Adel S. Sedra, Kenneth C. Smith Instructor's Solution Manual For Microelectronic Circuits, Inter
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Contents
Exercise Solutions (Chapters 1 - 16)
Problem Solutions (Chapters 1- 16)
Preface
This manual contains complete solutions for all exercises and end-of-chapter problems included in the book
Microelectronic Circuits, International Sixth Edition, by Adel S. Sedra and Kenneth C. Smith.
We are grateful to Mandana Amiri, Shahriar Mirabbasi, Roberto Rosales, Alok Berry, Norman Cox, John Wilson,
Clark Kinnaird, Roger King, Marc Cahay, Kathleen Muhonen, Angela Rasmussen, Mike Green, John Davis, Dan
Moore, and Bob Krueger, who assisted in the preparation of this manual. We also acknowledge the contribution of
Ralph Duncan and Brian Silveira to previous editions of this manual.
Communications concerning detected errors should be sent to the attention of the Engineering Editor, mail to
Oxford University Press, 198 Madison Avenue, New York, New York, USA 10016 or e-mail to
higher.education.us@oup.com. Needless to say, they would be greatly appreciated.
A website for the book is available at www.oup.com/sedra-xse
lV
Exercise 1--1
Ex: 1.1 When output terminals arc open circuited
For circuit a. n0 c = u,U)
For circuit b. l'oc
(,(t) X R,
When output tenninals arc short-circuited
.., CJrcmt
. . a. i ,.. "' l'~{f)
.·or
For circuit b. f.,.
v 0 ·~ 10 mY X
If R,
<'(J ·""
"~
.....M!Q_
100 + l
9.9 mY
10 kH
10 mV X _!Q_ - 9.1 mY
10 +I-
If Rt."" lk!l
i 5{ I)
For equivalt~ncy
t'o = 10 mV X _I_
I+ I
Rsis(t) ""' t•s D = 0001
1•.1
2 V => D "" 0010
15V=>D=IIII
(b) (i) +I V (ii) +2 V (iii} +4 V (iv) +8 V
(e) The closest discrete value rep.rcsented by
Dis 5 V; thus D ~" OIOL TI1c error is -0.2 V or
-0.21 5.2 X 100 •·~· -·4%
I
F..x: 1.6 (al T
(c) T ""
""
I<:x: 1.7 If 6 MHz is allocated for ench channel.
then 470 MHz to 806 MHz will accmnodate
806 - 470 '' 56 chnnnels
6
Since it starts with channel 14. it will go from
~:hannel 14 to ehanncl69
Ex: 1.10 Voltage gain '" 20 log 100 •·•· 40 dB
Current gain " 20 log 1000 "" 00 dB
Power gain ~" 10 log A,. "' 10 log lA, A,)
-~ 10 log )()5"' 50 dB
P,,
Ex: 1.11
15 X 8
120 mW
120 .... 18
102 mW
·r
-~~-
I ,. .!
.•• ~--dt
Ex: 1.8 I'
r.
120
R
"'.!.x.~xT
T
:cz
R
v.,"' l
PI+ P, + P, +
.
'"(~r ~ + c:~7J ~ (5·~~r k~ ...
y_: x
R
x_'
>: ( 1 +
r:.J
l ~
iJ
_I +
25
l't.
10
II)'·' V "' IOp. V
101' + 10 --
X
< 0.5
6.25 X 10
0.25 X 10 ''
Ex: 1.9 (a) lJ ean rept\~~ent 15 distinct values
hetween 0 anJ + 15 V. Thus.
l'.t = 0 V ~ D
0000
W
_-.-!!J_ X A X ~
+ R1
'"
Rl i· I(.
__
I - x I x ___
HL_.. ~ 0 . :25 V
p
Power nain ( 4 ) ~ _L
. 9l" 25
f
11
N;
Voltage gain
O.SI
10
I X
Fra<.;tion of energy in tirst live harmonics
····-~(!
;/
J()
I.
With the butTer amplifier:
l. + ... )
found from dirCl'! calculation.
Frat·tion of energy in fundamental
~
,.. ( 10 X 10 . )
IR
0
49
It can he shown by direct calculali<'ll that the
,_h
I?
cc·
infinite series in the parentheses h;ts a sum that
approaches
100 ·~ 15%
Ex: 1.12
Altt~mtively,
p
X
lO log A,
·.c
44 dB
0.5 p.A
0.25 p.W
0.25 V/V
Exercise 1-3
This figure belong,.Ho. ExerCise Ll5
r---1+
"- t
I()()K
-
lOt~,
Ex: U3 Open-circuit (no load) output voltage =
At'n''i
Output·voltnge with load connected
Rt
"'· .A~ II; - . - -
p
"'
L
11~
RL
100
'"'
v,
001!;2
-
lf,
fl,
11.
Jl,
v,
vi2
vil
Vs
_E ""· ._!.:~ X _,!,! X ...!.!
""' 90.9 X 9. 9 X 0.909
""8l8V/V
=
uh "'' 818 I's ., 818 X I
1-·- =» R0
R0 + I
"' -
Ex: 1.14 A,.,.
t·-
+
For t', .•., I mV
" R1_ + R"
OJ~
~
~
-=:-
+
lM~
lOOK
0.25 k.O
818 mV
250fl
40 dB ""' 100 V/V
"0.909 X 90.9 X 9.9 X 0.909::: 744
""
('A '" t'·-.!!.!:_)
rR
'RI. + Ro .
2
fur V,"' I mV
I.
2
=v~'X
(1oo x -I +1 -)
I IIJOO
' .
I
V.
·
n
2.:hl,.
744 xI mV
744 mV
t<:x: J.17 Usiilg voltage amplifier model.it can be
represented as
R,
IOlogA,,
Ex: 1.15 Without stage 3 (sec figure above)
~ "' (
IM
)< IO)( 100100
K
K+I
u5
100 K l- I M
. X
(100)(__Ji}Q_)
IOO+IK
~
K
R; "" I MH
)
R,. = IOU
A'", = A,. 1 X A,.2 = 1),9 X 90.9
·~ RL8 V
~, -~
R, + Rs
Vs
Vo
Vs
Ex: 1.16 Given v,
For R 1 ·~ 10 H
Oventll voltage gain
(0.909)(10)(0.9901 )( 100)(0.()<)09)
I mV
v,
X A
X
'"'
__!!_L__
RL + R,
·~ ...... !.!v.L~ X 900 X _J_O_
lf,
0
900 V!V
The overall voltage gain
I M 1 100 K
= 0.909 So
0.909 lis = 0.909 X I ·"'· 0.909 mV
10 ·' I()
409 V/V
For R, '' 1000 H
Overall voltage gain
.~
1'-_i_ _ X 900 X -
_ _!_.
I M ! 100 K
IOOO__ = R10 VIV
1000 + 10
.·. Range of voltage gain is from 409 to Ill 0 VIV
For u, ,. , 1 mV
1';2
9X
ll~ ·""· t) )(
1
9 mV
Exercise 1--4
Ex:l.l8
Ex:l.2l
FromnPdr
equllli(ln
atE
···-·--··0·--··----·-···-t'J,
G
= ihr" + (13 + I )ivR,
=i 6 {r.,+(~+
But vi> =
11,
l)R,I
and ip = i., thus
Ex:t.l9
R,
K": 1.22
f
Gain
10Hz
lOkHz
lOOkHz
I MHz
60dB
40dB
20dB
OdB
Gain !dB)
Rt
R1+ R,
,,, "'t's-tis ""
G,.t,1(R 0
::O.,vsRl
II
RE)
:~ Rs (Ro !1
RL)
111Ull.
to t<¥
R,
-tlo = O,.--.-(R
0
R1 + Rs
tis
II
R1)
Ex: 1.20 Using transresistance circuit model the
circuit will be
t04
103
...
3 dB
frequency
Ex.: 1.23
+
!1
~
is
-·
___!L
v)1 ~-'
;.x~
R
'"'
G,..V 1
..L + _!_ + .vC
R, -1· Rs
RI. ·+· R0
Ro
Rr.
Thus, Vo ""
v,
....i!L_
"'Rr + R0
Vo .""' R
i1
Now V 0 ~ Vo X !J. '~ R ....!!.L_ X ...i!.L
is
i 1 is
"'Rt. + Ra R1 + Rs
=R -.!!.L X _.!!.L..
"'Rs + R1
Rt + R0
I.
G,.
-~--~
- +-
R0
RL
I
sC
1
+ __.::._
_L + j_
Ro
which is of the STC LP type.
DC gain =
'
~ ?.!
..Lt..L
Rn
Rl.
100
RL
Exercise 1--5
..L
+ Rr_
J_$0,.
R
.100
0
"" 1Q.IOO - 0.1 ntA/V
I ··~. 0.0& mA/V
RI. .,:;·o·.I~ ~.O·.
t
R1. ~
·'
o.!>s kft ~..
Ex: 1.24 Refer to Fig. El.23
V2
V
=
s
RI
.
R.,+
.1..
+R
sC
'
R,
Rs +- R, +
s
s
which is a HP S'I'C function.
12.s
kn
1
hdn ··.. 2rrC(Rs
c~
+ Rl) 5
·-~I_ __
I
· -+-R,}
C(Rs
100Hz
2rr(l + 9) Hl X 100 '"' 0.16 ILF
Exercise 1--6
Ex: 1.25
b. 1imc tnken to cross 2 J.l.ffi
T= 50K
ni ""'· BT;ne ···1Jgl(2K11
length '=
::
..1.~;
6.75 X Hf
""' 7.3 X IOJ\50)JI1
.~ 9.6
x
e-1.12112xU2/IIJ-5
x~I!J
' qnJt,.E
= L6 X 10" 19
T"'"' 350K
n; ""· BTm e.
"" 7.3
c. In n-~i driff current density Jn in
J~
IO-:l9/cm3
X JO'~(JSQ)J/1 C·-l.la/tl X $.42
··-~
X I() • X .1~01
l V
2x
""' Aqn11n£
= 0.25 X 10·-B X 1.08 X 104
"" 27 11A
Note 0.25 11-m 2
From Exercise 3.1 n1 at
T ,.,. 350 K = 4.15
X
•
=
_
0.25 x 10· 8 cm2
dn(x)
Sx. 1.29/,. ··· qD,--
1011 /cnl
dx
Nv """ 10 11/cm 3
From Figure E 1 • 2 9
= 10 17/cm~ "" 10~ I (JJ.m) 3
llu
ni2
p,a!-N
'
Dn " 35 <-1n·/s
f)
X 10 11 ) 2
1011
dx-
J
106
-~-
lit
•
=
·-
qD dn(x)
,
n
Pp = N"
Want electron concentration
w
I
.5 X lO
>'I\
dn _ 105 -0 _
····· 10'1Lm~ 2
Ex: 1.27
At 300 K, n1 = 1.5 X IOH1/cm 3
TIP ""
4 2
35 x (10)
(f.tmr /s:
= 35 X 10~(!1-m) 2 /s
"' 1.72 x 106 /cnl
1.6 X 10~ 19 X 35 X 108 X 105
"" 56
X
10- 6
AI( 11-m )2
= 56 11 A/( 11-m) 2
"' 1.5 X 104 /cm)
For 1. = I mA ""1. X A
=-> .4
.2
;. N A = Pp = !!!..
=
I nv\ -· . !OJ IJ:A
::::.18
Jn
56 )J.AI(t.l.tn) 2
TIP
Ex;l.30
( 1.5 X 10 10{
1.5 X 104
Using equation 1 . 4 5
1.5 X 1016fcmJ
D.
llo
~ =
v,.
li-p
Dn = Jl,V,. = 1350X25.9X
Ex:l.28
a. u,·driff = -~J.-nE
Here negative sign indicates that electrons
move in a direction oppOSite toE
We use
'-'··drifT
$!;
Dp
X
35 cm /s
'
=
)LpVT "'
a.:
12.4 cm2/s
J<;x: 1. 31
--=---,
Equation 3 . 50
2 x Hr·4
•.
. 4
"" 6.75 X 10 cm/s "'""' 6.75 x 10 mls
W=
w->
2
'" -!J.-.E
·"-' 1350
w-·l
,,~ 1.08 X
A!cm2
d. Ddff current 1, = Aqnu,.-dritr
N 0 = 10 17/cm 3
(4.l5
1350 X
104
Ex: 1.26
=
X 1{) 1t, X
lit/l!IITI
= 4.15 X 10 11 /cml
It, ""'
30 ps
480 X 25.9 X 10
-;!
lltn~
Exercise 1--7
As one can sec from above equation, to increase
minority carrier-concentration (fin) by a factor of
2. olic mu1>tlowcr ND (= n,) by a factor of2.
J<:x: l. 34
Equntion
391,~
L
Aqn 2( . .!!.e._
1 L ,N
1 0
'"'
+
D, )
l ••N,,
D.,
D,
since ··"' and ·-·· here approximately
Ex: 1.32
t.,,
L,
In a p+ n diode N 1 >> ND
Aqu:(. 1._!!,/!._
+ Jl!c .. )
,N v
Equationl.. 5U(,
L,N,~
1
., lO"' X 1.6 X I()'
x[
v
w~-'---' --
Equation 1. 52 X J'
X
5
19 :><: (,
L5 X 10 1) 1
l
JO
+ _ ._ _
l!_l
16
--4
.·IS
10 X 10 X 10
10 -·1 X 1~
N_, + N 0
since N 1 >> Np
Nn
I
:::W·_;;;'
I,( e
\'/\' ..
1 ·-
I)
lY ,l
,.,
•.
r:quatJon
1
~
. .;- 3(. 1
<
)
-
.
tl > N1)
.
;V,
,
Ex:
l. 36
::. A,1,v"w
/l
1.()~ X
X
•J
1().
1.6 X 10- IQ
1.66 >: 10
11
(_!_-- + ___1_,)(0.814- 0.605)
ern
!(}IX
10 1"
0.166 IJ.rll
r;:.~-~--~~-
A .j2~sqN 0 V 0
Kx: 1. 37
[ " " v~
In
.-~ampk
N ~.>
1. 2;>
'f.
:
10' /em· and
V1
1
••,."
~~
" ' "~------,~-~"""
t~'(~; • ;~)v,. + vi?J
Ex: 1.33
to''inn·'
In the ll·Tt';!ion nf thi' pn jm:tinn diode
n,,
J' . ,
,\1 ,
II,
11,.
IO"'tnn'
ll-?
>-::
ll)'')'
10"'
(,J)l{
u~ing
Q1
)<
!()
Clll
equation 1- 53
' ·V V ,
Aq( -~~-"'-'·-~) W
iv_, 1 /v "·
0.601\
Jllll
Exercise 1--8
"·' I0- 4 xL6XIQ-ItJ ( IO
18 . 16)
XIO
X6,(}8XI0-Scm
1018
+ 1016
Ex: 1.40
r::quation 1 . 7 4
L2
::J!.
•"' 9.63 pC
D,.
I "" Is "" llqn/(..!!.JL..
+ [)~ )
l.,.N[) L.N.,
Reverse Current
10···!4 X 1.6 X 10-l? X ( 1.5 X 10111 ) 2
x( 5
X
10 · 10-4 X 10 16
+
18
)
10 X 10-~ X J0 1K
7.3 X }(}" 15 A
(5 X 10- 4/
5
--: 25 ns
Equation 1 . 81
cd ~ (;:) 1
In example 1.30 N,.. "'" l0 18/crn1 •
N0
""
l0 16/cm 3
A'>suming NA >> Np
'~'r::
-.fl ""
25 ns
:.CJ "" ( 25 X 10-~ ). O.J X 10-J
25.9 X 10 ~
( 10
18
10 16)(
X
%.5pF
!
+ 10!6 0.814
1018
= 3.2 pF
Bqualion 1 • 71
Ci""
C;o
R
3.2
X
10- 12
~
= 1.12pF
Ex: 1. 39
C1 =
'
1Sl.
""
dV
-!!...{-.r/)
dV
d
=-{-.rXls(e
dv
d
= Trl~(e
~ dV.
VN
T-J)]
I'll'
r- I)
)
Exercise 2--1
Ex: 2.1
Therefore:
The rrrinimum number of terrrrinals required by a
single op amp is five: two input terminals, one output
terrrrinal, one terminal for positive power supply and
one terminal for negative power supply.
The minimum number of terminals required by a
quad op amp is 14: each op amp requires two
input terminals and one output terminal (accounting for 12 terminals for the four op amps). In
addition, the four op amp can all share one terminal for positive power supply and one terminal for
negative power supply.
Ex: 2.2
v2
=
-
V 1)
isA = J.LGmR. For G"' = 10 mAN and
100 we have:
J.L
104 YN Or equiva-
100 X 10 X 10
A
lently 80dB
Ex: 2.4
The gain and input resistance of the inverting
amplifier circuit shown in Figure 2.5 are
- R2 andR 1 respectively. Therefore, we have:
Equation are v3 = A(v2
V;d
V 3 = J.LGmR(V 2
That is the open-loop gain of the op amp
v 1,
-
V;cm
-
R•
v 1 );
R 1 = 100 kfl and
1
2(v 1 + v2 )
=
a)
= 0 - .1_ = -0.02 Y = -2 mY
103
R 2 = -10::=>R 2
RJ
Thus:
R 2 = 10 X 100 kfl
IOR 1
IMfl
+ 0.002 y
0- ( -0.002)
Ex: 2.5
2mY
R= IOkfl
-1 mY
!(-2mY+O)
2
3
b) -10 = lO (5- v 1)=> v 1 = 5.01 Y
V;d
= v2
= 5- 5.01 = 0.01 Y = 10 mY
11 1
-
~(5.01
+ 5) = 5.005 y
.::5Y
1.1
From Table
c)
113
11;d
=
A(11 2 -11 1 )
=
v2
= 103 (0.998-1.002) =-4Y
~( 1.002 +
0.998) =
=
IY
I
V0
R,
d)
-3.6 = 103 [712- (-3.6)]
=>
V;d
J2
=
= v2
-
=
I
-(v 1
2
R; =
= -3.6036- (-3.6)
-0.0036 Y
=
IO\v2 + 3.6)
-3.6036 y
V1
=
+ Vo)
~
-3.6 mY
Ex: 2.3
From Figure E2.3 we have: V 3 = f.L V" and
-
G"'V 1)R
=
G,R(V 2
-
V 1)
0
V; - Ri;
0- Ri; = - Ri;
Ri;
= -R=>R,
i;
~'0 ~;" = o
-R
-IOkfl
v
__!
V;
thus R.I =
I
2[3.6 + (-3.6)]
-3.6 y
Vd = (G"'V 2
~ 0 I; ~ 0 , i.e., output is open circuit
The negative input terminal of the op amp, i.e.,
V; is a virtual ground, thus V; = 0
v 1 = 0.998 - 1.002 = -4 mY
-
R,
we have:
and V; is a virtual ground (V; = 0),
Q=
•
0 => R.I = 0 fl
I;
Since we are assuming that the op amp in this
transresistance amplifier is ideal. the op amp has
zero output resistance and therefore the output
resistance of this trans resistance amplifier is also
zero. That is R0 = 0
n.
Exercise 2-2
R =Jb1cll
~v.
Rt
0:5 rnA
C.onnectlng.the sig~IUI source shown in Figure.
e~:s to the inp!ll (If this. amplifier we have:
V; js a virtual .gruund that js V1 = o. thi,~s the ~r"
tent flowing through the to kll resistt)l' connected ·between V1 and ground is i.ero. Thercfui'e
V0
V,-Rx0.5mA,O-IOK'X05mA
""
=
-sv
For.the cifuuit shown above we have:
v (} = I
.:OW
A
If A ""
101
TXiO
then c; c., 9.901 and
E~e:l.14
(a) lo~d voltage
:·~ G-10 to X 100 "" -0.99%
......- . -I%
e
For V 1
'~
IV, V 0
G X V1
"'
,~,
I kH
x I V :::: I mV
I kH + l MH
9.901 V and
v,l = A(V~- V_)~ v+- v_ ""·
Vo"" 9.901
A
!000
(b) load voltage= IV
Ex! 2.15
:::: 9.9mV
If A = 104 then G ""' 9.99 and
E. ""
-0.1%
For V 1 = IV, V 0 = Gx V 1 = 9.99V,
therefore.
v • - v . . ,. vA0
""'·
?·~
104 "'
0.999 mV -I
- mV
If A ·"" 105 then 0 = 9.999 and
-0.01%
I V. V 0
£ ""
For V 1
v.- V_
""
9.999 thus.
A ~
"" Vo "" 9.999 = 0.09999 mV
JO'~
A
4
;:.:O.ImV
R1
V 11
2
jL ""' ~
I k!l
R.. ,on
v, "' I v,
;I "' OA, VI :::
""
,. R2 = 200 • IOOV/V
V,
\' 12 -
(b)Rid""2R 1 =2 X 2k!l=4k!l
Since we are assuming the op amp is ideal
Ex: 2.13
i1
R 1 =R3 =2 kH, R1 "'R4 = 200 kH
Since R 4/R 3 =RiR 1 we have:
(a)
G X V1
"'
I kU
(c) A,..,
=
=
~
V
101
I rnA
=
(....!!L.)(
I - ~ &)
R +R
R R
4
3
i 2 "" i 1 = I mA,
= (,: : }
1
4
-~~)
9kn
R4- R2
R~
R1
i.,_~
~--'-'"--+...-.()
v,
~+I
R3
The worst case C{)mirion-mode gain A,,m happens
wbenjA.,l bas its maximum value.
\1 0 "'
V1 +lzX9k!l
i ...
l'o
t. -
i0
v0
V;
I k.O - I kH
10 v
l V
I+IX9
-· lOV .. !OtttA,
""' it+ i2 =
""
=
~
IOV
If the resistors have I% tolerance, we
haveR.tnomO- O.OJ) :s;; R4 :s;; R4 nom(l + 0.01}
RJnmn(l + 0.01) RJ R:1"''"'( I - 0.01)
where R 1"'"'' and R411" 01 are nominal values for R3
and R.t respectively. We have:
R,311'"'' "' 2 kH and R~nom "" 200 kH, thus,
1.1 rnA
200 X 0.99 s ~. s ?00 X 1.01
2 X 1.01
R~
2 X 0.99
wYV
98.02 $
or 20 dB
~ .$
Rl
}{)2.02
Exercise 2--5
Simlarly, we can show that
-20 ==!X lOX I ms
CR
CR ·""
~~ X
'
I ms X 0.5 ms
Jt:x: 2.19
c
().OJ
-oV,(+)
IJ.F Is the input capacitance of this
ditl'crentiatoi:. We want CR "" !0- 2 s (the time
constant Of the differenliator), thus.
10-2
R= - -
= IMH
0.01 ILF
From equation (2.57 ), we know that the transfer
funcdon of the differentiator is of the form
V 0 (jw) _
. CR
- - - - -]W
V;(jw)
Thus, for w = 10 rad / s the diffcrentiator transfer function has magnitude
IVol ""
V;
+ ""
The input resistance of this inverting integrator is
R1, therefore, R = 10 kfi
Since the desired integration time constant
is 10-J s, we have: CR "" 10-:; s =>
•
10 -J s
C = JO k!l = 0.1 !LF
10 X 10- 2 ""' 0.1 VN and phase
-900
For w = 103 rad Is the diffcrentiator tnmsfer
function has magnitude
~~~~
""'
10~ X 10- 7
"'
10 VN and phase
::: 15Hz
9.9 .k!!
9.9 kH ::: 10 Hl
With this value of R:~ the new value of the output
de voltage (using equ:tti>fi, which
is the case .in this cxercis<~) and write:
Open-loop gain atf::: 20 log()")
vos, ··~ t•J
"• 0 .•... ,,r (/.\' ··r·' CR
. ""'" ~
::::} 1 """
0.16 Hz
Tht~rcli:m~:
12V··2mV
.,< I ms :::6D "'} 1
2 mV
Open-loop gain at 300 Hz ""·
"'
6D
20 Joe~ MHz
81) dB
300
~
Open-loop gain at 3kHz"'
20 Joo 3 MHz := 60 dB
"'3kHz
·
Open-loop gain at 12 kllz "'
201o<> 3 MHz = 48t!B
"'12kHz
·
Open-loop gain at 60kHz''
20log 3 MHz ···' J4dB
60kHz
Ex: 2.27
c
With the feedback resistor RF to haw at kast
• 10 V of output signal swing available. we have
to make sure that the output voltage due to l'os has
a magnitude of at most 2 V. From e(
!0 MH
beln'~:
v..---., \', 1+ 1
0 ········
1~F
l .•
~:R -~ ~ V, I
' )dt
ll1c wavefomls for one period nfthc input and tlw
We knnw RC ·.·· I ms ami
R
;.,., '
output sigmtls are >hown
10
7
-H--
.
1--I IllS .
_________,_.
2 Ill'
I ruo/ s
~-..._
__.·-JOV
Exercise 2--9
t ms
()
~----~1
2ms
slope-"'SR
1.6V
We have
-20 "'
=..!J' m• lO dt
CR
=}
0
CR =
t,.
=..!X
-20 "'
CR
;~ X
10
XI
ms
t = 0.9 X 1.6- 0.1 X 1.6
'
I ms "" 0.5 ms
Ex: 2.28
Since de gain of the op amp is much larger than
the de gain of the designed non-inverting
amplifier. we can use equation(2.35 ).
Therefore:
hdh ""'
__£___ and I
R1
I+ R2
R,
100 and
R,
J,"' 2MHz
Hence !Jdb
2MHz
100
20kHz
E.x: 2.29
For the input voltage step of magnitude V the output wavefonn will still be given by the exponential wavefonn of C{JUationf2.40)
lfw1 V sSR
That is V s SR ~ V s SR
w,
27Tf,
V:;:; 0.16 V, thus, the largest possible input voltage step is 0.16 V.
From Appendix F we know thal the 10% to 90%
rise time of the output waveform of the fonn of
equation (2.40) is 1r
:::
I VltJ.S
~I,
2.2 J..
w,
Thus. 1, ::: 0.35 JLS
If an input step of amplitude 1.6 V (I 0 times as
large compared to the previous ease} is applied.
the the output is slew-ntte limited and is linearly
rising with a slope equal to the slew-rule, as
~hown in the following fi.gure.
= 1.28 11S
Ex:2.30
From C{Jllation (2.41) we have:
f
--
M -
SR
2r.Vom"'
"' 15.915 kHz ::: 15.9 kHz
Using equation (2.42). lbr an input sinusoid with
frequency f
5 fM, the maximum 1>ossible
amplitude that can be accommodated at the output
without incurring SR distortion is:
V o "" V o ,..,( fM ) '"' I0
5 fM
X
! "-"
5
2 V (peak)
Exercise 3-1
Ex: 3 .1
(c)
Refer ro fiig3. 3(a). for V 1 ;:;-: 0, the diode q:mdu<:t!; and prcs.ents a ~ voltage drop. Thu:>
V 0 ..~ V1 For V1 < 0, the diOde is cut-off, zero
current flows thro1igh R:md V 0
""
0, The
res1dts is lhe tnmsfer characteristic in Fig E~ .1.
OAt
1~
Ex:3.2
see Figure). 3aand:3. 3b
f)uriilg the positive half (>f the sinusoid, the diotic
is Jnrward bia.~e !l${1.< ·' !II
V!V
"" I 5e
r '" 6.91 X 10· e
- 0.7V = V
·JJ
""54.6 rnA
R =
10-2.4
54.6 x to··l
" 139
I
2.5 k!l
1'
I '" '\'7
I ..,e
ln
:1
dioth:
\ ~ '/ \ '
i \ )
('
v
\':\·,
I
in
j \'J\.'T
;}):
ifi! t'
if)}
,)\
i,)j
-
c. If i 0
lmA
ai 0 (mA)
aip(mAl
'mall
expo.
model
·---·--·-<>·-··---·-
a
--10
-DA
-O..B
h
-·5
-0.2
-O.IH
c
r.'i
+0.2
+0.22
d
+10
t-0.4
;0.49
-~·-
Ex: 3. 15
v
:
mY
Ex: 3 .16
\! 11
lr
~3
.~.977
For a t.ener diode
I ().()1 X 50
\ 1 ,.,
--
,/,.7
+
\!,.,
\ 1 ,., -
--ilJo-
-4 mA
J
9.5 y
Fnr I z
~R
5
i,
0.7-143 v
Vohagc drop across 4 diodes
4 )f 0.7443 ..• 2.977 v
10
•
t'
5
'' 1 1n
\-'{1
+l:'i
rr~t\
)'. HI
. ('-·IIsJ)
Vn
25 mV
signal
"
tl7~·!1'3-
/\cross t'ach diode 1hc vollllgc drop is
thing equations ( I ) and 1~) r.:sulb ami u.,ing
jV(mV)
4.7 ·.r 10
I'.'\ I
I)
In I hi., problem i 1, 1
'
I'
i /)
II
\ l
in,(<'
V1
0.75
4
J
5 mA
9.5
1
(1.()05 X 50
~
9.75 V
Ex: 3. 17
-o V,,
t
15Y
I
~I/
a. In this problem __ _
V5 cos{1T- 0)- V 0 (n- 20)]
COS(1T- 0) :::-1
11'- 20 ::: 11'
vu .• ,g "'
A V0 -~"
AI,
-
Al;rz
I mA
_ 7 mV
mA
2Vs V 0
211' - T
-
Vs
Vo
---;-2
For V~"" 12./2 and V 0 = 0.7 V
t>o.av~ ·'"'
t2J20.7 = 5.05
1T
2
v
and
Exercise 3-5
c. The peak di 0
is
~'IT,... 2 fJ) X 100
PlV "" +Vs
::: 17V
2'lT
Ex: 3. 20
::: 97.4%
Average output voltage V 0 is
input
V.o""" 2Vs- Vn"" 2X 12./?:_- 0.7
1i
.
Peak diode current
'IT
in
is
, __ v5 - v 1)
In ·- ·--R--
J():JmA
F'IV '"' V.1. - Vn +
12./2 .... 0.7
a. As shown in the diagram the output is zero
between ( r. ~ H) to (r. + II)
20
•.• 33.2
V~
12Ji
v
Ex:3.21
Here 0 is the angle at which the input signal
reaches \1 0
:.
~'ssinO
o '"'
"" V 0
< \'
sin · i (
•
'/')
1- .\'
b. Average value of the omput signal is given by
v
z_5rr
.!_J2V~- 2V 0 (r.- 2H)j
1l'
Vn
"
But ens H ''" 1
cos( rr - U) "'' · · I
c.
Peak current occers when •f>
20
~- Tf
2V1.
Peak Cunent
~::.'!ll(:rgl_ ::-:_~!!
R
If 1\ is I 2 Vlrms)
then \/..:; "".
1T-
X
12 ~.· 12J2
:2Vn
1T
2 :X I2.J2
'' 1.4
<).4
v
10.1
v
Exercise 3--6
(b) Peak diode CliiTCI\t
''"'
Peak Volta~
R
" -~L=-~-~Q ,., 12J2- I~
R
lOO
~
(!)/(
156mA
PIV"" Vs-Vn = 12J:i-0.7
Ex:
16.3
v
3. 22
Full wave peak Rectifier:
.
Ve1T+I
't>.•v = wil1R
"
where wilt is the conduction angle.
Note the conduction angle is the same expression
as lbr the tmlf wave recti tier and is given in
EQt3.30
JivVI,
wili'"E· - '
(b)
Substituting for
J
~~,~ ~
tll ilt
we get:
if!;
')11
+I,
~-R
ve
Since the output is approximately held at v,..
~""I
R
,. Thus"
..
assumed the
/ l~eat diode ,
charges to V1,
""') j/),a>
$!!'
"' /1.[ I
1Tlt_!Fv; +fl.
;
'IT
jFv;]
Q.E.D
If t "' 0 is at the peak, the maximum diode current
occurs at the onset of conduction or at t = wilt.
During conduction, the diode current is given by:
i0 = ic
The ripple voltage is the amount of discharge that
occurs whel\ the diodes arc noi conducting. '111e
output voltage is given by:
V pf' -tfFIC
iv.mu =
"'
v l'e
RC
+-
'00%
lie
vI'(1-t! _ffi)
'
(t - l + T/2)
RC
v
"' _:_p__
2.{RC
.
/1.
lit
'
(a)
'lb find the avcmge current. note that the charge
supplied during condu<.1ion is equivalent to the
charge lost during discharge.
OsurvuEn "" Owsr
SUB (a)
jv
"
-Csin(-wilt) X wVr
+ lr
for a small conduction angle
sin(-wAt) =-wilt. Thus:
i0 _,. .. "' Ct<~At
Sub (b) to get:
for CR » T /2
--
+
-CsintlltX wV 1,+ 11•
"'*
.-:: V1
d;
c!!(v1 coswt) + ,,
discharge is only
half the period.
v'
cdtls
assuming iL is const. it=
·- 'lj£
vi' - v r
+ it
X
wVe + 11.
Exercise 3-7
The conduction angle wAt can be obtained using
equation 4.30
wAr =-
fi'v. =
rv;:
c=i.XT- = 0.36
.J 12J2 -
2 x o.8
rnd = 20.7e
The average and peak diode currents can bt~ calculatetl using equations P . 34 and P · 3 5
Ex: 3. 23
(}"
ac
and V1,
I
tJ
too n·
t(l + 21T~)
~~,
"" 2.74 A
PIV of the diodes
)
) 11
16.2 v
Vs- V 110 ' 12J2 ··· 0.8
To keep the ~afcty margin, sck>ct a diode capable
of a peak current of 3.5 to 4A and having a PIV
rating of 20 V.
• 11 RC
Ex: 3. 24
At the end of the discharge intcrv:1l
,,
·
1.45 A
+
(vr-~vvot'
l'o "" (1/ 1,
14.1JV
Here / 1 "' ----·-··-
12./2- 2 X 0.8, V, ""' 1 V
""'
it.•••
The output voltage, ''o. ean be t~xprcsS<.-'<1 us
llo
~, 11 ( 1
'"'".
-~-·]
line
voltage
~·)
· + 1T ..:.JJ....
2v,
1.1 •
2Vt>o- V,)
The discharge occurs almost over half of the time
period :: T/2
For time constant RC >>
,,·U/11'::
E »:
I -
2
~
I', r
1
__!__
RC
(\', I' ·· -'V {)() )X 2RC
_I_
cc
Hert• Vr
c.c
1/ /)1) =
0.8
11.['1 and Vr
7 ·'
The diode has 0.7 V dmp lll I mA curn:uL
v
c
···
·.o7 1.•11 ""
2
X
O.lO
X :::---:"::--'-:-::-:c:
.
2 X 60 X 100 X
{ 12.[2 ,. 1.6)
2:.~
6(lx-liio
1281
\Vithour com.idt'ring the ripple
otHput \ nl t;tg,•
C
pJ~
volt:r ''t ?: 5 V
Diode D2 conducts and
0V
.. 12 v
be;;ause it is ideal amplifier.
Po ;c"
I
+5 + Z(t•
1 •• 5)
Jt:x: 3. 25
=
>KJH--.0
t 1,,
(:ts+ ~)v
tl0
Ex: 3. 27
Reversing the diode results in the peak output
voltage be.ing clamped at 0 V:
v1 > 0 - diode is cutolf
110
""
0V
v1 < 0 - diode conduct> and opamp sinks load
current.
"o
=
"r
Here the de component of v0
Ex: 3. 26
IOkfi
+
+
D~
+
5V vo
,.,
IOk!l
Both diodes are cut-off
s +5
for ~ 5 s
v,
""
V0
"' ~ 5
V
Exercise 4--1
lwJ "" lsc ·""
'
. 13
to·•~
1()-16 "'
100
A
r.·l
t.~E "'· ' S(.. [.. , + !i. ~-= !(,..~~ x 100
'.<.H
"" 1.01 X HfHI A
n1JE1· ""700+251n. ( 0I• 1.)
V8 £
c]. .,
Vrln [ 1
Is
,,.
'"" 642 mV
v11 n
+ 251n
700
25ln [' 1 mA J•
·.10"16
· · 25 X 29.9336
'i
[ 0]
""· 748 mV
'' 758 mV
4.6
Ex:
Ex: 4. 2
Yn·=+5V
:.a= _L
Jill
.2Q_ <
Ct
50+ I
150 +I
< tl <
0.98
< __!1Q_
0.993
c
B
Ex: 4.3
fc·""'le-le
·:=
1.460 mA-0.01446 mA
=
1.446 mA
n ,.,.
!.r
"'
IE
l.440 ·"" 099
1.460
.
13 ""
!.r
"'
In
1.446 . "' 100
O.QI446
t•oe ·"' 690 mY
v/f£n'r
lc., I rnA
t5 e
lc "'
For active range V c '<:!: V 8
Is "" .....!s_ "" 1.446
'loE'vr
e100<~s
Rc(rnax) = Vee- 0.690
lc
e
=
1.446 A
=;.
10 -tsA
""5- 0.69
I
""'· 4.31 k!l
ell!
Ex: 4.4
fl .,. ~
I-«
and lc '"' 10 rnA
.
For u = 0.99
13 .,. ,. . 0.99
..
Foru ""0.98 13 "".
Ia "'
t "'· ~
,.,... 99
I -0.99
!.r13 "' 99
1!1 .,
1• .,
Ex:4.7
0 ltnA
f.~c ""' 100 X 15
().98 "" 49
1 -· 0.98
Ex: 4.8
.
'= 0.2mA
(~ e
11
lsf: e
~"Bt"' 11 r
(~ ""· w··tll A,
l3 = 100,
lc ""' 1 rnA
lsc + lsa ==· lsc[l +
~]
lsc"" e_ _
1
1{.
e"nc "r""'(!
~'nc1 YT
.... fsc e
0
1'nr'"r
Given:
=
VBE 11''r
for ic
10- 13 A
""
ic "" Is e
Ex: 4. 5
lsr:
.;:· w-•5 A
Areac = 100 X Areali
(~
fl.' HE- ~'ncl 1 "'r
nc' 11 t
Exercise 4--2
E~:
4.12
t
1.5
v
2mA
v
·-1.5
13 ·"·· 100,
Fig6.12
0.8 Vat lc
Vl!F
Vrlnllu11 1 d
V 1w~-V 11 t: 1
25 X 0.693 " 0.0173J
.\ \1 BEl
-
0.817 V
~_cr -::...~~r
lc
~
100
X
R1,
Ex: 4.10
I r . :-'~ ~~e FNF ~-.,
V
-
HE ••
~ lO
,,
~:li.£ - V !IX .~JL..
(13+1)
lc
1.5- 0.817 >< 100 kf!
2
101
({
2 mf\ "'
=soon
0.2219 '"' 22.2
t•/flf
338 H
t·r
50
1'\ !1
-- l
rLlO'z •
X
SO
5I
X
l'J' :j
' '
Ex: 4.13
tJOV
= 650 mY
_j~_ 1
p ' ; [
., 50
v
-~
5i . -
Ex: 4.11
t,.
1.5 A
:.Far ~· \', inl1.5 >< lfl ''I
(en mV
Fig 6.13
1:1 '·'
Vr
so. Vnr;
0.7
v
Fli ···· 0.7 V
I mA
Exercise 4-3
"" 0- 0.7
l. F-
~'
lr· =
.
= -0,7
v
E.'!.: 4.15
-0.7 + 10
10 K
0. 93 mA
~ /... '"' 0.91 mA
51 "
Vc "'' 10- 0.91 X 5
v
"·" 5.45
'" '"' !s:fl
= 0.91
50
0.0182 !.LA
-5Y
Ex: 4.14
V1Jr; < l I I = U:l99 mA
!01
E:x: 4.19
Fig 6.20
Fnr v118
111 = 0
0
Tr:m:;istor is OFF
.-. 1,. . 0
'
' 10- I)
; f()V
R,····JOk!LI3
(a) artiw
511
r, - 5 v
OJJ98 mA
1.68
v
Exercise 4-5
EX: 4.21
Ex: 4. 23
+IOV
ForV88 = 1.7V
18
.!.} -
'"'
10
0·7 ·""' 0.1 rnA
(a) edge of saturation ~rcc
""' 0.3 V
10- 0.3 '~ 1.94 kH
- .. ·-5·-
Vee · V cJi .
1.-
t'n: ~"' 0.2 V
(b) deep saturation
111
V 8 -0.7
""' 4- 0.7 "" 3.3
lc"" Pltl
"' 50 X 0. f """ 5 mA
Vc
10 - 5 X I kU
+5 V > Vm; (so Active)
Rc .
vR o"'
v
Rr. "' 3.3 V "" 6.6 k!l
·
0.5mA
Vc '"' V11 +2 V
.<'-c4+ 2"'·' ·l6V
Vee·-· Vr
Rc •<" ····--,-= 10-6 "'
g kH
0.5
0.1 rnA {unchanged)
lc ·- J3F.,«J/ 11 = 10 X 0.1 "' I mA
Nc =
10- 0.2
tm.A.
'"'
t<:x:
4. 24
+ IOV
9.8 kH
Jo:x: 4. 22
+lOY
t
~4.7 kH
+4';'
0.2
l3k!l
E__t,v
r-
(t
= I
flh>I'C<•l =
·nwn (
Fig4.22
At edge tJf ;;~uuration <'u: '· !D V
ldR( ·i R 1.i
;- 0.3
lO ·- 0.2
4.7 f 3.J
11R 1
Vn"
'
1.225 mA
0.7
- 1.225 >< 3.3
•c
4.7V
0.7
·.~.
5
5 111
1,:=61.
I = __
!!_)
-= (!:L _____
11
/( R, · 03 ' lFRr
Vn
-~,
ov
~3.3 kH
Io
v
v,
V,.,
j X
4.7
A
• 0.226 mA
61, .'< 3.3
·•··• '~A8V
V, I 0.7
"'· 5.18 v
6
X
3.3
Exercise 4-6
Exi
4. 25
Ex: 4. 27
+lOY
+IOV
~
~Rc
~
50513 ~£ 150
In active range
10 v
11, "= 5100
mA
- 0K
•7 ""' 0043
..
V. ·cc: I}
V,"" +0.7V
V, lowest for largest
!!L::::~U
0
2
···-·
0.99/,
.,. 1.5 kfl
10 - 0.4 i 0.7
Rc .. -·--··,··-~--,.
0.99 X 4.65
= 2.2 kH
0 ' 0.7- 0.4 "' >0..~ VJ
[V,{max)
Vee - 0.3
i5o
x o.(MJ
Rc · · ·
4.65 rnA
1..
13
1, ·""" J3 1. ' 150 X O.Cl43 A
Ex: 4. 26
Fur f3 '·" 50
1-',
10- 50 X 0.043 = (}.78 V
Forf3 = 150
v, "" 0.3 v
Ex 4. 28
+ 10 v
+15V
+
Rc
~ Rr
5kH
•0.7V~i
~---t:j
0
4V
'~·-I
~ Rc
~R,
I
(
!OV
I,
V,
15 X 50 "" S V
VJHI
= /,
9.~
v,~
150
kH
50i! 100
100/:~ k!l
4
--~11!! ..:: •....~.'.!!5........ .
--.w
10- 4
!
() k!!
1~in
_ ,____1 ()
~ 50kH
~
I mAt
T
1? 1
'
•
;
IR 1111 1W; l)l
I 100 .. 1
:1
1.1 X m;\
51
v
Exercise 4-7
¥f·
IE
Ex: 4. 31
1.15 rnA
co
1.28- 1.15
1.28
%change
+5V
Ex: 4. 29
Totnl current drnwn
+ 1.252 + 2.75 mA
"0.103
4.135mA
Power Consumed "'· V X 1
"" 15 X 4.!35
'" 62 mW
Ex: 4. 30
.
2.78
4. 32
Ex:
-~~-~-~---·--·"!--·-·------·
'15V
~2kn
t
+5
rnA,!
··-~( Qz
+t;IOV_ - .. +
.., 7- i
~-
J
::. T
IOkH
"NNr---< B
mA,T--KQ.I
CD
_, ON
(J:/1· _
"-
(j> +- I l/11
"" 7.()3 mAX Sat"
E
~
!
2.1 H!
v
~ ~~~~ n
I
lk!!
!
"·--··--~-~~.----.v....L-.,
{3
·c
J()()
IE: unchanged
lei unchanged
Vr; =·· V0 -· 0.7 V
11:,
=
> 5 :. Saturation
V ·"">lice-- 0.2
L
I
~~C::-
0.7
0.470
55V
101/u_;
IB
Vc2)
2.7 J
101[ 2.75
•
= 'LXV
.L!L=-~.:~
10
11.:••"> 4.8 mA
0.45 m/\
lienee
?..75
0.7
4.R
0.45
-Ll5mA
h.2 ... 0.7
Vn
047
")(
!00
!01
6.36
Ir
v
13.4 lll.-\
- ±.2?
0.45
9.6 << 30
-"4
4.8 mA
Exercise 4-8
Ex: 4.38
Ex: 4. 33
Vee- Vn; ~-- 10- Vee
l'r
0.025
A., •=
+IS
-320 VN
R,.
,, Vu: "'' 10- 8 -··· 2.0 V
IOk!l
JO-.1
8 kfl
2.0-0.3
1.7
lmA
10 ILA
I
l!a Swing
v
13
v
100
.LL
.l \f llt
1.7
5.3 mV
J26
J.....I:!IA
25 mV
~·
40 mA/V
Ex: 4.34
4{) X 10
·--400VN
fc "'·· Vn· ··-- lcRc
15 ... J
X
10
5V
N<{t) '"' Vc t· Nc(l)
.,., (Vee - lcRc) + Avvl,.(f)
= (! 5 - I 0) -· 4{){) X 0.005 sinwr
"" 5 -· 2 sin w1 (t)
;,,(l) "' lu +· 11lw(t)
•·• 10 I 2 sin wr (j.LA)
Ex: 4. 35
Ex: 4. 39
Ex: 4. 36
If· '" 0.5 m;\ ! cnnstan!)
I) · 50
0.5 mA
25 mV
-- 20 mAr V
II!
1,.
·- 20 mAN
0.5
50
II
0.5
200
2.5 mA
I 0 11.A
r4 ··"·
Ay() ""
':n · .· ·
Ov = JS(roft Rc R Rt) ,
x. 1000·
l5 -·- 049V
.-
"
5 kH
R1N
A11
n
zs n
·"' 40 X 5
R,.)=g,l?c
11
~·
200 VN
5
""
Gv =
Avo X .5~:5 ·-= 100 VN
~ · Av
Rs + I~ IN
"'0.5V/V
~-
:!£! • ! x 65.6 · -.12.8 VIV
2
!:1 "" ..2..... ""' ! "'* ~~ . "~
••s 5 + 5 2 "~
llro! "'
"' ,.~ ....
Avo"" g,.(r0
"o
tJs
= 100 "" 2.5 kfl
40
r' ""' ..2:.L
13+1 .... 25•
"'' 9.5 k
,..,
g.,
200 k!l
A. v _ -190..5 X5
5 + 9.5
= -65.6VN
G1•
4.44
.Ex:
r .~ A
5.0 kn
""· lO kl! 200 k
~'
10 nN
32.8 X 10 Ill ~. 0.33 V
Ex: 4. 43
"'* lc
gmRc
·i·
'·~
Av = 40 VN
'" 80
=
gmN;.
80!20 " -l kH
I lR f,'
r.,.
r~ i(
ri
I
P lnser130
I)
l+~.t~.A'QED
r,
!!.~
t>,
rc
I 00 ::~~ 1 + I() K + ...~!!.L
10
5!101
50
20 X 5 = 101.1 VN
Ex: 4. 46
r·~
~.,
"" 25/50 '~ 0.5 mA
= ~X
:. Rt:
'' I " I~ + _____!!.£___
X 100
son
"'* Av
;_!. •; + .CJL:
25
5000 + 25
Vr
mV
R 1N "" r, = = 25
- -....
.
lr
IF
Gv
vs
"'-
Ex:4.45
Rs""'
AwJ "'
~~~
+ 40.4
}c
I
g., "" Vr ""' 0.025 ~"' 40 ms
20mAIV
-20(2Ut'Jk n 10 k)
Rr-11
IOO.x 10
10
""'- l9.8VIV
"" -I ()();5 V/V
Rq
+ (~ + 1)0.35 ""40.4 k!l
R1N ,. 5 k
Rs1a + Rut
Fur lc=0.5-tnA and Rc ""' 10 kf!
r0
7 X 5 = _!- 0 3SkU "" 350 U
101
OJ
.
Vj
= A ....
,.., <~..• ""' .125••
r, "' ·~~
~. R""' -=
ib = ~-=:~VI.~·:,~~:: ih)!!J
l'w
Exercise 4-11
= O.S + IOIX. I ;;;: IOlSktl
CfvC~ = l(R1, mov¢d; t:o- = ,.;1
R_o = "'+-Rs
"" o;s101+ to ~-· .104 n
(i + I
Ov ,.,. ~ = ~
vs
Ro + Rs
=
l
O.H)4 + t
:tl
IE=
2.6
M95 mA
=
+ 3.3
..
l5J
%
= 0-995 - 0.984 X
IOO
1
"" 0.91 VN
Ex: 4. 48
Ex:: 4A7
+IOV
Vn:
R,
R£
lmA
lmA
Design I
13 .""' 100
Rc = 3 kfi
~~
R1111 == :
vliB
= 26.7 kn
v
1, = .!.=.Q1_ ·""- 1.01 mA
26~7
.
V
+3
101
Vt.
:. maximize Rc
I
= lcRc + 2 + 0.3 + I,RF.
VEE- 0.7
""
4.3
Re+ R 8 1(~ +I)
R~r+ R 8 !(f3
=
£
~Rr.
+ R11 /(f\ +
+ l)
-· l mA
I) = 4.3 Ul
For independence fnm1
.
f3, set R8 == 0 (OK for C8 )
c-=:>Rt: = 4:3 kH
ll""50
1£ "" ...2:.L..
.
26.7 + 3
·
/gg1ven
T
12 X 4() "" 4
80+40
""
lt.Rc
Av ""
""'
\!(.(min)
1.04 mA
= VE + 0.3 V
"" -0.3 V
VCQ = Vc(min) + 2V "" +10 V
151
% change = I.04 - 0. 937 x If}()
I
'z
10.3%
Rc ""' Vcc - V CQ
""
0.9
40
= ~sx+4
V B/1
""
"'·' 2.67 kfi
1.2 X 4 c= 4 V
8+4
13""'
fu· "' 4 ~ 0 ·7 "" 0.99 rnA
..
26.7 + 1 ~
101 ....)
'~ I 50
Vc
lr " _ll_ •· 0.984 rnA
2.6
51
13
::c:
kfi
+IOV
R 88
.
= 8.48
Ex: 4. 49
Design2
f\ ""· 100
R~::= 3.3kfl
f\
!!!.=___
lc
+
J.3
V,;
+ 0.4
l
100
2
+2.4 v
Rc '' ~cc- V r;: "" (()- 2.4 = 7.6 ~· 7.6 k!l
I
150
I
I __
Ill . . . . ._t:
f3+1
IOJ
mA
Exercise 4-12
{cont.)
R8 ·-- Vcl0l(2.4o;1) ,.. l7t7 kJt
.. JH Va ·~
..
+ luRtt + I eRe- V<·c
I
__
,., 0
lower· ::;) hr = 0.005 mA so V 6
10-0.7
ISO + 7.5(~ + I)
-OS- tM- 2
r0
It; = 101 X 111
'" L002 mA
v
= -2.9 V
""
100 ·""
lm
1
Vr
100 kO
1 .
25m
g- = ..£. "" .~ = 40 mA/V
V <: ""'· 10- 7.5(L002)
"" 2.5
-0.5
Va"" IOOV
9.92 J.1.
JH ''"
-t;96v
-L9(;V- 0.4-2 -4.4V
13 = 200 : upPei' still 8V,
Rc = 751dl R."" J80.kfl
().7
~
=
Using 5% resistors:
11 -
v
= SQ: upp.ih'lt111 8 V:.lower
~I,;= o:ol96mA so v 8 ""'
~wingf...,.:J.4 .;.2:= -3.4
~
·
v
r
Ex 4.50
"
J!. ""'
""
g,.
tOO ''' 2.5 kH
40m
r:,
r;. .,. (fH I) 1!!:! 25 .U
F..x: 4. 52
.I!Karilple 4 . 50
v
~~
R.
. , 100
~
R,
18
o.•
100 kH
7.5 k!!
""
~a~: r_ = 25k0
/1'!(!3 l· I)~~ IE! 100
lb
= O.ot mA
= 0- / 11 R0
\1~.
=
-I V
~=
X
R0
I X 7.5 ·""'-
IRm;
2.57 V
t
"' 19.3 kU
Avo '" -g,(ro
•
(,~.
=
V.-
10 - 11 k( I m) ·• ·1·2 V
\'"
100 k
\' 1,
··I·· 0.7 "·' --1.7 V
·ll.(l! m) ·
(3 ·"' HJO; upper limit=* V cc..
v;
V 11
-1.4V)
·····
II
Rc
R~_) "'' ,-40 X 3.5
= -g.,(ro
II
' 7.4 kH
Rc)
= -40 X 7.4 " -296 V/V
O.(H m;\
~
II
Ro "" roll Rc ~ 10011 8
lc ,. a( I m) ""'I mA
lower value
kn
Rc · 8 kH
·
Avo
Ex: 4. 51
Rd'er to Fig E4 . 51
X (
= 2.44
'""-119V/V
R"' Vcc-0.7+\!u= 19.3
•·
r, = 10011 2.5
Avo = ·-MmRc= --40 X 8 "" -320 V/V
v!1- 0.7 v "" .. 1.7 v
Vc .... \1<..<. -· ~.
!Hl
111
R,,u
RIN ""
~r/1
"
RIN
=- ----·
R~
+
/?IN
R(.
Av - Ru + Rr
~-(-216)--.1__
. 7.4 + 5
5 + 2.44
'"" --39.1 VIV
···IV
•
R;o;
+ /(IN
••
1'~ ·"'· - - - · \t ·.=
'
v,.
=
8V
V c ""' OA V (where
15 mV
RIN
lvo! "
Gv!;,~l
,,. 39.1 X 15 '"' 586 mV
Exercise 4-13
withRF.:
v,,, "" 5 +2020 = 4~~V-
"~
5 mY
wloR,:
~!!L '"' .5 + (2.5 8 100) = ~
5 mY
Av
v
I~ s
{2.5!1 100)
IV ol •• IVrl X
= 625mY
·
512.4 · 62 mV
Ex: 4. 54
g~ '"' 40 mA!V
r., '" IOOk!l
Example
4 . 50
r, ._,_ . 2.5 kH
g,, ····· 40 mAIV
1~,"" 100 k!l
V, "'" 100 V
R;,;g '"' 5 kO.
R,
1?, "" r" + ((~ + IJR,
R11•1
Rill! R 1
_, ••,
oc
25 ·- 2.5
l () l
A,,) '' 11··,~·-:;li;
0.22 kn
l
'" ii-4(3Tnij
R.,.1
·- 32 VIV
l?,;~+i!1+
OR G 1-
'"
___
l)(r,.
f!.t~-- x
R,,~
-12.4 VIV
25 ; 223
R,:
+ N1:-:
Note: wirh,,ut R,: :\ 1-
t
/? 1.)
A1
""
-
·
lr. '"' I mA
l'o)
'"
Rc II r0
,,.
II
ru)
!!__ "" __
,.,_
r, 1g
R, 18 + r,
·.· O.ll05 VJV
__?1___
+ 25
5000
Gv • o:(Rc.lR1.)
R,;~
+ r,
"'0.6V/V
°
2 K :<
25 K
12.4
R";;JJ;:;;..
C:_~Rc II RL)
R1 )
·- --123 V!V
¢1,
i
B
.
!
--i_~
nv
i~.
r
! .
1,
iH:
Rs
i
,-./VI/'---1
F,
0
51dl
. Ln
100 k)
= 40 X 3 "'' 120YIV
• -9.9 V!V
~.,,(R 1
II
7.4 k!l
A,, -'' +gm(Rc!l R,_
·--40(S)
100
0.99
'" 40 X 10-~ X (8 k
-296VIV
223 H
-8 K:i 5 K.'..
'"' -Ncii RL ____
r,
f3'"'
tX""
~
R,,., ""'8 k!l
;\,.
v,"' 100
Avo = +g..,(RcU
4 X R;;r. = 20 k!!
"'·'
"'!~,Rc
R."" 100 kH
Rk!l
R, "' 5 k!l
r.-"' 2.5 kU
r, = 25H
R,. '" r,-"" 25!1
= 15mY
c
Exercise 4-14
Ex: 4. 55
= ().5+ (101)(2011
= 96.7k.H
l)
R,N '"' R11 11 RH, ""'· 4-!lli 96.7
(•
__ V 0 ___ V 1
V0
V_1-
V1
1v ·- -
V5
~-
X ....______ _l~:+-
Rs+ R1N
r
.
28.3 kH
-·- X -
____RtN
(Rs II R8 )
'" 0.796VIV
,
_
40
.
( 'I'O
--· - . - - X
I0 + 40
~~I
=
ll R1,_)- - ._
I )(r, + (r 0 II) R1J
l)(r 0
+ (f3 +
. .
20 K
(.!.Q.JS.L
40 K) ('iF -t101
• •
.
•5
+ ?().. ·K·.)
~
lc
100(25 m)
---5 lll-----
ro
'-~:)
!00
5 nl
lc
Gvo "" 0.8 VIV
500 H
2() kH
R,,,, ""
r(,11
(r,
+ [R.dl R8 ])! (j3 +
I)
•• 20 II I 0.05 + 0.079] kH
--· s<~ n
· _ V, X (r 0 II r,) ......... 0.1)1
X 0.95
Vo .... ----------····--···
-·--.- 0,005
r,.
. . -
• , II, -
If;,
-
r.
-
i,(!3 ' I}( 'oil Rl)
--~~---~----~-~-~~-........,.""'"---
'•
Ex 4.56 blank
For R,(k{})
G, (V/V)
0.5
0.68
1.0
0.735
2.0
0.765
1.9V
Exercise 5--1
Ex:S.l
!,.·
~V 2
2 L
In =
34.5 pFl m "' 8:625 fF I (J.1m) 2
4 nm
~
for all 1los
o~t
~
= 0.25 rnA
V 01•
""
0.5 V.
I~x:.S.6
v.~
I
>.
IJ$
5.15
1 kn
I. "' 0.18 ~llll,
so
w
0.93
v > v0!'
~Saturation:
~un
10
F.x: 5.2
2.30 fF!
I
""'
"'-'
!
X
2
40 V,
....
0.()25 V ···I
v,,
v
l
50 X 0.8
VAL
200
v
0.5
'·"
I 11
X
J.ll
0.8
X
0.5 2( I
40
2 V
L
X. I)
78.4 k!l .:: 80 kH
0.51
SO K
!!::'
0.025
0.51 rnA
jU111
r ,,
0.2 mA.
!
~"
0.!!25 mA
Ex:5.7
= 20
:. V 0 r = 0.40 V.
0.40 V. for saturation
l<~x- S~3
h
I
•
!{
~
2 " !f
/. V ov
"'
/)
in samration
Change in /,, is:
(a) double L. 0.5
(b) double W, 2
tel double 1~,. 22 = 4
V,, no change (ignoring length
modulation)
(d) double
!:!:
,_,
I.
lO ~ k
,.. 600 I-LA I V~
I'
.
(e) ch;mgt~s fa) - (d), .f
J
ease (c) wnuld Cl!Usc leaving saturation if
v
l)s
~...:
+ -i v.
2 v(}\'
Ex: 5.4 In
~alunuion
<'ns
?
V,v,
~o
2 V or
fb) Triode region oc.:urs for
J'u/) <
or l'n > 1',; ·! I
Ex: 5.5 V 01•
gVS
k',
~
w\' (}\
7~~
(c\ Conversely. for saturution
0.5 V
1
kn
(d) (liven A. 20
I. W
0
.
2~kJ1 --I
L I· u\ .
For 1'ns
IV 01 !
"' 0.5 V
75 Jl!\
ll,"
v.
Exercise 5-2
/. 7i(; =
+ 35 V.
s
+ I = 45
Vo
Vt;
!! = .o.n v.m
L
V,
X ,, 0
<~) f('>rk ""' -0.02 V~ 1
10
75
.,.
(I) At V 0
~A-Aand
""
r,.
tind
_L ""
=
=
.o.s
667
kn.
IVovl
!21/ f)
V.
0.8 V."" 1.8 -1 0 R0
VD ""'
··
lo ""
3 V.
1k~ yiVolO + IA!Iriosl>
lo ""
4.0
0.18 p.m
I
2
t.s-
F..x: 5.10
At V 0 = OV,
1,1
"'
r,
~
kn
0•8 "' 13.9
72 ~A-A
:. R '""
"" 75 JtA( L04) "" 78 j.t.A
, W
i~A-nc ... y(V D- v,n) '"' 72 p.A
+
1.8V
75 JL.:\( l.i0) "' 82.5 t-LA
AYns
3V,
Aln
4.5 tA.A
Ex:5.8
lo ""'
1 . w 2
:2""11 C.,,TV 1,v => 0.3
!x l.lO
2
JO(l()
From Bxc. 5.9, V Gs
V ov "" 0.5 V ~ Vas ""' V(w
1.5
+
V, -
0.5
v
+
I
V 0 1' = 0.3 V.
1u
=
12 IJ. A (saturation)
At the triode/saturation boundary
Vn "" Vov = 0.3 V
- 1.5 - (- 2.5)
0.3
l.S V - OJ~
:. R• "'
72 ~LA
•
= 20.8
k!l
Ex:S.U
Rs "" 3.33 k{l
2.5- 0.4
0.3
R0
""
12.4 X 2 "" 24.8 kU
7 k!l
Vm ""' 5 V, Assume triode region:
Ex:5.9
I X (5- I )V /1.1. -- -2· .
24.8
~ V~ 5
-
8.08 Vus + 0.4 ·.-, 0
~ V ns "' 0.05 V
\/ Ut
0.5
v.
v;,s)
(
5- VM
0)
V 1 "'' 2.5. V:
t<:x: 5.lJ
t\lmed on. then
In "" 0.32 mA
"""> Vov ""'
V liS
symrnetricul V., = 0 ruid therefore Vos"' 0 which
0.8
""'
Vs
I DN
0.8 V
+
-
1 'w
ik• L ( vGS
v, )2
-
l "' 1.8 V
+ V < 0.1(}4
v
-· 2.5 V: Again if we as> tunc that
tttrncd nn. then
Or is
r., > -2.5 v anu VI;SJ <()which
implies the Nl'v!OS Q,,., is turned nff.
InN'" O
1.04
Exercise 5--4
!k w(v ·";
l !Jr'
"~ / 111 ,
IU04 mA
·-
Ex: 5.17
1v . ·)·
2 "L
·~
V0
- 10
0.104
I.C\4 V
7'(;,v
Ex: 5.16
0
\1
'
0.4
v.
L8
\i[)[l
V,;\-
v.
0.6 V.
k = 0.4 mA:
"
IV
Ui
\'/)/)
17.."
Rn
v
k,
v·
1
17.5
.~JJf
k -L v("'
IV
800 1.u\ ;
tl
v
lilr A 1
()
(t\)
k!!
Rn
kn
1 mA;
\
10
v.
0.4
'
I.
v
v'
I2.'i kH
CutntlJSaturation Boundary
V,
1.8
O..+V.,v..
v.
011 Saturation/Trindc Boundary
1',
1
,k,f v,;, -·
0.8
\'/).',
0.4V.
v.
!h) keep Rv
F,f Rn I
1
0.08 rnA.
0.6 V .• I"
17.'i k!!
-10
'
c.'}
/i,,.
."i71
~A
rV
J
<'us
! l.l\
0.143V.
:.\1 01
T',;s
lp
{JI}
,.{J)
0.54 V. I"
v 1_,,
'!0.7 p.;\
!U}4 mA.
1.1 V.
Ex: 5.18
O.:'IJ V.
(C) For "U(;s)(
v()\'
v.(j\
·=· 0.61.1 V .• 0.1 'iR.'i
v""
1.8 V .• triode.
u v.
( k, \ "')
Vos\r
17!) l!
18 mV.
~
I \'rw=5V'
Exercise 5--5
V, =
0.7 V,
kn =
I rnA I
i1
.o;·
£!+;=E!+g
tJ
_
·
. nr · l
r.o
v"
r,1
Ex: 5.20
2
( !k
2 ;; Vov)R
VN, '" 5 V.
l)
v(;s '" 2
·=
V,
v.
I V.
A= 0
and
VoF "' VIJn ~ V,- 10 /l, ~ 4.3 ··· 12.5 Vot·
k,. "
20 f!.A IV~
:. 11, 1 = 0.319 V.
:~19
!LA! V
78.5 kfl
2V
o"}
\1 01•
~"
I V.
V OS
200 JU\
o + 26 vi :'.=. v,
Vrw ·- I,Rn ··-
\-' {)S
27 mV.
·W
k ,, v '"'
L
('bJ g,,.l
"'
+ J V
400
j.Li\
If OS + 71J, ~ 2.2 :•;
Vn:;
I
v
Ex: 5.19
A,,
(c)
...;+..
0.2 sin
Rcq
-o.s
V..r,
Vv~
¢,
......; G
i _ '- - · · · ·
V.
(t)l
sin l•>t
v
(c) Using (5.43)
i l_! .
!k ( v '•' - \'
·-")
JJ.
fr.)
~~
!•
k--"( \'--·{tS- \''
ik·o1l,!,,.
2
1 !)(,':::-~ -+ 2
ts------
iu
200 p.A
t ( XO ,.u\) sin hll
+ (!{ fLAlsin: ..,l
ce; 3.R V.
Exercise 5--6
"' f2(){) .,_ 80 sinl»/
+
Ex: 5.24
(4- 4 CQSWt}liJ.A
1n shifts by 4 JLA
1 ·w
·:zk"L(Vm-
It> =
i2
2HD ""
X 60 X
J!i
0.8
X ( I 6 ·- I }~
.
216 iJ.A
Ex: 5.21
!2 X 60
.
2lo ·"· 2 x 216
1.6 ~ I
.100 p.A
0.72 mAIV
0.5
0.3 mA, Vur
v
(}.()4
"'* v_·,
~' J '""·
h
1.2 mA/V,
v
-~
In
720 JJ.AIV
g,. "·" Vov
X40X(!.5, 1) 7
rd ·-
1
jV,j)
I
0.()4
,~
25 V /n n1
....
25 X 0.8 '" 92.6 kO
0.216
15
0.3
)(} kf!
Ex: 5.25
,g,:nr.p .-.-.-.-.-
,,
'""'iH
':':,,
In
15
0.5
100 VIV
30 kl!
Ex: 5.26
Ex: 5.22
1,
,,
'~rol
0.8 J.Lm ''.}A,,
L
!.55 mAIV
(5.70) J\ 110
l mA/V, k,
0.1 mA, g.,
211)
'
- . - " " ' \• 0\
2
X 0.1
'''···-··
=
l
tt"ot
'.'
·""
(5.72) R, =
0.2
-
gm(RIJ
II
r0
)
Rl> II r 0
v
21/)
k,~ F~_~t
....2...:::_QJ_
100
-~E~•• o.2·
1000
Ex: 5.23
"
•'>g., ·""· 10 mA IV
200 k.H
""·
0
g,. = 11
Vol'
Gv ,. .
2/ n
- ·--=>lo
'''
0.2V.
·..•.•
rnA
TJv
l','\"i,\!
' UJno
R0
"'·
Rn ll r 0
:::.
RJ> "'" 20 k!l
.>ttl "" Gv "" - gm(/?1>
!I
r11 !I
Rl) :::
- g.,(Rn II Rt) ·.~ ~ 20 V!V
for
iJ~,
iJo
"' (IO'JH 2V 0 v ·""
( Avt'#'l
"'
{)J)5
mA I V)(2
kHl
"'+10
Ex: 5.30
CDamputicr
1
100 H =>g.,
~·
10 rnA IV
.~!..!!
211,
- - => 10
0.25V
•··•
1.25 mA
V.
l V.
Vov
Ex: 5.28
0.91
1}(J
,..,_
T.l.~-·-
....
.. ,
v.
91 mV.
fJs (!~ ···~····'-·--····
I +g..,R,
Rout·::::~.
IOOH
;-;;~~ T . ,~ 0.9V
A.,,.
I
~ RL
I
Assuming V ~ ·-t
x
From Ex 5.27
Ex: .5.31
CD (sou ret~ follower!
50 mV
_!_'J:'
---·-···- =>
vfh_,
200 mV
200!! "'
R'
· "' ··
'i
I
:-;~ ~
gm = 5 mA IV
,"'>,•tl
:. N, = 1.5 kH
-g.,( Rn
!!
~2
' g,.R\
···20
4
In -
Ex: 5.29
tkn
I ·IV
L
,2
-k,--- \· 1)\
2
0Ji25 mA
Exercise
Ex: 5.32
~8
V:s-
R ·' -
-2-(-5)
Vss
6kfi
0.5
lu
4Rs '"' 6.2 kH
lf We d1t1ose R., "" R, = 6.2 kfl then 1, will
slightly chl\1\ge:
I.
11,
I
2X
"'
2
I ) . Also
X (Vas -
Vas ~-" - V,f '''" 5 ··- Rsln
(4-
2/J) .,.
6.21,i
:} 38A417)- 51.6/~ + 16
=*.I0 = 0.49 mA, 0.86 rnA
1" =0.86 results in v~ > 0 or v. > v" which is not.
acceptable, therefore 1,. =0.49 mA
V,"" -5 + 6.2 X 0.49= -1.96 V
V, = 5 - 6.2 X 0.49 = + 1.96 V
g.. ra
+ gmr<)
I
0
R" .should be selected in the range of I MI 1 to
10 MU to have low current.
Ex: 5.35
11> "' 0.5 rnA
0.5 X 2
I
= 20 V.
frir tt,
v ov
+ .:..!lY
0.40
4(!\',
2
X I ( V 1,.
'"'
-
v.
o~
=> R"
6 k!l
0.5
6.2 kfi standard value. Forthis R,.
we have to recalculate / 1,:
Ex: 5.33
1 ·w .
,
In ... 2k,T.('-'Gs .. tt,r=>0.5 mA
_I_
5-2
"' v":} Rn
A 1~() = 0.99 = -~":-.- ~
"
I=
l )2
10 = ixlx(V 1;s-l) 2
"" lI ( Von
If
- Rnl n -
'
(V(;s = l'n ~ V 1>t>- Rf)IJ>)
l'r;s -" 2 V .
I
2(4- 6.21 1>)
Ill =
:.
'"* 11, :.s: 0.49 mA
IF V,::.: 1.5 V then;
1,
!
~.-
X
I X (2- 1.5) 2 '' 0.125 mA
d/ 0
0.5 - 0.1.25
1,.
0.5
::::J>-
v
[J
"'
0.75 = 75%
I
-"~
111
f<:x: 5.34
Rn ::;::-
\•
~
1:
~~~_._!2
ln
5
~·-
2
6 kH
0.5
=> 111 u
/llf"F
I
...
2
v
....
I ·'.
X .,~
y (/\'
.
5 -· 6.2 -~ 0.4? ~
1.96
v
Ex: 5.36
Using Eq. 3. 53
rrR: ~ ~~; ~ u '·'
I 11
"
=> Vu '·'·'·
X
~
O.l mA
0.1 . . .
(l.l .X 2
0.8
0.5
w) \! ,,.
1 ·( -·
~k,.
..
L
.1
.1
~ \·tit"
I
0.5
0.25 => V OV
3.5
v
v
Exercise 5-9
" = f~-~~~
""'
v, ""
V ns2 ?.:· V ov ~· \lDS,•i• "" Voy ""· 0.5 V
~
-45
vl>min "·'
= I mA/V
150 kfl
In
v
Ex:5.37
R;, .., Rc; '·" 4.7 M!l
v,"" 1,5 v
'w
.
l
kn- ""' I mA IV
R., 0 ,
VA = 75V.
10
""'
- t:.,RJ) = -15
APV
L
For r 0
0..5mA
1.0 V.
Rh ,,
,.,
R;,.
""
1.5 k!l
Auo
-g,(Rpll r 0 )
R.,.,,
Rp II
G~.
··"''
'" "'
R;,, +
R,;~
flo "'·
b. a 2.8 V 1• •
lf I>S +
,.
a 2.5V de voltage.
Ex: 5.39
-lOV
v(JS "" v,
II,.;
0
Yov
v~
~2.5
v.
lin
Vno
·~
2.5 v
lnRn ,.,
+ 2.5 V.
I mA' V
.Y..o!
//)
- Vx~t ~
Ex: 5.38
1~'.1
=
tso kn
4.0 v.
f)
4> (fR, '"'
gu/~!..'J~
i I)K
-
·""·
-13.6
13.6 k!l
-~A\'O
11 fiS '~
1.10
!5 k!l
150 kH, R 1•
4.7 M!l
R,,
R,_ + R,,,,
-7.0
li,J.,
sinusoid supcrimt)(Jsed upon
Exercise 5-1 0
Ex: SAl
_ _ _--:g"'roRt>
(Rti + rt)) ! R5 {1 i g,.r0 )
Rs
"~ ~(Rr:..!" r~ •""
1+
2.185
1-g,.roRD
3 R0 +-;;;
k!l
gmr(J
..
_l·
15 K. ru '" !50 K,
Ex: 5.40
g,..
I mAl V
r0
150 Ul
(a)
g,. =
50 H
gm
R;,
l kH
-~-----~-------~
oo
r,
=
150 k!l
4.7 Mn
4.7 M!!
1.0
0.993
----·-4------~--------
A.,
+ 15
gm! f< .. r/IF.mln}.
1,
~VIlE 1 VT
~'-,.
85.5
~tA
{'
J. I mA
- Q8 fi.A
3.1-mA)
V1. In ( - an d "•r
0.8 ~tA
(25) mV
Ex: 6A.2 f-or an NMOS Fabricated in th.: 0.5
= !(), we want to tim! the
L
tmnsconductancc and the intrinsic gain ohtaim~d
for the thllowin~ drain cum:nts: (/. 0.5 ~lin)
•
. . :tJ
f.l,C_,,"' ( 190}
J2 X
g..,.,=-
\1,
,.
_..2
J
r:
,q ...
J
I
___ v
0.57 mA
..r r .
,....__~-
2 rr (C,.,
t-
Cg. 1)
211' (8.3 + 2)
8.8 GHz
p.:;
v·
Ill --
190 X 10 X
V 'L
0.2 mA
v
20 >< 0.5
= --~1 -=----'" I MH
ID
10 f.U\
For/,= I011-A.
~" 0? mA
I 00
I MH ~ 200 ~
·- v
X
~li\ we
haw
v
g,., · ·
./2( 387 p.Aiv:~)('l())(~~
0.28 rnA/V
m,ing .:q.r6. 15
A,,
"
·"""'
A,.,
.Q;~
20 X
- 117 kH
85.5 fJ.A-
,~'='·'""""''""·-·---
/1, r.t c. (IV)
--I)·
L
"
lo
intrimic gain
..
fu
Ct,
~tm process, with !!~
1,, = (10)' JLA.I?
!'.~'.t:. · · ·.
"
"
A, = g'"r" "'' 66.7 V/V
= 207 mV
=> .G.\!liE
r
v, p;~;.~;~~f:D
·-·J
.Jili
0. 62 '.1_1:'~ - 0.6 m A
v -
v
20 X 05
100 fl ;\
.. ,. 0.62 mA >< 100 ldl
v
A,
100 ld!
50 V/V
Since g, varic~ with Jl~, and A 0 with
li2
vi v
.,!1;,
Exercise 6-2
Ex:6.3
For
I
lv
=
m(I~Ji
100 p.A =}g,, "' 0.28
.89mA/V
!.
A1, ,.
10 ) 1
so(-_
.I()(}
For / 0
""
rss VIV
···"'
I mA:
I
g,., "'
· 1 )2
.28 mA!Vh)iO
A{J ,.
so(-0: 0 Y= s VIV
"' 2.8 mA/V
I
I,.,
= I ,,,
J()O ~LA
In ..... O.I_J!.:A ~' 4 mA/V
Vr
25 mV
TJTJC"<
.1!_1_. "' __!QQ_
4 mA/V
g,. 1
' 25 k!}
~ "'"· .J~!..Y . soo w
I
0.1 rnA
IV,l '' . ~?.~~~- ..
500 kH
I
0.1 rnA
A,,= g~, r,, 1 (4 mA/V) (500 kHI ..,. 2000 V IV
A,······ g.,, !r" 1 II""")= ·· (4 mA/V)
(5()() k!1 !1500 k!l)"" 1000 V/V
Since all transistors have the same
!!:
I.
0.36
J,Ltll'
we have
lm~•ID2
/KEF
itil''
0.55 f-LIIl, and
?
f:x:6. 4lf Lis halved: L
= 1"-~~
l()()~tA
jVAI ''
jv,;j. L,
! V,J -~
5
VIJLITI-t9.:~;J: ()l) - 1• hnl. ··· ----...
2
1.55
(c} V 5 m
V 0 ~2
L8 - 0.2 ·-·
!
"
(().1 )
v
IV> r,) then,
R.,,
so that r0 t and r04 are >> r.,
Tb¢n; R,,n ""' (g..2r01 )r~: ""' ~ 2 rm
Al!ll1n ""
«.,. ·""' (g,.~r,12 )(r01 I! r..:~)
If Q1 and Q4 can be selecled and biased so lhat
R.,. "''·
Av_(lee~Wflen Qi and Q4 are selected and bias
v,1 ·"'
R,, ::: I0 k!l[ I
+ 40 1~(0.5
R.,::. 176.7 kfl
g.i!(~1rm !! f33r1n)
without R. (that is, R., = 0).
R, = r0
Ex: 6 · 12For the npn trasistors,
!Jd
jV 1!
0.2 mA ""ll mAIV
25'mv
100
"-' 12..5 k!l
=
JOkH
.f::X:6 . 14 Fig.G . 2l(a)
~'111
·"'
'o2"' 'u =
llmA/V
r
·-- r
o~
o1 ·
·- Ir '1I
--
I'd
5 v
0.2 mA
..
25 kfl
!'
~-
{)nt
1.67 MU
Mml"'
r"'r the pnp tmnsistors,
0.2mA
gm.1 ··- J.:m-4
r ,_1
'·"
,. 0.\ ·coo·
r ~• ""
S mAIV
25 rnV
1!. "·' ~::1 mA/V
g.,
W!
~
4V
~'o-• "~
·~·= ·-·;:;·-·ild
fL mA
R,.r " (,W"" 50f.l.ID
1 JJ.nl
So the dimensions of the matched transistors Q1
and Q2 should be changed to:
W •• 50 JJ.IU and L= 5 f.l.lll
I~x:6. 16 For the circuit Figure4 · 7we have:
(Wit.)~ I
/1.) 1 ' '
I
J,
•
and I
o----r
Since all channel lengths are equal
L 1 ~-- l-2 _, ••• =-' l. 5 ~' I J!.nl
)
(llld
Q!
.....,
/REI'
~
50kH
r~.>l
-~
4 rnA/V
R,.
~·
(I mA!V)(50 k!l)(4 mA!V)(50 kHi
(50 kH i! 25 kH)
I
1 .w~~w,
u~:~-w,-
·~
"
Ex: 6 . 1 51 n the current source nf Exampk 6 . 1 5
we have I 0 ~ !00 tti\ and we want tn reduce
80 JJ.A.
'"'
-- 60 --- 6
\()
/REF
w!
w, ,~ w~ ~ !2
,w" tV 4 J_,
I
- 1-l
2
r.
/l
20
/REF
10
~ =·
=
c ,.., (lV)
-.L ::! v 21)\' 1
10
~
( ~)
L c
v- -l..
~
120
200 >< (0.2) 2
b
l\' J
IV,
I '(W)
,1
-k
2 rl ~
L-:; \1 01' 2
-~·
IV,
Jllll . ......:: =
w,
6 c-7
2•;W_, ,, 2>:w,
w,
~
5p.m
}!O
ro~
X
! 00
1'01 ,_ 1,
~~;\
IV
Mn
I !'A
I--SO [.tA(IV)
--,w -- t0.2!.2 ""'
2
V\ 2:_!:
"}
I \H}
1
a~ tI1al
20 ;< /_
L.,
XO X (0.2/
lOUV
~0
v·
2 X SO
lOll 1u\
(I
ol.
4
In order to allow thl~ voltage at the drain of Q5 to
up 10 within 0.2 V nf positive supply we n<.'ed
That is !!.I,,
\Hll
2
~200 ~(~) (0.2)~ ~
60 v.A
the change in output current ,)./,) . .:nrrt~spomling
tn a l V change in output voltage, J. V,,, to I'k nf
.!2.
1
., -668 X 10' VIV
~·
20 IJ.A •
In order to allow tht~ voltage at the drain of Q 2 to
go down to within 0.2 V of the negative supply
voltage we n<.'cd V 011 "' 0.2 V
1?,, '" 167 lV!H
-·i mA/Vil67i\-Hl)
I
l,
/,
A"'' = -g, 1 R,.
w,
I IH'l'
1,
Prom Fig. 6 . 19 R~;
w,
w" ""} w~
I'
r,. ,_, 25 kH
r>~l
G_ :: M~t
/~ =~
1., " I, "-' 20 1.1.A and
,, -~ mtVV
g,. 2
10 11-A, 11 ·~ 60 p.. A, / 3
'
we have:
From part (a i.
g,. 1 - I mAiV
g,. 1
(WIL),
RHI'{W I L)r
e' I (WI L),
4(W /1.)4
s
I!;
I
«E!'(\V
50 p.m
V!p.rn
IV
.
-
n fE· xampI'~ 6 . 1 5 "i o! the trans1s!or
-;hnuld remain the sanw. Ther.:Jorc
•f ."' W.J ,_ 5() JLI!J
4
Exercise 6-7
Ex: 6.19
See next page:
J<:x:6 .17 From equation p. 72 we have:
'·
~
'"'(, +
~~· + v. ;,~")
r~
1, = I mA[·-I+L:f::J,
. I .)r I + ~ -I!)(Q_l.?_)·
1.02 mA
100
I" '"' 1.02 mA
R0
\!
100 v
=rm- "'-..:!
=---10
1.02mt\
1\x:6
,
=98 kH - lQO kH
-
.18
v,."
•'
From equation f . 7 4 we have:
In
~i-i~':, J1 J( I + ~~'~-~AVrn) ~
0.5 mA =
.. /R.·. a~--.
--.(.1 ; 2-500.7)=>
,
I ; (2! !001
0.497 mA
f Kl'F
R
.2=_ O.?_
OA97 111A
43 S.65 kH
0.497
1/""''" = k'u;sxr 0.3 V
F'or ~:, '" 5 V, Fmm equation P • 7 4 w(•lmvc:
t,
'··
j:/{~;>~~)( I
+
\/::.i:::;~~!f)
0.53 rnA
Exercise 6-8
Titus:
~Ef'_,
I+·~:! .. ~
Ignoring the effectl1f finite ()utput rcsistnnces, we
have
It "c. lz "" ... '"" l,v "" lq!II.H•
'CQREF
+r
I REFc~)
=:
I "" I fiQRU' + I Itt +· ...
13
'~REF
J
,e.
I.,
1.-.!n.r
i
-~(~
N + l
11
From{*) we have:
lcvREF +I ,., /RtF~
/Rf'!'
.
-·-·N· + 1 ?> I REF( I - 0.1)
1+-13
+ I !>N
leN
",,. (!f1
13
For an error not exceeding I O':f· we need:
j
~)
.-...
- , 1·
N
_!__ -~
' !;!.:±:_! .c.·c_. ---t\
0.9
>
,}·_! s
0.11
-~•:> N +
I
N i I
+ ··--·
I ,;; 0.1!
ll
~<;
1. I I
f) •~}
N + I cS II ~ N cS 10
111c maximum number of outputs for an error not
exceeding to be lessth:m I 0'.>;, then we need N< I 0.
In this case the maximum number of outputs for
anderroroflessthan 10'!{- i5N=9.
Exercise 6-9
whereas for the simple rniJTOr from equation
Ex:6 • 20Refering tQ Fig, 6 • 3 2
lm ""' lnz '"' lm '"' 104
'"", / 11 1!!'
""' 100 ll.A
p.69wcMve:
, t ,. (WJ'',v2
v "" ill.•"·"• L ,ov
_.1_
S . , 1,
•. mce
l
+ -~
f3
= 0.98
1/ - J REP!I
Hence "-"
X
100 ~· 2%
/l!lll'
For the Wilson current miJTOr we have
=023 v
'l1te minimum output voltage is
R. '"
~l
V,.+ 2V,. '" 0.5 V + 2(0.23 V) "" 096 V
Tb obtain the output resistance, Rrr we need g,.3.
g.,.;"'~ ~ =
Vm./2
2(0.1 rnA)''"'' 0.87 mA/V
0.23 V
,, ,. • ~= V,1 (L)
r
01
(?_Vi Jl.1U)(0.36~
0.1 rnA
,
1,
(h
1
·" 282 kH
Ex:
For the Wilson mhror from tht' equation
16. so·we lmve:
I
J
-i- - ..-.-:___ _
"" 0.9998
t!..::.- lnr.FI x
1()0
2
the simplcrnirror R., = r., ""· 100 k!l
]';x:6 • 22 For the two current sources designed in
Example 6 • 6
we have:
lc
lO f.!:A "" 0.4 mVA and
g~, ··=·
25 mY
Vr
For the currcnl source in FlgP · 3 7 awe have
It, "" r,.,_ '" r,. cc·., 10 MH
For the current source in Pig.6. 3'tb li·mn equation
6. 9S we have:
R,,::;! I+ g.,(Rrll r.)lr,.
In Example 6 · 6 Rr = RJ
11.5 kH,
therefore,
(3{(3..;.. 2)
'11\us
2
ro
18 k!l, From eq. 6. 77
R~,~=gm_;r,,1 r, 2 •·= (0.87 mA/V)(18kH)
~-S = 100 x 100 kn .,.,_ 5 M!l and for
=
OJ)2':-f·
r?,_-
[t , o.4 !!!~nu
v . kn
/REf'
=}
R., = 54 l\.28 V to 1.0 V
().4 mA( 2 ) ,~ 4 mA IV
Ex: 7. 2
~-o:zv-
(a) Tht~ vnlue of <~0 that caust~s Q1 to conduct the
entire current i~
·t
J2 X
2o v
J2 Vuv
0.3!6 • 0.45 V
A 4 = g,.(Rn Jl r 0 )
An·=- (4mAJV)(5KII50K)= l8.2V/V
0.4 X 2.5 "" 0.5 V
\I'm "' \! 00
+ 1.5 V
·"·
Ex: 7. 5
!b) For Q2 to nmduct the entire cun·cnt:
•·id
= ... J2
V 01•
-o.45 v
"'
With I '·" 200 JL<\ . for all transistors,
)/) "" {
2
then.
L
i !5V
~"m
- !.5 - 0.4 >-.. 2.5
~
Vn: -· 11 1>;: from 1.5 - 05
= ·- I
ro~
Since I Hl
Rder to answer table ft}f Exercisc7 . 3 where val·
ues were obtained in the following way:
JnK IV.· L --'> ~
.
L
=
2
0.36
jJ.Ill)
..
ru:;
"
(!!')
L
=
I
KV,w'
·•
(~)
·.o.
lo
f.llll)"
361d!
(".. ,,.
(w)
T 1.
'
'o\·'
J.L,~C,.,I V(!l.) 2
2( 100 p.Al
(!f)
!, . '
I~
v I
___2_!J,__
'"
L,
...
21 1u \
1m
~-tA
JLI11
rn1
U.!J-~:'J~Ill) (IU6
0.1 mA
V
Ex: 7. 3
V,il
-
rol
' l V
= 200 JLI\ "" 100
2(0.18
0.5 V
(c) Thus the differential output range ib:
to 0.5 · · 15
50 k!1
.c
OAmA
then.Vm""" Vrm-IXR 0
= L5 -
v
12.5
(!f) .
2
,L "
2( I 00 J.ll\ l
50
(l'fm 1v\ i v' i(0.2l'
Ex: 7. 4
I
O.S mA
2
2
l .('IV'
-k
• )1 V.
2 '·. L.
........~.i!."..
l\' 01 i;·,?-
0.4 mA
. ! I00 J.Lt\ )(_:!.,)
0.2V
I mi\; V.
so.
' So thm
1 )'
" .
An ""
~
gml(roJ
18 y;y
II
l"o:d
I( 111:\/ V)t36 K II .<6 Kl
Exercise 7-2
Ex 7.6
L
=
CMRR(d8)
2(0:18 JJ.!tl) = 0.36 Ji!tl
All r0
_
'!
IV,,i•L
'"
·--
~ ~'
20 log 10
(0~ I)
Ex: 7. 8
From Exercise 7 • 7
2002 ~ ""' lOO JJ.A
W II. '" 100 ,JL.C,,(0.2 rnA I V2),
fo ,, ~ ':" 0.8 ;nA ~ 0.4 rnA
36 k!l
Refering to Fig7 . 12 .. ~:· ~-
(h} .li,,,
-:=
5 -- 0.683
IC
:·c·
0.5 + 20
,, 0.5
I
_05
VT
4.317
i
·o;
v
~L~!
JlX}()t)
(d)
-
= (15 -- 0.5 X
feR;) - 0.1
c'(
ReSin( 2'lT
"n
10
+
X
10) · · 0.1 X 10Sin(2li f(){Klt)
"' 10- ISin(2'lT X IOOOt). V
0.25 kH
ll}() K I ((){) K - 200 VN
0.25 K
r,
0.1 Sin( 21T X 10001), mA
( v{ (.
25 mV
0.1 mA
/~'
IA,J! :::
in "' 0.5- 0.1 Sin(2'JT X 1000/), mA
t•n '
~I
Since R, >> r,.
v
x 0.005Sin(2'lT x
r,.
r,.2
20 '-'-~
0.025
50 kU
200 JLA
10001)
R;"
2r #
R;J
2(25 K)
r"
Ji
gm
···-·
100
4mAN
25 kU
50 K
If the totnlload rc~i~tam:e is ;~ssurncd to be
mismatched by I'll·.
IA.,.I
ARc
1Sin(2rr X \0001). V
(c) r•n ·· l'cl
(fl Voltage gain
cc
2 · Sin(2'lT :-.:: IOOOt), V
CMRR(dB)
·~
20 lm!
~-~d
I "' 20 lo"
~It! A,,,
el!l
1200 1
0.01
= 86 dB
Note: lf only tht~ load transistors are mismutchcd.
und since n::: I.
Exercise 7--4
( 100 K
II
100 KJV;,-.,
--2(5(fK-)-(}.25 K
Ex: 7.15
From Eqn.7 .127
,
v ""'
--- 0.4 99
\lp .. -[(R,~-~--~Rr Jil_'j:l\~'-'-"'
t
2REF + r,.
.t.
!t 1.01 l 101) Kl !I tOO K _ v,,,
2(50 KJ ; 0.25 K
C:\IRR
'lA,~~~!
•l·/~
25)10.021 2
(ILl (
2.5 mV
0501 li,.,.,
f3 ;
o so 1 -- o...J.!J'l
2 :.,-: 101
I)
:: 0.5 1-l r\
I ( ~13.)
fo,\
I 00, 000 --> I 00 dB
100
100
1/1
iA,J
, '~I ')2
'(if~,)'
\r,j_~····
"
Using eq. 7. 103
0.5
11
50 nA
0.1 J.LA
X
t<:x:7.16
20 m
0.2 Ill X I 00
'-0.8 mA .
From Exercise 7 . 4
v.,, 0.2 v
Using Equ. 7.108 we obtain\',... due to t:.R,, I R,,
as:
(~R'')
( v~").
. Rn
0· 2 ;< 0.02 -
i.e 2 mV
gm
J'!.
-
(0.8
mN2).
.
0.2 v
\im·
(Jm
g,,l
Ro
ru:t
ru,
' 0.002
--> 2 mV
~ Vt is obtained
from Eqn. 1 7 . 116
il
~-··-~-·-
20
t(Ul m/2)
50 kH
1/)2
~/..c'.L:
1/) . 1
(lUI m/2\
- -20- -
50 Ul
50
50 •·· 25 k l!
Ad
"'
·I '!!~
V
C1m R (}
Vu..,· :::-
CMRR
-
~---
J.46 mV
X
i (2 X !0
1)
Y.
25 kH
!00
~
Prnm Eqn.i? .148a
A.....:-: 2g~,',u:~
. . . . - - = :10; 2
- J3 (2
.)
v
ro.!
v. . ,
t:.Vr 2 mV
finally. from r::qn. 7. 117 the total input off,el is:
1
v
From Eqn. t7 . 141
'Inc offset voltage arising from
.j(2)< 10
·lm~
IIlii'>.
R0
0.02
4 mA;
...
v.\11
Wi/..
)( ~~v / _z.)
( .:.~~~.
2
w! I.
(~)t) X
thus.
~'o'J
To obtain V due to ~!Yc.L
"'
lbc Eqn.t7 .113
0.2 v
20 mA/V
From Eqn. ,7 .138
0.002 V
2
X
v
J
Ex: 7. 14
v,,s
-
Since all transistor' have the same drain curre!ll
(1121 and thi.' mun.~ product W II,>< ~-tC. then all
tnlllsL·nnductance~ g . arc identical.
R1, , : : 1.6:::\ MH
Vm
v
Ill~
'1()
(W/L)p X fJ-pC .. · OJ Ill X 200
~
tl005 V/V
100
0.005
20,000
Ex: 7.17
From &Jil. i7- 156 U. ~ g,.,
II1
g.,,- .,..;--
'r
(..QJ!__I!lA/2)
15 mV
From Eqn. ,7. 15 9
16 mA
v
Exercise 7-5
., v.\ 11
lc2
v ""
"' 100
Using Eqn. (7 • 17 9
v.~ _!~
lc4- 1 /!2
0.8mA
1 - (WJL),
1 - (W! L) (I)
~
12Hfi
=> 100 ·"" ( W !L)-, X
200
thus, (WI L).,"" 200
(b) For Q,,
From F..qn.l7 .162
N,,.,·2X r_,.
I .• I
-· 2Jl.p C"' (t LW) 1 "•m·i.,
~ 2 X _!L_p "" 2 X 25 m X 160
-
U; 2) ''
(O.ls m/2)
For a simple current mirror the output resistance
(thus Rr,) is r,
_v_a
tOO V
0.8mA
I
l"'n1m Eqn. !1 . 16 7
j=.::=~so············=:.
=> vov1 =
""20k!l
~
125 k!!
2
(c) g., ""
(J.J05 V
1.0125!
160. ()(}()
lo
\!·¥·
g,.,
Q,
501-lA
0.129V
0.775 rnA.V
Q,
501-lA
0.129V
0.775 rnA/V
Q•.
I001.tA
O.IOSV
1.90mA/V
104dB
(dJ r,,,
r..
Ex: 7.18
1 m:2 = 20m~
25
,.
~
1/2
() ~
-7
Ro4
' j),,r,"
Ill
v
IOOV_"'
200 kl!
0.5 mA
=
50 X 200 K "' 10 l\1H
!0/0.05
1;,. '"' !0/0.1 • lOOk!!
li ·o. 10/0.1" lO!lkH
(e)Eqn.l7 .176
A, ""
-g,.,
~ lr !1" ~"o-• H 13< ~~ l
i\.o
g., ~<
R,.
.
~
200 l
· 77.5
'!.v
.177
.:Zm() { r(J(J
n r(n)
Ex: 7. 20
'5 MH
20 >< 5000
i.e. !O()dB
-
l!
·.· ... 95VIV
Overall voltage gain is:
A, i" A, "' 775 >< 95 •··· 7361 V/V
From Eqn. 17. 174
riO!! 10) J\1!!
lro~ll rw)
= -0.775 CWO
/\::
"'
200 kH
= lO 10.05 ·· 200 kf!
Eqn.1?
R.,
?!.I~
v(,..
v
!20001
.,
ov~
·
-· .!ll25y
20 log 10( 1(;(). (ll)O)
v
0.129 V
100 '" .\ X 9() X 200 V
--2 X !15 K
CMRfl
"'
0.129
200
X
For Q,,
j"(;(i": VOYb
. - \ ( ). Flg.7 · 42 all/,. values an: rlw same.
If}' V/V
'0.
1',,.,.
V,
I /,J?,.
Using the equation developed in l.ht'
t~xt.
Ex: 7 .19
Refer 10 Fig r7 . 41
ra) IJsin~ Eqn. !7. 17 8
'"
=~
rlV.-·n.
--,·---'
r I ! 2l
rH·
!L),,
!00
(IV f.~:l._.>
thus. r W / /.) h
100
200
·-~
( @)
Y
\
)2o ···I}
50
Hl\0)(10 p.A) ...
5.:'7 kH
Exercise 7-6
v.,..,,"" v,.'S,l- v..
vetS.:= o:~.s + o.1 ,, o.ss v
+ I•Rn- v...
""0.85 + 0.15 -- 2.5
=- !.5 V
thus, V,;,.u = Vt;.t,!
Vov
11 ""
I"ov~l
~ Vr;s il ""
V1;
Ex 7. 21
0.3
""' 0.3 V
+ 0.7
''' IV
LS + I ,., ·· 0.5 V
11 "" ••·
Finally.
/1,""·90~A
Vn•""' Vm;-
J.t.C:., =
160 p.AIV 2
J~~
=0.6mAJV
""
~s +
101
Since!!.. of Q,., Q., and Q" are identical toN~ of
Q. and Q, then:
15.7
15.7 t 303.5
v;,... """
(t3 v
Thus. forQ ..
(0.3 )~ ""
4
2 X~~.!!:___
160 J.t tH·; L) 13
3
(W! L!u ,, 12.5
i.e. ( 10 I (J.!I)
Since Q, is 4 times as wide as Q,. then
( ~)
/_.
= 4 X 10 ,. 4()
2
,.--:-(\v) /2v-..( "' - - I
1
"'
-
J- .
2
·( f{W!L);; __ 1 )
~tw:np
214.8
100
N, + R2
=-
f3:
;,
'"' ().()126
0.8879
100
Thus the overall current gain i;:
"
_( /40 ! 0.!1 I )
tj 12.5 ....
.
2 X 160 f-1- X -- X 90 IJ·
•
f
R 1 -+ R2 ..!. Rr1
i .. ~
B
40-
13~ =
ihs =
0.8
0.8
l1
-t R 11
0.0492
~
= 101 X 0.()492 X 100 X 0.0126 X I OIL
jI
X 0.8879
0.8
X
I00
"" 55993 tVA
and the overall
1.67 k!!
The vollagc drop on /(, h:
1.67 H! X 90 ~LA • 150 mV
0.15
v
voltag~·
V0 =
!?_!~ . it-~
V;.J
R"
gain is
i1
. " __;~__ )( 55593
20.2
$256 V/V
Exercise 8-1
Ex:8.1
C8 ~
~X 10 X I X 3.45 + 1.72
=
10
XlXIOK
tO+ 0.1
;\., '' - 9,9 VN
Cgd
2
.f1',3
0.016lh
l
I
2;rX I J.L/2 m
,. - - - - " - · - - -
+ R0 )
2trCc2(RL
C,n, ·-
318Hz
27rl J.L X (10
+
10)
R
"''
40-m~ ~) l
r rr
=
2.5 kH = j[ •""> f;
r
'
"''
Km
25 mV
l mA
)
!K'(~)
2 " L "
40 m :-:. 25
=
Ill
1 mA
t
In
=
cJ, =
zs n
2) 1
1800
Ex: 8. 5
Jrm
100
= 5V
(V h-V )1
"'"· !2 X 20 X 20(5• · · ·
"
4.1 fF
2(::6 l
Ex: 8. 4
I J.LF
v
Jt ~~~·
n•
6.1
10
Peak <:urrent occurs At V1 = V,lt
i Peak""'
.h ::::.ln""' 318Hz
"'"
J;-·~· ~~~
C,u,
""'.8Hz
Ex: 8. 2
Cc1 "" Cj;; "'. Cc;
10
Vo
··-· -::------=-1-:-c----,.--o2tr X I t.t X ( 10 + OJ M)
24.72 fF
1.72 fF
Jt ~'Yss
l
2trCc 1(Ra +· R.1v)
2trCsl g.,.
= C0 v =
c.h = c,,.,
.f/'I = . .
f/'2 ""'
2
JWLC.,.• + C 0 v
""'
~f~ ~· A~~~
T.f·
·g..,
=
20
"'
40
11~/
x w- 12 x 40 x
Jo··3
'"' 0.8 pF
2 q,., = 2 X 20 Cz 40tF
C.,. ·" Cdc+ C)r = 0.84 pF
cje =
~ 2tr IJ!IIOO K II 2.5 1('!~5 KJ
frr
21.4 Hz
20 fP
'
, !OOK~l
2rr · I 11{25 .
.f..,
J
fr ~· --·~"!-2tr(Crr + CJ.L)
40 X ((J·.I
2rr(O.R4 + 0.012) X 10- 12
2.2 KHz
=
f 1, 3 = ')~Tr'
c·C2 . (R C ··!·· R1•. )
l
2rr · IJJ.(R K '5 K)
12.2 Hz
f~,
Ex:
JOI
8. 3
I" ·I - fr
f --~ 10
}t•
.'.45 x 10 ·lF / m2 = 3.45 fF 1 tJ..m~
C 0 1·
WLm.c .. ,
= 172 fF
10 x (J.05 x 3.45
7.47 GHz
-
.c.
Lt.
50
::.:> fr = 500 lVUiz
fr
·= 2:·1~ ..~. J.Q._
10 X 10·9
=
Ex: 8. 6
2rr(Crr
··II
t,.,
12 fF
Crr
f Cj!
+
CJ.t)
.. 40 X 10
2·;r X 500 X I 06
Crr=- !2.7 .. cfJ.
12.7 pF
12.7-2= 10.7pF
Ex:8. 7 Diffusion component of Crr at/, of I mA
'"· 10.7-2 R.7 pP
Since Cj, is propo11ionalto I c. then:
Exercise 8-2
ci" ·= 7 + r o + 40 .x J0-3 + 1.s + t03)
C4 (It:= 0.1 rnA)"" 0.87 pF
Orr Uc
= 0.1 rnA} = 2.87 pF
f 1·Uc
=
0.1 rnA) =
= 68pF
g,.
27T(C1t + C~-t)
:=:)
I .
= 1,42 MHz
211 68p · 1.65 K
fH ,.
-.,..-4...:..-~x I,..::.o_-.~~-
21r(2.87
·~·
+ 2) x w··ll
Ex:B .11 Using equations a. 61and a. 63 we can
write the general fonn of the transfer function of a
ditect-coupled amplifier as:
130.7 MHz:
Ex: 8. 8
A(s) = __A...,:v~c'--- where Af)(. is the DC gain
Rr;
A,., ·'·"· _____
., . Rt,
"'
R-+R-'""
·
(,
~Jg
1+-s211'/u/J
ofthi! :unplitier andf~8 is the upper 3dB
frt.'<]uency of the amplifier.
In this ca.o;e we have A,J(. "" 1000 and
f.ldn ""'· 100KHz= 105 Hz
1000
Therefore A{s) =
I+
s
R~. ~- 7.14 kH, g,.. = I mA/V
Rsig"" 10 kfl
AM""·- - 4.?Mfl
(4.7
+ O.ol)
MH
X l X 7.14
= -7.12 V /V
21tC 1 .(R,i~l !Ro)
C;,.
I
3.7 MHz
:r;;>< 4.26( w K n 4.7 M)
fn
21T X 105
4.26 pF
Ex: 8.12
For this amplifier we have:
i\M
H(s)
Ex:8. 9
=('I+ 2-)(1 + ....£)
Wp1
ex.• == I pF
C,.., "" (I + ,q 111 RL)CK,1 = (l + I X 7.14)
A;,
. . )!2
IH( .fuln
= 1
s.l4c~, 1
c~;~ =
Wp2
By definition at ro = lo, we have
=:)
"'
fr 2c. I MHz =:) - - -1- - - t?: I MHz
21l'C;n(R, 1 ~11 Ru)
C1n
= C~s
h(l
+ Ceq "" I pF .;. 8.14
f~,rm
I
+ 8.l4C11,1)pF(IOO Kl! 4.7 Ml
?
I MHz
I + 8.14C"''
C~~J ""' 0.077 pF or C711 "'f 77 tl'
1.63
=:)
?
[I+
c:;Jl' (:;:rJ
+
If ron"' Ktop 1 and uJ11
[ 1 + (·_o.9wp
(Dt\., = -39/2
-19.5 VN
- 100
2.5 X 40 · 10 'X R;_
100 -~· 5 . 2~5~T"()~{>:1 r ( I 00 !I 5)
r., !! He
R,~)
1.5 k!l ·· ( 100 ji S
..-> R1
(J)f 11
=
.....
If w11
=
2
= 0.1
k!!
( 1 + 0.99-) I
0
I
1.9 k!l
. · R,;, -· 1.05 kH
'
2r.C,., · R,,r
=:)
K ,
0.99 <.o>I'J, then :
R1
I? I.
· · - ·1
--
=
(o;r " ' ~.~ =(o;y
h>r 1
1.5 k!l
7.4 k
;r)
, +
2
}( WpJ
[ I + (~.99wpJ )2][ I + (
~·
K~"~~[I +(uly)ll[t
+(·<:!!1.)11
J
hll'l
I
w"'
I
0.64 ttl1., (exact value)
.,, ) we
(note that in this case the zeros are at S
have:
to>H "" I !
l~ I , +-:~-~
A
,.,.
1'1"
"'
l I. [ I
~(J)Pt2
~
+ }
K
{0/'12
"" 884.2 nsec
Rs,1CnJ ~- (R'
+ u;_ + g.,Rj_R')Cif,1
'T 8 ,, '""'
'Tgrt -~.-
803.4 nsec
Tt,1 ••
..
r-~.~7··- ~-
J >.~2 -
1
211' X 180KHz
80.8 nscc ~ T grl = 884.2 - 80.8
2
~
Wp2
J2hln
27T/n
wll
wn ..,., Kwp 1
R '" R;,. II
R,;~
., 80.8 kll, g, = 4
f'gd'"" I pF
1lnls
803.4 nsec .,., (80.8 kH
~ 324.2R1'
'
nt,
+ R;, + 323.2RiJ + I pF
S03.4 nsec- 80.8 k!l
I pF
""
::,; Rj_ ···"'· 722 ·6 ~!_! '"" 2.23 kfl
324.2
For the .:ase of K "" 2. the exact value of w" can
be found from the following equation:
[1 J][I (K:::JJ 2
+ (;;1
t
·=
Assuming "111
,,
Jhx
4 t (
X4 + (K2
K~)x 2
1+
+
I )X 2
-
+ 1
·
u•)
420
X 4 X 2.23 ·"" -7.2 VN
420 X 100
Thcrcfurc. the gain-bandwidth product is:
7.2 x 180KHz= 1.296 MHz:: 1.3 MHz
2 ==:>
=
R;n
R. + I?. (f<,."t
~~
"''r
M
AM =
X we have
b) I$!
Xl\
For this value of f?j_ we have
-'1
=
(I ! X 2 ) ( I + K:)
"'} R{ "" 2.23 kH
= 2
Ex: 8.15
~
IOkfl
K2 = 0 ~
F·or K ··· -) _,
._., wn
0 . 84
~ hlH
-
().•.8'"tl>Jp 1
Using Miller's theorem we have
(1)1'!
fn this case, the approximate value of (J), is:
. ~-,~-~
() 09
I' ., --~ '' .o blrt
..,.
K·
wp 1 1
4, using equation ( * ), the exnct value nf
ForK
wit is:
ltl 11
••
!{.
•
"'
R,"
'
\1" ~ ··AV,
Assuming V.,, = I V
A(V/V)
0.95
v. "· __!!.!_"-\/.
10 kH
A + I'
\VC
·l
have
V,-(mV)
R;n(!l)
\~,(V)
toJ,,,
t•>n
-~
~~----~
rl; i'
·~
K~
\Ve have A,
····10.8 V/V andJ~:: 128.3 KHz. therefore.. the
gain hand width product i~:
IO.R X l 2XJ •• I.J856 1\-lHz ;:_ l ..W Mllz
909
476
4.76
4.76
99
90
-9
•• <)
1000
9.99
9.9
-9.9
-9.9
10000
I
0.999
-9.99
-9.99
10
f-·
Ex: 8 .14
~/\lsi!!
(~)
In this ca'it\ tht approximate value of ''luis:
.
and
I kH "r
··-~,...,.·~-~
100
1----
L.-----.........
..i..
Exercise 8-4
Ex:S · 16 Refering tO the ~lution of Example
a .10 the value of/11 detennined by the exaCt
analysis is:
/ 11
=
R~
j X 10 K = 2.5 kO
=
!AMI = 8m • R[ = 2.5
J,., = 143.4 MHz
Also,
AM ""' -g,. ·R;_ = -1.25 X 10 = -12.5 VN
Therefore the gain-bandwidth produt.-t (/r) is:
f, "' 143.4 X 12.5 =- 1.79 GHz
Since fr is less than f,.! = 2.44 GHz and f ""
40 GHz, therefore it is a good :lpproximation of
the unity gain frC<)ttenl:y.
= 6.25 AlA
Using equation a . 9 3 and assuming/,,::::}~.
we have:
.Ifrt
.
+
2-tr{ (C"',
ro,ll.2 ' 02
R' ""
I.
!AMI ·""·
g,.!£!~ 'm ""'
° ""
1.25 X 12
+
CqJ(l +
.
g.,R/J] · R;ir; I
f
Sf( I
!·
VA~ "' 13() V
(CL
+ C~d)Ri.J
1.25 x 5)] · 10 K
+ ( 25f + 5j) X 5 K ]
g, 1
_ 2lvl
·•··
V
Ri. ""'
~6 k!l
I
C,n
~, 250 mA
v
r,
R,;~ ·+ r,
R;_
r0 u 1 i!
F
...!!
In
r 0 Q2 =
=
( r,)(JI
ij ~"oQl)
'""
t
=
=
175
~
v
C, + C"( I + g,Rj_)
···->
C; 11
448 pF
,/.,::::
21TC:,N,;~
10 k!l in example
increasing 1, by 4 reduces
Ill
=
-
i . roo:
I()_ K ~':QQ.!.
. , 4 "
4
+ 40
21ff'; ..
X
36)
~·
448 pF
(~~-r(R, 1 ~ I
r,)J
82.6 kHz
.!'lTJ<448 pFI5ll (36
I
0.2)]
~
::::.4.3 kn
Thus:
~"oQ1 "'
(" R')
v
V
both r001 and r002 by 4
~"o(ll I
+ r"
= 16 pF t 0.3 pF(I
9.10
Since r 0 -~
5 kH
(b) Using Miller's theorem \ve have:
(2Vuvl)
R;,,
""
v
J\.1,1 = -17;,-
,..,,~u
If
209-
40 rnA
--.- 5-_--(40 X 36 kH)
36 l 0.2 + 5
is 111 , is increased by 4 and V,,. by 2
(}\"'!
To calculate
130 kfl II 50 k!l
,1.11) From equation 13 . 97 we have:
''
0
""
I mA "" 40 mA
25 mV
V
g, = Vr
A,., = ~
then:
J.L
_lsomvA ·= -"iO k''
~"onpn !I ~'opnp
l?m
Ex:B .18Rcfcrring to the solution of
Example 8 . 1 0 ~
·-~
R; '""'
r, = ~- =
"'6.25 X 223= 1.4 GH1
130 k!l
I mA
1 -~-~Ae!
yopnp ~
};,::: 223 MHz
IAMI·f11
f X 2.5 I(
= 250 MHz
6.25 X 250 ~: 1.56 GHz
"'
I
I
fr =
f,. :::k
Ex:8 .19
I
In::: 21r[[20f
= 250 MHz =:>
/ 11
6.25 VN
Using equations. 92 and as~mning_t;, :::;J,.,.
we have:
f,,:;:
21T·HC.~,
! C8 ,.)Ri}
+ 5 f(l + 6.25)110 K + {25 + 5)
2-rr!T20 f
_{, : :. IA,,,J · f 11
and therefore
+ g.,R;)JR,,~ + (Cr
C~, 1 ( l
In ""
F..x:
Referring to the S())tttion of Example
8 • 1 b if a load resistor is connected at the output
halving the value of R~ , then we have
X 2.5
1
4
(c) Using t.he method of opcn"cin:uit time constants, from equation e , 10 0 we have:
7 11
=
CrrR,;f
I
C;.l (I
f
g,R; )R:;p +
f?!
I
t CtR/,
Exercise 8-5
Thus:
'fn '~ 16 X 4.3
+ 0.3[( I + 40 X 36)4:.3 + 36]
R, 18 = 10 kO,
+5 X 36
fn = --1-
.:::75.1 kHz
2rrr 11
(d) Using e"i:Jtiations s .102, a .103 and s .104
we have:
40 tnA
- ·v21T0.3 pF
/~,
.l
21.2 GHz
= 20 kfl.
I
500 K
+
1.25 m 1.25 X 20
g,.,r,
g,
=- 20Jlk0
~G =
R,.
•
R.;g +
To obtainfH:
Rii.•
500
II
R,;g
10
R;n
Rin
"" R1J R0
R0 = r 0 + R,;g
16.2 V /V
+ 20.8
10 K
II
r0)
·
21T[C,. + C 11 (1 + gmR/_)lR~i~
i
(C~_ + C11)R/.
75.1 kHz
+ (g
111
R,;t " 280 K
(same as in Eq. 8 . 12)
RM,, = 500 K II 280 K '~ 179.5 kO
cg.• · Rg,· + (Cg,1 +
T 11 ""'
!,2.::::
c,) · R#.~
+ (5 f + 15 fl X
+ 3.59 ns 3.72 ns
·"' 20 f X 6.75 K
...!_LC, ~E1,(1 + g,.R;)JR,, 8 + (C1_ + C~)!~
1C,.( CL
21T
i
6.75 kH
20.8 K
RRJ
=
= f, 1 ""
R~_
l
-+
-R,.-
R;n =
'fn = 2.12 nsec
15 fF.
Cg. = 20fF,Cifd"" 5fF CL
C") + CrC,_.]R;;gRt.
of,Tu
0.135 ns
Thus,
f 11
1""' - -
21TTn
f,,J "' 25.2 MHz
Sinct'1;, 1 << 2 GHz
ro)
CASCODE Amplifier:
R 0 = 2r 0
-
945 fF
ll
----7
I
I
A 1,"" --(o
= - -2X 40 '"" -20 V IV
2 ~"lm r,J)
~.
~Av '"' -g"'(R0
----:-::---='12::,:·::..5---::-:::-:-:- - 'i X fF
179.5 K
•c.
il
Rt)
+ (g..,r 0 ) · r 0
=
r 0 (2 f gmr 0 )
42r 0
~ .4 1.
'-'
...
g,.,(42r 0 !1 r 0
) -
-
g,, · r 0
Ex:8 · 21 For a CS amplifi('r fed with R,;g ~ 0
~.!'
we know that:
gm
I. .
. '
cc.
ln((\
CASCODE
=
c;,;)
f
-4Q , 2
-2:0
AFCS
~)) f,, : Neglect cornpnnents of T 11 that do Ililt
'·
0.25 c,,
include R, 1 ~; abo C,.t
CS ·Amplifier:
Tho;>rl·forc.
[
f,
L
f,
T }/ =
~- ... ~:~:!!!_(2nC,.~)
g,,! [2-rr(C 1 + C,J]
cI.
cgd
l
"'> L
/,
Ex:8. 22 R1
I
i
C,,, • R,;~ + Ced[ (I ·~ gm · R;)N,;(! + k',l
EL
(~utl
500 kH. am! from
Example 8 . 12
g, ·.·· 1.2:5mA,V,r,,
sine(' Cx.t
20 kH.
0.25 C_,, and g,,r 0
·"
40
Exercise 8-6
"* t If = C1, · R~it ·1- ~0:..25 X C~tj( 21 }( R•i•
=R~,.,,x c~~
x 6.25
CASCODE.- Amplifier:
Using P.A:J8 .i19and neglecting the tenns ~do
n~ inclu<,le R-.;1
T, = R.;,[e~, + c,lfr 11 '" 339
!A vi ·In
fT·CASCODF. """ (IAvlcA$CODf.) X(fH CASCODl!)
fiJ"" - 1 -
c) /r =
IAvlcs
In cs
= 2 x 3~6
~
Ex:S • 2 • Referring to the solution of
ExerciseS • 19 we bnvc:
"'" 40 mA
v and r 1t '"' 5 kH
Note that for the cascode amplifier considered in
this exercise:
r,, '"'· r.,2 = rrr;;; 5k!1
g,.,,
16 p)
ns
21fTn
=!AMI·
1
-469KHz
21r X 339 ns -
=
IH
= 242 X 469 KHi
=113.5 MHz
Compared to the CE amplifier in Exercise 8 . 19
lAM! has Increased from 175 VI V to
7.2
..
..,,.,
·
fr
R0 )
x $5 + (5 p + 0 + 0.3 p)(SO KJI
""' R•is · C$, X6;25 = 3 _6
R,;,. · C11 , X 1.75
.frc.
c 1.
=
+ 16 p X
26M)
35
159 ns
t.4 pf
+ 0.2 kH ''" 5.2 k!l
A., "" 8m · r., = 40 X 130
=
5200 VI V
Ex:B.25R' 1• = Rtll ro = 20K!I20K
= IOK
From Eq.a. ~2~ we have:
200 + I
X
,. AM "" _g,.R' t.
. '
I + g,.R' 1•
1.25 X 10 '" 0.93 y
I + 1.25 X 10
V
__!_X
130 +_50
130
2?T
+ J.Q.
·- 8 GHz
201
f
R;n: :: 35 f!
R" :: j3 2r 02
""
1.25 m
(20 f + 5 f)
200
X
130 kH "" 26 MH
__!_ · 1Im "" _!_ · L2 5...!E - 10 GHz
2?T CK,
2?T 20 f -
~
R!!d = R,;~ "" 10 K
R
=
R,;s + f!~
1 + g,., · R' 1•
~'
'0
1.48 kn
""
10 K + 10 K
1
l
1.25
X
10
Exercise 8--7
E;c;S • 28
lt "'·.
Using Eq f-.153 to f. .157 we have:
R'1. .,. . RtU R0 "' 20 Kll 60 K ·"" 15 kO
I
21r · Css · Rss
ll8 t1
.1
. = 15.9 MHz
l1r · (0.4 p)25 K
Ex:B • 29Por a loaded bipolar dilfereotlnt
·-
''
11•--
ro
"n. s Xr- -+· ·R-
I
,,,R,:;g
1! 2 = 0.5 tnA
Vr
25 mV
VA
100 v
0.5 mA
//2
=
7 11
2
+ f!.§
200kU
~O + .I "" 10.5 kkl
2
cg,Rg,. + cgt/ · R*' + c~.n·l.
""' 20 f( I0.5 K + 5 f X 235 K + 5 f
··~
l.46ns~f,"" -.. -1-
X
IS K)
= 109MHz
21TTII
20 mAX 2(Kl kH
X
20+ 20
'"
~· g~
1. = I kH ~ R,RJ
R.s
20 mAJV
T 11 ""
!
x.....l.Q_
2
where,
"Km
I+ 2
I+.<> R
Q
2_gm • ro
235 kH
""
R,1, +R:s
R _
amplifier:
t\,1 ~"
+ R' 1•
R.1s(1 1- Gm« t)
v
~-=
U/JP
10 X 109 MHz:::l.l GHz
'"' 2000 V IV
The dominant pole is set by !he output load
Ex:8. 31
capacitance
R;.,= {f1 1 ·l I )(r,. 1 + r, 2 )
f
I
' ·'" in-· Cr(ro2 II
Since in this case r,. 1
rm)
l
=
r,. and
13 we have
211" X 2 pF X (200 K I 200 K)!l
0.796 l\•1Hz. :: 0.8 MHz
2r,
= 2 ~\lr- 10 kH
lc
-
we have:
Ex:S. 30 (a)
Vo
=
AM"" -g 111 XR'~_
A,~~ =
-2
To calculate
°~~
T 11
X
-g.,(R~.II r")
(20 k!l!i 20 k!l)
I(
··20
=
Rin .. )
7 R·
Vs;g
-
in
>
+R Srg
. . .~ ...
R
L
~
using the method of open-circuit
we have:
time-constants we can employ Eq t8. 84
Tfl
""'
c~,.
R.,jg
+ c.,JR,,~(I + gmR'JJ + R'd
'C / , . R' L
T
20 f
Tu
X
20 K
l
5 f[20 K( I + 20)
I
10
Kl
fpt
= --·------.-:...-·-·---2...(6 iF+ 2 pF)(Io Kil 10 K)
!5fXlOK
T 11
''
2.6 ns => f
11
2·r.r n
The gain-bandwidtb product is:
GBP
(b)
61.2 MHz
lr1 = ~~·~1?
20 x 61.2 M ·•·• 1.22 GHz
With source degeneration of
-1f'--
2
gm
l~s
X
20 k!l = 60 kH
A(1·)
A.ll ,,,
--2
X
j.t
2·rr
1.
20
2 pF
X
10 K
- -.............AM. _ __
(I
v
.. 10-
l
i~~f':~ )( I • '2~>;.~)
I--;:A.,,I
v
2 mA
3
X
::8 MHz
Therefore. the transfer function of this CC ·CB
amplifier is:
R 0 = r 11 [ I + g,,Rd •• 3r0
R0 = 3
::6.4 MHz
we have:
I
v
J2
or IA ( s)i''·
·'
;. . //
Exercise 8--8
F..x:8 • 33To obtain Req:
Thus:
Req "' Rzll rol
II
r ..s
20k!l
JOOV
VA_
( 1+
f:t)(
ln
le2 ·-· 0.25 rnA
1+
~~) ""
2
Solving thls e C,. 5
4.8 pF
7. -
G,.1
211 · Cc
40 m
- 2 p ·· 14 pF
21T X 400 Ill
thus;
From Eqn. tS. 173
r · ·.
• ·
"
Ceq= 2pF+l4pF+2pF(I+40X3)
~
G,;
c~ f. "" 0.6 mA / V
.. z
211 X 4.8 pF
g, 6 =
0.6 mA/V
20 MHz
c•258 pF
Finally,
f,, ""'
·
From Eqn. 1 8. 177
0.6 mA/V '"' 48 MHz
21i X 2 pF
1
211·Req·Ceq
I
21T X 2.2 K X 258 p
·~
280KHz
Exercise 9--1
Ex:9 .l(c)A = lOOVIV andA1 = tO \ltv
_ . A
I
1
I
I
A f c.. 1 + BA => l':l
A! - A
I0 - t 00
= 0;09
Rx:Q • 4 For Example
A0
""'
6000, ~
(I +A~)
=
=
7 .1
10- 3
(I + (6 X 103 ) X 10-:l) "". 7
= fH( I+ A~)
/.fJJt
"' l X 7
=
7kHZ:
''"" 0.09
since: I
+ !'!1
1
Ex:9.
13
R1
V
R,
1
R1
().09
=> --= ""' -.- --· I
= 10. l J
I
+ AB ""· l + 100 X
vl
'~ I~·
v
v.I -"" ....!!
J\
V0
,,
,~-
-10 ..
0.09"' 10 which is 20 dB
i\1 Vs "'' to x 1 '" 10 V
x
0.09
!0 = 0.9
v
l(!()
80
10 -- 9.7561
Ex:9 • 2 (c) A ,_.
"' ().99
·---~
v
=
40 dB
li}x:9 • 6 Replacing the amplifier by its small
signal model
2.44% of A,'" 10
}
V,
I
Ai
I
I
I
10' - 104
A
v,
I
= --l =
15
110.1
Open~Joop
(d) The amount of feed hack is:
AJ3
«f+Vnf
+ 0.0099
=-9X!(f 4
l t-
V
I X V,.
l+(J00Xl)XI
an impr()vement of 20 log( 100 I I)
to4 V!V and A1 = 103
I ..;.. J3A
R1
+
AI
"I +A,A2~
9 7561
"'"
Al=-A-=>~
R,
=-'
+V
ArA2
+A,A1~
New SIN ratio = 100 /I
~" 0.1
-·"----------1
+go x oo9
ilAr
·'J
Thus, V,I"" I V and V,.,."" O.OIV
(I) If A decreases by 20%:
A 0.& X 100 SOV/V
A1
"" V
(I X 100) X Vs
l+(lOOXl)Xl
,
(d) The amount of ti.~cd-back is:
(c) For Vs "" 1 V: V0
"
5
I
+ l(f1 X
grounded:
9 X 10" 4
,--;
10 which
is 20 dB
i\ 1 • V 1
\1 1 ~
f3 · F 0
Vo =
V~_, =
(c) For \/5 ., 0.0! V:
V0
10 1
10
X {).()!
'
9 X
I(J .J X
-R,.v., ><
Rn
- \l~ ~A ~:: gmRv
Feed·back factor:
v
1.3
"
gain: without R2 and R 1 > Ru
0.390 maximum
-gmV.ts' Rn
Rrllo
R 1 + R:
1?2
+ 1? 1
Exercise 9-2
k} . vs.
<> R
~v.·.Jt~=gRV.
{
o
R+R
I
,..
2
I
Closed-loop gain: A J = Vo
Thus:
s
V;,
AI = --"·
+ g,.,RDR 1 !(R1 + R1 )
Vs
ifA•r3 >>l => (g.,R0 ) · R 1 I (R 1 + R2 ) >> l
10 = A 1gm2 V1 and V1 ""· Vs- VF
10 = A 1g;,:r£Vs- V 1 } = A 1g., 2{Vs-lo/1p}
~
loO + RFAlg,a) ""' A,g.,1. Vs
Ar =
~.
s
= 1 A 18"'2
which is the sante"
+ RFAIRml
asA =_A_
I+ ~A
.f
Ex:9 • 7 From Example 9 . :Z we know tha~:
-g,;,z.Ro where -gm RD .IS t he open2
! + _KmzRo
If AP >> I
A
JC:
~ RFA 1g, 2
A.g,.z.
R,,A1gm2
I
I+ RF
-')AJ - -
- RF
RM
loop gain
Ex:9 · 9
The equivalent small-signal circuit fhr
Fig 9.llb)
ifloop-gain A{3 >> I
Ex: 9 · 8The equivalent srnall-signal model for
Fig 9 . 1 0 b) is:
V
-~
V0
f:- +A
Il
-
>> I and
-
+
-
-A
I
v
'
-
oo
-v
-AV.-> V = _ _o
A
I
I
(I)
Is = I;+ l.r
~~·'
.
I
-
(2)
+
Opt~Jl·loop
gain: V, "" V;
=
-RrVo{ AR;r~
I A1 I }
I
t
··-·~--= R r-·--
( 1 1 _!.
A
+ RF )
AR;,1.
if A.>> I and AR;,1 >> R"
Exercise 9-3
Ex:9 .1.0
R;N = R((- Rs = 201 - lO
FrQm Example 7. l
Ao .... 600ll, ~ ""
x 103)
(! +AI3) ""' (I+ (6
:. f111
=
Ror .;':: {Rf'IU, II R,)
ro- 3
x 10~ 3 )
fufl +A~) "'~ I X 7
=
""
7
:::::} R,..,
~
Vc2
181
IK80
951
n
a) The loading effect ofthe feed-back network at
theinputis:R 1 11 R2
In == 0.5mA
0.7 ·-
lr;:J
= 5mA
r, 1
'"'
r, 2
~
Ex: 9 · 12The feed-back network is composed (}f
the voltage-driver resistors R1 and R2
= 10.7-0.5 x 20 = + 0.71·
v.. ,.
19.1
I+ Af3
191 kfi
7kHz
Ex:9 .11
1111
=
= - R.,
=
'"
b) The loading etTect of the feed-back network at
the output is: R 1 + R 2
v1!/!.J "' 0
The A circuit is:
VA I I ""' 50H , ~'e:1
5fl
A-Circuit:
Vo
1
'
Rtlt /?2
":"
"::"'
~·,
2K
A""' g,.lRDij{U 1 +R2)]
To obtain
R;
(20 II (1~ 2-t:" I )(rtJ
+ 2!1 10))
r + r ' + ~ + (I I! 9)
d
( 2 11 !0)
;." + (2 !! 10)
1?1
Rs
= 10 +
R0
For the CG amplifier:
"'
r•
.
:::: 85.7
~l
+
J
J31 +
PI!
9)
[r,., + ~~hJ
2 !! 10!!
-
/:;Q
X
I
v! v
+ ((3 + l )(rc~ + r,. 2 ) + Rr!l
101(50 +50)+
J3:
R~
21 kn
181 fl
;,' 1. =
B-Circuit:
v,.
= __!!.;__
V0
f<, + R~
A-I
+AI~
s· ub·
·
·
slltutm~:
•
Ar ~ _ _g,.,RI_,- -
I+ Rn(l-l g,R 1)
R 1 + R1
if R 1 + R2 >> R 1, we obtain the same result as in
Exercise !OJ)
From the A circuit:
/?2
~£3~
R
V;,
V)
1
..
· · ·~·
R;=
\! ...
v,
A
I
+A~
cc
_1_(1 •Af'.)
(Rv
0.1 V /V
91
lig,~R;,
t\ 5·7
-l i 85.7 X 0.1
~ X.96V / V
201 k!!
II R, + R2)
1 + t\f~
Exercise 9-4
(c)Rt
= Rsll
fl""
RF
Followin3 the procedure used inE~lilllpk 9 , 7
R·f "'·
'
--.i!.L_::::} j_ ~· l + ~
I + Af3
R1f
R;
R;
1 .. 1
R11
II R,)g;,(r0 II
(Rs
- = ·- +
.
R,
(Rs.ll R.,) · R:r
II
Ji. "' g,(r0
~
.~ or R 1
R1
R,
Substituting for R; '"' Rs
R;r ""' Rs
II
Rp) .
tfwecal
II
n,
II -:
fl
II
II
RF
n,. ""'
I
R;r ""·
-.!!J_
Rr)
II
Rr
10
=
6.67 kO
l
+ Af3
I
+ Af3
n
909 ''" 22s.6 n
4.(}3
6.67 K
4.03
' L66 kH
IOK
-1_+_5_•-n(.::,:I::..K::.::. .,.!1-I..,.O""'K'O:':') "' 1.66 kf!
II
r0
""'
•:c• cc r 01
""
1
-·· +
(Rs
R0
II
I KH R 1
.o,
JOKn
90k!lg, ..,. 5mA/V
r () '"' 20 k!l
Refer to Example 9 . 8
R1 =~ 100 K, unchanged
R
__
o_ ...;,. _!_
I + Af3 """' R., 1
R,,r
II
10 K
. "" 291
K H 10 K)
Following the procedure used in Example 9 . 7
"f
20
""'
Ex: 9.16 Ji ....~. 100
+ g,,(r0 !I
R
n
+ 5 m(20
R.,u, "" 20 K
Rr
. ·
(1 + Ji )
R2
R0
I + 3.03
Rs II R,. = l II 10 ,. 909
R0 ·"'· r 0
/(~
{d)
3.03
,, -3.03 K 0". -7.52 kU
AJ3
Ro
Since
I
=
-0.1 mA/V
Rr:
fl
=Rs
=
R;
Rr
II
Rf'
R;
··""·
'
l +
·
=
-1/lOK
Afl = -30.3 X -0.1
Ar ""' _A_.-
R")
= ,...t +
1
R1r
-c-
.
=
-1/Rp
=
_!_ + ~
R0
R0
R,)g"'(r0
II
R;
I! u~ '" - 100 .
A ,, - fl . Rl
I0
100
II 90
~.~ - L I I X I 0 3 A / A
Rr)
·~~--
R, · (r 0 1! Rr)
f3 ,,, --0.1 A I A. unchanged
if we call
Af3 = Ill
.\
A 1 '" -l.ll X lO
,,, -9.91 AlA
.l +Ill
Ro~ '~
R
()
II
.,.2_9K=900!1
R
Rr
"'
fl
1? 0
Substituting for
100
=
R,,"'
900 KSl, unchanged
=
(I 1- Ill). lJOO K
= 100 M!l
Rr
I+J.L
All of I 0 is fed-back.
=
r 0 •·H
(c) For g,.
R1
=
IOkH
R.1.
=
I kf!
if A!3 »
i\ 1 •= --.'.
r () =
f3
20 kfl
A = - ( l K i! 10 K) · 5 m ( 20 K
-30.3 kn
ideal
I A/A
R; "' R,, !1 R;,/
Ro
I --)
rm + (!?,
I
!?,
"'
!!
10 K l
II
0) I g,t·o}l?,
Replacing R~ for 0 in Eq 9 . 6 9
!i
':f~
R, .·.~ R,
i!
0) = ro2
Exercise 9-5
.... .,..R;roz
A
.l X r
g,,
02
A
= -.
Al
I + A!J
PromEq 9. 77
R.,u, "''
-fL. g,· R1
l + JLf(,. · R1'
8mVr
Rr
fL • R, · g,.rm ·
R 1 =o·
JLg,.r1>2 •
R1
To obtain R1":
•Ci2'0. A(JW)
... '""·(..hx-7.
10
.
_,q,
180 :::::> tan
(w 180ll04 )
J3 :::::> tu 1No
At. w.
18(1'
Amplifier stable if iAi31
4
+ jw/10
I
.. l
)1
4
•
((" 1110,/10 ) ·""' 60
J3 X 104 rad!s
at c" 111w
""
<.~IO~__,K,....__ _
(IOK+9K t--1 K
!I
114m)
(·_ _l_!L__) · 10 K 16.66
I K !I I /4 m
Compared to l7.39 obtained in Example 9 . 4
+ 105
X ().()J
=
fPt = J'p(l + A 0 ~)
For Closed loop gain
f.··~f(I+A
pi
tt
()
1001
.o
100
= I.
X
!OO.IkHz
100!
13 "" I
f3)
Ex:9. 22 From Eq. 9. 92 Poles will coincide
when
(wp 1
!•Jp 1 ) 1 -
+
4( I
i
A,,J:\) w 1, 1 oJp~ = 0
Using
+ JQO ~)X 10 10
(!OJ i (() 6 )2.
4(1
1+
1.0 1) 1 X !()()I
~;:}
J()(l
fl
f3 ''"' (
4
0.5
H'r maximally flat respon:;e Q
J.~ 1 +
I 00[3) X
=
0. 707 and
lOTI) ;;;-.) (3
JO'' + !!l''
CoJTC!iponding gain i~
A
5 m ;-< 20 K X 1 K
20 K + 10 K l l K
as compared to :un obtained in Exercise 9 . 15
Q
0.245
Corresponding Q
Ex9.19 V,
-co
lOO
I + 100 >< 0,5
0.5
L96 V/V
Exercise 9--6
ExEJ • 2 3CIO!!ed loop poleil are found usin$
I + A(s)(:\ = 0
(I + S/104)3
:/. + 3:P + 3St
10 12
108
'"" S 1
"
=
10'113
(I+ Sl4P +
c~
~=0
10)
1.+
+(I
H~
-
0
+ 100~)
""
o
104
+ JS., + 3S, + (I + 100~)
=0
+20dB/dec
for
s,.
OdB
Roots of this cubic equation are:
f
<- 1-10r~~~:\- t + s~uJ ± j 5./313' 13
Amplifier becomes UU!>tllblewhen complex p()Jes
are onjw a-.is ie. when f3 = f{,.,
1 ! .l
I013 cr
··•
.••
I
""' 2 ::::) f3~r
cos60;
0"
0.008
Rate of closure "" 20 dB/dec
Ex:9. 27 Must place new dominant pole at
.
A. ()
Ex: 9. 24 A
A.o
0 ·'
+ j f lf,,
jl;
fo ·"" -
t1
. 106
'"' -.
Ht
:.fn""' lOOHz
dB
+ J .fi 10
w' x 0.01
IA !3!
13 "" O. f =104 Hz
At f '' 10~ Hz
= -tan -
1(
IOJ/ 10) '"' -9()0
making phase margin 180- 90 '" 90°
.-E-
Ex:9. 25 From Eqn 9.105
. l;. = ,-·~·..!-,I ;r~
In.. .. (.'"''
I .
- '
i l + ('- i~~
For PM"' 30".0
I A 1Ut•J1 )! ! (I If~)
60".
= Iow .1.re·
f3
Ex: 9 · 28 The pole must be moved
1.93
fv where
f
!i1 1(}td,,,.) i i ( I!~~)
Ex:9.26 r~
1/SC
-.1 I , ( ..JC R)
Thus CR
2:
0.1595.
()
I +SCR
to
x w<•
I 04
~·····
1, 1
• Ar
f- (
100 dB- 20 db)
=
Hz
The capacitance at tiR~ controlling uode must be
increased by ~mnt· f<~ctor asfis lowered.
10 3
R + liS('
r-·:--.-.-1
=
A
10
= 0.707
f
~~ Frequency of_?!l~j)S!~
· u
~· LO
For P.'rl "'., 90". 0 = 90°
1
J
.r,t
tti)
150"
120°
(l =
IArU••'1) I / ( l! !3 l
~~~
102 JO.I 104 10~ 10~
10
d I
-
180° - Phase margin
quency gain 0
For Pi\·1
an
-40
:. c"""
·=
c."'' x tooo
Exercise 9-7
13.3
a. P1. , (V;..J./il
Ex 13.1
Ex
ForQ,
1 ,. Vee- Vcgsat
15 ·-0.2
. Rt.
J"k:fi
Rr..
(- V
-
rc>
-0.7-(-15)
14.8
""0.971c
">rA
"' • l - l · ~? X
. 100 ··~ I -~
In cxarnplel3 . 5 I,,
-~•
2
~ ~
<\,""
I
25 .• '" 6.25
+2
n
100
100 + 6.25
RL
1?1. + R.,,.,
v1
1 mA, and ftJr v_, '" 0
0.94
IOV
n -~
1U V
100
·"
,_ -
100 rnA
Again calculate i, (for lv "'" 2 mA) using equation
13.27 i, = 99.96 mA
0J)4 mA
(2)
~:.r_
is+ i,
25
99.6
o.25 n
+ 0.04
% Chanrre""
!_= I0 ·94 >< 100
,
=
6%
For Ia ~= 10 rnA, change is 1.2')}
l'ix /" ""· 2 mA, change is 6%
(c) The quiesent power dissipated in each transistor
1,, X V"
Totalpowcrdi~sipatcd • 2 x lOX 10
x 15
""300m\V
[
)()-.'
)
-
i.v(iN- 0.1)
(JO-D)2
i,,U, -· 0.1) = 9 X 10 ''
If i, is in mA, then
i,(i, - 100) =
I)
i/ - 100 i,- 9
Ex 13.6
From exampleD · 4 !',, ·- 15 V. R,
l 00 n
oc.· 10 ",>\ nnd 13
50.
Q, and Q,, malched and/,
/,,",
i,. = i.., ,. i 1
= 0. I mA
10 V and i 1
3 rnA
10
0. I A
For v., 10 V. f 1
100
As a first approximation i, :: 0. l A. i,.
0.1 A
"
j{)+l .:: .. rnA
=0
100.1 mA
., .. I 0
···O.l A
100
-IOOmA
0,
i,., : : .
A~
a lirst approximation assunw ;,
ix.:: 0 since i,
'"!
100 mA.
0. current through diodes
:I nli\
Exercise 9--9
3.x
:.V 11, - 2Yrln [ .
w-)]
Bul V89
"'
Vtln(
_i:v)+
-
ro·· u
Vrln
(m-l"'_)
_,_J.vnf·' '~r ·~ 1 X .,_!)05;o.O:l5
In "'· I X •
•
(4)
l
Equating t~quati(lllli 3 and 4
[ 3 >:
!
X 10
3
itir ·· ~-·!1
~o•J
1.~
""
~::
ln '" I X e 01 ·om~
=
54.60 mA
(10 !l)2
I= 54JJ{) +0.58 "' 55.18mA
For AV 1w ·~ -50 mV
I Cl
i" (i,- J{)()J = 81
i/ ~ 100 i"- 81 0
~ i 1•
100.8 mA
0.8 mA
V8 r
= 0.4 mA! m V
Thus, j_fc =• 0.4 X 2 X 5 "" 4 mA
X c-O.!IotOJl2>
VTJI' '~
I
I .0 V
R
In
Thus In -"' I mA
F:x 13 . 9
I
2.4
0.417 mA
X
111 n.i\25
e
= 0.018 mA
Using equation 13 . 4 3
C•.
1. 25 ~- 0.52 mA
,.,, _I_
2.4
I'' 0.43 mA
2.4
1.5 mA
0.13 mA
0.5 V
l'uE
Thus. I
'"
0.46 + 0.13
0.59 mA
For .1V 1111 , .. -200mV:
Reier to Fig.l 0. 14
Ia) To obtain ;lletminal voltage of I .2 V. and since
13 1 is Vt:ry large, it follows, that VIII .... VIa= 0.6 V.
l.l = 0.5 rnA
0.46 mA
~
Ex 13.8
v
0.37 mA
0.85 mA
1.1
2.4
'
1.2
"'
0.55 V
=
In '" l
I
!i!.1J!:~
£,---om5 iiHJJ.<
v·811 ----•l.l v / 11
where g., is in rnA I mV
25 mV
X
0.48 I 0.37
:lie = g., X 2 mV !"C X 5 "C.mA
=•
l
'"'
For .1 V 118 == - 100 mY:
Ex 13.7
"
1.15
2
VIlE ,.,
,, 0.575
Expressing currents in mr\
-~"'
IR "" l.l 5 ....,. 0.48 rnA
2.4
Vr111 "" 1.15 V
ip (i, ~ 0.1} ·~· 81 X 10 4
iv "" i, -- i, ""-
'' 0.58 mA
0.7 V
V 8E""
+ V T ln (l)
JO·l3
w--:Y '"
"'7.39
I"' 7.39 + 0.54"" 7.93 mA
For~ V 11fl "" +200 mV:
V8 n '" 1.4 V
l X .10 J.. -""
2Vrln [ -_·_-_--! X 10 D
J
'
V In (;,._- 0•1)
T
10 L\
= 065
. v
L3
2
1
= V-rln(_ ir(ip 1,)_·
to-ui ·)_ + Vrln (_,,-...
Here i, "" 0.1 A
0.54 mA
(3)
_
.! X J0- 11
3
.
IQ -
I
. (W!L),,
. n;,., (IV I L),
0.2 (_IV 1 Ll.,
I W !I.)"
( W i L),
(\Vi
Li,
5
J2
2~k·(!Y)t~'---v
11
' L
I'
(,.\
1FJ
,, !!?-~ :' {l_(_i)~
<'
2.72 mA
2.72 i 0.5'2
For :. l' u11
'
c1.24mA
100m V
0.2
I >< 0.250(._!\~)
2
,_,} ( T.W')' -
L
-10
(0.2)'
1
Exercise 9-1 0
""··,.~ k1.' ( !)2 ( VGS "'"' IV ,!F
Q,: lui a,,
Q,2
=!
2.
XO.IO(} X
(!f).
X (0.2)2
L z
Oain J!rror "" -0.035 x 1.00 = 3.5%
21
2 X I X J()·l
Cmo ""' g"'P
?"'' ""
=
0.14
= 14.14mAIV
R.
.....
= .
}l.(g,,
I
+ g,..,)
I
10 X (14.14
+ 14.14) X
::::3.50
Ex ;1.3 .12
See solution on next page
l k
. '("")
Q·
t ""'. 2.. -!V!)l
P• (I
p
f. I'(V (,S
I
~" !
2
( !!:,·') ,.
L.
X OJ()() X
(!f.)
.X
L.
0.22
p
500
P
Now F,,, "·"· V""
+ V""
+ V,)
= (V,,. 1
+ (V.,, 2 + !V,!)
"' (0.2 + 0.5) + (0.2 + 0.5)
= 1.4 v
Ex13.10
i1 ~,
l.v
10 rnA
!!') \-'1
. 10 = !k
' ('
2 ·n L
··
10 ""' !2
=}
.<:
nv
0.-..
"50 X 200
X
v.,,
0.63
X
V2
(~\'
v
Using equation 13 . 4 6
VDn- vm-lfhll'~-
\/oma:o.:::.:;.
• 0
vm-.. .
\!m·N
2.5 - 0.2 - 0.5 - 0.63
""J.l7V
Exl3.11
New values ofWn" arc
••
(!f)
L N
S~JO •• 400
4
I>:!()·'~.! XO.I
2
.=} v,.,,. '" 0.14 v
Gain Error
~"···~~~~~-
4ft! '.!Rt
··0.035
=
X 10
'x
IOOOX V;L
J()--~
Exercise 9-11
Ex 13.12
Need to prove when v.,2. = 4/r;/{L then vGSN2 = vtn
Assume Q,voff ( Vos.v *" V,..) so i/1.'2 =0 nod
ip2
=iu
(Von .-. V IJf)) "" - V SGPQ
[V(;sn ORlusing (I):
. (! ""
+ jV,,J ""' -· VsGPQ + IJ.(Voz- v,l)
+ 2Vsc;PQ- jV,Pj
+V , (Vs<;rq- jV,rJ)
V
+ V oVQ
- 2VsoPQ
!J.(V 12
41
VsGPl = -
+ IJ.(VQ? - Vi2)
v.SGPQ + !J.(Vm- Vii)
2lk"I• (W)
I
("
I
y
SOY~
.....
Vn····
-IV
j)l
l lp
-
Vw) "' - Vst;PQ
m•
en
IJ.
·IJ.
p
Plug this value for Vn into the value for V tun
and show VtiSN2 ,. V,,\'
r----------------
<.- 'J(s +
~l(l)pWml't -jV,;.D
2(V,HJPQ -jV, 1~) "' (I!SOP2- w,,,j)
2
Vs<;Pi "' 2Vwrv-· 21\1,~
"' 2VmPo
(
V 0 [)
·-
(l)
:
vsoro) + 1-t( v,, 2 ••
1J.(Vv2- V;2l ·""
~l(v)i-~- V;Q)""
Vovu '"' ( v(.iSNQ
-!V,,J
~' w•2
+
v(iN2-
VGSN1
where
+ jV,1,!
Find V 12 f{lr the gate voltage,
Vw, 2
VosNQ
Vt;s;vQ)
V
"'
v,.) ., O'st,'I'Q - !V,I,p
< Hf. 3 }( 100 K /2) = -50 VI V
Ex: 10.3
A 2 = -G,. 1 R 2
The small-signal equivalent circuit for the op-amp
in Fig. HI. 1 on page of the Text is redrawn below
for a unity-gain huller.
From Eq;o .s, lD. ·; ,le>. 14 ,lD. 1son page of the
Text:
(;,d
=
(2 X 10
1 )(r..
,,U
r,,1 )
""(2X to-")(40( Kl/2) = -·40V/V
A '"' A 1 • A 2
(-
50)( - 41)) •
+ 2000
gm 1 ~
2-rtCz
21TIO .. "
.,J\
IJ = tan
=tan _,100 X 10<> = 17.4.
fr
318 XJ06
+ 1.2 V
(c) ·- U V s (V reM )lio'ni :!i: + 1.2 Y
(d)- 2.9 V S ( V ICM ).,,.,., 011 :S + 2.9 V
Find SR lor f, = 100 MU:r.
0.2
""'
v
SR = 21Tj,V011 ~""" 2r. X 100 X 106 X 0.2
125~67 ~
=
+ 2.9V
(b)DySym;
- 2i~ V .$ (VicM)p S
Ex:10.5
v01'1
;;;: - 2S + 0.3 + 0.3 + 0.7 = - 1.2 v
:. - L2V :S WrcM)N :!i:
99 - a ::-: n.6~
t>M ·""·
:s + 2.5 - 6.3 + 0.7 = + 2.9V
~ ;.;.. vss + Vtw + Vov + Vr
vl('M(mlu)N
l .
.
2
11 = 2K(W/L)(V0 s 1 ~ Vr)
126 VI JLS
SR .,., cl ~ ../ = SR X
c
Ex:10.9
Cc
lz =
:. I "" 126 X 106 X 1.6 X 10"' 12
~K(W4/L)(V0 S2- V 1 ) 2
For 11 = / 1 :
~ 200 JLA
(Vc;~ 1 - V 1-)
2
""''
4(Vcm -- Vr)
2
i.e., Vos!- VT .... 2(V!JS2- Vr)
Ex: 10.6
or Vast = 2Vasi- l' 1·
+ 1.65 - 0.3 + 0.5 = + 1.85 v
VICM(mi•J 2 - V s.~ + Vovu + Vuvt + V,.
2 - 1.65 + 0.3 + 0.3 + 0.5 "" -0.55 v
$:
v.,(m•xl;;;;; Vrm--Wo~·9!-!Vov!
s + 1.65.-0.3-0.3 =
;;;;: - 1.66
Ex:10.10
m
2 X 100 X 10- 6
0.2
.
g,.r~[~o!l
,.,.
=
~'o
... VA
lt:x:
1.0 X 2lXi X I! 3 X 10~ ""' 13.33 M!l
X
X.
10-;
X
13.33
X
lO''
10:; V! V
Given: all V 01. = 0.3 V,
\1 1q
JVrl ·
fnr NMOS
VIOI(m"'l!l' s;
v/)(J -·
I
25 ~ 5 k!l
X
25mV
=
lmA
2s n
125V
JmA
125 kH
lc
10. 11
/~e
1'u: 1'r
""~ V 8 f: '" \!Tin(! .I 1s)
and 1·' ~~ 14 • I 1
""'
I~
Vov
0.7 V
vTin[.!..!.]
+ v 7 Jn[~J..J = v7Jnr '''4·]
1stIs~
~l\st~4
:. ln[___!l___J
"" ln[_1_]
l~• · I.Q
ls:•/.1·4.
2.5V
0 ,., 1
Vr
From cct ; V 8101 + Von = V llf'J + V 8 E4
..;,.; 10.8
Vm>
= 20015k "'40 mA/v
r .. ~~ f:J,. """ 200
2))
A '"" G,.R, '0 '· 1.0
25 mVInJ.Q._"' 633 mV
10-14
""
g.,= {3/rc
1.0 mA/V
Ro "' lg .. 4r,1(r,,2 il r,w>l li lg.,(,r,6r,51
(U)
50 V
-·3
V 91,
21
Vov
13.33
""'
+ 0.3 + 0.3 + 0.5 = -0.55 v
(,
V
r ·"' ...A ,. ?O X IO ~ 200 kH
"
I
100
=
125 V
"'
+ 1.05 v
IVAI "" 20 V. \1 0 v "" 0.2V. I = 100 p.A
C"<
I0- 14 A .(1 = 200,V11
pup: ls "" l0- 14 A,(3 = 50.V11
Ex:10.7
G
=
npn :Is
+
Ff
:. / 1 ·"'·
I s-1 S4 "~
/ 1[ -'-:L-j
I stls2
I
2
Exercise 10-3
Ex: 10 .12
VIlE""' 0.7Vfor/c
f.:"l::
Sec 1->ig.l o. 2 2 on page of the Text, Let R 1 = R,
l mAiorQil
R2 =R + ~R Assume f3 >> 1 and r.,s "'r,6 , then
Vn 5 = lln6 =i(re5 +R 1) = i(rc6 +R2 )
lc "" 10 ~tA for QJO
V nt:tt!
0.7 + V,lnl·to [LA] "'· 0.585 V
"'
. I mA.
Voltage across R4
rc~
.
.
r,.~
r.;+ R
+ R + tlR
r.~ 4
Ex: 10.13
"' + 15 - 0.6 + 14.7 v
For ~!i
R
+ 0.6
0.3
E
= .... Vss + V liE~
R
+
.
AR
1-----.
r,.~ + R + ~R
·-I
~R
0.02 :
0.02
().()2R
,,
R + 0.02R + r, 5
"'
+ 0.6 + 0.6 + 0.3
"' .... 15
+ R + AR
iu""' icb- i = i
0.115 V
:. It, '"' I
V /I.C.:: l_(J__~''J(2.(>3 1'- :x .U ..1 ''-)
X 1(} ' . 2.63 Ji ' .11.1 lA
5.5
;<.
10
Exercise 10-4
(A2)otc "" - G,.l Reo
1.68 x 10' or 104.5 dB
If the conunon-mode feedback is not present, as
explained in the text, coronion-mode transconductance and common-mode gain are both reduced
by a factor of Pr Hence,
CMRR ""
!·68 xJ_cf ·"'
50
"" -6.5
X
8.1 "" -526.5 VI V
Kx: 10.23
3360 or
CMRR "" 70.5 dB
Ex: 10.19
f:Jr6"' fJn = 200
r-'6 ""
25 mV
162 fLA
1.54 kU
25 mV
I61J.A
= 1.56 K
~
rj•t9
,.,n "" 25 mV "'' 45.5 H
0.55 IliA
R~ "" 100 H. R9 ·~ 50 kH
Substituting into Eq. «8. 77
R12 '' 201( L54 + 50 l! (201
r
,, _200
~lS
40X().J65
'' 30.3 k!l
X 0Ji455)]
'"' 4 MU
l':x: 10.2 0
.--1L.. . _v~>!.L
i.p
/3 + I r,. 17 +
.,
vb,,
--~
45.5 + 100
R~
··-· V ·[-1- I 6.6 X 0.917]
7 18.8 K
18.8 K .
I
:. r
145.5
V7 [0J)5 + 6.05]
VT
.
~ R 1 " " - "'- 163
needs Rm = (/3 l I )(r.-~,
lr
l
= 2011455 ; !00) = 2Y.2 kH
.·. V, 17
Ex:
··• V~e
Ex:
10.24
X 0.92
t
rl'
Ro
10.21
R,l
n
R8 )
~-~- ~'rl-t
+
R,.plll) R,r;
50V
0.55 mA
90.9 k!l
R c~ 0.0?2 +
0
0.005
'IS. F 1
f:lp +I
( 163 + 25 M + ~l K)
180 f.l
51
20 I
200 nnd
l25V
227.:> kH
0.55 mA
22 mA/V
I,'-' "
R0
200 = 9.()9 k!l
22
R, Thus
lOll H
R,,, ~ 722
I knee R,, 2
kfl
90.') !I 722/.. ·• ~~ kH
•
Ex: 10.22
!.~
V,
!fUl
+ (>.6
X 0.917
=fl. II
npctH:ircnil \ nlta)!c gnin
0.05 + 6.05
180 1.tA !rom Table
5 + ( 163 + UlJ -1 1588) ~' 5 +I.}·'
I'
'u=
201
·
...
14A H
Exercise 10-5
F-x: 10.25
Ex: 10.28
SR '" 0.63 VlttS
using eq.llO .129
using eq,
f ,
SR
n
2r.VO••••
0.63
""
10kHz
J "" Vrtn((n) where Vr ·"" 25 mV
Rz
ln
R3
R4 "" 0.2
R1
V rlnCs2)
10.26
Ex:
R2
A., "".
2.43147 X 105
(i I "
_!_X IIJ-J
5.26
"'
:. A., '"' Onot R
:. R
~"
""* 1279 MHz
A0 I Om 1
J<:x: 10.27
SR "' W, where
a
a "' G.11
2/
""* SR ~"'
Z/
G.,t
X
w
'
with R r inserted in emitters of Q3• Q4
G,.,,
= 2x
2
X
4r,.
I
+ 2Rr
0.025 mV + R,
1
"
for I ~ 95 X 10- 6 A
0 ·050
9.5 X 10- 6
Rr ""'
.
= 5.26 kn
X . .,,. , IR t: )
now,,''R ·"' -2/wr
1- • !- •'r .,.
nev"'
c·.:
.
G'
__.!.ti
"c.. ,
C,.
2C,
:. C,. must be reduce X factor of 2
30
2
Gain Au G, 1
:.
15 pF
A also hahcd
101.7 dB
{ 1,
lOJ.L
.·c
fJA
"'J; ha~ been halwd
I
"'
25m
20. ooo
.ls 1
--ln(2) "' I. 733 !l
lOtt
n
Exercise 10-6
Ex: 10.29
;,.
Find -:- for fVtp)
use RJ = R~ "" 20 k!l
and I
,~
For / 8
10
1st
froin Exercise 10.28
Assume i~r"" i0 and in ""' is2
10 tJ.A ·""'
""
For / 10
~tA
5 tJ.A "'
'·"'
~,then (!l:)
k
ha~
Since V s
l-.
Jl)
to equnl the original
(Vee- I "'RJ) = Vee- 0.2 soR8, R9, and R 10
can be found by
R8
""
R,, ,,
0 ·2 "" 20 kH
10 1L
0·2 '" 10 kH
20 fl.
0.2
40kH
5
Ex:
fl.
lb) i 11 J
10.30
(a) Find
!""
·.~ i~
(Assume j3,v '"' 40 l
I~ I,.
for (V11,;)
1/1;
.!.!L!1! ~ 6.25 tJ.A
(40) 2
~~
(Assume [3 1• = 10}
f3?f3s
10m
··---
( 1()) 2 • 40
Assume
in= iuv
( i.~)(')
·13v
!3,v
2.5 fLA
Exercise 11-1
Ex: 11.1
IT! (dB]
!T1 = I 0.99 0.9 0.8 0.7 0.5 0.1
A = -20 log
A::
0 O.t
3
2
6
20
Ex: 11.2
0
:. T(s) "' k - - -...,-~,;;_,__ __
(s + I) (.~· + s t s + 1)
Am.,·""· 20 log 1.05- 20 log 0.95 "" 0.9dB
sinceT(O) ·"" l, k = I
A.,.~~ 20 log [o.l~ 1] ·"' 40 dB
Ttmk T(.r) ··"" ---..:.:1- - {s
l)li+s+ I)
t
I
Ex: 11.3
+ j2)
T(s) : k--(s
(~·- j2)
(s + ~ + j }§) (s + ~- j A)
J( l -
i+.~+!+~
4
= k
I
JI -
""'k-
ji
(I
k
,,,2) + t/ (I + Ill~)
w4 ) (I -
+ ~t.l.ldll
(i + 4)
Ex: 11.6
4 s1 + s + I
''w.a:~~;
Ex: 11.4
(; =
T!.l')
!1'(Jul )I = ·
+ 0.1 + j8) (s + 0_1-jS)
+ 0.1 + jU) (s + O.l-jl.2)
~
4'r::'
A(w,)
=
"" 10 log I +
lllus, 10 log( I
J)lv
'
/~~-~,.L 30"
\
--
... ·1i
:. Use N
,..
rr
'y,
----~
\
'•(W--1.·)1~·1
f.-
I and Q
::Q
=
+
-
0.5888 1 ;..;
~=
1.5'Nl
>
30
II and nhtain
A."' " 32.87 dB
For A,., to be exactly 30 dfl
10 log(l l· ~.~ x 1.5'~ 2 !
I o:ns60 ~·· 1
2
0.3654 :::::> Ama<
0.54 dfl
E :::::>
1Cns60 ·-·· l/2
As shown. the pair of cmnputcx
hl 11 /
.h)p
..
.'lO
~~--~
('
rt.~,,I2Q ~
-~~)Jl\
{J)p
N"' 10 LHS = 29.35 dB
N ~ II LHS ~' 32.!\7 dB
I ---.... __ ,l
)(
0.5088
1
f '(
-
Ex: 11.5
- I
-20 log!T(jw.)l
r
+ 0.23 + 1.45)
~ 10 111
10 Iii -I =
(s
w,, -
28.6 dB
JI;3i'
4
.·. T(s)
(s
)
A(3) "" -20 log--1-
I
I rad Is
(>))dll
2
6
pnlc~ ha~
~.
l
20 Jng)l ! 0.3654 1
l<3x: 11.7
real pole i~ at s ""
Tht• .::omph:x conjugate pol~:~ an: at
s = 2 C{>~6(Y> :• .i sin 60"
Tht~
""
•m
0,5 ?:
i
~~
..... 2
Exercise 11-2
Valleys are obtained when
Cos2 [ NCos-
'C:,)J
= l
=
0, 1,.2
• = w .Cos
· (br}
5 k =
0, I, 2
5Cos··l(:,) "" hr.k
:. w
w,
pCosO
.;,_2
wpCos:?!'
5
·-----
-------~--"---------
(s I >(s + 0.5 j .R)(s + 0.5 i
i
;:;+ )}~7-,:-:;--,---1-l l~>r DC'"'" "'~ I
j
j~.)
O.Slwp
0.311up
t<;x: 11.9
~\mu,
E
...
10
0.5
'11i"
A(w:1) c.·, 10
Ex: 11.8
"" 10 logfl
·~~ N=5
Fur A ...,= I dB,
=
E
.
J10° 1 ,__ I
A(~o1 1 ) ·"-' 10 log( I +
'
l
u)iii - I
I
•
~ 0.5088
0.5088 2cosh 1 {7cosh- 12))
68.2 dB
This is an increase of 3.3 dB
Ex: 11.10
Jtoih~
for w
< wf'.
Peaks an: ob!ained when
1
cos 5
0.508&
(a) For the Chebyshev Filter:
~Ne(lS '(·~·-)1
L
A(t•>s) =
10 log! I
+ 0.50SS'cosh 2 (Ncosh - 1 1.5)1
:;:, 50 dB
0
lUp. _,'
N
7.4 :. choose N
Excess Attenuation
10 log( I+ 0.5088 1 cosh 2(8cosh
.icos '('-"-'-)- (2k + 1):?! k
hlp/
2.
·( 10")
"" 55- 50 = 5 dB
0, l, 2
(bJ For a Butterworth Filter
0, I, 2
.·.(J)
1 1.5)1-
E
0.5088
10 logf I
L
'
'(l•l )ZN-,.!
(. - -
s
(dp.
%
.)
h)p ( .,_.0!'. --.-
.~ 10 log(! + 0.5088 1( 1.5)zN I
;Vee
15.9 :. choose N
Execs' attenuation
• !,) -~-
16
2
50
50
Exercise 11-3
£x: 11.11
d '"'
I03 rad Is
selected to ytilld a centre. Frequency gam Qf 10.
Ex: 11.15
(a)
104
t
=..!..
R1
(Rt
IOkfl
= Ml~-tF
H.F. Gain "" - R 2
-10
Ri
R2 = 100 kn
Ex: 11.12
RefurtoFig.l1.14
..L '"" 10> rad f s
-•
CR
For Rntbitrttrily selected to be
17'(jw)l .,_
•·•-1 "'-'
1
'" 0.1 ~-tF
10:\ X 104
The two resistors labelled R, can- also be selected
to be 10 kO e.-tch.
10 kU C "$
w2~>)2
I+.
u
(l!lb- (ll2)2Q!
for any two frequencies m, and co, at wh.ich ITl is
the same
Ex: 11.13
•2
00
T(.r) ""
s
w1 w(J
lilfWo
2
+ sj2w0 + oo~1
(for de gain"' I)
(Ill;}- ~~~n2
.,.
"}
w 1(w;:,-w;) =
::::)(I)IW2""'
=
j;f + w4
1010£
WTJ
(I)
[t +--~~]~A
(wf, - wf)2Q2
1
WIWO
(l)b ::-;;~
=
2
1u 2(t1lo- lllf)
Now to obtain attenuation ""' A dB at m, and m,
where m, ···· t•l, = Bw,
2
-7w""o'-::-:=
At w
wi)l
(wi)-
'
.!. :;:;: JtOk' 10-
I SUB (I)
Q
w0 , !T1= _!_ which is 3 dB below the
./i
value at. de( unity) Q.E.D.
~ .!. ·:?. IJ(}I
IJW" Q -.~
Ex:11.14
j
=> Q :;; ·-·~-~-·--~"
T(.~) oc.·
IO's
-=
/H¥.,~0 ·····1
(b)
This
ln·::·l
l
tOtv
Q.E.D.
For A '' 3 dB
Q
IJW ,1
J10°'- I
OR BW, = m.JQ
BW,
Q.E.D.
Exercise 11-4
From Exercise 11. 16above 3dB bandwidth
= lilo/Q
2r.J0"~2'1T60/Q~Q
6
Q -""· w.,CR
6 -== 2n60 X C X 104 ~ C
10
1.6~F
Q ""' _!!._
w.,L
L
2'1160
4.42 H
6
X
Rx: 11.19
/, '"' lO kHz il.f 3rl 11
"'
500Hz
Q = _[_ = Hl" "" 20
ilf,Jn
500
Using the data at the top ufTable 11. 1
= 0.986 rod/s
CA ·'"
C6
1.2 nF
Rl .., Rz
R .!
2~1071 ";; 1.2 -;; 10- 9
"' 3.17
lnJx)! =
Q2
R6
.
0.69
1.44
~ QI w r
"
"'
13.26 kfl
20
2'1110" X L2 X IO .. q
=
265 kfl
Now using the data in Tablcll . 1 for the bandpass
c.:as-c
Ex:l1.17
Maximally flat ~ Q
oo,, "' 2o. X 100
>
=
K =centre-frequency gain"' 10
l l r 1 /r 1 =!0
1
J2
Arbilrarily selecting R ..,, 1 kH
Ex: 11.20
Eq fl6.25)-
1125 pF
T(s)
L
o------111"----,-:-o
I
;
!
'
C
~R
i
o~--------------L--1-o
R
Also Q
:. L
·co·
/{
bJ.,Q
90 k!!
Sl'lecting r 1 = 10 kH thenr 2
Hl
=
lu 1,
= 2-rr Hf"
-'
w'
8.1408(s + 0.2895toJ 1.)
x
--------'---------eX
2
(/ + s0.4684(v, +
0.4293u> 1,
)
(~' + s0.1789'''r + 0.9883111/l
The circuit consists of 3 sections in l'asc;u.k:
(a l First order .;cction
2.25 mil
Ex: 11.18
L
'T~r--r.
o-------·1
---o
the number cod1icient was set so that the de gain =I .
T\.')
-0.2S9:'it~
s + 0.2S95t,,,.,
Exercise 11-5
Ex: t1 .. 21
LetR 1 = JQkfl
=
_,. !)0
R "'·-1-.""
IT(}t~t)l ~ 0;2985we
·· ·
coC:R 1..
w
I
C""
0.2895 X 2-tt 104 X: 104
(b)SeOOlld•Order l!ection with transfer (unction:
T(.v} =
•.
.
0.4295w~
15.9 kfl
""
2
where the numera«>r cQiiflkient was selected io
yiclcl a de gilin.ofunity
10 kU
=
""
R2(2Q- I) ""' 10(2
x 2- I)
= 30 k!l
a" .
High .freuency gain = K "" 2 -
s· + 0.461J-1w,. + 0.4293t~t;
·=
R1
Ri "" lO kO
Using Eq.l1.63 and setting R2 .~ 10 Hl
R3
. . . .·
1 . _9
l1rl0 X 10
UsingEq.li.62 and selecting R1
= 5SnP
.
4
la),;C
1
1.5 V I V
T!w tf$i~1er function to the output ofthe Jirst inte!rarotis
sKI(C!!L_
2
w.,
2
s+sQ+w.,
Tims the. centre-frequency gain
"' K J1
CRw.,
""'
KQ
lJ/Z
1.5 X 2 '"' 3 VI V
·~: 11.22
Select R 1
=> C ·""
C4
=
""'
R2 = R 3
""
R5 =
10 kfl
1
J0.4293 X 2nl04 X 104
"" 2.43 nF
C6 = C ,., 2.43nF
Q = ..}i)A2'93w,, = 1.4 => R6 =
0.4684wP
-'L
= 14 kfl
C
(c) Sccond·Order Section with Transfer-function;
to) 0
~ '
0.988..Jw11
~-· + s0.1789w 1, + 0.9883w 211
The circuit is similar to that in (b) above but with
T(s) "'
,
R1
""
"'
R2
R3
·"'
R5
""
10 kH
given C "" lnF RL = 10 kH
R ~,. _1_ "'
I
2n5 x 10~ x 10··~
w.,c
R1
"" 1.6 nF
Q ·""" Jo.~ss3 =
O.lr89
5_56
Thus Rt. = Q t 111 0 C ""' 55.6 kH
Placing the three sections in cascade, i.e. connecting the output of the first-order section to the
input of the second-order section in (b) and the
output of of section (b) to the input of (c) results
in the I)V.:ralltmnsfcr function in eq. Ill . 2 5
"'
Ulkfl~Rf,
31.83 kH
IOk!l
R2 = 10 k!l ""} RJ "" R1 (2Q- I)
= 10(10 -· I) '"' 90 kl!
811 w 2
R 1, "
''"
1,)
n
OC gain""
2 ~ R11
·'-"
2
10(~)
5
""·
x::. "' ( b):: '"
Rf. '"' 3 X to
2- 115
2-
'" 16.7 kfl
25.6 k!l
3
Exercise 11-6
Ex: 11.23
.l·+J{ 10-'J
RefertoFig.16 .2S(b)
CR =
.!. => C
""-
L59 nF
2'lT 10• X l0 4
Wo
RJ '"' QR "" 20 X 10 "" 200 kfl
Centre frequency gain= KQ ·"" I
2
2 X 105
X
i
+ s(3 X 104 ) + 1011
s
=
-3
x to4 ±
+10~
9 X
l
5
X
0
=
J9x-108-:..·4-x.lo8
2
= -'0.382 X 104 and -2.618 x 104 rud
s
·K=l=l.
..
Q
20
R~
)+ to- l 10
104
Ex:11.27
200 kH
'" R!K"" 20R
Rju
l<::t: 11.24
Refer to Figll. 26and Table 11. 2
C "" lO nF
R ""· _l_ "' -
._, 10 kH
104x 10 >< 10-~
lo.,c
QR ""' 5 '< 10
,~
50 k!l
C X 11m gain "' In '/ I
R
10 nF
~I
R! I • 10 k!l
gain
,. " 10 k!1
I?;
R~ ,,. Q.' = 5 X 10
+
50 kfl
v,- v,,
~--~ "" 2 X
I()~
C\
·~
C1
t~V 0
+
10\
SC;R.Jl?.,
Sf~ 1 t 1/ R ~ ~- -
.
4
ZOO "'' 50 kH
4
Ex:11.26
~g
F,
R., • 5 Y.
i
I
ClC 2 R,I(,
.
C 1 ~ 10 ..., F. R, .... 2 < !0' H.
w' n.
I ---
SC 1 1?,!? 1
I
~
C 1C2 N,R,
'I11is is a bandpass function whose poles arc idcn~
tkal to the zeros of l(s) in Fig. 111. 29a).
ForC, c.- C,"' lO''ER,"' 2 X Hl' 0 & H.• = 5 X
IO'H
'l11c tr:msfcr function of the fcedb:Kk network i>
giwn in Fig. 111 . 2 8a .The pole'\ arc the roob of
thl' denominator pnlynmnial,
"'
s(-1- +_I_)+
c,u:, c,R,
S' +
s(·-1_ ; _I_ ' . I )
C,N,
C,R,
C,R,.
1-
C;./?1
-S{J(/(C 1 R~)
Prom eq. ill. 75
R
M
=0
-n!
Thus R, = 200 k !1
for C 1
,\
0
In!-'
I? "' l X !(~=-~ = 200 kfl
Io··>
s2 f
.c
-ll
R,./(1-a)
l'o +SCI v() + £<:.1 v, +(I- u)Vo + ciVj
R1
SC 2 R 3
SC 1 R 1 R 4
1(,
Fmm eq 111.76
w,,
A
R,itY
Ex: 11.25
CR "" ?_~
o
'
I
gam
For C
at (A)
v
_.!! + sc (v - v ) +
··-S X 2 X J()J X«
,
s"i-r. S X 10;1 + JIJo
For unity centre-frequency gain
2 X HI' X
0
Thus~(~
"
=
n
=
10·1 =:}
100 kH:
rt '~
0.5
R,
··~
I ·- ''
100 k!!
18
10
=0
Exercise 11-7
Ex: 11.28
S~"
=
S~
'-"'
_!::::) Awo
•
'
., -I%
w0
2
-!::::) ~ """ -! X
Q
2
2
2 = -I%
(c) Combining the results in (a) & (b)
~Ci)Q
-
""' ·- I -· I "" -2%
ulo
~"'
~
I- I
0%
(d) using the results in (c) for both resistors being
2% high we have:
·
.:\u)o "" ••"'oACt ·!· •.-·.. AC2
"''CJ-.>-J("..,--
c,
1<1()
'"' -!(-2)
2
. Cz
+ ::l(-2)2
2
.
2
"" 2-2 = 0%
V,
V., I SC, V,, R,
= V,,(l +SC,R,)
:£ I nt (;\)
AQ
Q
V. Vo Vo
,,<'('·', V o....! + -R + -SC.R,
.
R
R 1 ••• •
1
1
-!-·
0(-2)
Vo SC·4 (•SC· ~ R2 )
vo[S2C,C<~R2 + SC3R2 + SCl + 1.]
R1
-
Rr-
IIC~C4R1Rz
st!
ACt
•0 .:\C2
="' • nc + sn Cz + o
+ (0)(···2)
I 0 '
0%
)<;x:ll. 31
From Et]lll. 96 & rll. 97
Cl ""' C4 '~ w0 TcC
= 21T 10 4 X - - -1---- X ?()
200 X
~=
10~
-
6.283 pF
From Eq. 111.99
c\ _,_
C4 = 6.283
Q
20
0.314 pF
From Eq tll. 100
Centre-frequency ~2ain "' C 6 .,. 1
.
Ex: 11.29
Frnm Figll. 34(c)
R,,R,=R IOkil
CR -~ 2QioJ,,
Ex: 11.32
c"'2Q..=
R ~ R1.il r 0
w0 R
111
C
4Q~
.""
Rr ~ woLQv ... 2r.10 6 X 3.2 X JO-" X 150
2i.fi
2 ;;(4 X J03) J04
5.63 nF
Q '"
R,. ~ 2 k!l=::::'> R 1_
(R 1
I!
R;,.) ( b) 0 L
1(}1 i! 1()-'
(21TX455X lf)J)X5X 10·<•
::!.81 nf
Refer to the rcsu.lt~ in Exmnpk 11.2
(a) :!J.RJR. ·· +2%
='> j,Q I
/3.R., I R,
II
Ex: 11.33
4( D"' 2
Ex: 11.30
!h)
c5
Q
=
BW
~.
J,/Q •·
(2n
~X 2
X
455
I J KHz
455/35 ·
X
X
5
X. 10
24.47 nf
c,
24.47 - 0.2
24.27 nl'
••
15 kH
3kH
Exercise 11-8
Ex: 11.34
To just meet specification!>
fo ." 455 "' 45 _5
HW
10
Q =
· ~.JL 112 R"~
"' 72.8 pF
R"" _J__ =
CB,
72.8
,., 45.5
m,>L
1? 1 1! nlR;,. -"" 45.5 X 455 X HP Y 5 X 10-u
650
n 2N1,.
n
12 X
141.4
C -~· __1_""'
R '"
24.47
24.36 nF
IR!n &
40 -~-Q,~)?
1.36
I
74.7 pF
I ___
CB2 ... 74.7 X 10
I
12
X 141.4 X 21T
Ex: 11.37
191~
.A
). X 15.5 X 15.1 "" 117
Gain or ~ynchronous-tuncd amplifier at to
Ex: 11.35
Lf2
Q
,fzt .
! -
I
tl
Eq ti6.1H)J
"' 4!D
Ratio "" 1 17 "" 2.42
I
48.3
10.7 / 10!') 2 >< 3
X
7.1.7 pF
c I?
73.7 ·-..: 10 "X 2rr X .HO.ll X 10 3
Ex: 11.36
r
- "'
=
r _,_ ~1rB
?.,fi
Eq(l6.! 15)
- (!-
10.7 Mil' ; ZOO kHz
10.77 Mllz
2
Eqnn.llt>J
10.7 ;\Ill!- ~f)f)fl./2
200
.J2
For stage I
R"!itil~t: I X R'!.Ug.t~
6.95 X 6.95
= 3!0.!lkHz
c
1-ll..t Kl11
10~
Gain of stagger-tuned 11mplilier atf;J is proptlr·
tional to
I
I
J2R'"'t<
16 ><
to ''
I kHz
Ex: 12.5
Working from the uutput bm::k to the input and
continuing the equatimh \Vt get I
~! + -~'L-
R1
( "". .
.. R, '
SCR 1 R
V.,
SCR 1R
SCN
Vo )
SCR!"i?
v
(I
SCR,f?
\Vhc"r<.'
R
10 k!i and C
~
16 nF
Thu'
Us)
S X 16/ lO '
The clo><'d lnnp p,>ks are found hy
Usl., I. thai is. they arc th.:: v:tlucs
.~
I S X 16 /
10 ' '
~\·tting
of~-
satitiying
)
sc
Exercise 12-2
Ex: 12.8
c
--
--~-1~--1
v,.
SL2
-0
+
gmV·+1
v,
L2
- -
-
L1
-
I
V
v,. '"' v" + S/.
-.....!!. · · SC
Thus:
2
-SCR;
Vo
Node equation at collector:
v,
v
v,.
·:-'L + g,. v + 1:
.Sl-2
vo .
v,
R
v,.
()
+ -·
S/. 1
-(JW)
+ V"( I
Ex: 12.6
s.
The circuit will oscillmc at the value of w that
v
v,.
l ::2-(!~L···)
0
,, -· '
Since V, t- D, (oscillations have started) it can
makes ~(;'w} a rt~al number.
be eliminated
It follows that w, is obtained from
S 1L 1 L2 C(g,,. +
~) + S2(L,C +
l-1
R
Substituting S
jw
3
CR
~
.WI)-
_L_
w 11 CR
'!1ms. I,
~ -=
I
:=) "'··
.
J'J.CR
2rrJ3 X 16 /. J(f'' X 10 X 103
~
574.3 Hz
re.~ulting
in
L2C)
+S- +I = 0
11 ·- ~./cu_,
L"} 1+ jw
1
For oscillations to begin. the magnitude of
= 0
v.
~(j1o1} should equal to tor greater than) unity.
v,
that is
o1~C 1 RN, ;_, I
4
Thus the minimum value nf R, is
4
R, ·"· - ,-,h)~C" R
4 R
4 R
I
w~C2 1?:
= /;.!_ + L~
L!
c..}
for
- 12 H or !20 kH
f:,.,R
/. 1 ( 1.1
o~cillatinus
to stan
g,.R --, L,! 1-:• Q.E.D.
Ex: 12-7
..,,.,- L .) CR
CR
l•:x: 12.9
2;rl0 '
R
For C ,_ !6 nF R ,, 10 kH
.·. the output is twi.:c a<. larg<: as the Yoltagt•
=·----l_(~_J-li 2X IO'JI IOOx 10 5
IO'ix 10 ,,,
anm,, the resonator. the 1"-'ak-w-peak amplitude
lOij 2!1
b
-1
v
1T
-H2
X 1.4)
--- ., •
·-··-~··•·-~~v-"'-
3.6
v
I.M kl!
!00
40
X
1.64
'~'
65.6
Exercise 12-3
C1 = 65.6 x 0.01 "" 0.66 p.F
Ex:l2 .13
L-= - - " - -
z clc1
C1 + c2
10
wo-.- -
-----=-----:
5
6
lOll X 0.()1 X 0.66 X 10
100 p.A
$!
+ 0.6C,
fl.OI
-5
J<:x: 12.10
from Eq 112 . 24
f~
-10
= _ I _ ""
21T .{l:C,
--;:::=:=====;:;:
21rJ0.52 X (},012 X 10· 12
10
'"' 2.015 MHz
from Eqll2 .25
1
ft• "''
2r.
rz-.
tci'J+c"
-10
CsC"
Time delay =
I
001'>
r" ..
J
2'1T 0.52
X .'
- X
. I )·-12
4X
0.012
-!
ms '"0.125 ms
_L_.
+4
2.018 MHz
=
Q
r
r
21T X 2.015 X 106 X 0.52
12
120
"" 55, 000
~-----~
Ex:l2.11
~·~~.<
'-"'v••·'~-·• -•-••,...
J
111
-12
1!1 -
1.6
R!
lfi kH
Rr
A comparator with a threshold of J V and output
lewis of :+ 12 V
Ex:l2.15
Ex: 12 .12
Vm- Vn "''
~W
Wri= 1 ~=50mV
.
5
R:
· 1Rr
Pt.ls,ihk du•icc R 1
Rz
10
R,
0.()6
Rz
.200 R1
ti)r R1
lkH R2
20() ld 1
Exercise 12-4
Ex~
12.16
~ = _R_t_
Rl +Rl
r
=
100
100
1-(3
'~OJI0365
"
t
2"rtn!...±.l!
2 X 0.01 X IOwt> X
j . ""
0.01)1
+ 1000
1(}6
X In(
LOI)I )
I - 0.~)1
s
!T ""'
274 Hz
l)uring T,
J;;x: 12 .17
V11 (f) ""' 12- ( 12 + v,)e··llS
R,
V8
~-~-~-f
Vn uti"" T/2
z
V 0 -~ 12- (12
IOk!l
12 ···
"'' 2
f
+VI>)
v,>
21'ln(l2
T
+ V 0 )e-T 12 ~
>c
0.1 X JO-t> X J(} X I()J X
1
=
?)
tn( 12 + V 1
.12- Vn
>)
T
111 (12 + v1
12- Vn
500
ln(l2.7)
IJ.J
-12
Ir-
T/2
-
AI ObC. V n
T ..,_,
II,.
=4281Hz
0.7 + .05 ,. 0.75 V
·c-
500
In( 12.?~) = 3, 995Hz
=
11.25
At 50¢C. \'/) = 0.7 - 0.05 = 0.65 V
=
fj"'c
..
-
500
= ''· 611 Hz
1n ( ·12.65.)
-·
11.35
At IOO'C. l'v ~-
f1
•
0.7-0.15 '"' 0.55 V
500
<. ,.
ll>l'.
In
T/2
(I'>~.:l.-'i)
··· 5. 451 liz .
11.-15
Ex: 12. 18
To obtain a triangular wavc:fonn with 10-V peakIn-peak amplitutc we ~hould have
VHI
--\'11
5 V
But \'.,. 1
Tim~
- 5
-~
--·
l.
I
R1
R,
···10:< 10
R::_
Exercise 12-5
Por I kHz frequency, T "" 1ms
Thus,
=
T/2
05
. X
10"'3
To ()})t<
10
R1
6171 fl
6
0.1 Xw·"
'""
X R1
In(:~:~)
3 + 8.7
1.25
R~
5
:::.>Rj '""' 1.2.5 Ul
Ex: 12.20
T ,,
R3
= ·~ + 8 -
6.4
""
R2
To obtain a perfect nmt1.~h at 11"' 8 V we must. have
to select R., so that i = 6.4 mA,
I.ICR~R
At
T! l.IC "" 9.1 kH
t'"" 3V;the circuit provides
.
~ · = 0.6 mA while ideally
5
i '"' OJ X 9 ·""· 0.9 mA . 'lltus the error is
-0.3 IliA.
*At V .., 5V; the circuit provides
i "'
Ex:12.2l
T "" 0.69CU(l
+ 2R8 )
I
J-{){) ~l~
:::.> R.1 + 2R8 = . .. I
4
0.69 X 10-
.,.
R,~
A + Rn
R 1 + 2R 11
+ R11
(I)- (2)
0.75 X 14.44
~
10.88 kH
(2)
Rll =' 3.61 k!l
Nnw. substituting inio (2l
R,1
7.27 kf!
Use 7.2 kfl and 3.6 k!!. standard 5% resistors.
Ex: 12.22
l
5- 3
2.6 mA. while ideally
1.25
i ·"" 0.1 x 25 '"" 2.5 mA . Thus the error is
+0.1 mA.
*AI t! .,., 7 V, the circuit provides
Using Eq (13.45)
0.75 o~
~
5
14.49 kfl (I)
i =
Z+ 71.25
- 3
5
"' 4.6 mA , while ideally. Thus
.
the error is -0.3 mA
*AI v "" I()V; t.he circuit provides,
j
~~
.!Q +
10-1 + !0- 7
i(mA)
i
I
(}----;..
Slope
+
4
•.c.
0.1
At "
(]
,r
2 V. i
0.4 mA
2
5kH
=R 1/fR .m?,
~
, '-....__,..,... 'I I
,.,._;'~---..
i = O.li.:'
6
;
10 lilA, while
6
1.25
1.25
ideally i = 10 rnA. ll1us the error is 0 A.
Exercise 12-6
Kx: 12.23
= 0.5 V
so
VA · ·"' Vo.
+ lin
0.51 V
""
= 1V
111 ""
l V~
i 1)
I mA, v0
u1
-
-
110
=
0.7 V ";\
=
1.7 V
I V -The negative feedback loop is not
-
opcr.:ttive.
ll,, '" ()
~-
. 242 Vr
Thus, t'n
v
l',1"" -
.12 v
········-M/'R
-l
~'
1, 1
~-=
I+ {2.42 V,); I?
1[ I + ~.42 Vt]
II?
.
t[t
~' /(I
Ex: 12.25
I• [ """ I( I
'·' 2:!.5
. . 42.·)
1,., s: '(·,·'I -
?;42 )
•.5
lc~R,.)
{ V,.,. - l.lR,) - 1V,, 1,. 2 )R,
·- (/,. 1 -
-· IR
-
X , . · }.4]
r
;:._ /-_
0.25
2.5
2
X J() X
X
l.4l
2..5
4.H4 V
For the diode to conduct and close the negative
Ex: 12.24
feedback loop,
, :~~=t>~-;.::·
!
'f
:. -""~'-·~·-· ·"-'"~···~~~~~-·~m-~····l-~-~--~0
.:~R t .J,...'
R
l'·'o
·. The opamp i:;, id.:al ''", '·' ;•1 h.•r v
"Ut
in
Given
10 mV
lO mV
N
·.~
i,,
Pn
~.
> 0.
0.7
v
11,
= 111
The diode will be cut otr and
In ~ummary
F,.,,
0 for v, ?
()
l-'n
,,,
0
For positive v1 • the
for
l'J
~;
vr> ·""'
0.
Ex: 12.26
For ;•1
!0 ,.li\
ntA
the op amp and thus
Refer to Figtl2 . 34
!0 mV
'i't.i
must be negative, in which
op amp saturates in the positive saturation level.
R"' tkn
..L
P,
.:ase. the negative feedback causes a virtual short
circuit to appear between the input terminals of
+IV:
D, will conduct and close the negative feedback
O.l rnA
0.6
v
10 p.A
0.5
v
loop around the op amp. u _
0, t.hc current
through R, and D, will he 1 mA. Thus the vnhagc
--'--< 10 kfl '" 10 V
10 + ''m
10.7
v
Ex: 12.27
-1•,,:;:;::
fJV
VI>()·· Equivalant Circuit
- D,, on . IJ, off
15
R
o-- ..
''t > ()
Current llnws from
111
through R,. R,. D, into the
output terminal of the opamp. '\• goes negative
R
and is thus off. The fnllowing drcuit results:
I ""' 15
.'R
A~
v1
f '.]
goe~
R
negative. tlw above dreuit holds so
that i'o = 0. "lltis n<:~:ur~ as the 15 V supply
sources the eurn:nt I ewn for small negative r 1 •
This
~ituation
.!~+:!.
R
rernaim the case until I ·- 0
0
3R
l't!
~
J'
R.
R,
....._.._
l';
< --· 5 V
/)~
nil /J,
·Oil
Exercise 12-8
b) t'l "" I V - similar to the cin:·uit in (a) but
with all of the undergrounded opamp input tcmtinals at
r11 -~
I V
t•u = I V
I = I I 10 k!l
.1)
lo<> ( 0
"' I
+ 0.1
"' I ·! 0.7
''ll "' 0 -· IR
0.1 mA
1.6 v
(c)
~'
v1
"'-v1 - 5
note
11,1 '~
:. tim
r•11 + 0.7 "' -
0-
1!1 -
4.3
10 V- similar to (a) & (b)
-an input terminals (not grounded) of opamps is
equal to I0 V.
>0
'111;
v_, < 0- {)2 off!
""
10 V
I •., 10 == I mA- diode voltages "" 0.7 V
10
V,, •• vii··'· Um
d)
v, .•
II 11
·co·
10
+ 0.7
.···• 10.7 V
v
V0 + V 111
••
= 0.1
0.1
+ 0.7 + 0.1 log ( 0 ·~12 )
"' 0.63
v
Ex: 12.29
l't
_f + /~"' 0.02 mA
Vo=·· ,.,
.. o.tv
~'11 ~
_,...J), i O.IV"'7h
o I
IOk~tO.OimA ''I
"(.l.l + o. 7 + 0.1 log
lei 111
~'
0.1
\I )
·· I V · ll"-:' cin:nit in !til
0.1 rnA
;•1
fl
'"
I '1/
I.
0.1 m/\
1,,
I ' /. .... 0.1mi\
1',.... V. ..... \',.,
v
NB
1\)r all o.:in:uits. cmTenls are giVt~ll in mA.
tano.:e in k!! & vnltages in \'.
0J.O::!~
,. 0.6JV
,.
a) ,.,
\'n + l 1m
re~i>
.. I l 0.7; O.llog( 0 ·2 )
. 1 .
- 1.63
v
Exercise 12-9
I<::t: 12.31
t'ft is ti sinusoid of 5-V rms (peak voltage of
(f) 111 = -IOV' use circuit (d)
/= .OmA
tt0 "". -rJ1
=
10 V
5../2JThe average current through the meter will
1, ""· J.O rnA
be
1.,. "" 2mA
10-,-0.7
+ 0.1
log(D
""10.73V
~ x5
f. To oblain toll-scale leading. This
currC!Jt must be equal to I mA. lltus
.~
rr
x
sji "''
R
I n1A; which leads to R "" 4.5 kll
v..
l!:x: 12.30
will be mnximum when V,1 is at its positive
peak, i.e. v.~
"'
5../2 V. At this value of VA, we
obtain
''r ~~ V m + V M + V tn + V 11 where
Vm = Vm "'' 0.7 V and
V,11
For u1 '·'" {j • i.e.
v2
v0
''''·
lv11•
1:1
! ·- 2 X
-jv11
-I t•J!
-!x- jll1!!
For ,,, os: 0, i.e. ,,, ·"'
11~
=
= 0, t•n "'
S../2 X 0.05 "' 0.08 V
4.5
Thus
V,!m.s "" 0.7
··I n1! and
·- I r>1
=
+ jt•l!
Thus. the block diagram implements the absolute
value opemtion.
Using the circuits of Figt12 · 34 a). with the diodes
reversed, to implement the half-wave rectifier, and
a weighted summer results in the drcuit shown
below.
+ 0.8 + 0.7 + 5J2
Similarly we can cakuhue :
ti,.Jn>in '=
-ll.55 V
8.55
v
Ex 13.3
Ex 13.1
ForQ,
I = V CC
a. PL = (Vol ./2)2 = (8/ ./2)2 = 0.32 W
RL
100
Ps = 2 V'ee X I= 2 X lOX 100 X 10- 3
=2W
15- 0.2
1k!1
V CEsat
RL
-
I= 14.8 mA
V D - ( - V cd
= -
- 0.7- ( -15)
14.8
14.8
= 0.97k!1
vomax =Vee- VCEsat
= 15-0.2
= 14.8V
v•.ru. =-Vee+ VCE"''
=-15+0.2
= -14.8
Output signal swing is from 14.8 V to -14.8 V
Maximum emitter current = 2/ = 2 X 14.8
= 29.6mA
R
Efficiency Tl = PL X 100
Ps
= 0.3 2 X 100
2
= 16%
Ex 13.4
2.53
w
Ex 13.2
At v0 = -10 V, the load current is -10 rnA and
the emitter current of Q 1 is 14.8 -10 = 4.8 rnA.
Thus, v 8 E 1 = 0.6 + 0.025
'TT
At
V0
iL
4
'TT
I
2.53 X 100
2.15
lJ
(d)
Peak input currents = - 1-
= Oand i£ 1 = 14.8 rnA
Thus, v 8 E 1 = 0.6 + 0.025 In 4 ·8
PL X 100
(c)
v1 = - 10 + 0.64 = -9.36 V
= 0 V,
RL
= 6 X .!_ X 4 ·5 = 2.15 W
In(I~· 8 )
= 0.64 v
Thus,
I Vo
p + = p- = V cc X - -
(b)
2
Ps
=59%
X
J3 +
= 0.67V
v 1 =+0.67V
=
V"
1 RL
_!_X 4.5
4
51
=
At v0 =+lOV, iL =lOrnA and iEI = 24.8 rnA
Thus, vBEI = 0.6 + 0.025 In (24.8)
= 0.68V
v 1 = 10.68V
(e)
22.1 rnA
UsingEq.U.0.22
p DNmax = p DPmax
To calculate the incremental voltage gain we usc
~=~
RL
V;
+
rei
At u0 =- 10 V, iEI =4.8 rnA and
rei
=
25
4.8
Thus.~ =
V;
Similarly, at
and,
5.2
=
Ex 13.5
(a) The quiescent power dissipated in each transistor= lg X Vee
Total power dissipated in the two transistors
n
1
I + 0.0052
0.995 V /V
=
~
V0
14.8
I + 0.0017
=
1.1
= 0.998 V /V
Atv0 =+10V,iE 1 =24.8mAandr" 1 = 1!1
Thus,
I+ 0.001
n
2/Q X Vee
15
=60mW
(b) IQ is increased to 10 rnA
At V,, = 0, iN = i" = 10 rnA
From equation 13 . 31
= 2 X 2 X IO-' X
R
OUI
= 0.999 V /V
= ~
ip+iN
25
JO+JO
100
100+ 1.25
Exercise 13--1
1.25
n
v
...!!.
=
0.988 at v.
i 0 = Ia;.. - iaN = 3 - 2 = 1 rnA
=0V
V;
V88 = 2 Vrln [
At v. = 10V,
iL =
1~ ~
=
I0-3
!
)
X I0-13
3
0.1 A = 100 rnA
This
use equation 13 · 2 7to calculate iN
ii - iNiL - fa2 = 0
iN2 - 100iN-1Q2 = 0
~iN = 99.99 rnA
~
is because biasing diodes have
~
area of
the output devices.
+Vee
using equation 13.26
[2
ip = :Q -1rnA
IN-
R
out
=~
iN + ip
25
99 _99 + 1
=o.2475 n
100
-1
100 + 0.2475 %change= 1 - ~- 988
100 = 1.2%
X
In exampleD . Sla = 2 rnA, and for v. = 0
R
out
'!!!!=
V;
=~
~-6.250
2+ 2
iN+ ip
RL +Rout
v. = IOV
IL =
IOV = lOOrnA
too n
Again calculate iN (for Ia = 2 rnA) using equation
13.27 iN= 99.96 rnA
.
[2
22
tp = :g = - - = 0.04 rnA
IN
99.96
25
99.6
o.25
+ 0.04
n
But v•• = Vam
V 1 ln
I
= Vrln
From example 13 . 4 Vee = 15 V, R,_ = I00 0
QN and Q,. matched and I, = J0-' 3 A and ~ = 50,
/ 8 ;,. = 3 rnA
A
100
As a first approximation iN= 0.1 A, i,. = 0, i 8N=
0.1 A -2 rnA
50+ I -
(2)
[l ~~~~,) ~
V,ln ('•
~
;,)
6%
Ex 13. 6
...!.Q. = 0.1
eN(i~~ iL)]
Equating equations I and 2
For Ia = 10 rnA, change is 1.2%
For Ill= 2 rnA, change is 6%
(c) The quiesent power dissipated in each transistor= IQ X Vee
Total power dissipated = 2 X I 0 X 10-3 X 15
= 300mW
For v. = 10 V, I L =
(I)
(~) + V 1ln eN~ iL)
2V,In
%Change= 1 - 0 ·94 X 100
+ V8~:p =
[
!
10-3 ]
X 10-13
2
= iN(iN- 0.1)
( 10-13)2
3
iN(i.. - o.1) = 9 x w-•
If iN is in rnA, then
iN(iN - 100) = 9
i,.'-IOOi,.-9=0
~iN = 100.1 rnA
i,, = iN - it. = 0.1 rnA
For
110
= -10 V and iL
-10
100
-0.1 A
= -IOOmA
As a first approximation assume ip
= 100 rnA,
iN= 0 since i .. = 0, current through diodes = 3 rnA
Exercise 13-2
:. V 88
-
2VTin ( 3 X lQ-3]
!
0.54 rnA
(3)
X lQ-13
1.3
2
3
But V 88
let
1X
=
0.65
v
/vBE 1 vT =
1X
eo.osto.ozs
= 7.39
I= 7.39 + 0.54 = 7.93 rnA
For 11V88 = +200 mV:
Here i1• = 0.1 A
Equating equations 3 and 4
2VTln( 3
!
x 10 - 3 ]
3x
w-3]2_ ip(ip- o.I)
~X
lQ-13
1 X e 0' 1 10'025
=
I eJ
VBB =
(I0-13)2
IR
l.15V
]CI
1.1
2.4
lei
where 8m is in rnA I mV
=
I X
Thus, Me = 0.4 X 2 X 5 = 4 rnA
V 88 =l.OV
Ex 13. 8
VBE =
Refer to Fig.l 0 · 14
(a) To obtain a terminal voltage of 1.2 V, and since
!3 1 is very large, it follows, that VR 1 = VR 2 = 0.6 V.
Thus /CJ = 1 rnA
Thus,
I
=
ICI
+
IR =
0.5 rnA
=
I
vBE. =
1.25
2
0.625
I eJ =
I X e
R
!J.VBE 1 VT
·
=
=
I
2.4
0.417 rnA
-
0.018 rnA
Ex 13. 9
Using equation 13.43
=
I .
I
(WI L),
B•a'(W/L)I
Q
=
0.2 (WI L),
(W!L)p
0.52 rnA
v
e
0.13 rnA
I= 0.43 rnA
I
1·25
2.4
IR
0.5V
I c1 -_ l X e -0.110.025
1.5 rnA
(b) For 11V 88 = +50mV:
V88 = 1.25 V
e-oo5t0.025 =
I= 0.46 + 0.13 = 0.59 rnA
For 11V 88 = -200 mV:
10 rnA = 0.4 rnA/mY
25 mV
2.4
0.46 rnA
V8 E = 0.55V
Ale= g,X2mV/°CX5°C,mA
= ~ = !.1
0.48 rnA
For !1V 88 = -100mV:
Ex 13. 7
R 1 + R2
1.15
2.4
I = 0.48 + 0.37 = 0.85 rnA
iN = i, - iL = 0.8 rnA
IR
=
1.15
2
= 0.575
= 1 X e-0.025/0.025 = 0.37 rnA
u, -
"'
54.60 rnA
For L\V 88 = -50mV
i,
0.1) = 81 x w- 6
Expressing currents in rnA
i, (i, - 100) = 81
i / - 100 i,- 81 = 0
~ ip = 100.8 rnA
=
=
I= 54.60 +0.58 = 55.18 rnA
ipip- )
V In ( -0.1)
- + V T In ( T
lQ-13
J0-13
g
0.58 rnA
V8 E = 0.7V
=
X lQ-13
3
(
- -1.4
IR
2.4
V88 = 1.4 V
002510.025
(WIL),= 5
(WI L) 1
Q,: /Bias =
= 2.72 rnA
I = 2.72 + 0.52 = 3.24 rnA
Fori1V 88 = +IOOmV
0.2
=
~ k~
V GS- V,Y
! x o.25o(!!')
2
L
40
Exercise 13--3
(!J:1 (
(0.2)2
I
Qz:
/Bias =
~
0.2 =
!
kp'(-I1
Gain Error= -0.035 X 100 = 3.5%
(V GS-
jV,i) 2
X0.100 X(~)
X(0.2) 2
2
L
2
_
_ ?:.!...fJ _ 2 X 1 X lQ-3
8mn - 8mp - V
O 14
Ott
=
14.14 mAIV
1
Rout=-----
~(~)2
= 100
!2 k n '(~)
L N (V GS - v )2
QN: Iq =
t
!
1 =
2
~(~t
Q,: lq
J-l(8mp
X 0.250 X
(~)
L
0.22
N
10 X (14.14
Ex 13.12
See solution on next page
=!2 k/(~)
(Vas-IV,I> 2
L p
1= !2 X0.100 X(~)
X0.22
L
P
+ Va,
Now Vee = Va,,
= (Vovl + V,) + (Vov2 + jV,!)
(0.2
=
= 1.4
+ 0.5) + (0.2 + 0.5)
v
Ex13.10
IN= iLm" =lOrnA
:. 10 = !kll'
2
I0 =
!
2
(~) v~v
L
II
X 0.250 X 200 X
V~v
~ vov = 0.63 v
Using equation 13.46
vomax =
Vvv-Vovlsias-vtn-vovN
= 2.5 - 0.2 - 0.5 - 0.63
=
1.17V
Ex13.11
New values of WIL are
2000 = 1000
2
( ~)
L N
= 800 = 400
lq =
! kp'(~) v~v
I X
2
L
2
IQ-3
P
2
= !2 X 0 . I X I0- 3 X 1000 X V ov
~V 0 ,=0.14V
Gain Error=
~=
=
4tJ.lqRt.
-0.()35
14.14) X lQ-3
::3.5 n
200
=
+
+ 8m11)
0.14
4X lOX I X I0- 3 X 100
Exercise 13--4
•
Ex 13.12
Need to prove when V0 z = 4 1i?L then VGSN2
Assume QN off (VGSN = V111 ) SO im = 0 and
= V,n
- 2VsGPQ
.
IL2
4/Q =
= -Vo2 =
RL
~k; (~)
41
Q
(VSGP2
-
+
VsaPQ
J.L(Vo2- Vd
[VGSP2 OR]- VsGP2 = - VsGPQ
+
~-t(Vo2- V;2)
using (1):
ipz = iLz
ip2
(VaP2- Vvo) =
+ jV,pj
J.L(V;2- Vo2) =
V i2 -_
-!V,PI) 2
+
V
02
+
- VsaPQ
=
-
VsGPQ
+
~-t(Vo2- Vi2)
+ 2VsGPQ -!V,PI
(VsaPQ -jV,P!) = V
02
+
VovQ
IL
p
J.1
Plug this value for V; 2 into the value for V GNl
and ShOW V GSN2 = V,N
(- o/.s
~;(f) p(V SGP2 -
IVtpD
2
v GSNQ +A~
2(VsGPQ -jV,P!) = (VsGP2 -jV,p!)
VsGP2 = 2VsGPQ- 2jV,P!
vi2
v ovQ
(1)
for the gate voltage,
V GP2 = (V DD- V SGPQ)
v GP2:
+ JJ.(V 02-
J.1(Vo2- V;2) = V GN2- (- 'l{s)
- ~ -v
;Q) =
v GSN2
where
+ IV,P!
= 2VSGPQ -!V,PI
Find
+ V GSNQ) +
=
(V asNQ -
V g,6Q- V c/tvQ
v,,.)
+ V 111
=
= V GSNl Q.E.D.
Same proof for p transistor.
V;2)
Exercise 13--5
C
+I.OV
Ex13.14
(a} From symmetry we see that all tnmsistors will
cmiduct eqttnl cumntund have equitl Vne's Thus.
t•"=OV
+l5V
~~....t5k!l
-15V
Negleetiiig .I83 ,
lc1 :::: IE! ::;J 1 = 0.86 rnA
But at this current
V/.IEI
= V1
ov
ln(1Isc•)
3)
x 10
3.3X 10- 14
"" 0.0251n{0.86
""0.6V
'Thus, V£ 1 = + 10.6 V and 11 "'0.88 rnA No fur-
ther iterations arc required and let :: 0.88 mA.
To find lc:z we use an identical prot'edure:
-ISV
V11£2::0.7V
V£2-~
If VuE =-0.7 V lllen
Vet"" 0.7Vand / 1
/2 "" 9.3- (-15) ·=
""
5
I (4.86
V BEl. "' 0025
• • n
At !his current, VnE i~ given by
X~~~~} :::0.63 V
3.3 X 10
Thus V,,.1 ~' 0.63 V and / 1 '" 2.87 rnA
No more iterations arc ~1uired and
10-.')
-H
+ 1.0 V:
To start the iterations Jet VBEl
"'0.643V
V~-z "'' 10 --.643 ""+9.357
12 "' 4.87 mA
10 =. 4.87 mA
Finally,
,.
lo
In
Where
iCI ""· in "" ;0 ·~, in:::: 2~7 mA
TilliS,
X
3.3 X 10
lc 1 ::: 2.86 mA
V 11 E = Cl0251n(2· 86
4.86 mA
5
15 - 0 ·7 = 2.86 rnA
If we neglect 183 then
(b) For v1 "'
10-0.7=+9.3V
V 11t:"" Vn;
:: 0. 7
3.3 X IO·IJ e
'"::~ .,
0.6:! V
V
Thus. / 0 "" In ,._, 1.95 mA
Exercise 13-6
. !'t',
i'h
1
The :.-ylll!Jletry of the circuit enableo; us to lind tbe
values for v1 = -10 Vas follows:
ln "" 4.87. IllA lcz""- 0.88mA
10 "" Ic4 "" 1.95 rnA
For v1 ,. + I() V, tt" =·Vet ..... Vnt:.l
"" 10.6 .... 0.62 = +9.98V
For tl1 "' - 10 V, v., =V t;;1- l 181,3
"'-9.357 - o:62 "" -9.98 v
(c)For v1 "" ·l·IO
t'l)::
v:
3
· In ( ·.100
= 0.025
_ .X 10_ 14 )
Vstl
3.3 X 10
= 0,72V
Thus, 110 = V£!- VRE.l
19.86 v
" 1.0.58- 0.72
Vo£4
= v,, -
V£2
"" 9.86 - 9.36 = 0;5 v
'fh·.
US, 1c 4 "' •3 •3 X
.J()-!4 eO.$dl.OZ5
.-: :. 0.02ll'IA
JOV
For sylnmetry we lind the vahte for the case
1•1 ""- lOVas.
ft:::. HlOmA
lcr= lllO rnA
10 ·"" 4Jf7 rnA 10 . "-'0.38 rnA
J0 =0.02mA fc"4=lOOrnA
I{)()
lin "" 201
tJ"=-9.86V.
::0.5 rnA
'1<::.:13 .15
For Q1 :
t9t'."Jl! VT
ic .::: 13,.., l_~p e
Thus. Elfective seale current
= !3N fsp
ic
iu=---
pl,(f3N, !)
Assuming that V8 £ 1 has not changed much from
0.6V. then
-
ic
"' /jl'f3N
Vm :::.10.6V
/ 1 =
irj(f{N+-1)
~-==-~Q;_~
.)
lEI= 1 1 -IR~
" • 0.88 mA
0.88- 0.5 = 0.38mA
In :::038 mA
V/11'1.
c•
(b) EtTective current gain "''·
1 _ l (0.38 X 10-:\)
00
• ~.) ll.
··P
3.3 X 10
Vn ·~ !0.88
v
•.. !5 - 10.58
5
I
0.88 mA
50
X
-t•
PFBifW25
10 • <' .
=
= 0Jl25 In (2 X 10 11 l
0.651 v
Ks 13 .16
See Figure 13. 34
When 1',, " 150 X l!l ' X
0.643 V
10 - 0.643
Vr2
12
jH
JO()()
''HI
'llw$. I 0 ·- 0.30 mA
Now for Q:! we h;m::
\!/IF';' =
=
IOOX Hf 1
0.58V
I
"'c 2() X 5()
;(.
9.357
4.lP mA
R,.,. then /, .,
2 mA
vi!E5
"
VIIn
('f)
ln~::O
lcz
:~
4.87 rnA (as in (h))
""25:>: 10
Assuming tha.t I 0 :: I00 mA.
'" !Hi51 V
Exercise 13-7
1
tn( 2 xJ() l~l-~.'_)
I~
!.,,...
150 X 10''R., ""Q.65J
Rn
=:
II' peak output current
V..,. '' Rn X lOQmA
"-'.0.434V
irs """ Is e
4.34X 100 X JO·'
v- - :v
Jl£5'
RL
I.-
= 100 lllA
c,
tal current out of mode B
A
v
.r
"" 10- 14 e<~.4JM 2s x w- 3
!
8.2
Thus
·"" - 3 -- ]
;!'r
Q.E.D
-·6 mV /"C
hut FromEq.llO. 58
iiV..(;g
.JT
(l + R~·)av/JE6
iiT
""
4
R
( I + ~)
Thus - 6
~~'
X ... 2
2
R~
Ex 13.22
Refer to Figure 13 . 4 4
ForAR >> R 2
,,
2R,
_!,!- ----==
I DN
-·-
100
X
10
"" !2 """ c ~<'V.
L ~ (,S·I
J
e
"'
·~X 2CIVd-
=
-
v >1
j
3J'
I!Vd .... vy
VGJ = 3.32 V
0.1 "
:=)
Exl3.18
I fJP
VL{; ""
R
=
2
v,, "" 6.64 v
vr;o ~. __(~.£~---
20
~ ..32
20 >: 10 '
111/\
n
Using equation
Vc;c;
..:xl3.19
For Fig. 1 1J, 3:.: Wt~ set: ihat for f\!i~'!.ip;tt!tm to be .Jess
than 2.9 W, a rnaxin1um ~upply voltage of 20V is
called for. Tht~ 20-V-supply curve intersc,·ts the
3'Yc distonllion line at a point for which the output
power is 4.2 W. Since
(I i t,)v11f.,, (.I
6.64 == ( l +
Exercise 13-8
I
:n )(
0. 7
i
(
.
I
I
~!)VuE'·-
i
'!..! .) /
R,
41/111.
0.7 -- 4
X
0.7
Exercise 14-1
Ex: 14 .1
Noise margins stay unchanged, because
rn the low-ouput state, the transistor is on ami
R,,
V ot.,V011 • V JL· V 111 only depend on V 110, V,, and
I ___
= rns:::: ----,-,.,.-,..:....
V,. Since V, bas not changed, noise margins
c !!.(v <;s -vi•·
J.ln ''·' L
stay the same.
Inotder to calculate the power dissipation, we
need to llrst recalculate
Tht:refl)re, the current drawn fmm the supply in
this ~tate can be cah::uh1ted as:
f f>l>
.,_,,
I
\l
_J!Q_ ,,,, 50 l.tA "'~ RJ} + r 1,s
Ro .,.. ~'~>s
2.5 v
50 11A
·I
JJn·
"" Vnn- Vot
nn
Rn
=
L 8 - O.IZ ""' 168 uA
10 kfi
'"'
1.8 X 168 tLA ' 302.4 IL w
50 kH
151 p..W
Note that keeping V, unchanged resulted in
higher power consumption. but noise margins
stayed the same.
Substituting for
~') l
R o + ros:,.
.,
-·--soriJ:>kH "'¥rf)s
':> <;
.... 2 '"'1
"'
Ex: 14.4
and hence:Rn '·'· 48 k!l
To determine V, , we use:
.,. I . W
we usc
o n Jlalll
7: ,
and with Rn ·•··• 10 kH and K,
v,
unchangt~d:
V ,
Rm
vt!
Rm
..2.:: '" -
1 X 101 X 125
. !:1::
.. L
=
'5
= 0.089 X~ "" 0.22V
10
'
To calculate the new noise margins. we hnve to
!O·t• X (2.5- 0.5)
X
=:> V
lind V 011• Vm. V 11.. Vt.n ..
?
1-' 011
-
V nn '" l.ll V unchanged
-co
V, ' V, ....- 0.5 1 0.22 ..., 0.72 V
V 11
\\'hen the switch i~ closed or in low-output state.
the current drawn from the supply is 50 ~lA.
v/II
v1+
=
1.63 jff;;;;fl~
- vI
Pm, -· l'm,l rm "" 2.5 X 50 X 10 ''
125{111'
When the swiich is npen, Ill) current i., drawn
from the
~upply:
P 0 [)
=
UIV
0
___
VDD ___
llur =
-
-
I + Vm>- V,
v,
Ex: 14.2
\Vhcn input is low. the output is high and equal to
V011 . In this
R c~ .
ca~.:\
thcr~~fore
lienee,
1.8
I + 1.8-0.5
V,·,
~
NM 1
The power dissiption
::;
\Vhen the input is high. the output is loR' and equal
to V m . The switch is cmmcctcd to R,-, . Hcn~c
.5
2
;<
l
Voo --~·££
Rn
3V
P,
. .•
" ""''"''
Ex: 14 .. 3
.
, o (IetcrnlJlit."
II'
stitute \',
O.OWIV. R,
E, we usc:
L'
J\ .,
R
become~:
1 v[)fl 1fJ[} -... !v
2 /)/)
5V.
T
0.22
NM 11
the- switch is conncetcd to
the .::un-.,'nt through R n is zero.
~-'uu ·•
! :-:
Nntc that ke.::ping
0
I.S ;< 1_,8 ·- 0.2fl '·' I J9 ll W
2
l0 kH
~;:
unchanged resulted in lower
noi'e margins and highcr power dissipation.
k,:
~oo 1 ~.A · v' ·
J()() X !()
/
W <. 10 / !()
L
0.26
3.75
v
Exercise 14-2
Ex: 14.5
~ Jg~~~j:
k,
=
Ex: 14.7
FtolrtEq.14.36wehavc:
!t ~J
0>•
3
From Eq. 14 · 20: V 011 -~~ Vm>- V,
::: 32.4 X 10-11
""'-
32.4
JJ..W
ci V 2 ,,m
v,i
C.11' 2 nm
1k2r(Vm)- 2V,)
v
66.73 :::66.8
From Eq.l4 . .?.2 V 1c.
0.5
v
unch:mgcd.
From Eq. 14.23
VM "" V f>l)
1
..:x: 14. 8
(VM-
"" 0.12
•
,, fCV 01,
6 XI00XIO··Ux!8 2
Pi(>nl. = 100XI0
•
1.3 V
unchanged
From F-t1.·14. is
V ..Ot ="·
P,~y.
Jtx: 14 . 9
+ (~c_!,~~ --~ 1.8 + (3 - 1)0.5
k,
+I
...
---Ttl·--
l',.(r)
=
V,('<)- (V,(:c)- 11.,(0'))e- 1 1T
"" 0.7 v
From E(j;14 · 26 t.()gethcr with St~tting
.. I:
fon 1,w = .10 pscc. with C
1
= ! .S . If~(· 0.69
lOP
:. 32 [0.12- VIII+ 0.5 + 0.12] = L8- 0.5- 0.12
9[0.74
~
V111 1
l.IS~ Vm ·"'· 0.61V
Nlvl 11
v,m -
NM 1
Vn- V0 1. ~~ .5
V 111
=
1.3 - .61 = .69 V
.12 ''·' .38 V
Ex: 14.6
The inverter area is approximately
IV
A ~' W 1 L, + W •,L,• Since
___! = K, and
.
L,
W,
-~ =
L2
I
·:·.we hav<: IV,
II,
L, ·= k, W 2
d and
•
Asstuning k,
~>
I. we have
w..
Thus:
A
k,L,/. 1
W 2 kr1V,. -~
k,t/, + k,.W' 22
=
l 0 IF and Von= 1.8 V
1.2 mA
Exercise 14-3
Ex:14 .. 10
throu$11 the Pu whiclj is 20 1¢0:
V(l(t) ""' V;i(i») ~ (V 0(~f- V 0 (d'))e"''1'
z<1 V:vn + v. wJ
l3S pSec
For trHL• the output starts ai il08 and goes to VOL
through Pd which is 10 kO .
fpUI"'" 0.69R.C=0.69 (20K) (!Of)=
FQr ,,.Liflbe:ou~p!Jl SIUtu ai: vQL and goes to Voll
.. ,
""'
vOil- ( ·v(!II -
1
z(Voii
·
+ Vm)""'
.
..
Vol.- (Vm.- Voy)e
··-•,.u~.lr
. . -•t•t;t/T
Vm)e
(~~Van+ ~Vr)tJ
- (il,;,} + Vin.
T(-··ln(~))
= tptu
""
Inn "" 0.69.R.C "" 0.69 X 10 K X IOJ "" 69
tp "'
~(t 1,nr + tl'l.ll)
~( 138 p + 69 p) "'"
p!ici;
H)'' pset
Exercise 14-4
F..x: 14 .11
V 0 (t)
V0 (x)- {V 0 (i.,)- V 0(0+)]e-' 1'
Vo(t)
() .~
rv[)n-
tris when wltngc is
,
, IVm> = - V 00(~
O}e~I!(21C' I!.IOh ~ ~
= .l Vt>o
··(0·· 1/)17
-ln(.l) "" tf_t
r
2.3 · • ;;~ 'r
tr"" 2.3 · 2 K · IOO I
0.46 nsec
Exercise 14-5
Ex: 14.12
k.'(r)J(v;- v,n>Vo-~v,1]
~ 0.2 X 10·~ = 50 X 10'" 6 X (r).
;I)=
a)FromEq,l4.S8
V
,....jV,IJ) + V,,.
... r(V,,p
··r+l
M""
0.6 = r(l.2 ~ 0.4)
+r
=0.8r + Q.4 ~ r
r
·.or
+ M ~0;6
[(s- 1)o.2- ~ x o.22]~(it ~ 5
""" I
= r;;;;;,.··. ·""· U·~-~!
~'j;o1wn.
+ tl.6r
,.)4-
'"'·
I~~
From abt.we:
4
tVt,
·wEt
EX!
ov
~(5 X
= ft(SVM - 2V,)
Vm
·~~ ~(3Vm>+2V,)
:::: 0.:55
~(3
1.2- 2X 0.4)
X 1.2
~(~)
= 10
L I'
14.14
UsingP..qsJ4.63 to H.67wehave:
u.c
I Pill, "''
.
+ 2 X 0.4}
U.tt
.
k.'(z). von
=0.6SV
Vn
= 2w
II
~ (w1, "" 4 X 0.13 1un "" 0.52 J.tlll)
b) Vvu "' vllf> ··= 1.2V, V(n '"
wl'
'
z7__3v,.1
~::::
4
Vnn
( ~~·· )2
/)/)
v
NM 11
V 011
-
V 111 =•= 1.2 ·· 0.65
0.55
v
Noting that
c) The output resistance of the invc1ter in the low-
t l'l!t
jV,"j. then :x.
_.:...:;:e;;_;~~::_:_:.::__ _
300 X 10'" 6 X 1.5 X 1.8
1ru1
6
C'O:
1.99 X 10 X 10' 1 ~
=
output state is:
430 X l 0
v,n
= -
a,C
k/(.~)
\1 0 /)
1. I'
"" 24.7ps
or
X I ( 1.2 -· 0.4)
24.6
X
4
X 1. 5
""2.9 kH
Since QN and Qp are mawhcd, the output resistance in the high-output state
the same:
"' 49.4 ps
2.9 k!1
rusr "' rosN
d) For
i~
(!Y)
(!!)
L
L r
I
t,. -"' 7,(11'1.11
+
=
O.S( 1.2- 0.4) + OA
I + 0.5
M
37ps
From Eq.H . 6 Bwt~ haw:
0.69RNC and if we substitute for RN
tPlll.
hence: V
~(24.7 + 49.4)
Ex: 14.15
-
= O.:l,
X
1/'Hl)
4
1.0. we have
n
J4.rr--I
3
0.53
v
from Eq.l4. 70 i.cRN =
12·5 kn then:
(w)
L"
Ex:14.13
ll>ing Eq. 1.4. 51l ·3nd 14. 59
0.69 X 12 ·5 :X: 10 1 XC
8625C or
(IJ,.
r( 5 -
I) ;.
J
~····················~/'
r+l
/t-tp~~-:1.' :::::> ~-l_:r
·~Jl"lt'"
When V 1
H',
\inn and \1 0
=
0.2V.Q.vnpcrates
in triode rt•gion and hence the circuit is given as:
:.(. l.Y)
L_
=
3.5
,1
Similarly. using Eqs.l4. 69and 14.71 we obtain:
Exercise 14-6
.
0.69Rp C
Ex: 11!1.19
= .0.69 X 30 X lu -.-.c~
a)As trientioned on page of the Text, C1n1 is the
t;ontribution of intrinsic capacitances of QN and
Qp .. Therefore,
(t)l'
or
Cine =
2egdl
+ 2C~tJ 2 +
c,,, + Cnhl
+ 2 X 0.3375 + I + I ""' 2.*.) fF
CgJ + Cg4 + c ... ""' 0.7875 + 2.3625 + 0.2
.·. Cint "" 1 X 0.1125
CUI ·""'
= 3.35fF
b) From Eq.l4. 79 we have:
8.3
t1,
Ex: 14.16
t 1wt and 'NR arc pmportionaltq C
t Pfll. - .orginal
"'
·.. ·..
""
0.69(R.qc., +
part of1P is: 0.69;R..,C.,, and in order to reduce
c~
•.,
Cn~d
the e>ttrinsic part by a factor of 2. S has to be
increased by a factor of 2. Note that S '"' R.9
and
R.~
TI1crcfore,
c..hl + Jp ·.,
I
2{1nn
ex.
28p
11
!· 'nu) '"
* .l0625P.
625f
'poLo·~
Rc-q
has to be reduced by a factor of 2
or cquivalenlly (. !!') and{!!') have to be
6.25f + .tp
L.,.
..10625 pF
I
~R.qC:•.,) The extrinsic
im:reascd by factor of 2.
c) From Bq.H. 79
C
·d:,::
"" 0.69( R.,,0 C;, 14, + ;R.,1,1,c...}Itence,
476 psec
Ex: 14 .1 7
2.9 + 3.35i2
W1, is reduced from I 125 jlm 10 0.375ttm
375 X
. .0·1.125
.L . . 1,
2.9
100 = 33% reduction
X
+ 3.35
28 p~
·"' 20.5 ps
d) A ·"" W X Land since
(z) is doubled and L
is constant, then A or area is a!so doubled.
C
(4 X 0.3.l75f) +If+ If+ (2 X .78751') + .2f
Ex: 14.20
4.225 tV
I Pill, =
24.6
X I() 12 ;,.:
(~-225()
6.25f'
Using &J. 14. 35
16.6 psec
PdP'
I
fCV 01 / =
X
109
X
6.25 >< 10- 15
=J9.1
p.w
1l1e maximum possible opentling frequency is:
!c 16.6p + 21.3!') ~"
2
19 psec
/,.,,. '~ ;}- . Hence.
~tr
PDP
Ex:
14 .18
!7.9 Glh:
The minimum period at whid1 the .inverter can
rdiability opcra.t.:: is
Tmin
r PilL f t rur· Thus..
2 >< 28
"' !7.96 Hz
· ·· ('. ·x··. l'. 2 [){) X t,,
"' ...I-. ·"
2r 1,
6.25 X 10,
2
15
X
2.5 1
19.5 fF 1 J
X
2.5 2
Exercise 14-7
Ex: 14 .21
.. ~ . 1.e.,
.
' seale.d
The power u<:ns1ty,
d .PJ..,
1s
eJVearea
a) Pot NMOS devices:
~ = ll
L
=
0.27
OJ8
0.18 X 1.5
0.18
For PMOS device~:
0.18 X .,;.·_,
,
.~ ...
,., 2.16
4 p~,-L
0.18
O.IS
b) For NMOS device~:
W
-'~'
!!{.o4n.,.,O.ISX4XL5
/,
0.18
For PMOS dcvi;;:.:s:
}! ,, I' '"' QJ.S.
L
0.18
y
1
.
Using Eq.14 , 94 we have:
0.25 X 10
LOS
0.18
i~
Ex: 14.26
Por the NMOS transistor, Vos "" 1.2 V results
l.
Ex: 14.23
.
. power
· d't!>stpatmn
. . ts
. sea Ic db. y -;;
I
,S mcc
t Iynam~e
S"
and propagation delay is scaled by
~,hence. PDP
Ex:l4.24
\Gtlc
by
~ . thu~
CV~n1>. i-> st:alcd by
Cl
s
X 1.5 X 0.34( 1.2 - 0.4
(I f 0.1 X 1.2) "" 154.7 I-LA
if velocity-saturation-were absent, the current
would be:
hoth larger than
a remains
~cab! by ~.and
linergy/Switching c~dc. i
I
b
v{J/)
scaled bv S. therefore t
•
greater than V n.m• ~ 0.34 V. Also,
1\' Gsl
any·
mnrc. They are kept constant 1
-..:
1.2 ·- 0.4 ""· 0.8 V which is
in the ab,encc nf velocity saturation.
For the PMOS transistor. we see that since
If VJ)D and Vt <~re kept eonst<~nl. the entries in
Table 14 . 2 that ehangc arc llows:
I'
v,. '"'
X (I+ 0.1 X 1.2) = 231.2 J-lA
Satumtiun is obtained over the range
SnPDP
decreases by a factor nf S.
1
in Vr;s-
in "'' 430 X 10
Obviou$ly. Vnn and V, do not
101 X 10.. ,
""0.63V
0.54
0.18
s'
X
4oo x ur· 4
Ex: 14.22
(a) ·ntc minimum current available to charage a
load capacitatlt.'e is that provided by tl ~inglc
PMOS device. ·n1e maximum current available to
charge a load .;apacitancc is that provided by t()Uf
PMOS transistor~. ·nms. the ratin is 4.
!b) Th.:rc is only one possible configuration (or
path) for capacitor discharge. Thus the minimum
and maximum cuntnt~ arc th<~ same
=> ratio
-6
''
~
X
11 ()
v
~aturation
X
1.5 X 0.6(1.2
0.4 )"(J + 0.1 X 1.2)
59.! (.t/\
1',.\ <
v,,5 C',
1.2
v
or ()J:{V
~~
lins
<,
1.2V
Note that the velodty saturation reduces the
NMOS ClllTt'nt by JJ% and the PMOS curn:nt by
-~ 7CfL
Exercise 14--8
Ex:l4.27
a) Using F..q.14, 102 we have
in
l,;v'(;sl nVr
·
Thus, log 'n
~-' Jog(~+
Vus log(e)
-V
It
T
Therefore, the slope of the straight line represent-
ing subthreshold conduction is. given by:
11V 1
--""' 2.3nVr
Jog(e)
b)Vr = 25 mV for i 0
IOOnA
atVca = .21V
lOOn =
/se·l!IU2i2Sml
f., ··· .InA
in = .lne
'• ol'
r "" .InA
c) For Vcs ""' 0, i0 = .InA
Jtmul
500 X 106 X .I X 10- 9 = 50 rnA
l,.,a1 X Von
= 50 m
X 1.2 = 60 mW
Exercise 15-1
15.1
(W/L)3
fnn '~
1.5
""
0.32
(WIL)P =
0.5 ns
since \1ou """ Von- \1,.
tn.H ""' 0.03 ns
Substituting values, we. gl.'t
15.2
0.5
V,
Using eq. i1 S .11
(
(2.5 ... 0.5) :<
0.27 V
""'
ll:!
03V
1.8 V -- V, I 0.85 V-
J. . . . . . . . . ...
O.R5 V)
+ £J.3V 112
V, "'' 0.5
Squaring both sides yields
v;- 0.446V, + 0.05 "" ().(}9(2.65- V,)
using eq. u.s .13nnd t15 .14
V,- (Vnv- V,)r I·-
NM 1•
·I
./ .................- ........................... -
J2.6sv- vi- o.:w' n JO:s·!fv
v,- o.223, o.3J2.65-=v,
[t- Jl- ~]
V 0 ,_
v,., + 1 (Ji'rm -=·v, + 2:[;~- Ji;J;;)
v,
TilE NOISE MARGINS WILL NOT CHANGE
L
r;=·I'Y __ .
~•
1 --]
/y( 'I + I ).
..
v; -- 0.356 v, -" o. 189 '"' o
$n that.
Solving this quadratic equation, yields nne
0.7 v
NM1
practical value for V, :
\1 1
~'
NM 11
(I
(Vno ·-
~ ~) -=
..,13.4
v,>(t ·- ~)
t2\ X
0.648 V
0'
Vrm- V, .... l.X V --- 0.648 V
\' 011 "'
"" 1.15
v
0.85 V
15.4
. (w)
I: "
ll.,f",
'I
oc·
'
. (w)
!.lt.c.,.
T
.
.-.(T)i'
/)
(a) Referring to Fig 15. 12 withnut loading,
115 J.l cl.375)
0.25
4
cv) ,.
t_I
Jo 1
5 V
V011
-7
\! 01
-)()
V
(bl Refening 10 Fig.ls 12 (a).
.
1.44
'n.v(o) ....
I (··w)·
:::,!.:,
-L·· (
-
v,,- \i.,,)
J
Jj
<
using cq. 115. 12
I
">l~l
P0
"" !c;o
1.4-1)(2.5 ··· 0.5) 1 = Sh..f ru\
2 ," •tl!
t'
.
I,.,.,Frm = 86.4 J.L2.5 = 0.22 mW
"'
using cq:tiS. 15 and
as. 16
'~
1.68
·(w·)
.
=
.
I
:::,kr•
(linn- \',.,) 2
J. {'
lpp(o) =
~(20tu\! V2{1J_;~}_~
. . ''·/. (.375).
......... .< 2.;,.25 .
I I'\ 'C I!)
.
+ 320 JLA
= 1120 JLA
To ubtain iv.vUnu). we note that this siwation is
idcntio:altn that in Exumplel5.?. (s·. 2'v)· _2!(s
V) J..
2
.. 2
2
icl.,,
=.325 p.A
&o1vb1g for
~(li~Op;A + 3l5,.r;A) = 7t2:5 p.A
=
c(v;'')
.
clap
tJs v -lv:;r+·o.6 v - Jif6V 1
JV,~- o.6l3V = o.sv'' 2 js.6v -- w.p]
7225( l0:- 6)A
squaring both sides and setting one side equal to
zero. we have the quadratic equaiion,
== (t24 Its
(c) Referring to Fig.ts .12(b);
i,>N(o)
""lk'(.~)
(Vllh2 n.L,. ··
~(50~tA IVz)(i)( 2y)· - !(
V)
'.2
2 2
5
2
]
V 00 which corresponds to V ,.0
2
""'
Rr6 =
T
-y[ff+~- ~]
, r /5 V
r;;-;-;-,]
; o.5 v -lJT + o.6 v- .,o.6 v ,
J
i
I
C
U1'
l
1.49 v
L
I'
Rrc "" 12 ·5 kH =
' (i},
g:~ k!!
r,
21) fl A . 70ll JLA
"~
0.69J(C., 0 , 1 + Crr; 1 )R 1
X IR 1 +
91·1 Jt A
12.5 kH
I
IC;.~ ·I C 71 ;~)
R~)
"' 0.69(( !OfF+ SfF)( 15kf!) + f !OfF
X (IS kn + 125 k!l) 1
c( v~'>)
=
(I}
Using Eq.llS. 38
So.
'I'll I.
lU kH
with Eq. jl. 5 · 3 6 we see thut
1.49Vr"' 20 tA-A
!_!,~>.E:!l 2+ 708 1~!~
!~Si :.:
30
30 .
R,. 1 -= (~) k!l -, ( 2 ) kH = 15 kH
icUrm) ": ii>N(IP/1!) I· im,UrmJ
= 688 JLA
~~~·~)"
15.7
rp
J-
1.5
lhingEq.tl4. 71
= VIO +
~(20~tAi V1 )(i)[ 5
=
Von
·n1us, 'nrUPiu.
.
>= 2
Jk·(w)·
·,. L P [v/)n
2 - ·v
! 1].2
=
(I). (it
Using Eq.a.s. 36
::: 688/iA
To find im.( tpm). we first determine V, when
"' 1 v
= 0.69RC = 0.69(5.5kfl)(70)( I0- 1 ~)F
15.6
5
1
.
liS
tpLJI ·'~ 0.27ns which is close to the- value of
0.24ns obtained in Exercise 15 . 14
f)/)
IV.,.I
~(0.24 ns + 0.1.9 ns)
= 5.5k(l
.,voo _ !.(vm>·)~
[( v _ v"'2
22J
=
=
15.5
ic{o) "" i 1m(o) ~· lm,(o) '" 800p.A
r•,
= 1.6 V
I V)2 = 320p.A
= ll20~tA
= SO
.
IV,A
= ~(tpu1 + tPF11)
:: 0.22
~k~{~)1,( VIJD- V,,i
ftw({)) '
jV,~,
jV,~ "" JV + o.sv'' 2
7!J< to-" 15 )'(~)
t;.w. "" -i.-- =
jV,J
v,.. + 'Y[JV,, -jV,,,] + 2, - .firl
w.~ =
::275 p.A
Tbuli;i,;(tpui) = SQJ.LA + 27!ip.A
<2,;\11/iJHumoffwben V., =
where
t,. "" 0.64 ns
0.19
ns
I
5fF)
Exercise 15-3
"'lxsox
2
1(5- t) 2
= 400~-tA
= 50 X
I[ ( 5 - I ) I -
i
X
I]
"'115J.LA
=
iodv,. ·"" 400 ·; 175
(c)illl
L.,,at c:·.
288p.A
c,_, Av,,
AI= Cu(Vrm- V,)"" 40 X Hf 15 X 4
imlu,,
288 X w··fi
""0.56 ns
(d) f'OIIowing the hint we nssume lhnt Q"''
remains satumtcd during At .
imj 11 ,,
. 1>"' '"
11)2(
=
'(W)
L
--
lI
im!,... ••
:
I
3V)"'" :;k~
(3- I)
f'¥(2
.
'
X 50 X 1(3 ---If
= l()(lJ.LA
15.9
. . ., 1 1 _ 1( c·
S.mcc
1n( • Ill> ·· ::; .J.Ln "'
J( Lw) (.•,.tw -
...
doubling
'"')_will double( Lw)
(z
so i 0 (VM)"' 2(76.1 J.L;\)
This new(
v' -)2 ·
i'lj
"'I
100 X 10. 6 X 0.56 X 10
and i 0 ( V v!J)
152.2 J.LA
z} will also double i (V;") :
0
4
;,f';&)""' 2(68.9J.LA)"" 137.8 J.LA
Thisdoublesl,,.to 2(72.5 JLA)
the new tpm_ is
o_·v)
145 J.LA
= - 1.4 v
Thus, "r2 decrease lo 3.6 Y.
15.11
0
Vou
V 01_
=~
-0.8:! V
SHOULD BE SHIFI'ED B '( ·- lUiS V
V 011 =
0.88V AVrER SHIFfiNG
·
11 01 •= - 1.76V AFTER SliiFrJNG
l'
C··(· \' /J{J - -:;-<
=
40
9
X 10- 15
-
15.12
Refer to Fig. 15 · 12 Neglecting the base currelll
of Q 1 , the cnrrentthwugh R 1•
n,. JJ2 and R 2 is
P-5( lO '')A
15.10
_ 5.2- vm- Vm
I-·············----
Refer to Fig. El5 . 10
-~(-·
R,
!!:)
=
~(---~~)
__ f.
0.907
'
R1
5.2- 0.75- 0.75
"- ._I.
(. ~-)
L."''
f
_I X ~ "'
2
2
~-
4.98
l~,
~'
0.6285 mA
-0.57
v
1.32V
Exercise 15-4
15.13
r•011
Referto Fig.1s .26
-
/E.. ·"" -
E
0.779
Vdok"" -yX4XRd:"'-4X0.245 "~ -1 V·
Vdo.1 (!8 "'
OV (because the current through Re~
is zero}
15.14
v
::::0.818
vIll
4.018
=·
1.205
-
Vc!un""
4.12 mA. The
3.98 mA
11
o.?s + input rc~i.stancc into the hasc of Q_1 is
1.205 . 0.788 .,_ 5.2
Thus. ~'ciux
0.788 V
-0.88-0.788 + 5.] '' 4.53 rnA
0.779
V ud Q:l = 0. 75
v.,. 011
0.783 v'
and correspondingly
=
= -0.22 X 4.53 ~· -I V
VN(Jll
- 1.205 - 0.75 + 5.2
··----·
. ·"' 4.166 rnA
0.779
Thu~ a
0.906- 0.798
E "'" - - ·-
-= 0.767 v
IQ.I
6.88 mA
A better estimate !'or V nF.Iv:~ is
= 0.785
and eoncspondingly,
.
x 4.12
0.9()6 -· 0.75 "-" .,., 1.656 V
656 ±.1 '"
=-1·0.05
,.,,
(JJ
-0.22
v
"" -0.906
liNOR ''" ••·
·nms a better estimate of V 1u:j 011 is
F1/tlo"
'
The value of IE we found in Exercise 15 . 14to be
Assume FnEiaR .., 0.75V, IF. '"' 4.018 mA
X
''oR is
0.49- 0.8.18 "" -1.31 V
15.15
··- - 1.32 - \1 lWQN + 5.2
0.779
0.99
15.2 mA
REFER TO FIG.lS. 32for
I£ -
''"
+2 =
Thus a better estimate for
V(IR "'" '-·
Refer to Fig.lS. 28
For\/1 '"" V,1;, 1{>8 "'' 9910 ,,.
I Qll
0.05
v1u:IQ2 = o.1s + o.o2stn(1;·2)
m
""
1.24
A better estimate for VnHif!2 is
+ 5.2 . 4 A
..., 1.32- 0.75
0.49- 0.75""' -1.24 V
-
11E Q2 = -
VII- VsEIQR -. (-V,;e)
-R
""
X
0.245 = -0.49 V
I
5.55k!1
Exercise 15-5
Th. f) ...
US,
v, + 0.79- OJ
0.22
.
,. 0..99
X
v,- ().79 +sa
0.779
·=>
v$ ... -t>.ss v
15.16
Refeno .Fig,1s. 26 For the referenceclrcuii, the
current through R1, 0 1, D2• and R2
d) See figure below. Assume V»t: '"" 0.79V
.is 5.2- 2 X 0.15 = 0 6··2.9 A
4.98 + 0.907
• · · '"
VBI(It ""'
v It
-0.57
v
= - 057 - 0;75 ""' ~ 1.32
I I! !·Ql.
·""' -
v
1.32 + 5.2 ""' () 636 rnA
6.1
•
Tbus the· refereru:;e circuit !!raws a current of
(0.621J + 0.636) = L265 rnA from the 5.2 V
v<>--i
s
supply. h foliO\Vs that the power dissipated in the
reference Cit"C\d.t is 1..265 x 5.2 = 6.6 mW. Since
the reference circuit supplies four gates; the dissi-
'
-5.2V
(because the current will be 4 to 5 rnA). At the
range of saturation,
lc = -·
·)2
v,.
2
" ~(·~oo
P:)(L 0~)[(1.8 -- os1(·!&·)- !(.9)~]
.• 4
0.18.
2
2
.
l
(12 X 10
b)(.!!:')
L
~
172 X Hf~
=
(·!!'t "·) "" 2.5
a
14.3 "" 2 ·6
(US
Ex 16.2
BiiS for row
=::.
~
fJ.nl
Ex 16.5
·C,, X .6. V :
I
addr<~ss:
1,_ "' !(,
2 .-n C. "'
2M~·· 1,024
log 1 (2M) ···-~ log 2 ( I. 024)
.. (IV)
fl.
~ ~~1f:t_L~.024l
10
>("")(V
-I' - V
L
fJ/}
. "'
Q·)
2
a
~X 300
I,
.
2.5
Ill) . -
L
·r·o hn
. d I ,. we usc
~
X
10
"X
2-5
X (
Ul- 0.5 -- 0.5)1
lng 2 ! 2)
240 f.LA
Bils for column address:
ut "'
7
(bl
Bits for block address:
2
10- 12
X
X
0.2
(Tw)"
=
1.s
15 = !x300X 10
J
Bits
1.7 ns
240 >< 10
,;X
1.5X(I.l\-0.5···0.5) 2
log_.(:'\2 i
I"H J.lf\
logJ(2)
Ex 16.3
ut
or:
1 X 10· 11 >< 0.2
14"1 J-LA
2.8 ns
----~~····-~--·~~----
~~ n
J..
!,
o ---1.....
('.!:)
therefor<~.
L "
.ilr
1.7 ns >< 2·5
1.5
2.~
ns
Exercise 16-2
Ex16.11
Ex 16.6
~ (!f) X 1Lnf I (!f)
L "
· L. ., IJ.,,L
V,.
(I -
V m> -
~ (!) X 4 J< f I - (1 (!)
L
L "
L
P
·)z]
v ...
)2]
0.5
I ,,.
L8 - 0.5 .
""2.5 x 2.5
~
Cf)
• .,,
1
--or
I
cvi>Q
5()~'F X L8
At
0.5 ns
v
I'
en"
cv /)/)
From Eq.l6 .18 At
180 v.A
1.8 X 180 fLA
/)/)1
, 324 1-1 W
I<::X 16 .12
Refer to Fig. 13.26
Our decoder is an extension of that show:
6.25
J<:x16.7
F-rom F.qs.lG. H andl6 .15 we have:
We have M hits is the address (as opposed to 3)
and correspondingly there will be zM word lines.
Now, cuch of the 2M word li.nes is connected to M
NMOS dev.ices und to one PMOS trunsistor. Thus
the total number of devices required is
AVio:::
M2M(NMOS)
For minimum area: sck-ct
wn '"" w,, "" w., =
0.18v.m
;;ox w··l-' . 1.2
- - - - - - - - X -~
0.3 X 10 12
2
I)
Refer to Fig.13 · 2 B Our tree decoder will hav,~ 2N
bit .lines. Jltt1s it will have N levels: At the llrst
levels there will be 2 transistors. ut the second
- Cs X l!ml
C8
2
22 ••• ., at the Nth level there will be 2N transis·
tors. llms the total number of transistors, can be
lind as
,. -60 mV
Ex 16.8
Area of the storuge array
X
+
Ex 16 .13
AVw;::
64
~' 2M(M
6.0 mV
+ 2M(PMOS)
1024
X
... ; 2'!'{
Number
1024
X
2
"' 2
"" 134.2 mm1 or equivalently
(I r 2
·!
22 + · ·
Geometric series r
I
2N- 1 )
2
11.6 mm X 11.6 mm
Total chip area
=
174.46 llllll~
1.3 X 134.2
v
2' ··· I
Thus,
f~X 16 . 9
Refer to Example 16 . 2
Number'~
c
Since !:..1 is proponionalto r
reduce
t:.r
by a f>tt:!Pr of 2 by
- . \VC
G,,
decr(~asin)!
can
r by the
2(2·"'·· I)
Ex16 .14
f
=
same factor. C:.t " r ,.. •L.
c·'m
Hence. G,., has to he doubled. G.,
and h1>th Kmn and g,. 1, have 10 be increased by a
factor of 2. The itKt<:as<~ in g.,, can lx· nchkvcd by
.mcreasml!
.
. thc con-e,pondng w
-.
c
••
L thus:
(rt
2
(TJ,.
1X
'"lo.:
0.54
0.18
2.16
ll.IS
6
24
100 Mllz
2
X
5 X 10 ''
Exercise 16-3
Exl6 .15
(a)
1,.
=
(c) In one time-constant the voltage reached is
k; (f)
.P
[< J()
IZ X
J ;< lff'
2.9 ns
Chapter 1--1
1.1
.!: "" .~ "" 10 mA
R
nn
f "" lQ V "" tO Ul
(a)/ "'
(b) R. .,.
l
I m.-1
(c) V = /R "' 10 mA >t; UHU ... 100 V
t'd)l=·y""
..!2..Y.
"'OIA
R
lOOfi
.
Ntut: \\ill,$, rnillillmps, and ldlo-oltms ~tillite a
~t ~~Ct of lft\itll,
1.5
v1, = v 110 ~
R, + R
1
To fi~ B0 , · we ;hort cirwit VPt> and IOI.1k !:mclc
.i.ntn liOde'X•
Rn .,. R•. U Rl ""
1. 6 •
RiR:
R1 + R:
Use vol• divider to find Y.,
v• "" .;-22"""·· ·.•··•
. l.l + 6.8 ,.., "'·""'
f.~l\'ill~toorputresi~ Rn i~
1.2
{I} V "' IR •~ fOmA X I kJl "" lO V
I' "'.jlR "" (10 rnA)1 X I k{l '"' WO rnW
{bJR"" 'VJI"" 10 VII IliA .. lQkll
P "" VI "' 10 V X l mA "' 10 tuW
(c)/"" .NV" I W/lOV=O.I A
R() ... <3~~ ~en
a 6,a w> .,
.2.~ ~c.o
The extreme Ylllues t)f V(, for ::!:!i % tolemnce
nm11toran:
V •.
.., 9.-.
),~(1 - O.O!ij
.
o ..,.
3.:tt 1 - o.os> + un + QM)
"' :t7S V
R "' V t I "" 10 VIO.l A ''" 100 {}
(d} V "'PI/ '"'OJ W/IOmA
"' 100 mWIIO!rtA.;. IOV
R"" VIJ- l!lVIIOirtA"'I\fi
{e) P ..., 11 R,. t = JPlR
I ""
./tooo mW/1 kll
... 31.6 mA
V "" IR •~ 31.6 mAX I !ell
= 31.6 V
NCIU!: V. lrtA, Ul. and mW conmltute a cunMS··
tenlllel of llllits.
1.3
Thus. there ate 17 ~ble resistaw.:e value,,
This figure is for 1.3
10
46.7
1.4
Shtullbig the 10 Ul by ll miliu u!'vatue of
R rcsuti in the wmbtfllllioo having a mistance
R•••
R "' lO.R
"1
R + 10
Thus, fur a 1~ redu.ttioo.
.....!!_ "' 0.99 • R "" 99o kU
R + 10
For a 5'.(, teduclioo.
R : IO "" 0.95 ~ R .. 190 kO.
for a 10% n:duetioo •
U)
lQ
~6.1
10
·~-r-~L.
- L.W.-1 -
"*
10.
·~·
__JL_
= IQk{l
R + lf) ·~ 050-,R
.•
Ul k:O by
(Jl) I Mfi mult in
:..>lUX IQQ2"" J£L,; 9\:tkO
"' . UJOO + 10
tot
. ' .· '
a 1% R!duclion;
(b) 100 k!l rosuli~ in
R
R~., "'
Ml?.tW
roo+ to ""' .ill
u "" 9Wt~O
· · ·
11 9.1% rt®ciion:
(u)
to kfi tesultll hi
.. ~
10
"••
- iif+:1'0
u50%~ioo.
"' ",.,..
" .....
?0
8.0
......1L._ "' 0.90 R ,.. 90 l / 1
+ /JRJ
I,:::. 1.06 rnA
I,:::. U 3 rnA
11 :::. 1.19 rnA
V :::.. 2.38V
(b) A node equation at the common node can be
written in tenns of V a.~
15 R1
v + 10 - v = .Y
R2
Thus,
.15 - v + 10 ~
tO
5
v =
=>0.8
""HJ/1 + (/ 1 +12 ) X 2
R3
v - !:
2
3.5
=> v = 2.375 v
Now, / 1, 12, and /3 can be easily found as
"" IU 1 + 2/ 2
I, ,
15- V = 15-2.375
10
10
= 1.0625 rnA :: 1.06 rnA
+l5V
{, = 10- v = 10- 2.375
+IOV
"
5
5
"" 1.125 rnA :.: 1.13 mA
/J·
= RX"_
3
= 2·375
2 = 1.1875 rnA
. -- U9 mA
Method (b) is much preferred; faster, more
insightful and less prone to errors. In ~neral, one
attempts to identify the least pos.~ible number of
variables and write the corresponding minimum
number of equations.
1.13
That ls,
121, + 212 = 15
(l)
Similarly, the vol!age drops across R 2 and R, add
11p to I0 V, thus
lO '~' 11R 1 + 13 R3
511 + (1 1 + /2) X 2
which yields
21, + 712 "' 10
(2)
Equalions (I) and (2) can be solved together by
multiplying (2) by 6,
=
12/! + 4212 "'' 60
Now. subtracting 0) t1·om (3} yields
40/?
45
1.125 mA
en
From the symmetry of the cin:uit, there will
be no current in Rj. (Otherl•lise the symmetry
would be violated.) Thus c.ach branch will carry a
current V.f2 k!} and l,. will be the sum of lhe iwo
current,
I = 2V,
'
2 kfi
V,
I k!l
-
1,=2VJ2 kH
V,/2 kfl
fr
Chapter 1-5
1.14
(a)
v
R ,..J =I kH
I,
If
Now. if!(.~ is raised to 1.21dl the symmetry will
be broken. To find 1; we use Thevenin's theorem
as follows:
v1
+ 1 + 0.545
At
f
l/jw
C
.
= -~2"1f'j X
= too kH1~ z
""
2
1 GHz,
z
1 = 60 Hz,
R,
Z = -j0.8 MO
At f = 1 GHz,
Z == - j79.6 0
':'
lk!l
At
f ""
=
_,
j2-rrf X 10 X lO .
Z = j3.77 0
At J = 100kHz.
RJ
on
= - jt.33
At/= lOOkHz,
At I = 60Hz,
J;t
lkfl
-it 59 n
=
Z = - j0.016 H
(d) Z = jwL"" j21fjL
p;
I
lOX 10- 9
Z = - j265k0
= 60 Hz,
At 1
At f
At
v, + 0.022 v, x o.s
"'
z=
= I kO at all frequencies
(c)Z = 1/jwC = -jZ"'f'/X;Xl0-12
£5 = 0.545V,- 0.5V, = 0.022Yx
0.5
(b)
z
Z "" j6.28 kO
t GHz,
Z
=
]62.8 0
z "' jwL = j211'fl· "' j211'j(l X 10 -9 )
f = 60Hz. Z = j3.17 X 10-1 = ]0.311 il-0
(e)
v,
Vz
R.
J=
R4
l.2k0
lkfl
lOOkHz,
Z = j6.28 X 10- 4
f = 16 Hz. Z
=j0.628 mO
= j62.8
0
1.15
(a) Z = R
= IOJ +
""' (I -
lb) Y ""
+ - 1-
jwC
l
_9
j21f X 10 X 103 X 10 X 10
jl.59) kO
! + jwC
R
1
-6
l
= --;_ + j2'1f X 10 X 10· X 0.01 X 10 .
10
=
z
=
+ J0.628J
1o·\1
n
1000
I + j0.62S
y
1000(1 -· j0.628}
I+ 0.628 2
= !717:!- J45.04)
(c)
Y
=
-~
R
n
jwC
t
--=----:, +
--j'~<
1
j21T X 10 X 10 X 100 X 10 '
100 X 10
10- 'n
l
+ j0.628 J
~
7~
10
I + }0.628
(71.72 .... }450.4) kO
05\f, X Ul22 '·" 0511 V,
F1 +
1
F' ,, I':
I,
/l ' 1,
v
100 + j6.2S X 100
.c
0.467V,
·
0.956 V,
. H. ··-~ "" -'
I, ~- LO:'\ Ul
:-q
~
100 + j2r. X J0 X J0 X 10 X 10
l'
"'l"k?l
v_
· · v2
I
'··-···"!Hi.
R+J(oL
(dJZ
I,R~
(
100 + J628J
n
J
Chapter 1-6
1.16
1.18
Norton
Equivalent
'rbevcnlo
F.quivnleot
R,
··C .
v0 ·= v, - i 0 R,
''o
Voc "" rJs
=
(i, - i 11 )R,
isc "" is
·= i,R, ·- i 11 R,
vs = isRs
v0 = v,- i 0 R,
Thus.
'~-'oc
Open-circuit
(i0 "" 0) --+ v,
Rs"" -..
isc
(a) ''s ""· vor
=
voltage
10 V
is =· (sc = 100 p.A
R· "" !!££ ""' _!.Q..Y_ = 0.1 Mf!
5
isc
tOO p...<\
(b) vs
is
=
=
= \00 ldl
0
voc = OJ V
isc "' 10 1-k"-
voc
Rs""" -.1sc
~ = 0.01 Mn
tOn:\
=
10 kO
Sbott-drcuit (t10
""
0) current
r
1.17
1.19
(I) 1.26
(2)
R 1 represents the input resistance of the processor
Thus.
For ''o = 0.9
__!L "' 30
R
0,9 ''"' - -1-·- =>I?,
Rt. + Rs
I+_&
For i, "'0.9 i,
100
and
_3!_L
'-'r
0.9 =,
\0
I+~
10
Dividing (I) by (2) give~
l + (R./lOl "· 1
1 + (Rs/100)
·
~ l?s = 28.6 kH
~=>N
Ns + R,
L
1. 20
~---.------~-------~.-------~
Case
w (rud/s)
f
(Hz)
7' (s}
! a
6.28 X HF
l X 10''
I X !()''
----,r------+-------L~----1
b
I X 10'
1.59 X !0'
6.28 X I(P
60
X
Suhstituting in t2) giws
11, "' 38.6 mV
The Norton curwnt i, can be found 1"~
10;
Chapter 1-7
1. 26
1.21
v,.,t
= 117 x
(c) V pc two harmonics will have the ratio 7 to 9.
which is confirmed by the measurement reported.
Thus the fundamental will have a frequency of98/
7 or 14kHz and peak amplitude of63 x 7 ""441
mV. The rrns value of the fundamental will be
441 ;J2 = 312 mV.Toflndthepeak-to-peak
amplitude of the square wave we note that
0
l
I
I
+7
4V !1t = 441 mV. Thus,
I
0
0
0
-0
I
0
0
I
-l
I
0
1
0
-2
1
0
1
I
-3
I
l
0
0
Peak-to-pe.al.. amplitude
=
2V
Period T
441 X E.'
1
693 mV
I
71.4 !J.S
f
-4
"
1.24
To he barely audible by a relatively young
listener. the 5th haml(>nic must be limited to 20
kHz: thus the fundamental will he 4kHz. At the
low end, hearing e11tends duwn to about 20 Hz.
For tht> flfth and higher to be audible the 11fth
mu!\t be no lower than 20 Hz. Corresponillngty.
the t\mdamental will be at 4Hz.
1.25
\:;,! 1?.
If this power is 10
equal that delivered by n sine w;;vc ofpe;ik amlitude
l' tb~n
1
0
l
-5
1
I
I
0
-6
I
I
1
I
-7
Note that there are two possible representation of
zero: 0000 and I 000. For a 0.5-V step size. analog
signals in the range ± 3 ..5 V can be represented
Input
lf the amplitude of the square wave is V..,
then the power delivt.>red by :he ;;quare wave to a
re~iitptcn~y.
.,~·~·-~-~"~
Chapter 1-8
1.28
(a~Foi'Nbits~wm bell"~blc Jc:v~
1.31
el$. ftoD:I O.to Vn• Thus there wilt be (2'<- thfls~ stf:Pa. from 0 to v.!$wid! the step. sit,.C.given
by
'
'
A,. = uoio = .!Q X !!?
v111
u1 11
= 2 X 10' X 2000
=4 X lO'W/W
or 10 log.A~ = 86 dS
'
Step size = · V F,t
zN -t
Tbl& Is too ililalog cbailge t;Otresponding to a
change in tbe.LSB. It is the vulue of the reSOlution
..
OftheADC.
='
2
2H=I
i1
I mA
~ = lOOOAIA
ttlil
i,
"'
= 10 X 1000 = JC}'W/W
or 10 log., A; = 40 dB
Vrs
22V=l
This is known as the quantization error.
(c) 10 V
v
u0 i 0
u0 X i-0
Ap=-~-
ll
1
I mA
or, 20 log 1000 "' 60 dB
step
step size =
=- lO V = JO V /V
i;
i_
X
v,
or, 20 log 10 = 20 dB
A, = !g ,.. ''ol Rt. == 10 V /10 Q
{b} Til¢ DI#Ximllm error in conversion.~
when the analog signal Vt\llle is at lite middle tlf a
step. ThuS the ma:dmum error i~
!
.!!:!
(c) A =
1. 32
+3V
:s 5 niV
1 rnA
lY /'\
\r\"',
2N-l;;;o.2()00
= 11,
2N~2(J()l=:::)N
2.2¥
120 mA
t(average)
ForN= 11
Resolution = - 1110
= 4.9 mV
Quantization error
...-
= ...42.·9__=_ 2.4 mV
2 -1
1.29
There will be 44,100 samples per second
with. each sample represented by 16 bits. Thus the
through-put or speed will be 44,100 x 16 =
7.056 X Ill" bits per second.
C\{1" "'
-3 v
A, = "o .. 2.2
'';
0.2
= llVN
or 20 log II = 20.8 dB
A;
1.30
(a)
=
'
~
A = ~ "" _!.Q_Y_ = 100 V /V
~
v1
= •'o I R,,
ii
i1
tltJ~o
""
Tltlt
ltnA
or, 20 log A,
,. JO VI 100 0 _ 0.1 A
tOO J.L11
- tOO p..A
!11 X ~
= 100 X 1000
I;
VI
= IO'Wtw
.,. ~ ""
1 V
A,, = Po ·""- (2.2/ ,/2)1 /t(.)Q
P;
0.2 X to··3
"" 2 X 10~ V ! V
t10
!R1 = 2 V/tOk!l
!flO ni\ ·
= 0·2 mA ""' (},2 X IO . ~. '"' 2000 A i A
100 nA
HlU X 10
or 20 log A, = 66 dB
-~
,/2
n "'
('>2/ /?}2
;~ = ~ioo
Input power=·
i;
./2
or. 10 log A,"" 23.8 dB
Supply power: 2 X 3 V X 20 mA"' 120 mW
Output power '"
v1
" t t1
IOJLV
or, 20 log 2 X Ill"""' 106 dB
A.""~""
'
i1
= 26.8 dB
= 242W!W
or 10 log HY ""' 50 dB
(b) A
2.2 VI 100 {}
=
=~""22AIA
I mA
100 mV
""'JOOOA/A
or, 20 log .1 000 = 60 dB
Ap =
~
I;
or, 20 log 100 = 40 dB
A·
=
~~
24.2 mW
=" 0.1 mW (negligible)
Amplifier dissipation:::: Supply power ... Output
power
"'' 120 .... 24.2 "" 95.8 mW
Amplifier efficiency "" !]~t£Ut power X 100
Supply power
"" ~X
ll)(l
120
. == 20.2%
Chapter 1-9
(a)
IORs
x A X
IOR 0
IORs + Rs
"" lORo + Ro
V(l
vs
- r - - - - Voo - 1.2
= !Q X 10 X !Q
II
= 8.26 VI V
11
or. 20 Jog 8.26 = 18.3 dB
(b)
_ _ ___,_-- - Vnu
!:Q "' __&_ X A. X .....!!.rL_
vJ
Rs + Rs
" Ro + Ro
= 0.5 X 10 X 0.5 =2.SV!V
or. 20 log 2.5 = 8 dB
+ 1.2
(c)
Rs/10
XA X
R 0 !l0
(Rs /10) + Rs
•'u (R 0 /10) + Ro
For V00 = I5 V, the largest undistorttld sinewave OU!pllt is of 13.8-V peak amplitude or9.8
= _!_ X 10 X
V.,... Thelnputneededis9.8V/500= t9.6mV_.
Il
or 20 log 0.083
=
1.. = 0.083
11
VI V
-21.6 dB
1. 35
1. 33
IOfi
(a) For an output whose extremes are just at
the edge of dipping. i.e., an output of 9- V,.... the
input must htwe 9 V/1 000 = 9 mV-·
+
•
I
20log A.,0 = 40 dB
(b) For an output that is clipping 90% of the time,
0 = 0.1 X 9~;
'"·' 0.5
won
w
9V
... 57J V and the input rnust be 573 V/
1000 or 0.573 V,,,.
1. 34
T
lkH
IC~HH
AM-o--,
P,
10
!00 mV
"':"
H1t';. f.
I'
~l~-~.
OOP,
":"'
-
'
.
IOOfi1· ''·
-
::(!
__!Q_~_fi__
'\
10 k!l + 100 k!l
.!.(~.
110
X 1000 ;x:
X l ()(Xl X
.! ()() "
1100
I 00 {}
100 f!
8.26 \ 1 ! V
·t
I kU
Chapter 1-10
This figure is for 1.37
tMn
IOOkll
1.36
The signal loses nboot 90% of its strength when
connected to the amplifier input (because R, = R_,l
Total gain
10). Also. the output signal of the runpl iller loses
approximately 90% of its strength when the load
is oonnected (becauseR, = ~/10). Not a good
= 99.9 X 9.09 X 0.909 = 825.5 V/V
The voltage gain from source to lood is
= r'r. ""' A., 1 X Av:z X A.3
flil
!1: =!!!:_X~= A.,·
11s
vi 1
"'s
design! Ncvertheles.~ if the sourt'C were connected directly to the load,
1111
115
= 825.5 X 0.5
"!!.!!,.,~
"s
= A.,
= 412.7VIV
Tbe overall voltage has rediiCild appreciably. It is
due to the reason because the input impedance of
the first stage, R... is comparable to !he source resisrance R,. In example 1.3 the input impedance of the
first stage is much larger than the source resistance
R1_+ Rs
100 !l
too n + too k!l
:::_O.OOlV!V
R9 = 100 k!l
1. 38
a. Ca'\C S-A-B-L
V0
vs
+
v,
""
v,b v..
(1X
Vo
which is dearly a much worse situation. Indeed
in.~rting the amplifier increases the gain by a factor 8.3/0.001 "' 8300.
Vs
V.s
IOO ) X (100
100 + 100
=
X
10 )
___!QQ_)
X(
100 + 10
100 + 10
4.13 V /V and gain in dB 20 log4.1 =
12.32 dB (See figure below)
1. 37
b. Ca~e S-B-A"L
In exrunple 1.3 when the first and the second stages are interchanged, the circuit looks like
the figure above
100 k!l + 100 k!l -
Ad = ''12 = 100 X
V; 1
·-
V0
_
V5
-v1.·vi"
V 0 V1.,. V10
v~
= (100- X 100 IOO
) X (I
+ 10 K
_ (} 'i vI v
100 kil
l/11
"s
Vox V 1bx Vi•
'
(
1 Mil
I Mil+ I kil
""99.9V!V
X
lO K . ) X
10 K + 100
lOOK
)
100 K + 100
Vo "'-'· 0.49 V / S and gain in dB is 20 log 0.49 =
lOX
v5
lO kH
-6.19 dB case a is prefcrTed <~~it provides higher
voltage gain.
10 k!l +I k!l
= 9.09V!V
l X
JO() !l
won
+-
ton
~· 0.909 V!V
This figure is for 1.38(a)
;
... ~-~-~
!1)<>!1~
..
·0;
,'
This figure is for 1. 38 (b)
I{IHU'l
w.
l
v.cb
0
I:-.. Y~,
!Ul
•. l
\'
"
~"'Hl
w.
<:>>IX<·\·,.
I
'''"'~ :.
Chapter 1-11
200
IK!l
+
IOKO
tlf2
This figure is for 1.39
1. 39
Deliver O.SW ton 1000 load
R,
Source is 30mV RMS with 0.5Mil source resistance. Choose from 3 amplifiers types
-D>--1>--D>Rp IMU
Rp tokn
R;= !OkU
,'\,,"" 10 v/v
.4,,= 100 v/v
A,r I v/v
RJ: lOkU
R,,lkil
R":20!1
Choose order to eliminate loading on input and
output
A -I st-to minimize loading on 0.5 Mn source
B - 2nd-to boost 2ain
c . 3rd - to minit;lize loading at 1oon output.
(See figure below)
~ = .2 v . =
vs
1 .,.,
235.7 < (.
30m V
0.5t,L
Co~ to)oool(w1 ~
)no)
+ lt,L
RL
=X-Rt + R
300V/V ""_2Q_XA
90 + 10
667 0
··,
I
l + 0.667
X
=>A,0 = S55.7V/V
= 100 kfi (I X 10' 0)
Rv = 100 !1 (I X l()l U)
(e) R,
300 = _!QQ_ X A
too+ to
~A,
"•
X
1000
1000
+
too
""363 V /V
1.41
R,= IOOkU
w
p
-
1.40
(a) Required voltage gain
1-
,-~l'•
R,
IOmV
""' '!!.'!
(a)
'~
J03 mV
__jQ__
10 + 100
Ol
.. IL''• = -IOmV
- . - ~ R, + R ~c 100 !d1
/?8 + R,
·
'
(b)~ "" 303 mV
Thus R; '·" 90 kn.
h:1r R,. = 90 kD. ( = 0.1 t,LA peak. and
(c).~
v,iR
0 wrall current gain '"' ~
-.- - =, 3 X 104 A. A
0.1 f,L/1
~::)'
____ ::{}:
_____
' l()(j()
..·) (0.1 xii;~;;:
(. J!L?<_!i>.~
J2 'X -J2 ·-)
C)XI(YW/W
(This litkcs_intn an~t the power dissipated in lhe
mttornal rest stance of t:hc !iourct•, 1
fc) If (A'',, ,,) hrt~ irs peak value llntit(•d ro:;
largest value of N, . is fnuntl from
''s
(d)
I,
J mA
v. the
~
-
10 mV X
(b) The smallest R. allowed is ohrui ned from
+
tl-
lOkfl
~
= 300 V i V
(
=
0
0.58
-.
3 '
(d) For R, = 90 k!1 and Ro = 667 0. the required
value A.., can be found from
1
1)(1 )(2o -~oo)
n0 = (253.6)(30mV) = 7.6! v RMS
3V
""
(lfR., were greater than this value, the output voltage across Rt would be less than 3 V.)
235.7 < 253.6
= (l.OI V
. 2
-R1
'"' 3 => R0
0
..-omy
1000 X
X
1000 X
100
100 + 200
= 303 V 1 V
100
100 + 2fJO = 3.H3 V !V
Chapter 1-12
Connett 11 resistance R, in parallel with the input
and select it~ value from
(R,. II Rl)
- 1 R,
(Rp H R,) + R~ - 2RI + R,
~ I + ~ ""' 2.,::::} R u R
Rp H
~
Rl
'
21 2
100
2i - 2t
Overall voltage gain
"" .!!.!!
•• J!s-_
::::}l + .!_
R,.
p il
Rs _
= 4020 ~
1·9 .·o··'"' v /V
21
R1
WO
1.44
R1. "" .
I
"" 9.1 lr:U
0.21 -· 0.1
1.42
R,
+
(a) Current gain =
i, = it + j1
i 1 = tJ1 /R 1
~I;
=A-~
i11
"Ro + Rl
= 8m . !-lfj _,
Vi
VX
"'· 10010
11
=
(b) Voltage gain =
90.9~ =
39.2 dB
, - - - + - - o Out
::'!!
t:5
=
= 90.9 X..!_
101
'-" 0.9 VN ""'· -0.9 dB
(c) P!Jwer gain = A
JOOk!l
~!L_f!L_
i1 R~ + R,
r ,~ ''oi.o
nsli
,,., 0.9 X 90.9
= 81.8 W/W ~' 19.1 dB
1.45
Tmnsresistance amplilier
To limit a,,, to 10% corresponding toR, varying
in the range l to 10 kO. we select R, sufficiently
low;
R _.:::; R .nniu
1.43
I
10
Tim~. R1
"'
100 !l
To limit ~"-• to 10% while R,varies over the
range I to 10 Ul. we select K,sullicicntly low;
RI > ...s Rt_.~-~
10
G.,.= 40mA/V
R,. , 20 k!l
R, • l kfl
ThtiS, R0
=
100 0
Now. for is = 10 !LA,
R,
¥1-.
R.,i.
-~;:·_·r~-I to
v,
IOkO
7:Jo
-
--
Chapter 1-13
Rt""l"
= 10 -~ R- R
R,.,;,. + R1 •Rt.mln + Ro
ttomirt
= 10_,
1
=;)
1.47
Current Amplifier
1000 R
1000
1000 + 100 '"1000 + 100
+
R., = L21 X 105
121 k:fl
lkto
ln:-1;;
1
toon
JOk.U
toon
For R_, varying in the range I kil to 10 kil range
and load voltage variation limited totO%. select
R, to he sufficiently IQw:
-
R ~ Rsmin
I()
I
1.46
=
R'
Voltage Amplifier
l k!l = 100 H = I X 102 H
10
For R, varying in the range I kfl to 10 kfl and
load voltage variation limited to 10%, R., is
selected sufficiently large:
R,
I to IOkfl
R"
vs
l',
L-.----l.-..,...........J.....---_j
Rt
1 k to
10 kfl
R0
2:
10 R 1_mu
= 10 X IOkfl
= 100 Hl
""' 1 X 105 U
Now we lind A,,
V Omin = 10 J.l A X
For R, varying in the range I K to 10 k!l and
!J.i<' variation limited I<> I0%. select R, to he suffi-
R;
IO X 10 -6
10 Rsmo.
2:
10 kfl = 100 kfl = 1 X 10' H
X
For R1 varying in the range I ro l 0 k!l. the load
current variation limited to !O'Jl.•, select R, sufficiently low:
~A;, =
l(smin
/(~min
+ R,
R_~miu
+ Ri
10 X 10-6
ciently large:
R, "" 10
X
X
• · X R0
~ nH
.
Ro
=
'''(Ro
I kH
I kfl + 100 fl
x
10
Now find A,.,
+
I X
tw n
Rt
1x
"100 K
~"- 10 X 10· X
IO(J kf!
Hi '· 10 kf!
I
X A", X 100 fl -'· 10 kfl
I X 10 ) = 10 X 10-' X !OO X A ... X ~l_110
•·
tton
I K
111.1 A/A
+
121 V!V
R,,
A, i,
I
X
102
n.
A 1,
'0 '
111.1 Ns.
1.48
R,. .,. Open-circuit o_~!.[Hll voltage
Short-circuit output current
=I kH
1+4
8V
R.,' lkH
r---¥/'lror
.
Vollage amplitier equivalent circuir i>c
l X 10" !!, A,.,,
R11 ~- f X 101 H
121 V!V and
v,,
R 0 =I X 10' fl
10/
R,
+
Current amplifier equivalent circuit is
w" n
_ R,
X A ... ----·j __ _
·+- R,,·m~ro
... Ro + R,!!l:.lx
100
A,.,,
Rlmin
i(!
...,...
i, mio '" 10 mV X
J0- 3
R l-tnin
i\ 100 K X I K
:-~"
l k!l
II
A- RoRtmin
Rstl\in
R
Ro:s;
RL
~
' lfl \/
I
11,.
n--
lOY
lO mA
Chapter 1-14
A =vo=
8
I X 10""'3 X (1000 10) X 103
• v,
fur(a) V0
=888 VN or58.9 dB
i
'!11 =
A, ""
3
= 81(4 X 10)
R1
110 /
10-:; X _j.QQ._
100 + 10
V~jl Rt.
Vn
V1
ltO
wherek=l
;fRI
I
(
10
1 + SCR
w,. = ..!_ from table 1.2 it is low pa~s.
"
8 2 /(4 X 10))
=
v{ 1 /;~~ R)
= _._1_
tOO
10- 3 X -
"' 2200 NA or 66.8 dB
A. .,.
""
····~X ~)2 10 X
100 + 10
103
RC
V v,[-R1)
R+-
for(b)
0
·=
SC
19.36 X 105 W/W or 62.9 dB
,
Overall current gatn '"' 1
io
fU\
V 11
_ tJa!Rr _ 8/(4X 103)
-
=
I
fU\ -
=
S
v,
10-3
s+..L
RC
2000NAor66dB
where k =I
=
w6
1.49
..Lrrom table 1.2 it is high pass.
RC
1. 51
Using the voltage divider rule
" = 100mA/
...-y
R1.= 5k!1
lim
a. i(?
::-:..~ .l:mi!t -- 8m"P~
''o = i,.R1. "' g,.R~_(v 1
h.
l'J
=
Fl .'.PI! ""
·-
v2) = ''n
__R_,__
~
OV
V~
=
(R 5
+
R,
R;)+ sC,R,R~
I +
+ R,)
s(' 5:-;:!R,R~)
.R1
IOV
\Vh~rc K ·~
1. 50
(R~
N,+ R,
<•J "" (.'1? R.
"f
(
I
!?,
R,
(i(,+R,)
from tahlc 1.2 low pass for given
·'
values w,, = J2.5 Mllz
Chapter 1-15
1. 52
1. 54
The given measured data indicate that this
amplifier has a low-pass STC frequency response
with a low-frequency gain of 40 dB, and a 3-dB
frequency of 10" Hz. From our knowledge of the
Bode plots for low-pass STC networks (Figure
1.23a) we can complete the Table entries and
sketch the amplifier frequency respn ~ - - - - - . . . . . . ; . . - - - - 2;;- - 21T X 0.1 X 10 6 {10 + 40) X !OJ
31.!1 liz
K
0.57
J2
I
I
I
I
I
I
I
I
102
103
104
106 j(Ht)
.ftHz)
IT!(dB)
LT(")
0
40
0
100
40
0
1000
40
0
to•
37
-45"
10'
20
-90"
10'·
0
-90"
T(.Y)""' R2/(R 1 + R2 ).Aiso,fromthecircuit
observe aJI s ~ cc , ( I I sC) ~ 0 and
I
I
1
I
vrv
1. 55
1.53
Using the voltage divider rule.
c
R,
v.r:.-~~
*
~
RI.
··
T(s) =
where
R 1 + R, + ..L
·
Since the overall transfer function is that of
three identical STC LP circuit.~ in cascade (but
with no loading effects since the b1llTer amplifiers
have input and zero output rei>istances) the overall
gain will drop by 3 dB below the value at de at the
frequency for which the gain of each STC circuit
is l dB down. This frL'qUency is found as follows:
The transfer function of each STC ci rcuil is
sC
<•111
= I !CR
Thus.
Rr
R~_
s
+ Rs,
which is of the high-pa.~s STC type 1 ~ec Table 1.2)
with
K '·--~L...
Rr t· R..:;
For/,.~<:
"'-">
····I
h
i~ ('
0.64 ~tF
w 1d 11
051<<~ 0
Rv2JI R;3 =
= 27rCl R
6 MHz
100 kfl .
Thus R, 1 = 1.07 k!l
Similarly, for node B,
27r X 15 X 10·1 X I X 10" 9
1 5
10.6 k!l
21t X 6 X 106 X 105
""
11.9 Jc!l
She should connect a capacitor of value (.~to
node B where C" can he found from,
""0.26 pF
1. 57
10kHz
Since wben Cis connected tbe 3-dB frequency is
reduced by a large factor, tbe value of C must be
much larger than whatever parasitic capacitance
originally existed at node A (i.e., between A and
ground). Furthermore, it must be that Cis now the
dominant detenninant of the amplifier 3-dB frequency (i.e., it. is dominating over whatever may
he happening at node B or anywhere else in tbe
amplifier). Thus. we can write
!50 kHz
R,, 1
= 2r.C(R,,
. 1 II,
1
R,2 )
~c;.·
I
27TC,r dJe input circuit, the corner frequency f.., is
found from
,
f.J
2rr X 150 X 10~ X I X 10-'l
R.
""'L06k!l
!Okfl C1
P-'Wv-J~+
\i
·'
"'::"
R;
~
IOOkO
":"
FDr f." .:s.; 100 Hz ,
1- - , - ---···--2r.C;(10 + 100) x 101
~;
100
..
Chapter 1-17
"* Ct ~
. I
. . = 1.4 X 10-R
2'11' X 110 X 10' X 102
Thus we select C 1 = I X w·· 1 F •"" OJ p.F.
The actual comer frequency l'e!>1llling from C,
will be
··' I
"' 14.5 Hz
2r. X tO . X Ito X IOJ
For the output circuit,
fl/ 1
.f,1
""·
-
Better approximation
(3,dB fmquencies)
Bandwidth= 10'- 10" =9900Hz
1. 60
TM)
llsC1 + R 1
LP
I
-1T C·z( R o + R1J
-
.,
For f,, 1 $100Hz.
I
2wC,( I
"* C, ~
•
+
< 100
l) X !03 -
.
·· I
· 2 = 0.8 X lO
2;r X 2 X 103 X 10
Sele<:t C 2 "" I X 10-• = I ~F
This will place the comer frequency ar
fo2 =
_J
,
2r. X 10 . X 2 X HY
T(.s) ""' 100.
= 80 Hz
s
(l + 2;>,J( I + 2;~/'J
1. 59
The LP factor l/{1 + jJllcY) result~ in a Bode
plot like that in Fig, L23(a) with the 3dB fre-
qucn<-)' fo
=
104 Hz_ The high-pa~s factor
11(1 + 104/jJ) results in a Bode plot like that in
Fig.l.24(a) with the 3dl:l frequency
lo
=
= 53 Hz
2;rl00 X 10-Q X 30 X 10'
:. T{S)
T 1(S)T 0 (S)
104 Ht,.
The Bode plot for the overall transfer fur)(.:tion
can be obtained by summing the dB values of lhe
two individual plot~ and ihen mising the resulting
plot vertically by 40 dB (corresponding to the factor I 00 in the numerator). The result is as follows:
~1 •. 1.
3 dB frequency
I
1+
X -666.7 X
S
2rr X 15.9 X 103
-20 dB/decade
t 20 dB/decade
dB
t
1- 20 dB/decade
- 20 dll/deeade
Bandwidth= 16kHz-53 Hz:::; 16Hz
1. 61
!?I
vi ""· v,f? .\.+ R'
a\ To satisfy constmint (I), namely
.ffHz.)
V12o(1
+
,;~))v1
\Ve substitute in Eq.( I) to obtain
r = 10
W 2 10; HY1 j(}i 106 HP (Hz)
1.·1,1 ::20 40 40 40 20 0 -20 tdH)
-~;:,
Rs
+
R1
1 ...L
100
S
S + (27T X 53)
Chapter 1-18
Thus
The firsl factor on the LHS is (from constaint ( 1))
greater l?t equal to (t -.r/100). lltus
RR+RI..::: __l _
Rt
- 1-....!..
G,. 2::.
100
Ao
(3)
(t - I~)) using
8 '" 7.3
10 1 ~ (.'m 'K " 2 :
X
k ., 11.62 X 10 \:VtK: E.,
1.12 V .we
hav.::
T
7(l"'C '" 203 K:
rt,
or
C 1R 1 "" C2Rt
When this condition applies. the nttcnuator is said
to be compensated. and its transfer function is
given by
V0
c,
_
v, -
!l"C ' 27:1 K
T
1.52
!1,
T
C 1 + C2
which. using F.q. (I) can be expressed in the alternate foon
Vo
V;
Th;~ti5.uneou!uf.:very 5.13 :< 10 1' silitnn
atnm' is ioni7ed at thi;; temperature.
!+&
R1
Thus when the attenuatot is com!JI!nsatcd (C,R, =
C,R,) its tran.>~nission can he dctennined either by
it.> two resistQili R,. R, or by its two capacitors. C,.
C,. and the transmission is nm a functitm of frequcncy.
)(
20"C
'"':
=
II,
..
~
N
'
105 X. 10
H
:'9J K.
KfiOX to"
c::;-~
r
J0 " i.'fll
llln"c
t'Jll
' '!J
l
N
nx
10. ll
C\?:l K:
~
2.87 X 10
c·m : ~
9.45 X H!
I'
l ..Ux !0t'l!l
T
U:I'C ... WS K:
'{I
·1.72X 10 12
';
"'
,\'
!l
:I
1. 65
Hnk coilt.:t~ntm1hm in intrinsic S,.
II;
1. 63
The HP STC circuit whose response deter·
mines the frequency response of the ampli!1er in
the low-freqvcncy range has a phase angle of
11.4" at/= 100Hz. Using the equation for
hdow intrinsic k\'d hy n fm:tnr uf 1o't
L T(jw) from Table 1.2 we obtain
:. f Jot~ c\mc·cnlrull<\1' in I' doJ'c'd Si h
tan-! /,, = 11.4'';=> (.,
100
.
1.5
hokslcm
Y
hnl~
ln pbn,;p.bnrus ./,.
n,· 'p.,
"·
!.:" /
1. 66
l\tf = 100Hz the drop in gain i~ due to the HP
STC nctwork. and thus its value is
20
lng-;:::::::::~=1~ ""
I, + (;o.l6r
'-'
T
=
27'C = 271 + 27 '"
-t\.!7 dB
At 300 K. II;
Phosphorous doped Si
100 .
Similarly. at f = l kHz. the flrnp in gain is caused
by the LP STC nelw(lrk. The drop in gain is
n, = :V 0
o·•
10 16 /cm'
0 17 dB
llok C<>tH:cntration
The gain drops hy J dB at the e<>llKf frequt'IKirs
of the tW(l STC nt•tw;wks. that i•. atf 10.16 fl;.
and( 4959 A fl1
T
!25'T
···" p.,
~00
K
Chapter 1-20
= 7.3 xto'sx(l9Slae-t.I21(lxs.6zxto -5 xml
2
= No
n1
p,.--
10 nm
~
= 4.72 x 10 11/cm~
tV
E=_!_Y_=
Need }drift -~ I mAI~tm>
9
- 2 .23. X 10 / em-~
10" 3 A
-
_8
_
l
-
,
4
lOX 10- em
1.6 X 10
= 10' Vlcm
= q Nn Jln E
--19
Nn X 1350 X 10
)
10 Clll
At 398 K. hole concentration
~Nn
"
p;, = 2.23 X 10 fern ~
= 4.63 X
10 17 /cnl
1. 67
(a) The resistivity of silicon is given
For intrinsic silicon.
p "" 1i "" n1 "" 1.5 X 1010 cm-' 3
1. 69
Using 11. = l350 cm21Vsand
From Figure p3.IO
l Op.v-Pno
8
__
= 480 em2/Vs, we have:
l1p
=
p
- n/
P.~ - No
w
2.28 X 105 O•cm.
=3x
(b) "• = N 0
2
p. =
~
!!L
"•
= 10
R
lOp
••
o.t x
w- 4
x to- 4 em
X 2.25 X 104
0.1 X 10- 4
3
dx
10-s cm 2 • we have
= 2.25 X 1017
Hence
0.
R = 7.59 X 109
-
since OJ j.Lm = 0.1
=
Using R = p • ~ with L 0.01 em and
A
"" (1.5 X J0 11lf
1016
= 1016 em -J:
1 = -qD ~
Pdx
P
= 2.25 X 104 cm' 3
=
Usingj.L. = !Jl0cm2!Vsand
-1.6 X 10- 19 X 12 X ( -2.25 X 10 17 )
= 0.432 A/cm 2
11.1, ""'400cm~/Vs,wehave:
1. 70
p "" 056 0-cm: R = 18.8 kO.
1018
(c) "• = Nn ""
P. =
n/
"•
cm- 3 ;
N, = N" ~~ 10"' em' and
n; = 15 X 1010 cnr-' we have V0 =< 695 111 V•
. 2 ··3
2.25 X 10 em
~
A
Using (3.26) and E,"" 11.7 X 8.85 X 10" F/cm.
we ha~e W = 4.24 X 10' em= 0.424 Jtm. The
ext~mst~ o! the depletion width into then and I
rcgwns 1s gJVen in (3• ·?7)
d 0.28) respectively:7
k
• nn
Using fl.."" 1110cm2Nsand
f.lt,
2
= 400 em !Vs, we have:
5.63 X 10_, 0-em:R"' 188 0.
x.
= w-~
N,.l + Nf)
·-
·"' ~ IJ.m
As expected, since ND is increased by I 00, the
resi~tivity decreases by the same factor.
xr
'~ w-~
NA + N[) ·-""
0"1"
·- ~ IJ.lll
p ...
(d) p = N 4
r
·
"" 2.25 x
~
.,
10 1& Clll
tt
P
'"'
·
',Since bmh
. re
. gwns
are doped equally. the deple-
~
n'fl
uon regton ts symmetric.
w·i em-·'
Using (3.2(J) a~d A "" I o·'' ern' the <·hargt~ magnitude nn each stde of the junction is:
p '" 156 fl.-em; R = 521 kH
(e) Since pis given to be 2.8
X
!0 -~H-em. we
direct! y calculate R "" 9.33 X 10- '
n.
1. 68
J~rifl ~"' q (IIJ.ln + P~lf,) E
Here 11 ,.,_, N~ and since it i~ 11 - Si. one nm assume
I'<< 11 and ignore the term p!J, Also
·-0"1'
01
=
3.39 X !O"C.
.
Chapter 1-21
1.73
1.71
Vr a.t 300" K = 25.8 mV
built in vobage V,
n, .
JfNA·orNoisi~byafactorof 10, then
new value ofV0 will be
·
ln(NA~») = 25.8 X 10-u
V0 = Vr
Vo = Vr ln(N4~»)
.
fit
1
5
( to fl X 10'
lio = Vr
)
In ( 1.5 X to'6}1
tnCO n,N~N")
The change in the valueofVois
= o.633V
,..
ln(lO)
ln(NA~D)
n,
1. 74
with N,. =
to 16 cm·3•
1016 cm·3,
N0 =
and n1 = 1.5 x 1010, we have
Vo= 635 mV.
and V.~t= 5 V, we have
W = 2.83 X 10'4 em = 2.83 Jlm.
with A= 4 X l~ em:~, we have
QJ = 4.12 X 10'" C.
Depletion with
w=
~1'::'·-Es-(~1-.-+---c-1 -::-)-v-0 +-equation 3.26
4q
No
1. 75
J? X l.04 X 10.:.17(.l +-'-)X 0.633
=
W
Nil
1.6 X l0-19
"j
lOti
1015
= 0.951 X I0- 4 cm = 0.951 J.Lffi
1o find
=
X.
Xn and XP
N
W
N 11
16
+" Nll = 0.951
X
IO
10 t6
+ 10 ts
= 0.8642 p.m
=
X
W~ = 0.951
NA +No
P
X
IOI$
1019 +lOIS
= 0.8642 J.lffi
to calculate charge stored on
either side
Q1
=
Aq (
~" N W where junction area
NA +No
°)
"" 400 J.Lm2
= 400 x
I0- 8cm 2
= o.95t
x to·•
Hence.
Q, ., 5.53
x w· 14 c
1. 72
1. 76
Charge srored Q1
""
q AXN
Here X"' O.l flm ", OJ x 10 'em
.k" 10 fl.m X 10 t-tm "'" 10 x 10' em
X 10 X 10 'em
'' 100 X ]() 'cm 2
I, "·" Aqni(_!!L + /)" )
t 1•• N 0 l..,N,~
A "" 200 p.m 1
= 200 x
I, "·' 200 X 10" 8 X 1.6 X 10 !'!X ( 1.5 X lOw)!
+
So. Q1 ''' L6 X to l'l X 100 X 10 8 X O.l
-
X!O 'X 10 16
5 X 10- 4 X 10 17
"l6fC
10 .g cm 2
10
""l.44Xl0
6
A
18
10 X 10 ·!4 X 10 16
Chapter 1-22
I~ I,e
VtV
cJ""'
T
cJ,
,t.44 x to··l6 x e'llt,,z.w
( I+
79~tA
s
c1 =
ForV.= I v,
= BT312e·lioHlKTJ
""'0.45 pF
For
At300K,
II;
= 7.3
X
10
I$
J il
.
X
(300) .
X
e
-1.121(2
X
V,.= lOV. cj =
n12 (at 300 K)
At305K,
=
7.3 X
10
X
I
JQ..)j
0.75
=0.25 pF
= 2.232 X
ts
0.6 pF
(1+
8.62 X 10 -S )( 3().1)
=1.4939 X 1010 /cm 2
11;
1
)'i
0.75
( I+ 1
1. 77
nl
0·6 PF
VR)'"
Vo
(305)
1. 80
10""
312
X
e
-uv(zx8.62>' w- 01 ,oos)
=2.t5i X 101(1
c4
GJ
7
=
10 pF "" (
'Tr
_·)_ X 1 X 10- 3
25.9 X 10
n12 (at 305 K) = 4.631 X 1011.1
3·
'Tr "" 10 X 10- 12 X 25.9
n/(at 305 K)
so .
. = 2.152
= 259 pS
n/(at 300 K)
Fori= 0.1 rnA
1. 78
c,
=
G:)x
1
== e59 X 10-12) X 0.1 X to·J
25.9 X 10· 3
= l pF
1. 81
_ !::1 ""'
for p + -n junction N4 >> N 0
:. I :::: 11,
IT(c
·• D
Aqn,·
=
p'
71' -
V.'l'r
I)
,!!-{f-n ~
10 4 .><
Ll-~ cm1 x
Q,. =
1.1\" l'l · Fl
~r~
X ( 1.5 X
4)1
71'/1'
100 X to-'l X 0.2 X 10 '
10 1") 2
.
(10 X 10
10
note 1t.un = 10 · 4cm
= 100 ns
{1
For this lease
l, ::: Aqn/
Dl'
IO
10 X
J(J'
4
2oxw·'~c
X 10 16
C
"'3.6 X 10- 15 A
.J
= (_
'r)l
Vr
F \' ..
= l,.(e
3.6
X
1 -
!} = 05 X lO ·l
10 ''( ,v:(ZS!h
""} \'
0.6645
v
9
( loox 10 - )xo.2x 10
25.9 X
l!·· ') ··
1)
0.5 X 10
1
- 772 pF
1()-'
Chapter 1-23
1. 82
a.
X,.
0
Depletion . . I
Region ....,
Length of p and n region
=- 11, + I,
Find current component 11'
b. The current I
QP -- Aq X 'I2 [p.(X") - p 00 ](W. -X.)
t . [
= :zAq Pnot!
wv,
I
= 2Aqp. 0 +(e
I
nr'
-
D
P •• HW.- X.)
-
Vll'r
-l)(W.-X.)
:;Aq-N· (W.-X.)(e
> Xp
dl
1 1'd\1
But l = l~ (e 1'/V,- 1)
dl
d\1
lse
VN.
'
Vr
f
-~'r
. c···d ·= 'TT ·
SO
(e
fI
~~
V,l'V•.
' .. 1)
w!
= 1
1 x 10 1
'
2 10 25.9 X 10·'
Solve for W,
d.
The excess ch~mge. Q1, can b<: obtained by multiplying the area of the sheded triangk of the
p,.(x) distibnti<1n ),!J:tph by Aq
c,.
w. '"'
63.25 lllll
'-1)
Chapter 2-1
2.4
2.1
2.2
ik. vot~c a..t -t\u.. pos\~,tt 'lnpu!: h.Q.b-\::.o
be. - ~ .~ooY .
~ •-3.0.2.6\f"
,A. t...\
, =.--3.
(U'+ -Lt.)
=\00
- 3.o26 ..(·3)
2.5
''"" = I u Sin (27160)r "" ~( ''' + v2}
,,, = 0.01 Sin (21TIOOO)t
2.3
#
t.l
v,
1
'I?,J=
u.
vr/t•,
vl- ,,,
1
0.00
o.oo
0.00
0.00
-
2
1.00 1.00
0.00
0.00
-
(b)
1.00
1.00
2.6
(a)
4
1.00 1.10
0.10
10.1
I0 I
Circuit
5
2.01
-O.ot
-0.99
99
a
1.99 2.00
6
~0
(c)
0.01
1.00
(d)
-5.10
Fo
·- 001
V
(h') v i i = )
\ = 1.00
100..
T' 4 "'' l.OlJ- 0.01 ~" 0.99V
The missing entries for experiment #7:
(a) v, = " ' -
1', =
"
.:· 5 · 10
100
= ··0.051
v
u,+ ,,,=5.10-0.051 =5.049V
All the resluls seen to be reastmable.
(C)n,
t>/.,,(VIV)
-100
lO
100
experiments 4.5.6 show that the gain is app!'O)ci·
mately 100 VN. The miasing entry for expemnent
#3 can be predk1ed as follows:
(.d)
v,= ''""- u,/2 ""Sill(l207l')t- 0.005Sin
201:l01rt
,.. , -= 1• ''" + '',I 2 = Sin 120m + 0.005 Sin 2000111
3
2.00
= 1•,- "'
"" -10
R.,.(kil)
10
b
-10
c
-10
10
d
-10
10
10
virtual ground no current in I 0 kil
Chapter 2-2
2.10
2.8
=-~a
-=>- R..2. = 5~1
R,'+ R.2 ... 1.2.o.t..n. =>- SR, +R, .=-l.loK..Q~
R, .. !tet/<..Jt ->- R.,_::: roo ~Jl
~
::: - 5
1):.
There.are four possibilities:
,~
f , 1(/l.
~ "' -2 VIV Rin
lf;
=
10 kJl
lOkfl
2.11
lOkfi
lOkfl
~
= -0.5 V/V
ll'i
R,. "'
t
lo~IG.(= ~ JB ~ u..,.-fq. qs- ~v- ~ ~ "' ..
:=::)- ' \ . = I i . 't S" R, { 10 H Jl
For IQr~esl-
10 kfi
R..2.
poss:.lole i"'~
-= 1o HJl. --7
R, c::. Soo KJl.
R;"' ::: Soo ~o
.A,
I
~ c.lo.s...d toop
~a.:_ .( C:r}.
G-:
2.16
~ :s..SD~
£?
If
v.·c-*>
" A,
-=
2.15
2.17
""
-R,
R2
5
G=-"=-·~-=v1
R1
R1
15
v, = OV,
1• 1 = l', = 5V
For ±I% onR.. R,: R1 = 15 ± 0.!5 .kll
R2
-R,
v, = ll;~ =
s "" s !5 X
= 4.9 v $
For
$
v; "'' -
(Jp:5
=:> 4.85
1_R2
'R;
.tsx4.95
~
15.15
-
~4~~~5
''"
$5.1
110:5
P..
I); -
....
~
t'i - ~
RJJ.
. - Ro/R.,
5.15
(1+~)/A
_ -
100
1--t
1,~bo
... _
.9o.g Vt
~
::::.)'-.!&,. ::::lJO, g ~(T
v
+ .
· V 14 85 X 4 · 95
15-0.15
· · · 15.15
:5
«>"'; -
Oj "" 1 ~
";·
~ lf.ssvMe
- 15 X 5.05
l ).
•
14.85
l'
5 ::!:. 0.05 kll
"'
v
R; .. RJ( (/ R,
wh..en ~. = 11($1..
YR. ...~1oo V'jv
V,·
!J.· -IJ;
~
,.,,
:;
tJ, - tJD :arR' a R .. ~- :!P)i~-~)
f?
'
J ·1oo 1oooV\ tooo
-~
R.' _ J-O.:.!
, -
""'o. sqeu~..n.
J.oot
__., ~ =~. 9 I< .n.
!:!
5.
;;: ~::: .....&sf<,-t ~ I +f\..c.
al- I<.Jl t ,z
Chapter 2-4
2.18
~t~lb:.t3c ~f. +4
c.Ji(t
-tk_..
';l
• ..
1ttlltll"h'rt1
v.= -At•. =II.- i,R,
i,R, = 11 (1 +A)
ijl~ Wwt~•,.eJ.
=
TJ•..
(
lroM
Vil"l ~ j"'lllfl
Vof~3t c.F ~f!ro
:Joll t.o ~ . TJu.JS
d.· (ooo
Will
d
tooo ·
tpoA.t-
.il..
i-t"PIM "U'\e
"J a, M"">C;,,u,,.,
i; Rl
I+A
Again v,=i,R, +
• ,_ 1
lq&lj
.R +
-1, · I .
ctP .t 10 ,.,v:
~ =
SoRID =
11
2.19
11
.
R2
11--
J+A
R +..!!J_
1
A+ 1
2.21
G'
VJ
=
-R21Rt'
t+l+R 2 /R 1'
A
il
,.__-+-....:.._-l----4·
In order for G' = G:
G =
=
-RziRI'
I+ l +R2 lR 1'
A
-R2
R1
a) For A "" oo: v,"' 0
b) For A-finite: u1
-tl
:::::l> 1!0
""
-•-
A
""
-~. tJ0 = v1 TJ
(Rl
i 1Rr
+ Rc)A
R,A = R,
Rr
= ARc+ Rc
+ ~(R 1 + R,.)
+ GR, + GR.
R" _A- G
i 1Rr= R,. "" .-!" = - - - ·
''
1 + .!_
A
:R;-I+G
2.22
2.20
(; = -R.,_ /R.,
I+
E
I-+ lt1/R1
p..
::I !;-~........:........!!
"'l v.Gt........-.....1..
~.. ~
lb.- G_.....,...J • ..1oc
'tv
f>. = (I Hoi>) ( .L -1}
C>·(
ih•'.s ;, &
~J
-'\
e- .. (o'l.:::: o./
='lo, V{v
,W......:.wJM ..,, ~J..iN!d
vtJ.u..o. tnr ~ •
Chapter 2-5
2.23
(a) A!Gl!IOI "" I+ R1 1 R1
AAIA
A
{b)
A~l ~
0..5%.
o.oos _
1+
A:= 50%.=~= tOO
too
S--A-
A = tOI(.5)
(.005)
= 10.1 k
2.27
a)R,
=8
l!
82 = (R
R3 ·= (Rz
II
R)
+~
R) +
R4 ""
=
1+1 =
+~
""'
11
+ / 12
= (
P..,
<:.l ", = I, R = -/8
"' = ···l,R = -2/R
", = ... J,
I? == -41R
- I 8 + T, 1i
'
.2
• · · · 4/R
···· 81!5 •
2
-818
Chapter2-6
2.30
R1 = R, = lOOkO
!e ,. -RuR• + R. +
v1
R4
R1-\R3
v
= O=:~o...!!
v
R2
-R
= - 1 = -1 =:~oR1 = IOOkU
R1
1
R4
""
t)
10 kO =:~o.!!! .., - JO
v,
= -l X
(.!!!.!!!
+ .]!!!!_ + t)
R
100 kO
3
+10 =
(1~ + u)=:~oRj =
1.12 kO
Potentiometer in the middle:
!?t
'~';
=
5- + i )
-1(5 + R:1 100 + I
~"" -1.87VN
V;
2.31
2.29
.
~
..
t>
\t. =.0
..-!)- t..
=
0
~~:JMA
fr.o..v.
po.l"t"Q..:
.
"1..="" "''1:
~
.::t
~0 Wt~
Chapter2-7
2.32
cltoou ikt 14)~"'-t-~d S~--tr Co"kdura4"
Wt
tt-·--
2.35
v.= v, + 2v,·-3tJ,-4tJ.:
R
For a weighled summer
~__...,,__...~
).
circuit:
assume:
R0 = IOkO
2.33
v. = -(2l', + 4v, + 8v,)
R,R,,R,;;.IOitfi
R.
R..
R
-!; "" 2, -!; = 4 ....[ = !!
R1
R2
R4 = 10 kO => R<
R3
R1
=
Rb
I ~ X :!Q = I
R2
R,
Rb
=
R,
=
R.-= IOkfi
~ .= ..3.s;-.,. c.t.n~<&c t:)- o. ol s.-... ( :trr ,.,ooot>
sl\.o o\c\. 'bt..:
c.Pt
= _5s.:""wt. -5
fl.Sf>VI'II\..t.:
I)":
5S•".ut}
~:: ?..v
~=
-v, -+~S'{)
! .... a.. wual\..t:-c.d .ScJW\~f' ~\dll~~"":
Jk. :: .<. .s
JSL : .....,,
R,
R.1.
p_
:li>K..Il.. ......,.. •.,.
&
3
L',::: 3 S.n.(ltJJC6ot.).,.. o. ofS,·" ( .:z n Ylflttot)
"'lhe.. ov....tp.....t "~d~\
R2
40
2.36
2.34
oH.
:!Q
R.""' 40kH,
R, = 20kfi
R, "'40kfi
~~
40 kfi => R3 =
= 13.3 kO
Ra x
R.1 "" 10 kH =:) Rr = 80 kfi
I.T0
=
"'1c:;K
:R,
q
v)~f' Co
ltcwe :
~ .= 10 ~ _ 10~
Chapter2-8
2.37
This is aweigbled 8UIIltller Cireutt:
V.,
·
2.40
R,
. R11 . . R, . R,.. )
=-( -. V0 +-t!t+..o-11:1+~t1J
. R0 · R1
R,_
R3
•
wemay~:
"'I=
Vt.
SvXao
til= 5v>kfl
0.1 mA
As indicated, i only depends on Rand v and tbe
meter resistance does not affect i.
2
F
v., = - R1;
(20a0 + 2'a 1 + :22a2 + 2la3)
2.41
tJ.=
fl 11
+3n.,-2(v,+3v,.)
RR,N3 = 2 I·r RN)
= IO.kfl=>R,""' 20kfl
RRrN4 = 6=>RN4
,. -6· = 3.3 kO
RN
=
20
Rm II RN4 = 10 K H 3.3 k
(I+Rr)Rp
R,v Ro
=
l=>(t+12.)R,.
2.48 R,. 1
= I =>9.06 R,.
Rt' "" R,.t
II
Rn
= 2.48 kO
II
= R,. 1
R,.l => R,.
2.39
..!_+...!_+...!_
Rl't
Rn Rl'll
(l + -Rr)Rr
-
RN Rl'!
R,,
R
1•
= 3 => 9.06- =3
R,.2
=> Rn:::.3R,
X 3Rr
II Rn = 9
""9+'") = 2.25R,..
R,. "" 2.25R,
II
R,.D
=> R,. + Rro = 2.25 R, 0 =>Rro = 1.8 Rr
Rp "" 10 k!l => R,. 0 ""' 18 k!l
~=2
u,
Short•circuit R,;
R., = 9 X IOk = 90kfl
R., = 3 X 10 k = 30 kfl
Rr=20kH
IOkfl
IOkfl
IOk!l
--
Chapter 2-9
2.42
2.45
tokO
2.43
Setting
output component due to
tJ,
11,
= o, we obtain the
as:
110
=
t•1...J.Q__
l
v.,=-20v
Setting t•, = o, we obtain the output component
dueio t•,as:
v02 = u7( I +
2~X 20~~
R) = 20
(t + !.Q)
I
2.46
t'2
lOOkQ
The total output voltage is:
tt.= n,., + ""' = 20(r•,- tt 1)
Fort>,"' 10Sin21TX 60t- 0.1 Sin(21TX lOOOt)
u, = IOSin 211" X 60t + 0.1 Sin (211" X 1000 t)
tt, =
+ lO
+
lOY
4Sin (211" X 1000 I)
a) Source is connected directly.
110
2.44
\)"o
0,-
"''+..&.-R,
~>( ?<. 4 \ _.,.
=\+~-=- \+L-t=:.L
:x.
".X.
I {. QA. -<..
-~
DO;)
= 10 X _I_
=
(}.()99 V
101
i1 = ~ = 0 ·099 ·= 0.()99 mA
.
I kfl
I
Current supplied by the source is 0Jl99 mA.
b) inserting a buffer
;Jt w~ ~d. o.r-6t.$\-or o" -tloo..l. arou"c\. to:th,
~=I-t lt-?C.ltiOK
q.
.?t.artOlt + R.
Go,,~y....., 2.\ wh~"'
-x.... o
-=*>'" !U :::\ -t
-'r
10'<.
R. = .!B.!S...
!tc:>
R.
.=
0· 5 \C. ..t)_
110
.
11
=
IOV
JO V
= lK =
10 mA
current supplied by the source is 0.
The load current i1. comes fmm the power supply
of !h.? op-amp.
Chapter 2-10
90k0
~:
-
eq,
lOkfl
IfA= lOOtben:
a_..., __,_A =
••.-....
I+ 2_2
to
tO
9·- v...,
1 + 1 + 90/10 = iJ ""' '"" '"
100
I +90+90
G"""'l'"•••lod = _ _.:.:10:.._.::.36~
1+
ertorof dam magnitude
f!it- ,I
l!LJ
1
~)
=
= --'A+l
1000
100
125
'i:i25
I +90+2Q
. 10 36
too
.
= 11.1 V IV
10
2.49
Vo(Y).
v
0.999
Oain
error
-0.1%
0.990
0.909
_§L• G0 -GX tOO= GofA ~tOO sx
I+ Go
~';
-I%
A
-9.1%
:::} A 2: G0 F where F = 100 - 1 .:: 100
X
A= 50V/V
I+~=
R,
G = !:Q =
IOV/V
Thus for:
Rl
Rt
l
t'1
x=O.Ol:
+ t + R.yll 1
A
1 + 98/10
I+ I
+ 90/10
0 0 (V/V)
A(V/V)
x=O.l:
x= 1:
G0 (V/V)
(510R,h + 90R,h + 900)
= 50 X ( JOR,b + 90R,~o
3()()();:}R,b ""
+ t)(J())
36 kfl
x= 10:
10;
G 0 (V/V)
A(VIV)
IOOR,, ""'
tol
J()l
105
tot"
tO' J(}ll
Hfl
too hlgb to
be practical
A(V/V)
X
to4
10
_ 10 _
- 1.2 - 8.33 V/V
50
In order to compensate the gain drop,
we can shunt a resistor with R1•
Compensated:
10
X
= lOk{l:::}Rl = 90kfi
1+
G =
l+ Go
Go
A
2.48
ifR1
G"'
G0 (V/V)
102
lO
to2 tol lo4
to4
HP
loti
101
10
102
tal
104
1(}1
104
lOS
J()<\
10
to2
Chapter 2-11
2.50
2.53
for oon-:inverting amplifier
lf we assume R3 = R 1• R4 = R1., then
G =Go
-l + G(J
A
,e = Go- G X 100
Gt,
Rtd
for inverting amplifier
Ga
,. Go-G X 100
G=
,E
(!
It)
I+ 1-tio
i\(VIV)
R,
R1 = R2
Go
{VIV)
a
e%
G
b) A,1 =
(VIV)
-I
b
10
-0.83
16
10
0.91
9
R1
= R3
c
-I
100
-0.98
2
R. 1 = R3
10
10
5
50
d) AJ =
e
-10
100
-9
10
f
-10
1000
-9.89
Ll
g
+I
1
0.67
33
-
30X20
20 + 100 + 20
-
- 15
·
+
+ 20
20
2.52
~ ""
Rl
=
1.()7
10.714 V
=
v+ :
onlv:
R1 = "1!1
.
I "" R
R
V
Noh(#. f{..c...t
...u hi:a.ll'e. ; & "" .&.... _too
{0
f{3
~
--
R.,.J. :: 2 R, :: .2o~.n.
p.)iN
20Ul"" R4
0.5 V/V => R2 = 5 kfi = R4
I'() :S;
.:1Po "" 2 X 10.714
=
= 10 kfl
= -10.714V
+ 10.714
pot has 20 tum, each tum:
-f0.7)4 :S;
= !Ok!l
= lOk!l
Considering that u _
••1
30 X 20
20 + 100
= 2 V/V =>R2
2.54
when set to the top:
t.o -
= 10 k!l
R4 = 10 kfi
""'
R 1 = R 3 = JOkfl
when potentiometer is set to
- l'i+
='>
~
R3
z:
c) Ad= R2 = 100 VIV=>R2 = 1 Mfi= R4
Rl
the bottom:
,
""
d
2.51
=
a) Ad= Hz = I V/V=>R 2
A
case
= 2RI => R.
t:l;threovf hy Rs = (R 1 + R2) II (R3 + R4)
"'0
II
R
2.56
A,M .. ~ .. A_ (•- ~.!!..)
t_, i ltlf
UiU"f
1l..c. wot-StR
" 3 connected
to both
f<,
~
\~ w~t'." ~- t.....~ \\-S.
c.o.JJ.A-
R
lT
1&
~<--. = \
" 2:
.&1...
R1 =.!!"" R
I
t \-
·H
B:.a.. !3..)
~~
ct,
R-...,
M~ ~..... :=:...,..
v~ O..f\d
h
hAs to l.:.c oJ:
o..\SoR'-'
~
ho.J>
~{.&. ~ ~.
R
\OO't:X.
R_'1
R
bt..
""''"''~.
\00 -t.ll.
R. 1
\oo..:x.,
1oo-~
\00 t)(.
{{....
l'f~
to
t~,{h-1:.. ~
IOo-.::X.
Jb......:
R
.Ba.
\t'£ ~i~
-
~ :: 2.o tod l ~ \ . No.-\ ~t. h.A.II'C to c.a.lc:u.ltJQ.
~ \!>~ o""' ~ ~ c_h,ose fa.,- f\-1\
C.t'\
.,....\ft.
"""~
V<;.
'\...,. Ko.,c. •
(:(1.:::.R.~-::1oo-.,;
~::. ~~ + '-"o2.
1'
~
::
-.,.., qppty \II\~ .su~r pas:Ho""
-~_l._ " ; + '-'2. ~ li-t fh. )
"1r,"
R.)+~
~
'{"' _
l)O
R,:=:~..:\00-t-:X.
~oo-X.
o- +
'-"'2-
•
=-12o10.01"'-~ {)- -+ &
I
2.
1 aot;t.
(I-t
:J.,bo
\00-t-X..
j}-
wt_
C..O"'!h'd.H
~ ~~
jOO-tk,
(oo-;x.)
1 oo-tX.
Chapter 2-13
CMRR
CMRR =
lk
20 log
for~' << L
2.57
A
""
if k "" AJ,.,.,.1
...J!.L.(l~~)
R + R.;
R R
=
3
1
=
vu 1
+ tlo:!
=
-R1 v 1 + 1':!~(1
R1
R., + R4
+ ~)
R1
"'·
~ ~~.
100,
= 20 log 0.04
lQ!
CMRR
4
In order to calculate Ad, we use Superposition
principle:
v0
+'I
4~
0.01
"' 68 db
2.58
then replace v 1 =
'=
100 (1 - 100.100)
JOO:'ti)O
I00 + I00 .
A,m =o
Referto 2.17:
~
~
R)
R1
R
=> A,t = -1 "'
Rl
li) Since A~ = o.
IOOkQ
Ad
CMRR
ClvtRR
then if we apply
v"" 10 v, and
Therefore, VA
I,+!~
CMRR
20 lng
I
II
·•
!~!
-:R;I
2 R, ___
~-~~
R, R,
I
fm worst case, minimum CMRR we haw ~o maximizc the dcn.mtin-
~i5
v,,, ''., = o.
·-~lQQ_____
"" 100 + 100
11
5 '$
I.'...: Ill ~7
5
~
2.5
Chapter 2-14
c)we.aplythe Supeqmition principle to calculate
A,.
tl •• is the oulptlt voltage when 11 ,. = o
= t'IcM ( -
""" i.s the output. voltnge when u 11 = o
...3z.. = Aer•
I 00 kfi II 10 k!l
)
I + 12(100 kU 11 10 k!l) + 100 kfi
=
0
f~(;M
Now we calculate ~~.,. range:
- 25 s
-2.5
119
s 2.5 ~
100 kfi U 10 kfl
< 2.5
< I'JCM X (tOO kH II 10 k!l) + 100 kH
-30V,.; t'R"',.;30V
lOOk!l
2.59
+
11" 2 =
(l +
11,;,1
too ~c.n II 10 ~c.n
!l 10 kfi + 100
t'u 100 kO
. 100 kfi
)
too kn II to ld1
= tJI.:: X
~
v,=
11"
so we can consider v , . v a virtual
1
u,} ""- u., 1 + u.,z
= -
1~ 1
+ l'lz~A<~""
I
Now we calculate A,~:
then:
()~ l'AII
l.-'tl
.-::::
·~
j
' ""
v, =
t'ICM
kf!
lkM- II,;
100 Hl
T!A -·
i1
lt
10() kO !1 10 kH
100 kH il 10
+ !()() kfl .
100 kfl X i,and
.
·•
i, '= t 1
-
FA
!(} kfl
", ·- I 00 kH X i,
+
lO X ,, ,
·-[i;R! +VIlA+ i,Rl]
11'1
- [2i-;R2
+ 11o,d
Chapter 2-15
2Skll
25kll
2.60
a)
and 0 together
a)
v,
iv).!!! ""
l'+
=
b)~ = -1 V/V
v,
+! V/V
2 ""
2
v.
Ttl
i)
25k.O
25k0
2.61
v,
ii)
!2
= +I VIV
'l.li
3-o.a'ISi,.,.,t V
251t.O
25kD
o.oc,S:nwt ::f3-t1.o~S,-,.wb) .: o.olS.~wl;)'•A
'" + Sotr.wt.:
. 3+ 't.olf s,·,. wt, v'
.. 3.,.o.o~S;"'wt
1.. =
~
3+
V, .. 3-C'#.ol.tS,~~t:-51>/C-" i =3-4.ol.fs:..,..,t, v
"t ='b =t ~ . . '· s _ .2 .o 2. s:;'l ..1 t:: ~ If
The circuit on the lefl ideally has infinite input
resistance
iii)~ = +2 V/V
l';
\{ ::: ..,.13 -~ ... - B.oB S,'A.wl; , r/
Chapter 2-16
2. 62
.a.
(t + ~) ""· 101. If
The gain ofthefmt stage is:
the opamps of the first stage saturate at :t: 14 V :
-14 V
~
:S
v 1 S + 14 V
-0.14 V :S
viplained in the teJ>t, the disadvantage of drcuit in Fig. 2.20a is that v ~is amplified by a
gain equal to v"' (I +
~:)
in the first stage and
2.64
therefore a very small tl ...,. range is acceptable to
avoid saturation.
b) In Pig. 2.20b. when v...,isapplied. f'. for both
A, & A, is the same and therefore no current
flows through 2R1• This means voltage at the out·
put of A, and A1 is the same as "'~·
-14 :S v, :S 14~ -14 :S l';cm:S 14
Aa(2) of the second stage is
&
R)
=
0.5
R, = 100k0,R,"" 200k!l
we use a series configuration of R,. and
R, (Pot): R, = 100 kOPot
(Fixed)
Minimum gain =
This circuit allow for bigger range of n •~·
(I
+
t,) ~ 05(
I
+ 100
:'+
R,
2
I :s A.t s 100 => I =
o.s(t
+
l
2R2
RIF+ IOOk!l
=> R1r + 100
2.63
A., "" Jb_ {_1-t- Ra.. )
...
f\,....:oo
~j
c
R,
.!.fle!
(l ,_
1<» 14
J!Yl.!:) ,. .21 \//
!i',._
lti
,..d r:: 21
(t + 2L)=>
Rlr/1
ZR, = 199 Rw
(2)
R,
= o.so5 k!l:: o.:nn
= 50.25 kfl:: 50 kfl
2.65
a)
!]!
1~
=
I +
otpplyVc.., t.o
c-d note. tt....ed: \lew.. w ~ \\ o..pptiM"'
t:..._ ordtr to c.c...\UA.t.oli. Ac.,.....,
b:>'tk ;.... ~..J:.s
loo'\-h o "'kp....t "'\er-..l"''''h a f. 1"W.. ~:r,.h~.
to..,.. lt."Cli.lt.lok
W4.
:;.c..(O~ S~f.
Cod
t.'o
':.t
±;x..(
.t..:::;J
Ac__
=,..
..ll
~y~:"'a-- 'iWt.
do""- \..._ f'f'6\o\~ '2..6!.),
Wo.b
I... P2 . 6 'S ,...a .s ho...xd 1i-..J i ~
I.._
(1)
Maximum gain= 100"' 0.5
!>0
-±ll•
•
U:. a.\1 rtl\st-o.-s: on.
No...>
2R1
o). <2) => R1r
C.HRR = 2.o~~ ~~~ :::
cJ;
=
=~ . lh~"~6t"f. ii.Q
o-"""
i~ 41S6~.
Ac..w..-:L
c.-~ R R. ,. !Z..o
l>o
c
e.'\
!:o
So
,::;:-o.a'L
$<)
..2.1..._ ,. 6 0 Jh
=
z,_;,,
we use supe.rposition:
Vl
~
J11- V:-
t•, only:
V 11 = V 0
2
Zt.i.,t
ul---2-
z j
..1.:...£!.-i
ot (•
.(.J_ •·R)
,
2
----~--
RI
i 1 R~it'"~
o
R
•'<;(\-')
...
15 ~
z~,i.,t
2
r)
Now if only (-
11 ,)
is applied:
tllJ
l'A
The total current due to l>oth sources is:
1.5
:.1--~=~
R
R
R
has ideally inlinitt·
The circuit
input re&istant-'e, and 11 requires that both terminals of Z, be available, while the other cir<:uit has
finite input rcsistnnce with one side of Z,
groumkd.
Chapter 2-18
2.67
2.69
ITI
= - 1- if ITI = 100 VN for/= I kHz.
wRC
!TI
then for
= 1 VN, jhas to be I kHz X 100
=
lOOkHz.
Also
RC ·"" j_
wT
1.59 p.s
2'11' X I kHz X 100
2.70
b)
+
~,a..,4J qt/' ioD.
J
-d
if reS«Jlts
b~ ~i.. u
R., = R, ThusR = IOOkO.
ih::.!...,
.
•
·~~ ou.t'p~ ~..ek"J -Hu.. ,-,f~b'/ '10
!6 iF .Fre1vl!"'GV
;F- /o,.Jt!rt:J /,'/ ~tfb.tfor
JW
J
ol- tb, t'h.e111. #.c ~tA.l-rJ 'th>old..t.., Cf'(!ASt! IDy t:4h:ctlbr
C.) L-'0 .:: 1):•
.,,. (0,
ITI = - 1-
wRC
RC
..L
~ C = -:-::-::-:---=--,-,...,..,
RC
1000 X 1002
with a 2V-2ms pulse at the input. the output falls
linearly until f = 2ms at which
-
ft.,& D~
I at w = j_ .
w = 1000 rad I s =
1' " -
cl; Th.t P~e doun.t~f: ~e. d!VIJ
S*l'fl le..ds 1-4 ~ l..y 'fo"
=
v,.
=
t'o
-1
C
-2
t = RC t = -2t Volls
where 1 in ms
TilUS 11 1 "'
-4 V
2.68
R,·t'l =R. ~ 1ooK..n.
c R "'r s ->-c.
2!------,
_1_ •
to 6ltlcl>
'OJf.F-
"7
0
r (ms)
2
2
-t.h.a ~H·""t'~ ..-~c:.w
o.I'\J n- r-e.~ t..S
to v- tJ:
a.t t~ tofU•. :
e "' :M s "'"
olf
10
-lo
.-
...,.t(msl
los
'
f:
with V, ·•·• 2sin 10001 applied at the~ input.
2 X __
I .--·······
100{) X 10 '
l•.,(l)
P ,.(I)
o•.•
2sin ( lOOOr + 91Y')
~in( IOOOr
..
f
90'').
IOmF
Chapter 2-19
2.71
R,=R,.,20ill
a) when noR,
1- = t a t
111=wRC
w = 21T X 10
I
kHz~
C "' --......,....,-!.,,----:--:-:
u.,(t) ""
2w X 10kHz X 20 K
;(U
l.dl
-62.8t o""
r"" 0.1 ms
C=0.796n.F
~=
RFI R and dte finite de gain is -R"·
u1
I.+ S CRr
R
There fore for 40db gain or equivalently I00 VIV
-Rr
we have: R = -100 V!V
~ Rf
= 100 X 20 k
b) with R.: !t0 (1) ""
(Refcrto pg. I 12)
110 (oo)( j ·--
t>,("") = -I X R., "" -tJo(t) = - 100( I -
e
11CRI')
2~ ~ X 2 M
-100 V
r•t !.~)
= 2 Mfl
The comer frequency _l_R is:
c
0.796mX 2M
I
F
=628Hz
T~(ll)
----4-------L----•H(ms}
0.1
2.72
Each pulse Jowers the output voltage by:
j..,
"
= _!__. 1·· l.dt = 10 !}:S =
RC
RC
~
I rm
= 10 rnV
Therefore a total of I 00 pulses arc required to
cause n change of I \'in vJt).
l
<\(1)
0.1
~-_._____0
D
Chapter 2-20
2.73
__g.._
~=
"''
,·s.
VR,
~:a. .otSC.
y~
-a.
1>1 k.:c:k
:h.
LP dr~ w;T\.. o. de.~
Stc.
Oo.lfl
•-
0.5ms
of- ~ owl· c. ~-~ID ~ ~ ~::: t~:
~ {~ rt~'"~:~Mc.t.. ~~ t-o~~ . S6 S!or:
R;. t l<
R, ... \\C.J"'. o-d ~t~~r clc.~}~ .ofJ.O~r
=>
-2,____.
....., Ra_=\0 ~
;..Ba..::r\0
R,
\0
~
~clb ~~~ <~~~ L(~'*4!o-! W.C, ... 1tt~~
a
{h.c. 11>'1.4 '11-
(.0 d. b)
is 4 0 \4- ~ 2:
it>
=
-CRd";
''o = -I rns x _!__Y_
10
and
110
2.76
2.74
~(s)
= -sRC
-s X O.Ql X 10'
6
X 10 X 10 1
V;
~ 1110
1 "'
'l1;
I when w = 10\ad ! s
orf= L59kHz
for au input 10 times this frequency, the output
will he 10 times a~ large as the input: IOV peakto-peak The (·j) indicate~ that the output lags the
input by 90" . Thus
110
(1)
d1
therefore:
For 0 ,s t ,s 0.5 :
=
·-·5 sin ( to'r + 90") Volts
2.75
)p
0
0.'\m~
=
0.5 ms
0 otherwise
-2 V
Chapter 2-21
+""- oJ~
1'hl.t4 -t\.4 f>CA t. vcJ.vA,. of.
it. O•'tMI.xto"-A... li\1" • '1ha.
l$. ~ ~~ D.'
$,Wif\.L W4IAI
w.e, ""*"1 b~ \-WI o~t.rl:
tk .... ~~ (.lIt\+'!.) •
'"'~
R
\u4.A· \o \oc.
w~
Q;
0"
~
'"'c.n.tMMl
-t'\..A.
\-'K\t-1...
I
IV fl"K
+\..4
t.o
tl\pr.J::
0 ·
0~
\0~
hl.
J..Q_""
~ .s" \..e.1'1'!'-
'*lil\.&
~o..vt is ec.~t!
4
tt ~
ba,.
~ oJ:. t.W. ot..t'pt.J: •
.:s:a-t
~r~d "t
M&.t:~~ s \
Of'.L 0 ..
tt..... .f.&~~v.JkCf
.
~\%<.t>.o: -RC.
,_\kol'l.
Slot'
Slbpt
o~ lft~ *.vt)
oc..c»rs ·~ tk
oukt.,J: c )."C\)II.\oo6,.. ~c. • 6-:Ul"ll"
"l1\.& Ma.J:i."'JJ& ptA"- of.G..'J r'J: cr-t. wt•o, .111, •••
=
RC
to- 3s when
C = tomF~R
100 k!l
=
..3
Jc.f;· ,. _oJ,.fo-~/OJ __ ,o J«?;·
~c
4t
~l.t) ::- _,;3,. .t.rt ,.1tuto , Co., tlltP 106 ot:.)
Jt
q> ""
I ~~
'{.,1:)::- :trt C,.s (.2.rt,..!D<:OC't)
Tf;
-90" always
=
I
~w
""
1
= lkrad/s
unity RC
Gain is lO times the unity gain, when == the frequency is JO times the unity gain frequency. Similarly for w
= .l
krad!s.
10
gain is 0.1 V N . (for
w = 10 krad!s, gain = 10 VN)
for high frequency C is short circuited,
-R = -100~R 1
110"'
R1
u1
!::!!
~ »'Jdb
hJh
I. kO
-Res = -HfJs
R1Cs + I
10-5$ + 1
=
v1
1::
= 100 krad / s or
= !5.9 kHz
R
-tk ov.tp.....t.
.W..
to
t.-j>
i>
tn\/~rt~
"' - 6. !li
S.·t1
S•"' t
•
:trt.c tooot +'ftJ)
(,2 rf• 'CDc:> t. +
q/)
'-'ot:t1 ::-6.28 Co.a<..:>.rt>;.s:"'as, L~s vc..\.141.. \s .2.1"\~~<1000 .""'t\~ts. "'""' ttA"-
2.77
"{ .. >.'yt(..tt\ll:.\000 b}
o.. !>lM!..$oi tlo..l
wo.;.\M ~rM o.c<
ou..tp....t
~C.~"MA.ot;~
equal t.o
s~u. t.W. ~
1M.. ~c. ~ ol- tkt o~p.J: ;s.
,.
t.) 1M pt.Qks o~ t\...t.
for unity gain:
ilo 'sl
if w
=
=
Ito-\ + 11 ==< "'u
10.1 krad! s:
~~~~
F;j
~ i""frtA~U. ~~t'QCJI.d\.tS.
t\. (. a..s. ~ \tcc...~ -a-'''o~b\;t J-.tt.ll} S6
uu. e.w... ,._\ul:- R, .. \6\l-J\.
a) for w
To ~'oi-cU-11\. "W.a"1.-~~ ~~ of. LtO!\.
l t. t.. too) ~ .B.!: .. \OO -""")- (\.:\.. • \ M..S\. •
R\
b)
~&r o.
'3.d\, ~\1&\.C.~
..L- ,. .2 "'- 'tt\Ooo
Et,c
"""Y
.. ~
c.. ..
1~!10 t\1:- ~
\5. ~"" F
\00 \ '("'C;.4~u.s
l.h
"o(. )
-RziRt
-R2R1 .
.Raw
- JW - - - - - j-TJi
+ ~)- wtfjw R 1 w1
(t
JW
forw, << w<< w2
"o ·
-(pv):::
-R2
k,
1';
c) for w >> w 2 and w2 >> w1:
~ uM.t-1- ak
-R2R1
!1!(jw)I
11;
~-. ~ ..e lr•de _J,·~l'"eu,...... \,e\o~, u.>t. se.e.. t"ka..b
= w1 << w 2
- -RtRI
+ jwfwl jwlw 2
""
j(RR2t)(l:z.)
..
from the results of a), b) and c) we can draw the
Bode-plot:
=
~ o.o\ ~>
~
h
C1
"'·
w1
1.59 IJ.F
'~ 10Hz.~ w1
""
21T X 10 X 10 1
'~ ~
R1C1
Chapter 2-23
2.80
= v05(t + ~ + R2 tm)
= ::!::2 mV
t'o == 0,01 sinult X 200 + ''os X 200
''os
=
9.09= v0 s X 101
2 sinwt ± 0.4 V
=> l00v08
2.81
Ou.t:-t">u.t Pc. of-fs~ ~ , '-'os
.s:""'oiJ :s
"'
oC:
tlvor.J: ruulhs ....._
t>M
-l"
tw "'-f\i~u
~
-0.22 => ''os
-2.2 mV
A.., ; .. p~
Qp,.
IOMO
.:>..J:.puJ::
13. 3 =to\}".,q 11::: JPI
I 181
~~f-ore 'f"hc.
+ 100001111
(I). (2)
•
IO
-
M\1
is CAp~'"* toup\~,~:
R= IOM.O
= i.L,.
t3-""
\600
c) In this case. Since R is too large, we may
ignore u05 compare to lhe voltage drop across R.
''os << R18 , Also Eq 2.46 holds:
2.82
R3 = Rdl R2
therefore from Eq. 2.47:
''o = los
1.4 fit!
ttos
X
R~ => 10 s
=
10°~~.0
10 s = -80 nA
= :~ = 1.4 mV
2.84
2.83
IOM.Q
p
a) In
(lfJ!
+ flr;J/2
R2
open input:
t•o ""'
t'+
+ Rzl Ill
l'os
9.31 = "os + J{J{J{){J/1!1
input conneded to ground:
+ R2/lll
""
cl
•
R 3 = JOOkO
1+!!2=200
Nl
(l)
"f>l
=·
100
199k
'\()? fl
...
-
RlCl = 27T X 100 ~ Cl o•
3.18 ftF
_I_ "·' 211' X 10 c7 C.,
RJC~
-
.,.• 0.16 ttF
SClO x
I
27T X toO
Chapter 2-24
2.85
\t, • "as(. f+ .!.!!)
"ol ,.4(1+100): l.f(llf,..l{
IOJI.
llt~ o...t"f'.J"oo-ft'~ --"'---L
:£8 or 'i"t....t" .
c,\)p\ed
="cs
V+= "..
d44A- Co VOs is :
dt.U.. -to
CAfo..u 'h._,.\~
h,..
"'TII.t ol.l.t'p....l:" c."" pt l'ltl'\ t
v,. • Vos
Voa
o\.
"A'+,,.,' .YJu_
"
Ill\.
VI)= IOOI \lOs -.!'"Uoi.t\1'
~Of ~hV(.~ teup\t..cl
\1'-t..v- • "o~
b:a.t C..UO"'\td- i"'
~f.....t
C.C...St.·
2.87
AI O"C , we expect
:': 10 X 25 X 1000 1-L = ::+:250 mV
"'fl,...t
w.,rst 0..0 t Llo.~.s t) DC. o~f..H.t- 1.)1)11--~
td:" -tl-o. tw.!p.,J; i &
"fO~ + 6'o ::: Lf 5l( """V'
At 75"C, we expect
:': 10 X 50 X 1000 1L = ::+:500 mV
\~e
expect these quantities to have opposite polar-
Jlles.
2.88
~ =F\IIR:t=.,_qKS>.
(Rf!f.e.r to .z ·lf6)
Vo::.
'lo.s R.:l
~ • 2.."11
Va ::.o.l.t,. :tos ~<- tM =>
lf. ~s: IMV
'(,.,. -~((3~~s
lb!> .. o.Z.\ ~"
Chapter 2-25
..!..M!!)
v.0
= 3 mV(l. + 10 kfi + 10 nA
""" 0.303
X
I Mfi
v + 0.01 v
· V 0 = 0.313 V
..L=- ..1..
R3 fl.,
'T
-1r
=-,. r\l • - re, .. ! ~~ (,t-1"}->
"2.
I
... ,.. ras. :: :t I"'"
"l."\l'o
lr VJe ·Clft..,
o\\.Qf'
"'* f,
~ s~ tA.l~M- c.~ ~ S; +o t~
CM.~
i l-
~~\\ ~u.\.
e F-ht.l- o"' ~
2.89
.
2.90
w, = Aawh
,. ± o.t )A"
e.w::l cJ. R_, , "\'h.t"'
oH·S-Q.\o
,.
av.k"
ou.t;>tA.t: . !
a) To compensate for lhe effe~t of de bms curren
I B• we can consider the followmg model
t
"kt.
0. !JIA
"'
Aofh
Ao
fb(Hz)
J;(Hz)
lOS
lo2
to'
Jo6
lo<'
lo5
J0.1
J08
1[}7
J0-1
1()6
2X
10~
10
2 X 106
1Mn
2.91
10 nF
A.... ~..
I tJ' w;..llo
=::>
IAI- \"-•\
J. . . ~z
>A.=: LiOdb@ j:., IOoii;.He
R3
·~·
R1,I[
Rr
IOkn!.i I Mfl~R,
the. de output
b)
.
voltage of the integrator when the mput JS
grounded is:
v,,
Vo.{ I +
~tJ + 1nsl?t
9.9k0
Chapter 2-26
2.92
d) A 0 = JO X 100 = 1000 VN
A
11 + fbif.l =
lolXU 1--~
10 ::::> O.J X 109 =
!&
Jijij => fb = 10 MHz
/, = 1000 X 10 .MHz = 10 GHz
e)A = 2S V/mV X 10 = 25 X 104 VN
11 +ill=
fd 10~l.SkHz
,, = ../99~/, = 2.51 kHz
UlOk:Hz
10kHz
A.0 = 8.3 X leY V/V
2.., A
Eq.2;.,:
[, = Aofb = 25 X 104 X 2.51 X 103
=-L
2.95
l+J
h
= _R2
GNom
[, = Avfb
5. 1 XlOJ ""
8.3 X 101 ..
J~ + ecx;:sr
~ 1 + (100 kHz)2
(4
w3dh ""
hdb "" 47.6 kHz
!!Q _
2.93
= 20 dB +A(db)
20 db= 20 logiO => A 0
""
lOA
a) A 0
= 10 X 3
A =
=It+ j.ll'
"" Ao
=·
I + jflfh
fb
A
X 105 =
Ao
l.
6
_2UQ_ =
fo
1=
3 X 106 Hz V/V
-20
-R2/Rt
1+
1't
we have:
I
s
w{t+~)
0· 11Jdb=>/!!Q/=
f""
l0/ldb=>l 1101 =
111
10 ::::>
l+~ =roo~,,.
R1
v
,.,.~
A = 50 X 105 X 10 VIV ::::> A 0 = 10 X 50 X 10~
=SOX 104 VIV
c) A
~~
h
"'
.,/99=>J~o
= l H1
= too J
= 19.9vtv
= 19.9VIV
=
.4 0
Cf.,._,.,
=>;>
""'
1500 VIV
QL~!~ "" 199 ==> / 1,
r"
f~r "" 15000 X !0 K .,=- 150 MHz
10 kllz
, f.~ :2-oMtt-a
fc.t-i
a-
_.,. Cf= - t~
/9-3<1. ._
L ,.
f.Hto
J.,.5-.).U,.1C~{, 0 ,.
Cf., -t'4•--.> J:.,. ~.u..
= 50 MHl
lO =>
Gg·ul}
I +j
"" 10::::> lOHz =
fh
= !500 VIV
It + jf.l'
~
./J + 100
f~l. .. -b.:_ ::: 2oo
= Aofb
- 20
JJ + (0.J )2
2.96
b)
f,
215
J99
f, "" Aof& = 3 X 106 X 60.3 = 180.9 MHz
h
+ j[l =
t
.fb
+
2vXJ06
V;
=>ft. "" 60.3 Hz
r
w,
= 2r. X 10~
I+ R2 R 1
I+ 20
"' 2v X 47.6 kHz
A 0 fb == 8.3 X 10~ X 60.7kHz = 503 MHz
A 0 (db)
-20 Ao = 104 VN
= 106 Hz
fb
[, = 60.7 kHz
""
R,
= 2,65
f, =
= 627.5 MFlz
Ao
w
~ rtt.t" =
2-IKI-l:!
1.'l Mffe
Chapter 2-27
2.99
a) Assume two identical stagett. each with a gain
2.96
function: G =
a)!!l <= -IOOVN.!Jdt. = lOOkHz
Rt
Rt
W)d"\_
w,
10.1 MHZ
overall gain of the cascade is
(t
"" 10 MH1.
f ., = f 3db + ~)
Rt
r
= 2 VN.
/Jdb
= 10kHz
10 MHz X2 = 20MHz
d) -~ = -2 V/V, /_~db = 10kHz
F1JJ2-
=
fJdb
b)40db=-20Jog
Rt
= 10 MHz(l + 2) = 30 MHz
fJdh ""
-lOOOV!V. hd~"' 20kHz
=
_Rl
e)
R,
.'
=
_r,
I
1 MHz
f,
~
"" - L
=
I+
!!J
Rt
1 MHz "' 10kHz
100
Rl
6
""
10
TO
= 10 5 Hz-
2.100
+
e.,.
~o ....
~
,..J o o
Vflf'
+~
R.,
f-1:::-:.: 8 x 100:: a-oo t:.t+e.>'.)b
= '-"'f-H:c :
:
f, ""' 100 x 5
'tr \C.\-\ 1:
0"1),• ~
2.e.
::::.40
\!~,..
f'v
IAI = :!Q
I MHz
"/l + f
gain K. then its 3-db frequency will
fin MH1.
I
IGI
2
!GI
0.'->9 "~> f
0 1,12 MH;
The fullnwcr hehs STC dn:uit
with a rime con,tant
T ····
= 8 vrv
To obtain a overall gain of I 00, three such amplifier cas~:aded, would be required. Now, if each of
the J stages, has a low-frequency (d} closed }{l(>p
2.98
(.
= 500 MH1. if single op-amp is
u:;ed.
with op--amp that has only f, "" 40 MHt.,
the possible closed 100 p gain at 5 MHz is:
5
fldh
=
2 MHz
2.97
I
= 100
The overall hd~ "" 10~ J./2- I "" 64.35 kHz
which is 6 time greater than the bandwidth
achieved using single op amp.
(case b above)
f >dl> = 1 MHZ
= I MI. I + l)
G0 ~G0
+ ~ "" 10 and therefore a 3db frequency of:
.f1
-
== !MXI "'lMHz
g) -
I
R,
20.02 MHz
R2 = I VN • };db
·
f) 1 +Rt
/,
1 +~
= 201og.fi
c) Each stage should have 20db gain or
= 20kHz( 1 + 100)
f
f(f.Y
(f;~b = .fi. Note 3db
1+
f,
d0
The gain will drop by 3db when:
+~
f,""
+ jflft
I
Rt
c) 1
Gt1
I
= 100 VN. fH~t = tOO ldlt.
+ Ra
b) l
t + j!!
(l + R~)-:::. f = tOOk X 101
w, "'
__!!_g_
1- ,
--
2-rr X tO~>
:" _!_
211'
JJ.S
'!2
Ml-lz.k
Chapter 2-28
Thus for each stage the dosed loop gain is:
I~=
J':H
~ -
1Q
which atf
IG~Mtbl
J:J:...Af~1.
We
d v't.,S
i1a.. - ___, .,"i - I+ S/WJ)
=
Jt + (~Y
The overall gain of 100: 100 ""
I-tS
'\) ~t,k.
k
5 MHz be<:omes:
k
""
- ~IR.ta.
'~•/2. -
[frrr]
+UY
I•
I
..,J:s..
.
3
S.·-iiM- r.t.!l.u(b c- he, c\.+~
l.ZA..
3
~
t
2.103
k=5.1
Thus for each cascade stage:
hdh
_7
= 540
lhe peak value of the largest possible sine wave
that can be applied at the input without output
iJJh"" 7 MHz
· 1s:
' "100
±llV
c rtppmg
The 3-db frequency of the overall amplifier!1, can
be calculated as:
value
""' 0 .PV
~
= PO
~ m V nns
= 120 = 85 mV
J2
3.6 MHz
2.104
~
~ o~p....t '' -h-f4\Jh1"" wa-1.\. ~ s~e..V
2.101
o f- Ao ~.> . L. order b:.. -r-~os.L.
GBP"' Gain
GBP
h) I
+
X
f
l
ht~
I +k
R,
R=
I.
"
f
'•lb
- f,
-
l
GBP
k
k
The non-inverting amplifier reali7..es a higher
GBP and it's independent of k.
2.102
~3.11..
~(.1::
V1..
t»t
I.AOA.
::::o
I
N?w vs;.,~ ~t:Vt~i-1-'
~ot"e.;'l-1,..
c_,-,..~
kh
""
''d.!= f,
10 P..·...,.,I
)J.S o::O·I'>)AS:::: r!)O>'I.S •
~f'fhN ~ .......A.""'-~ pv\s.~ .A.:cl~\... \.s
k....b__
=
'3> \J", ,·'" t'-o.I'U
Co s....... ptd·;r tkR.. i~
r(!svlts ;..,_:
f\$,
l':)o
~~.
Chapter 2-29
2.105
w
p
0.9
~
[/
0.5
I
0.1
-
"""}!
I
~t.W= 2~J.S
t, + t, = 0.2 W = OA JtS
t, = r, = 0.2 J.LS
= (0.9- O.I)P
SR
t,
= Q&2UQ
\
1\.
II'
~·
= 40 V/p.s
0.2
2.106
Slope of the triangle wave = 20 V = SR
T !2
Thus
¥!X 2 ""
~ T ""
4 J.LS or
For a sine wave
t<"
f ""
f=
1 P.,
1T
mu
~ ;,.,
.
=
250 kHz
= l' ,sin (27T X 250 X
.
= 2 X 250 X 10
du.,
--...!'
dt
10 VIJ.LS
=
l dll,, "' !Otucoswt ~ d 1
dt
dt
'"1
.m..
=lOw
The highest fre.quency lit which this output is possihle is that for which:
du,,,
dt
m~:t.
= SR= lOw,,,.,= f>OX
= 6 X 10'
~
fow •·• 455 kHz
I0· 6 ~(1J.,."'
Chapter 3-1
3.1
1!1
I.5V
(b)
~
LJ
vl)
------()
The diode can be reverse-biased and thus no current would flow, or forward-biased where current
would flow.
(a) Rc.verse biased J = OA.. V0 "' 15 V
(b) Foward biased J"" L5 A V0 = 0 V
J= 1 kH3
3.2
(c)
V11*:::::0V
Vp-=-IOV
(a)
+I V o---Dof----,
+ 2 v o----1'>1-----+--o V "" 2V
;f
Conducting
2k!1t
u0 =0V
Neither D 1 nor D2 conducts so there is no output
(d)
t
= 3.5 mA
-5V
(b)
5V
1
~2kH
,
I
Conduc~ nl
J
T
5 -(+1)
/=~2= 2 mA
+IV~V=lV
Vp+"' IOV Vp- =OV f""- I kH 3
Both D 1 and D~ conduct when V1 > 0
(e)
+2V~
[)., 'c·
.tllnrt·
"
3.3
(a)
VI'·•"' IOV
V1,_=--IOV f"" I kH 3
when l~ > 0 ami Dz conducts when ''I
< 0. Thus the output follows the input.
[) 1 conducts
V1,.,"' IOV
f''
I kll 1
V11 ..
,,..
OV
Chapter3-2
0)
(f)
Vp~-= lOV V,_=OV /= lkH3
-D1 i$ cutoff when VJ < o
Vp~-=lOV V,_=-SV f= l kH3
When ''1 > 0, the output follows the input as D 1 is
conducting ~
When l'J < 0, D1 is cut off and the circuit becomes
a voltage divider.
(k)
IOV
Vp~-=OV V,_=-'-lOV /=J kH3
D 1 shorts to ground when v1 > 0 and ls cut off
when "' < 0 whereby the output follows "r
(b)
"o = 0 V -The outpUt is always shorted to ground
as D1 conducts when fiJ > 0 and D2 conduct\ when
t'f 0, D1 is cutoff and D2 is
conducting. The output becomes IV.
When ''t < 0, D1 is conducting and D2 is cutoff.
The outpUt becomes:t.r0 = v1 + lV
3.4
Vp- = -5 V f
=1 kH3
When ''I> 0, D 1 is cutoff and v0 follows :•1
When I'J < 0, D 1 is conducting and the cin:uit
becomes a voltage divider where the negative
peak is
lkH
•-IOV=-5V
I Ul +I kH
Chapter3-3
J:~ J-s: \!!o rtt!uc~A. \o~ \O•fo ~
pJtAk ..,o.\"4t. .o~ i.~ r.oMo.\f\s "\'h~
~----~----- eonlu~~o~
~\@. ::.
11"-2.&
- &0k"
c:Jtl'\d
\):r
1),
/
\..,
lT:c
1>?.
4· S \1
<
·,$
''* o~~
4 ·S V
'}),
ccnduets.
c.uT~ so t~;tO~. !="or
l:h. Cb'\duc.+$ a."' d.
\:u.st
""'4...
Cor
c.o~cluc. ~\olo\
~Y"M\-;tH\ o~ ~-t. c~c:Jc..
1'"-Z.& :
1l' - ?. S,n.'("'•S'I
I'\ )
21t"
z.ll'
= .J_
3
T\.o:l>:
ie.o."~ =- ~ [•oo . f 1
i-h""' d.\s. c.o"""ec..\-;~ th"f.
'"~ot ~ . ~\\ o~ th~ c:.un"tU\t
t\..._V\ ~lev:>& Thro"~\,.. tJ...c.. bo.-t\Q.t ~ •
toS\n
e
e
~ ,_,.~
=.
c:..or.t\"c:.ho"'
f""MLT\o"'
oy
i5o ~·"v.o."'
x::.
C.~V\~~ S.
:.
~3·~
1'\'\f!.t
V
(:!·'"fro)
ca.~\t. = 1\- z.&
Sll'\ •I
c.'jc..\Q..
3.5
5-0
R.
\,'\1 ~~
Tr-
z.e
R.
21\
3.6
The maximum input current occurs when one
input is low and the other two are high.
~q sO.l mA
R
R ~50 kl!
Chapter 3-4
3.7
sv
(~)
If·
(b)
5"
~ V= 5/z.- CJ/z.. = -2V
v
+-
101/IO::.Sit..n.. c.ufofi
s t•o). T
101"10
= 2·5V -::-
-sv
(b)
-tSV
I 5-C-S}- .L
~
IS
~
-
3.9
R~~~3.4kfi
Ml\
50
The largest reverse voltage appearing across the
diode is equal to the peak input voltage
'Pz..
•
C..Cl'\c'-U<.tm~
V=
- 5+ 4/!> (s)
S~.n..
120./2 == 169.7
v
3.10
; - 5f& \}
-sv
3.8
V
I=
'-
'7/,----
\.!_0112.0)+1..0
::.
D .
_!_
2.. 25- A
.. ,TT
D starts to conduct when
1~
>0
Chapter3-5
3.11
::··::.
.
~·
... ...
••
lo; o.v8
= i.,, pak/z.
.:
Z6mA
··\10
.-··
3.13
I
uo,
0'0,
~" ·-=
.,..~
:
:ov
+S
v
RED
GREEN
3V
ON
OFF
• D, conducts
0
OFF
OFF
OFF
ON
· No current IJows
-'3V
3.14
l1'o J. t
: t [3 ~
J
• D2 conducts
.
11
=
3 /2
o.11vr
= 1, e
•
v
=
1.0
_3
0.5/VT
'2 "'
•
1., e
,
~ =
...!1... .. e
il
10-3
O.S-0-7
IHll.~
i; = 0 ..335 p.A
3.15
.: _ ..,trf".rr
3/z.
100
.
IS
, -
-'-$
e.
-
0·11
3.12
v
0·"1"
o."~V
D·C,V
Z•... . --
D
.z.T
==--
:;::l\
=~J
.::r; = s (to-') e.-o·"'lo·ot.~ ~ a·Lt'x'asA
3 V
2.:5Y
•·""Jb·o~s
= .:r. e.
0
tPo
r·J.fJ, ""A
273·21rtotA
3·!>~
rvt.A
tfl· 'S.,....u.A
lO 1,
-r1.1i/o·ot.s
,:--se
Chapter3-6
3.16
eaJcJ:~I-.t.t
To
X$
t.ti..IL
-V/n.;,.
x6 • X e
:::
:r
-Y'fo~~,,.o'Z.S
e.
'To c.alcul~k +A..c. vol'h!3Jt.
"} tli~
i,.:::
4
m..t~~esVI".ftl 2.
It=/
/1::.2.
CA]
[A]
Cv]
f_V
t·l~1
ro·•
0·1
O·SB
Z.•O'i~
t•t.fl•rO
0·81.
1\ O·l
t>·~
t ·OI.f2.
2·15Kt0
O·"t
'"'~
&·IIX ro-•S 2·24XIO-'t 0·53S O·'-ilO
5·(11(10
0·1-
a,.cuxro-'' t'·~1.x to~'
'2.·2.,)(10
0·5~5
0·1.12.0
o•S
0·1
3.18
0·,50 lOJtA
10-1
o·s
IOf\
0
l·t~"l
10
0
•1\
(~~l
1·131-
0·1
ltw\A
\0
(A]
Is
11.
0•"
O•(,
-t!
~~
~·Cfl XlO
8·~Zl(t0
D"585 O•Lf70
~I}
10...1\
v'l..
[v'] [v]
.,... 11
1
b::J =- v~ = n. ~ l~t. ~~~
:I:
y,_
qt IOI
t!alt:.v laf.f.. +J..e. .I..~J-4... vof/-ryj.e
° TJ.f.. vo 1~-t.
-to
...
....
o.5J«f 0·1.170
3.17
t
.t..
.f.-t.
"T"'
X, :: ..t...s
I
z.
e." ' Itt vr
= Is e
v.. fntlr
,.
x, /!o
2·81
'"A
Iz.
-:r, = e.
It
'"'A
ltlh-
Vz. - V, ]
II\
:r.!A;
:r. .
::.
2·SJ.
:::
-T,
Vr.. -V,
= -'- [
Vr
:r=3·et .. A
Vo
~
+
r:::.
v=
v2. ~ 2.
e~~Vt))?
D•o\.
{yL- %)/fie
3·Br
-:'.':
0·'1
~c;k J.;oJ..a..
across.
Vt,/3
t$
O·B
e.
o-oz.s
- 2?.. ·BmV
Chapter 3-7
3.19
~e..
c:."tf't:t\1\t
~ro"(}"'- lh.
.YJ_
Is
et\Vr
I. s
::
O·Ot-
\S.
r2.
~
=(o ·O\- r2.) e."
w:t\..
two J,~,e~ ;1'\ po.ra\\-c.l J the..
CIJIT'tt\t sp(,ts.
h... \wll..e.f\ eAch d.;od. ~
&c ""'a.t ~e. clt;Ju.
eC\c.l,. h.o.O- 58 0
=:
S,;.e4 t~t; J.,~.,. 'lol~
'-r
Yf"'" ' :'Afc{y
1/tU.ftUt.fl.
II\
.J.t,..putt/..,rL.
1~
'Z.
M'/ r:,..
.,....,...1-A~/...,f'.tl.. ~
wtvst ltQ.\"-
:t
120MV
ro 't
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3.29
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Chapter 3-12
3.32
-CONSTANT VOLTAGE
DROP MODEL
(b)
(a)
3V
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~
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v "' -3 + 0.353(5)
l
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s kn ®-o.7; w
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-3V
-IOV
(b)
3.34
+IOV
(a)
OO!I20)kfl
5kfl
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ti=O
1
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102
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15
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v
(b)
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3.33
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2
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·0 ..';
v
G)
Chapter 3-13
3.35
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0
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1f1.pc4k
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R
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50
=
n==-'2.
~or c:.. ~ ~ \i..i,\....J.~.:!\0\
3.38ldl
"" 120./2 "' 169.7 V.
The design is essentially !he same since the supply voltage>> 0.7 V
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Chapter 3-18
3.42
11()(11)
ii;
(mil.)
;, ·'"' i,
;. =i,
(rnA)
(rnA)
'"
~~""
(mV)
v.= v,
(mV)
I
~3
iljf_
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+
-- vl
v~-
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112 -
=
,,1
and v 1
0.7 V, 11
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= 700
vT
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I mA
tn(D
0
0.5
0.5
683
683
0
+l
0;1
0..55
0.45
685
680
1.005
+2
0.2
0.6
0.4
-687
677
2.010
0.25
-693
665
5.028
+5
0.5
0.75
+9
0.9
0.95
0.05
-699
-62..'i
9.074
+9.9
0.99
0.995
0.005
-700
568
10.09
9.99
0.999 0.9995 0.0005 -700
510
10.18
10
1
0
10.7
vo ., o, io ""' O, the current 1 = I mA,
divides equally in D3, D.1 side and Dt• D2 side.
=
11 =
· -::::
l:t
·
l_1
=
·
l4
xr;:::;
2I =
05
·~
JUt<\
700+ 25ln(0.5) ;;:683mV
v = P 1 = "" ""' 683 mV
From circuit
v1 c.c - 111 + v3 + u0 = ·-- 683 + 683 + 0 = 0 V
For r•o '"' I V, i 0
'~
1
10 K '" 0.1 mA
Because of symmetry of the circuit
!.
2
and i 4
i1
+ t.f!
2
co
()5
+ 0.05
0.55 mA
= 0.45
t''
v1 = 7(X) +
v4
~'J '"
700
_,,1 + liz+ t•o = -0.680
+ 0.685 + 1 "" 1.005 v
Similarly other values are ca.lculated in the table
for both positive and negative values of v.,
The hugest values of v0 on positive and negative
side are +10 V and -to V respectively. This
restriction is imposed by the current J == I mA
A similar table can be generated for the negative
values. It is symmetrical.
For V, > 10, ''o will be saturated at 10 V and it is
because I = 1 mA.
For I "' 0.5 rnA, will saturate at 0.5 mA X I 0 K =
t\,(V)
Calculation for different values of vo
·
/1
=
0
5V
+ 25ln(i)
25!n(f)
7(HJ + 251n{i1 )
685 mV
=
680 mV
...
0
tt1
I""'· l mA
Each diode exhibits 0.7 V drop at I rnA current.
using diode exponential model we have
~~,
I= l mA
~v,
+
~+tt.lnV
Chapter 3-19
::
= oVr
"VT 1- 1:. t{
+
J:(.
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t+ scrcl
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Chapter 3-21
3.46
+5V
R,=IOO!l:
l,=ll=l5mA
.
100
The diode current reduced by= 15-JO = 5 rnA
.·.AVo= -5mAXrd = -1.7rnV
R,
h
._-=-=--...---o ,fO
=75 n:
h
= 1.5 = 20 rnA
75
Diode current reduced by 20-10 = IO rnA
:.Avo= -lOrd= -IOX0.34 = -3.4mV
R, = 50 !l:
IL =
~i
= 30 rnA
Diode current reduced by 30-10 = 20 rnA
AVo = -20 rJ = 6.8 mV
Diode has 0.7 V drop at 10 rnA current
"• =1.5 V when R, = 150 n
3.47
VJVT
10 = lse
10 X 10-3
=
lse0.1I0.025
~Is = 6.91
X 10-IS
::r:..
A
/
Voltagedropacrosseachdiode= 1:} = 0.75V
:.fo = fse
VIV
T = 6.91 X 10-IS X e0.1SI0.015
= 73.9 rnA
IL = 1.5/150 = 10 rnA
s"'a., 1oMt.
or IS-'~
f
I= 10 +/L = 73.9mA+ lOrnA
= 83.9 rnA
:. R = 5 - t.s = 41.7 n
83.9mA
Use small signal model to find voltage ''" when
load resistor, R,, has lower values
rd = Vr = 0.025 = 0.34 n
10
73.9
When load is disconnected all the current/ flows .
through the diode.
:. 10 = I = 83.9 rnA
110
=
V·1n(!.Jl)
Is
1
= 0.025 X ln( 83 ·9 X IO~l)
6.91 X 10 IS
~~o=0.753V
So No load, v0 = 2 V0 = 2 X 0.753 = 1.506 V.
Increase in voltage = 1.506 - 1.5 = 0.006 V
Now load is changed
For ~
mua.t'
t"-...
W.t
loa!
~ o~
'l..
~ -:J."'tt 1 'J:.
bt.. ~r-~ ~II'\ 1- """'. 1'1w!!>
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.kod.c..::, wil\ WA.~ ~.,.._
f:t,_~·,~
"tiv-evj"- ~M. -.t ..ll ~-s.
Chapter 3-22
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~ ~\o.\.t~ '.(. r:,o .. ~
ll~• ~I>Z.. ::. e.. W/2..., n\h-
L,"e ~~~
<&MOll s~ ~
~-..d,o~.u
L,..
b.\J/o~ 1..
o
IAfa;J~ ~
o\
"', e..
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= - no""\1
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b
iij.
Chapter 3-23
3.48
SO now
+5V
zvo
=
= 1.401
v
Yo==
z[o.7 +2.3Yrl(ige~~]
b. As found earlier. with no load v., = L434 V
+
v
"
tSI)fi
c. With tso n load connected and v., is lOWered
by 0.1 V ofits nominal value.
V 0 = 1.401-0.1 = 1.301 V
Voltage aero.~& each diode = I .301/2 == 0.6505 V
= 10 X 10-)t
In
AVIV
7
where
i\V .., 1.401; 1.301 .,. 0.05 V
Both diodes are 0.7 V, 10 rnA diodes
First fiild
v., with no lood, I.e lr. = 0 and r = 1,.
... 7.4 rnA
= 1.301 V = 8 7 rnA
150 0
.
Use iteration to find Vo and lo
I ,. 5 - 0.1 X 2 .,. 20 rnA
I
V1 - V, = 2.3Vr lot{~)
:. New value of SV supply-=
:.I= 10 +/t"" 16.1 rnA
t8o n
»
V2 =
t
o.1 + 2.3 x zs .x to- 3 x IogG~)
= o.717 v
ln = 5-0.717 X 2 = 23.79 rnA
180
180 0 X 16.1. rnA + V 0 E11< 4.2 V
So the S V supply can be lowered to .... 4.2 V
=
d. New value of the voltage supply 5. +
(5-4.2) = 5.8 V. Now do the problem again a.\
done in the begining and in parts a and b.
I
v1 = o.1 + z.Jvr tog(2~·:)
f)
V2 = V1 + 2.3Vr
= 0.7216 v
l
= 5 - 2 X 0.7216 "" 19.76 rnA
v2
= 0.7
19.76)
+ 2.3Vr log( W
= 0.717
= 0.722V
I
v
_ 5.8 - 2 X 0.722
180
.,. 24.2mA
DQing one nmre iteration, almost same value is
obtained
:. V0 "" 0.722 V, In = 24.2mA
Now when 150 !l load is present
11 -
= 5-2 X 0 •71 7 = 19.81 mA i1ii I
/)
180
It is almost similar to earlier result, we stop itera-
I
tion here
V0
0.717 V and l/) ""' 19.81 rnA
=
So V 11
log(~)
0.7 + 2.3 vr log(z;"i,4)
=
180
p
= 5.8 -· 2 X 0.7 = 24.4 rnA
180
= 2 X 0.717
= 1.434 V
I
"" 2 X 0.722
=
9.6 rnA
150
So lv = 24.2-9.6 "" 14.6 mA
L
a. Load of !SO 0 is con.nected
I
I.
... v, "" o.1 + 2.3 x vi'
= 0.7!7 X 2 = 9.56 rnA
150
:. I v
= lr 9.56
= 19.81 - 9.56
=
10.15 mA
0
logC: 6 )
= 0.7095 v
V 0 "" 2 V 0 "" 1.42 V
:. Loaded output voltage ""' 1.42 V
e. Percentage change in output voltage
= 1.42 ~ 1.~<2.!_ X 100
5.8 -· 4.2
'""'7.4'*'
Chapter 3-24
3.49
~)
v.,. = Vao
..,." +
'tt-'2-
l t:> :::.
r~
+
=-
r.,
_..,.
r-a-T
r..,.
@.) r ·6
"so~tP
'V?:o
~ \J~o
\Jc ::. '/"2.0 + rr \C 2l"eT
-=- t\ ·4 .... ~" 1&01C lo-'!::>
::: IO•q\}
l ·~ ~ LfSO)( I D '"3 .:
r· 'i
p .:
-3.
= z·2..\1
V"'t :;. ;. •'2 t'PouJ.d" Ollh~!
+\ ·S X ?.00~10
'":1•1 V
x t~CO )( to-!o =- 3·12W
3.50
(a) Three 6.8 V zeners provide 3 X 6.8 = 20..4 V
with 3 X 10 == 30 0 resistance. Neglecting R, we
have
Load Regulation = -30 mv/mA.
(b) For 5.1 V zeners we use 4 diodes to provide
20.4 V with 4 X 30 =< 120 fl resistance.
Load regulation = -120 mV/mA
(b)
3.51
~ W14JI 1!o~nAI 'M'U>th.f .ft:N" J,:..C. ,.., ulcl,;.,!
~..11..
(C.) (,·~ -::. (0 ~
:r~:r
+
).;'""--~
u;y
Z lC :C;,r
=lOolllll\t-
~ ::
(cl)
1i .::.
r~
:::.
~~·2..
I fC •i
t:.U"~
-3.
lfctO..Q...
V?. -;: . J'f. 'lp ::.
+-re- "'Sllto
X
.....
~'-lo
,,0"' "'lliC(~
l0
J(.
t t'>- 3.
.::
5
::. \ g
·F:"
5+&2.
= ..!L
8'1
::.
l8 ~ M\10
s
3''1-
.::
)(
~lis
X I·~
1-41·1-
m\J
Chapter 3-25
'*" 05 Vo-= +7. 1.1 V 7o + 7.' V
3.52
f'\Q ':
WITf.l
30.n.
Vo
Izt< = o.s MA
~ Vz.0 + 11-\lo
'l(
0
o. 03
==/' V0 =- 7. ss V
-
II\}
1:1 MA
+I d. ~ 3D ~ ,0 - 3
7. S -= Vz.0
A'-'n IJ,..::
0·'-5
v].. = 1.s v
rz ':
v+ = 1/ v
"'
=-7 Vt...o -= 7. lt.f V
12. :::
7.5 :.
S£UC.T
:r ':
J. ~
S'O '1#41'
\J
'.is ,., A
1'-7.J55
{0 ,..,
-0·'-'
c:. 7. "'Jf ,..A
A
I..z -.:. 3. 7,.., A
~' c.~-t rs
'/
IzK
R::. 10-7.5
/0
tpJ
-
7.l'l+o.o3"'
-::. ?.tsSV
5"" rt
2_ ,..., -V·
o.s
~,___ _ __
J
--
d-_50 .D..
Rl...MI"' - 7. 155
7. J'i -0.5
= /. 04- k. ..a
F'<)" lsv -t
= ~ 1v
/1~ : "!: I X .
J.:J.fo. 03
c. ~ 50 +(J.1/j o.o3)
::::.!O.IV
Chapter 3-26
3.53
l35V
15V ;t 10%
300kfi
R
..--------<>
lOrnA~
Rt. tlllr
03mA
8.83 v
t
lkil
= V,, + ri,
V,., + 30({W09)
v"" -""' s.s3v
V,
Lowest output voltage= 8.83 V
9.1 =
Vz = 8.83+30(0.01)
= 9.13/1 kfi =
[IlL
IR
=
10
+ 9.13
= 9.13V
306.8
=
19.13
""'300
Vz
=
=L - . 30
R + rz ... 300 + 30
""90mv
v
9.13 rnA
Load Regulation
= 19.13 mA
ts- 9.13
.·• R "'
Line Regulation
=
-(rz II R) = -29.1 mY
rnA
n
n
8.83
+ 3oC 5 3~Vz_ 1 ~)
3.54
"" 10.33- Vz _ _ L
10 100 Vz
=
9.14 V
vz ""·
8.83 +
Vz
=
3oC 5 =!~- vz _ ~~~)
1.113[8.83
+ L5 ± 0.15] = 9.14 ± 0.13 v
:.±0.13 V variation in output voltage Halving
the load current "'
v.z =
R~_
8.83+3o( 15
doubling
··Vz ..
300
1033 = 9 '6
= 1.115
.. .;.
v~_-_)
2000
v.
:. 9.26 -· 9.14
0.12 V increase in output
voltage.
At the edge of the breakdown region
Vz"" Vzo ""' 8.83 V lzK = fU lllA
R _
I. -
-:::':8:-'.:
. 8:::..3_ __
13 ..5 - S.83
0.0003
30()
" 578
n
!Lu.l
~r
VPo
::O•rV
tl
fj, ..o
..,
t~; vo
+
Chapter 3-27
=- 0'":. +t>'':f \1. 1
\Te> ,..o,
(-..)
U"'eoo
~ tr$ •':f. .. Q5 A
l6o:
· •·s-xtt>s - ,. •
(e) 1'1tJ
OQ!OI'S
+!A.A. fUAL
(b)
..ol.a~
A,J.
V. is. af
c.To .::0 •
'PtV ~ C"""""""
tSV
3.55
D
+
R
(c.)l\....~ ~
& ::.
s,t\-t
¥s- -:.
~+
Z·b~o t s-lops
o.t
~ T-& ... r:rT-·'33•
"llw.~ fu ~~ ~\ ..•: \~ 11"-2&
~ 11-4 .,
'o
or 3·05 ~ ·
\1T-&o
--i;r J
U., • .,~-=
~
~
~
t 11 = lse
vpl~T
iu
.., /''o
inO rnA)
''tia~lmA))IVr
vn- tln(at lmA) ·""-
Vlln~
I rnA
"tiR]
t'o"" r•D(atlrnA)+Vrln [~
10 .
t& s •.,;- a ·1- ~'(>
-Jr L'r,c..,., 4> -o·t-~ :-&
= l's- t'o(at lmA)-V.,.In(7f)
where R i~ in kH
.
tfi.
Chapter 3-28
3.57
3.56
ideal 0.7V
Po = 12./2 - 0.7 = 16.27 V
Conduetlon begins at
vs "" 12./isin6 = 0.7
0 = sin -t(.~}z)
= 0.0412 rad
Conduction ends at 'IT - 0
:. Conducti(IJI angle =
The diode conduCts for
'1;ncl
..
2·.S""
,.
""7'f
}.,_:::::;
-'!
3
2
~. ~ h .. \,\\'•
(I •
V o. .., =
Tf~- o •Ofi.ST:::.. 0·43'01
= 5.06 mA
o.""' = vo,,..,
R
(z·s -o. 's-) (=i -·~
1000
to'Z-'D
( 0 . :,
.-,;.;l)
'#
~335"
=
3.06 rad
2~ J (12./isin~ - 0.7)d~
= 5.06V
;
20
X 100 = 48.7% of the cycle
"'$"'
.--....
t l
'If -
Chapter 3-29
Peak voltage across R
=
12./2 - 2 V0
12./2-1.4
15.57 v
0
=
sin
-I
lA "" 0.0826 rad
12./2
Fraction of cycle that D 1 & D1 conduct is
20 X 100 = 47.4%
21f
_'If-
Note D3 & D4 conduct in lhe olhet half cydeso that
there is 2(47.4) = 94.8% conduction interval,
1'f ·-
n
2(12J2cos9) _ L4(v- 20)
1T
9.44
v
9.44
3.58
9.44mA
3.59
12oi2 ± 10%: 24J2 :':: JO%
=>turns Ratio"' 5: J
.+
II
12Vm,
1>.-., =
24 J2
2
:'::
W%•
P[V = 2Vslm., -· V 00
Vn =
o.?v
"4Js
2 X • 2 "' X 1.1 - 0.7
J6.6
v
using a factor of 1.5 for safety we select a diode
having a PIV rating of 55 V
Chapter 3-30
3.61
3.60
.....
The circuit is a full wave rectifier witb centre
tapped secondary winding. The circuit can be ana·
+
lyzcd by looking al tJ,:) and vi) seperately.
12Vnns =
(i)
v,"" (V p - V00 ) ;;R
R
v:~
Eq. (4.28)
T
O.l(Vp- Voo) -- (\'p- Voo)CR
C
=
I
0.1 X 60 X 103
= 166.7 ~tF
(iiJ for
11,
I
,--', /1'-<
'
,,
II
= 0.01 (V1'- Vn 0 )
(a)
'
\'1,- V no-
'
'\
2l Vr
= 12./i- 0.7-
~(12./2 -· 0.7)0.1
(12./2- 0.7)( I - 0{)
15.5
v
... ''o. "8 = ( ,..,-·J12 -· 0.7)( I
I!)
=
ZVs_o7
15
= 16.19
1T
assumed Vs >> 0. 7 V
v'
= !2_+ 0.7 7r = 24.66 v
s
2
Thus voltage across secondary winding
= 2V,,. = 49.32 V
Looking at D 4
PlY
~-
V, -- V 0
\'_~
·o
+ ( v,
2V1
-
-- 0.7)
0.7
48.6 v
If choosing a di<'{k. al!uw
1.5PTV
7:1 V
8
Sltldy margin nf
v
-~
2
001)
Chapter 3-31
(b)
we have the conduction angle =
i)
wilt~- J2V/(V1, - Vt;o)
2 X O.J(Vp ·- 0.7)
(Vp-0.7)
J[2
0.447 rad
..
f-raction of cycle for
conduction "" OA47 X 100
211'
= 7.1%
(ii)
wilt:::.
2x
O.o!(~: =~:~)
=
0.141 rad
Fraction of cycle"' 0 · 141 X !00 "" 2.25%
211'
tc)(i)
1r(1 +n,F~)
if>,avg
VoR·••·•(t + ;r f 2(\'p- ~)
•JO.l(Vr- v, 0 )
=
15.5(1 +
IOJ
f2)
1T
•JO.l
233 mA
+ ;r./200)
(ii) io . .-g = 16.19(1
!OJ
'
·.· 735 mA
NB next user
h
z Vr!R
v p - v fJ(J. - !v
2 "
hut here are used i D.,,~
wiJkh is
mon~
15
-;l(t
10
Vr ... vL~
R
R
accurate.
i
2r.
[2 )
'. ,.1
(ii)
iv. ••·•
(d)(i)
(ii)
=
100 = 4.5%
i
./it.)
= 119 mA
+ 'lTIJOJSi)
""' 356 mA
14 79( I
15·49 (1
l
= 8.84 v
+ 'IT
(b)
(t + 21T 4'~)
0:2 = 223 mA
15 .49 (1 + 21T II) = 704 mA
I
~Ml
10 =
10 =
""
VD - Vno- 1/2 ttr
::;: 9.3-2I 0.93
(ii) Fraction of cycle
=-
:. v 1·= 9.3- 9.3(1- T /CR)
l4.79
1
At=>
3.63
tlt => ..!Q..."' 0.93
T 14
+
R
V"
IOOH
At
At:: 0.02325T
= 0.02325 ms
(c}:. Charge gained during conduction=
Charge lost during discharge
i,, :J~& .6-t-= Cvr
100 X 10
"I
6
X 0.93
().02325 X 10
= 4.0 A
,..._---T=lms·~i
.
-~
Chapter 3-33
R=
c ihlt
=
at
= JOO X 10-6 X 40 X !OJ
11=
0.2
75
n
C = 1722 j.~>F
Consider Ds when looking at PIV
:::: 4A
= 4 + t/0 , 0,../IOO
= 4 + 9.3/100
PlVoverD,
::: 4.09A.
Note that in this case ''""'' = i"- during to the linear input (it is constant and ic is approximately
constant).
3 •64
let capacitor C be connected acros8c each of the .load resistor& R. The two
supplies, ttt and t-1}- are identical. Each is a fullwave rectifier similar to that.based on the centertapped-transformer circuit for each supply, tbe de
output is 15 V and the ripple is 1 V peak-to-peak.
Thus W''' 15 :t !12 V. It follows that tbe peak·
value of v,,musl be 15.5 + 0.7 = 16.2 V.
PlV=
= 15.5 + 16.2= 31.7 v.
Allowing for 50 % safety margin
PIV= 1.5 X 31.7= 47.6 V
use Eq( 4.34) to find
iv.••:"" J,(l + 1TjVp12Vr}
:: 0.2(1 +
v
RMS across secondary
.
Turns Ratio = 120
= 32,.:4 =
.Jz
I 2)
1r Jt5.5
= 1.95 A
. . Voltage across SCL-ondary =2(16.2)
"' 32.4
tiQ + l;s
3.65
\.Jo = tT:t:· ( 1:1:-
r:?.fR)
::. 2. U r.
~
22.9V rms
IA:J
fu
d.1 od-e
I!::.
vr.h~-
= 5.24: I
22.9
Usc Eq.(4.35) to find
;,_.,,..= T1.n + ztrj\i1,i2Vr)
"' 0.2( I + 2rrJJ5.5/2)
= 3.70A
Vr
~'r"' 2fCR= l
Eq (4.28)
DISCfli\RGE OCCURS OVER T 12= 21f
~· C= _..cl:..::5.;c5:_._
2 X 6Q X 75
where
:wo mA "" ~
R
u;.=m
U'o "l.f\)
=
9
& :::-tv
(JA""
~V -' J_,_~
tf'o=~
0\}
v:,.
=
tt.
c.v.f
oJj
C.onJ -
Chapter 3-34
d)
\li: =-2.. \1 U"'" =.. 1'2..\1
IJo ~R.lt
0:. ~~
r--.,.•
-sr
3.66
(Go)
3.67
a·~"
+
...:Z·f~Ui ~ 2.·S"
·~·~'ti-......
c.owluc:fs.
"''*
D---4\~"""--r---o \To
lMPf &*)~
Ub =o·":l-'1/
-bo:\\o. '>, z
Cl~
-\.1. ::. \,}:t
(1,)
w
~ -;q. :l· s..J ""D, o t'\.
U~>r :::-o-r oJ: t1>1 ~ 1rvdr
~'ll
= 'J·":J-\1
._t. U';t:.::: ?.·1-+ t~'
----
: 3·2.V
1
Chapter 3-35
3.69
~
fbr ..tZ&tck
~=
H:.r
~ "i!:.~NI.r~
b-{~ r:~~os;
____ __.,.. __
... "t,..•'5""'
-~--· ---~. -~·T
I
I
't>t.
~
I
« -t>t
t• 2. :::.\leo,_ tO'JC 10'!. x"Z.O
V"l:C = 'i' ·OV
~ t>z. ~,-_.,1-J.,..io.l), ofll
1k. \~M·,i(r "'llvu..\..n\L,. OJtiL
± ( z 1CO ·4lG +t·o) :: ± q.3 v
For t1"1:? tot·~ (•s~cw ~ \Ji: <- ~t-'~)
~~ :. r.,, ...,. t"a.,. r1>t- ::. 3(to) ..· ~ o-os? v
l\l.fi t r"' tft'tr.z.
tkJtt3(tc)
\1
\To(V)
I
I
I $1Df"'.
t-5
Z.·5
I •D-f>"
I
I
-q ·?.
- - - : - - - - - • -'1-'t I
--- - -
I
Dt 0 FJ"
"Dt.. OtV
b,"\ 'Vl,.
OF,::.
t\ •?~...
oFF I
'1..
1:\,l>t,
I
I
I
flrlL I)to IH!'!S
I
1 t--ow~ AU OFF -....,
~)
t>L;& OtJ
1'>,5 1 l>tc- oFF
Chapter 3-36
3.70
forb,
G,~
.k_
,.,
IMh
-....,
(trt.-o·"+) ~ 1\V;-t,.,(i'j)f,rv-.t.:)
: ;:; o · r to"' t:
<1- .:CA" ~~
lb_)
•
• • i D =- 10 ·~
:::: 10
IJ'o (\1)
o·~
0·1-
t>·+
,. T-
O·f
10·1-
0
0
-o·5
-0·
-o.,
)C
c.T.x. c..v)
O·Sr'O
51
-o. '1-
-0·7:1-
-t·+
- 0·8'
-10·1-
o
r
>t
_3
D• t"
..J \.to·3
tro -o·':l-
tO
o ·•
tb(IJ'"~o-o·r)
Uf>&ll'w\ LJ>
Tt_ \;~ ,.\u- ·,!. Ce.:.n·\~ \.ox!,
J<.
v,;,'"'
~
l
L-t ~ O·Y;V '
L- ::! - o ~tv
a.
0~
Chapter 3-37
3.71
(b) ltt< lli: <.(0 't>!- 1 "\)q O!) =:;>
I=- - -.r:,
\Js -=- - ( ..,...,. 't"()pt.)
VA :; - (v&
~~ lt6+
ef
ve =-0
po'"""~
.,_5X.)
\1" "70
fo..l
I;_
(Mfr)
\llt!o, ~ Oe~U:+Vr.~t-~
c·bool
tl i)ll()l
O•'f
O•bO I
O•eet
0·~
1•00
l . t) l
0·01
O·l
O·~
I • 2.\
1·2.e,.
o ·1
o ·I
O·f-
•·So
l·C\0
b·'Z-
O•l.
O·'T3
l•l>'-
2.•€1"-
0·3
0·}
0·15
1·('0
5-~o
O·'i
O·f:h
0·-.:rf
,.qz.
i>(\2.
o·e
O·lf
O•S
'2·0l.f
lf·5'f
0·~
O·f.o
2.·' ~
5·1 ~
lh..
(v)
(~)
0· ?t"
(.\/)
0-t'
""
N)
0·~
Chapter 3-38
(F>) Ls.+
"1
""Po,W\k {o,r
:r,
Oo,,Ott...
(•l)
(v)
Ue.
N)
" ' <.'0
"'"'
{v
-o-to
-t•OO
.. o.to
-t·Ul
-f ·l!>
b•IO
C•lt
0·':1
-l·ttb
O•'tO
0. l-l
-l·tf{,
_,.qo
-z.q"
0·~0
o·7'S
-l·SO
-~·00
D•ftt>
0·1-k
·6·52..
b·St>
0·11-
-·~
1·«3"\
0·''0
o. r.r
-1-5.{:.
-Lt~
D·l-0
0·~
-1·51-
-f>·Ol-
o•oool
O·'f
t>•bO f
0·5
&01
a
-l·ot
a-...
-4·oq
-s
-If
~_l
-z..
"A
'""'
lT"
(V)
-ID''f
·Oi.
q
3-
-01
./ue.
-J•f)
-I•'L
.,..,
-··"
t;;
3.72
3kJL
lYo
<-;.
j
~
i .l:J)\
I .
i.l>"l-
tv
-2J
ftl:Wrt'btf!> i,,) /waft 1 Vor tt O· +V
kt tJ:~u =- o ·-:r-1'1 o; .,. s·
:rv
Chapter 3-39
Vo :. 1·1-\ +
=
1•1-\ "-
l1>1
1
D.~clwh·d
tkSL
(\)%'- -~·1-1) 1f'
Loa
u.,
10..0
o·l.4\1-
H-1i'-
1·'2.1!1- 7!~ -.t-
to·•
0·52-l-
1·521'
I·S&l3
'"''
t"'
\To)'~
10
D·Sf'f
I· 5'15
(0 -'f
DVl/
•. ,lS
t..O&f'Z.
-'"at.
t.,,~o
1·1'12..
-5
:o~ lJi >S'J '61~ lro{IJi.
lf"J:~ "2
'It-
Y«t-
LroCv2
i1>z-=- 0
10-~
tJi :('lk)iDI ~0"..
V
+I
2.·1-
s·:J-
5·f
'Z: 1-?.'Z.S
O·l'ICIO
D·1-'i
,.o
~·1lf
J'Z,• rY
Z·l-i"t5
ro·'l.
O·lSt'
11·'?5
Ltl·'fS
7·0
!.·D~'Z.S
~.0
.s. 'l.t'Z...S
't·O
~·S32&
IO·O
·l't>
0
-'L
"'·-
3·1-'t'Z.'b
O"o (\I)
~,.
Yw.::.
+2.
.::Lse.
lJ),&
=-
ii>'Z.
:0
\ipJ.fo.o:t.s
®
t,..,. ... ""--.'t''2..
®
l>tn. .. D•D't61t'\(~llfiiiO'iJ)
ITo - i1>2.
Vi=
X3
o..
'Z.
.,.,,.·'"
-··
-2.·01
~ ~ ~·'-!"~ .u-~ ~"
9A.L.
cole··~
....,.,
v
~ ~ :
-'l.·IO
-z:zo
-l.-5
-l·'
-Z·~
10
3·'1'1110
•.a.
to·•
(f)
\c. SL
(l/)
-2.·001
@)
v-.
(\I
D·SU -t.·'SU'
ao·S
to·"
-t·~I\S
D• ..&fl. ·l.· 'll.JI
-!.. O'l.lf
0·5tS
-~
-?.·51\®
-z·,'L5
~1110
o·t.'IIID
CHo\'\ •t•n,
- '5·'{SC\
~~~A-
o•S'IUO•'S
O•,tz. -3·1U'
-'I·G•U ~
Dof.JCIOJ
t>•(Jol?
-?t.z.n-
-5·01'-'f
1•1Jl ....~
o•lt~
-3·'11.1
-1·5'1~
O·'l2.0
-~·If!
- If>
,,,.... "'
-2·'f1
'"' z.·z..,A
1"5-.A
® (w
~ Lot, 'D.!. i~"" ~ 1>z i. 00\
•·• \.,.
~'-s. ~..~~ tk.n. ......-;~'-"
c.o..rr.
Chapter 3-40
(!}
O·S
\1 ~r
'Da; fo
©
UA
....uwt.s
l>&
CAU&..t.s
6~ \-o ~
=' .. 2.·1-V
VD~ ~ l)~ ~5
~ R'- :s.o ~
"D!, u,:M.~el Tlt.. ~ lCt'l..
1lut.
O•'!J
""'*- vol...,._
Average (de) value of output
;
14.14
v
= 10J2
Chapter4-1
4.1
4.3
C!~U 1!1
A~r2 "" 10- 6Am ~
-wlot>lt'
I
141
d/l.l:fn~A.
= m·• ln
0 BEl
1 VT
= ls,.e "111J1
1vr
2.
$a:lvr~ f-ie#'\
ir: 1 == ls1t
3
a.c:/-;,JJL
~lvYit.k~
ic:J.
't
5
,;.v1r+ul
.
'"
f.ne
tA.C.f...~
Wlod.L
ls1t
~~-.v;.A.
":/
e
I
)·-l6 i\
t·'· ··· .. ~Is ·~' 6273 x <
For
· -.0.907 mA
1
Chapter4-2
rAI'l'J•U
For
ic ""
e
to p,A => 10 x to- 6
""
6.273 x
lo
w··• 6
(rowt.
15· oslo~ A
''se.UO:.t1
:.
v111,
HM Po........,-
0.587 V
""'
Alternate way-- without calculating Is
_is:_
0.7 •. 0.76
e't'i:iu3-
=
10mA
:. ic
= 0.907
4.9
rnA
•
'c:
For ic = 10 p,A
vBbJVT
= lse
is=~
USE-Q16
tox w- 6 "" e~
.;...;..____._
13
.
10 X 10- 3
.
'E' ""
= 0.587 V
t'BE
'I X .Ic.~Jit =if '.II'""
= tas,.,w
= 0.7 V
For v1111
:a
f!...±:_! .
1c
13
ic = (5
X
10- 1')e 0 MO/R025 = 977 p,A
i c is .constant and independent of 13
.
.
4.7
.
•
z, :: f'Lea
4l>b::
;;t
to
p-1-·5
:
o( • .}_
x to- 6
50
19.6 p.A
-6
!£'
= 977 X tO
13
200
= 4.89 p,A
iE ranges from
':ll2J2 ::
I
ic _ 977
's ranges from ~ -
f!...±:_!
'f·!"
13
::
.
=~
977 X
50
10- 6 ., 998 fLA
tolL±_! ic = 201 977 X 10' 6 = 983 fLA
13
200
t> ·1K2.
P.tl
ir
4.10
if: =
4.8
1 mA
Qls!U: 111
50 p.A
""
=
ic = it- i 11
J3 = ~
Tc. =f?> I e.
YIU"J-'S
:=. "ox-;;;"'~A
=
3mA-
!YoM
+o
&;oxS"g.,A
1o 15h1A-
a
=
950
=
1 X 10--:1
X
50 X 10- 6 ""' 950 p.A
lO. 6 = 19
iy
50 X 10"6
__IL_
13 + l
=
~:ifl
-
19 = 0 95
20
= IOp.A
-'\
ir = i£ - i 11 ""' I X 10 . - 10 X 10
990 j.tA
•
~ ~'
i11
_lL..
!3 + 1
?90 X
--6
10
10 X 10- 6
= 99
~!.. "" o.99
100
-·r).
1\
Chapter4-3
~:i 11 =
2511-A
=
= IE- ;11
i(:
11.3/VT
I X 10
-3 .
A-25 X 10
-6
inc = f.,c e
A
For Vee= 0.3 V
= 9751J.A
!£· = 975 X 10
=
ia
iuc
-~
.
J!
= 39
~-
40
•
toe
tI e
v
VIIEIVT
'
=> V uc "' \1 r In
= 25ln[·5 X1010. !>_lJ
3
liE
18
= I cll25
"' 10001125
=
[jj'JI]
T;
-U l4
-1' 2R
i11 "" i 81, +inc = 14.5 + 0.02 = 14.52 11-A
ic
p.A
=
= 1.45 mA
1.45 mA/14.52 J.I.A = 100
ice- i11c == 1.45- 0
icli 8
"""
V8 c,0.4V
ForVce"'0.3V
+SV
+SV
ib = 14.5 + 0.089
=
145.89 11-A
~-:::::
=
1.4.5 mA
ic "" 1.45 -
iclih
c
B
= 2.65 mA
= 10 ·e = 1.45 mA
V8 c=0.3 V
lsc e
ForVce=0.4 V
650mV
=8
= 0.7-0.1 = 0.6V
= 10 · e
0.1/VT
ice
= 0.089 J.i.A
= 10 · e
0.61VT
=
= om6s J.i.A
= (l7- 0.3 = 0.4 V
·-U 16
OAIVT
I sc e
For Vll£:: 0.7 V
"" 0.975
4.11
I8 =
lsc: e
V ac
ForYc£=0.1V Y 8 r
25 X 10" 6
_jL
p +I
=
U 12
10- e ·
=
= 1.45 mAJ146 11-A
For Vce=O.J V
'• = 14.5
= 9·9
V8 c= 0.6V
+ 2.65
"' 4.1 mA
ic "" 1.45-2.65 = -1.2 mA
1.-'cE loo low for model
Rn =
\lllil-
vllh
~--,--
Is
Vu:-·Va
fc
5 .. 0.65 = 544 k!l
0Xl08
5 ... I
I
4.13
given: ic
4kn
4.12
~
~~
0
• "V
vcr 1v1
e ,. sc•
. r l e · __,.,
I .v
se
IV
llr
lsr·
l."c1f,;l
,
r fe
thai
lsc
- 13-/- f.IT:~
-~
X
r + !Hsclld
~"''"""
lsc[~i + B ~~<><~•<~ll
Is J:l-13 ~ton-ed J
~
25ln TOO
For va ~' 0.4 v
v !IC
fsc[ I + ~~f.,t.:«l
0.7
l
QED
T; .T-llrot<'d/13 J
115J.nV
0.4 '' O.J V
For f3tnrwl ~' 50
-
.
·'
.-,-.--cl\c:-,- - - -
I+ SO
-~
'5 Itt 1_ HKJ
. ·.·."···'_--·
-·
. ·I :'iH/!00.
Chapter4-4
= :25 ln[i02.00] = 230,8 mV
f1or J'io~cM, =
4.15
10
= lS ~n[too· 1 :1o~~ooJ
Vctw.
= 2S tn(t22.l} = t71.7 mV
Pr"'"" = S
Vc,... -= ~Jn[lOO·l :;,~ooJ
For
= 2S tn(631.61
1
·: ic: .,. 13 e"£s Yr
= 161.2 mV
For Pro..u = I
...
V C!>ait = 2S In
= 2S
•Es -·U.8
i
Use -.!:
= to.w
lA
[·100
t+t ]
to calculate v68 for a particular ic
• 1 - 1/100
For ic =
= 132.7 mV
ln{202]
to lilA
Forie= 5A
4.14
tl611
= 0.685 v
veil= 0.840 V
4.16
ic.,. ~ i.e ~ lQ..
,, ~ 1o -:. 'f.Df,.lt
=
~
'
'-
l
La ::. Le-- k:=· o •'t I an'~~••IVr
i., ... T.s e.
,,. '/ca/o· o?r
'f·D7110 -:s = 10- e
-lOV
f3
Vc =Vtt&
= 40
O.f ""'
40
41
1.~ = 10'" 13 A
'". "" 0Is (Vz,;s'"r-1.~
I
s(
/t'BtVr(! - l) =
Vnt"r + 002 X 10-} A
.
0.02 X 10·-3 A
10 -n /£8 10'02~(:~- I) = 0.02
V EB
.
11. ""
=
Is
0
0.570 V ~
e
11t:n 1"r
=
··IJ .!!Jl.
= !Q__
cum;
40
V c ,. - 10 + air X 10
=--2V
for
f3 = 15
iP.
= (13 + I )i11
10 ""
ai£~
'"' - 10 + :!Q
41
10-J A
-0570 V
VB ""
=0.82 mA
ic
X
4.17
X
0.82
X
10
(J3 + 1 )iB
=-
o· roo\}
Chapter4-5
4.19
(a) 11 = 10.7-0.7 = 1 mA
10
(b)
• -- lO
18
16 -- 0625
· • A
Calculating lsi
12V
rt
~ i~
it· =
·
!HI
c
1.5 x· l(J _ 1
j6
~ Sl
/ 31
=
1~ 1
=
'
''~rat•r
e ·
,;l.H~JtL;)l;l
£.
1.608 X 10~ 14 A
V1 "" 12 - 5.6 X 2.5 = -2V
Compare this to
. -'-'li»lt~r
lsl ""' 'c e
"' 10~) e-0.7/0.0l.~
= 6.914
X
-2.7V
-4+ 10
--~
10- 16
2.4
=2.5mA
·:Isnarea
Areal ~ ls1 _ 1.608 X 10- 14
Area2 - fs2 - 6.914 X 10··16
-IOV
{c)
"' 23.3 times larger
IV
4.18
15k!l
OY
0·-( -10) = I rnA
10
-lOY
E
It
{d)
ir
I De t lrc
::Z u-
+lOY
+
¥'u•
-~ .
¢''
i
i,
u
-lOY
Chapter4-6
1SV6 + 139.5 =SO- SV6
·;J
('~
= 3 rnA
v6 =-4.475V
16 =
1 _ V6 - 0.1 + 10
IE = I,+ 18 = 3 rnA
s
5 -
= - 4.415 -
s
4.20
0.1
+ 10 =
V, = 3 rn(l K) = 3 V
0.965 rnA
!1
1
=
p+
I =
8
SV
(a)
Is = 6.3- 3 = 33 f.LA
lOOK
1..!!!.
33 fJ.
= 90.9
p = 89.9
4.21
Vee=+ 9V
+2V
!£ = p = (. 2K)
(_Q_)
18
-1..!L
=
= 93
0.0215 m
200 K
SV
(b)
For F.S.D i6 = SO
~~oA
ic = 1000 - SO = 950
~~oA
SinceR. = 00 VcE=VBE=0.1V
:. active mode
R .. Vec- Ve • 9- 0.7 .., 8 3 kO
e
lc
lmA
·
I c = PI,..•
(1,
+ 1,) =
I, =
!.£
11
ii
e·~-K2·3 )
950
P = 10
= 19
ForFSD/5: i 6 = 10 f.LA, ic = 990
= 10 rnA
~~oA
=tj3=99
= 0.1 rnA
= (10 m-0.1 rn) =
0.1 m
For FSD/10: i6 = 5 f.LA, ic = 995 f.LA
p = 99
=P ..
4.22
199
+I.SV
(c)
IOV
+7V
VE =-0.1V
-l.SV
Chapter4-7
a""1
It= V,r;- Vet: = 0<8 = 80 p.A
RE
lQK
lc = _1_ Ir.
Ji + 1' .
I~- I = (- 0 ·7 - ( -lS))
= 50)(
llQ = 78 v.A
51
"
= 80 = 1.6 v.A
p + 1 51
=
6.8 kO
= 2•1 rnA
v. = 15- SJ K(2.l m)
Iii ,.,. . lt
Vf = V•r
"
Vc =
4.3 V
4.25
-I,Rc = l.SV- O.Q78X 5 V
t.n v
+t.5V
4.23
+SV
0.1 mA
lmA
o.sv
-tv
p = 100
-t.5V
-sv
A.VsE = Vr In [1C7]
In
V Be(lmA)- V BE(O.tmA}
=
:.
= 251n IO.ll =
2.'itn[.l.]
0.1
V11E(O.H
Vuf:(lmA/ "" 640 mV
Rr·
.
!~ 16
R _ - 0.742 + 1.5
""0.99mA
"-
Rc·= -l-(-S)=4041t!l
.
0.99
.
1!..±1
13
lc
= 100. 0.758
101
Re = 5 - ~-698 = 4.3 k!l
Vc can be rai!led untill
=
1.5 - 0.5 "" 10 k:fl
0.1
V8 = -0.742V
+ 57.9 mV
=698mV
lc""'
-57.6 mA
= 742 mV
0.1
=
7 5 k!l
.
= +0.4 V
Rc ""' 5 : : / = 5.45 kCl
4.24
4.26
(A.)
Va:::~
y6 ..,. 'le. -o ·"f
+ISV
IIi'~
-==- -
-0·1-.,...!?
"2.·2-
·-
3-
I • o 'Z. ~ 7. • '2- .: 0 •
rS" \]
X~.,::: oi.IE' • -\} )f 4· ,_. 't :::: if· 73~ f.t
Vc. ... Cf- I lC X c.::: Cf- 4.1 • rl ~ _U·U U..,
:J:"y,_ -=- ~~ - -=-
::t'a ....
4.27
Vao;.~
v6 v..... o·1t
::::- .Q·
I11 :: '!-'/e- :::. 3 ... c·1- .,. ?t$MA
l
r
l
~~.
!.\
~ ,o·c.. ~
'2.·2.'~-t A
... z.
r&·l.~
t"D
}(
:= I 2i"O ~t /+
....La - ::z:c... ~ z.. 23 ~ o · t>9·'+3~ttA
- fO
~)
4.28
lena
30
ic
-c
'/&,;.;~
Ve: Ve. t-0·7- ~ .3· 'XV
.,. ,_J.l
ti-Ve
I .
I•I
:::: .1U?,. )(
'51
="'·f2~ttA
E
4. f'l. = ~·' ' .... t\
0)
for [:l opendrcnited. in"" 0 and ( l) gives
pe
1, •·1n; 1 "r
Ve'=
3-o·"f;:,
= fuw=:>l'
''tn 1 vT
.
suhstitute into (2) & 13) =:>
~
Ic::: VtFfo·4r;. '2·>
B:' MA
.
e) -n..
~a.s.e - t~,\\:.t.r \1c \~.._ ~,. C.~
~t£ro-- c ·-=1--V
tA.t ~tu. ~~
At: w·c...
<10-.""
l{20
lt!O
30
3.)(tD-l
v
;;::
3·3~J"L
:::.f:>·'CN
Chapter 4--1 0
4.35
f..,,,
4.33
6,.
,.J.
h -
r.,"' "" fr.c..
ic. =
f.a
P#f ,...
~
ts-r J) c.
t
1•2.Y¥1Ao
"ir,/f.-\
O•Jmlt. :: ~
O•yh
~ g~·3 kSL
4.34
= 0:72V -ic
VBE
= UlmA Vel!= 2V
ic = 2.4mA Vcc
=
~U"c..s:..
14V
ro
r 0 = &Va = ~ = 20k!1
Me
2.4 - 1.8
:r.
Near satumtion Vn, "' 0.3V
:. &
v.a
= 0.3 - 2 = 20 k!l
ic- 1.8
Ate
ic
=
L72mA
Jttx
;uA
-t-
I• 'LNIA- _,... A l~
••• 'k_.-:.
C-alculating VcE for ic = 2.0 mA
AV
---.fE:
Me
= ro
4.36
VeE- 2
2 _ LS = 20==>Va
= 6V
Take the ratio of currents to find the early voltage
(with Eq 5.36)
V/1/i'- VIlE
e~v-·1'- (I+ 14/VA)
2.4 =
1.8
~
=
2.4
+ 4.8 =
VA
1+2/V,1
I
1.8
.
B =so
+ 25.2
VA
VA=34V
v
ro ""' ~
(a) active region
lc
where l (: is the current near saturation •·~ active
boundary. As calculated above l c "" 1_72 mA
ro =- 1 7~ VA
.... m
=
19.8 kf! compared to the
above calculation of 20 kU.
-· Vee1,- .R,'i ·- 1
·"" lK
In =
\1118
v,.
= 4mA
~ ~" ~~
= 0.7
2fn•3k.T2. : :.
0.08 mA
+ 20:::':..:!
= +2.:\V
50
=
0
·Zf-1/ttAit
Chapter 4-11
(b) edge of saturation
I
c---.--.
@)
"c "" 0.3 V
lc:IP = 4.7150 = 0.094 rnA
""
V1111
+ 0.7
"" 0.094 X 20
O'IJ
-
eu+o«-
'IE-::.. .Q..lL
Ie-=ott..
Yc.= IDL
- 5-0.3- 47mA
l8
V& :::
= 2.58 V
(c)drepsaturation t'c = 0.2V fir= 10
lc = (5 - 0.2)/1 = 4.8 rnA
ln ""' lciProrced
V811 "' 0.48
= 4.8110 =
x 20 + 0.7
4.39
0.48 mA
"" +10.3 V
ftN
4.37
Assume active:
VE == 3 V, V 11 ""' 2.3 (Assume V8 E"" 0.7 V)
18
""
Ve = Ye.
2·3 = 2.3 rnA
IOK
1. = 2.3 m(50) = 115 rnA
V< = ll5m(l K) = 115 V,
Vc < Vo(not true!)
saturation. Use VECSAT = 0.2 V
+ 3 - Vl£SAT - Vr "" 0
v,. = 3 - 0.2 = 2.8 v
\18 = 2.3
Vr,:= 3V
I
_ lcs~~.I -
.
!lr,mJ -
I;;" -
'2-
::. O·&t'l, H6) l·G>S,.,A
- wo..rtf' Vs
2.8 rn
0.3 rn
. R
II
{ 11
3
Ill
'/e.-::. 2V
'Ve:::
z..-o ·+::1·3Y
'ie:., = I ·3"' J\
l
Ic.
O·IIS"mA
"' VB "" 2.8 "' 93 kH
~\·oM~
(b)
r
~ ,.h...,_ ~
a>~
At~~~ e~ ~Ml.
VSE =-C>·S v
4.38
...LG' ::o
:rI!! = -fo x 1·I -;;:A1 A
Ve- = -3 + {).liS" K '2... -= -2.. "1-1- v
VP.:. = Ve-+ 'D·1- ::. - '2..·c-:r v
- w,_J-
( o..)
v.:> J...u,
=
= 9.:n
Transistor will openlle at edge of saturation when
v,. '" v8 ,, 2.8 V
"
-t>·,.
:. -1·1-
=> - '
~ I:e = \1 E' 1'"?,
v. > v8 < VE .·.SATURATED
(2.8)
~ =
( 1 ~ K)
-o:=1
Va -:-:.IV
-
VG% 1 -o ·"'f= O·~Y.
\t&
• .. Va - o · 5 - Z.Ie -t?>-:..0
VB"" -2.·6\J
~Xe-:::0
Af~.
efCD..Jc,chcm
Chapter 4-12
4.40
sv
a.,
.~~ ~
\Jur • 0·1..'1
\lee, =-o J0'\1
__,
0
: • 'Ia + '2.· a -:.: . 4-·5-'1•
v.. :.
~·\'J
+'all 'J!t =
(JO
-~•" '
~. +fe.l.-
#.
=>
:r:CM..t:
\Ie.- 0•1:t'c.s,~.~~.'t' =
(le.z
S :::.2
~.~z..
Rr., +-2•1.- "=' ~co \c. ...sL-{'.e.
=
XC&At:-
.,- IGcAJ:
Ea2
)!.
'2.5
'Ra.,__ =
'JE
t
= "2-
:::Ca
:reF' ::. .:Ce. -r
S
5 = 2..
to k. J2...
R •' -=- _!P »&..
=
-3 T Z:I'E
::.. -3 + arCU!.'i::'2.•
3
'\-'i&
3.
Xe... lie(.- l..'la -o· s}
:::
X(.CA:f:
2.
~·<.>-Va-- ~t-z.·~) -r 7\J~
Ve. :::
fco, G)
2-
'fJ:r) -o·-r -.14(•)
to I
Te • I·ZtfwtA
=0
Chapter 4-13
tft.;''!j 5 •to ,t.t.Ais.W ~
"j2_. :::s
'Rc.- '2:1-kSL
"3·'\k.J2..
Ier -::. 'l- O· +- :\ z.. 1 z.~h3·tf
4.41
'leo
\V
'="
-
't-t-
..
'Z.·I'Z;,c '2.· 2..
: () '-J '@, (., -=
=- -14• oV
!::!. .~\1
4.43
\fc.
:\S+o>1CS"(t;)
=-H>":J-6'1
=re-re.
:O•f-0·015
-5v ::. 0·7SELt1A
®
f6~~=
::r:e.
D·Dio
!) o(:::. I.:.. =
:r"'
O·'"f8'5 ::: 52.·3
0. 1=8"5
-;::. 0 • -l f'
O•'r
4.42
IE : ;:. 2. "!. '2.4 ~ Mlr
Ve. ==- 'R& :rt!' fa, :::
V~; :: 'f- ~ ·1- Ie-::::.
I<
- 21f'L i- 't :.0
"·11-Re-flc. .-:o
LeT Ve.:::.
Vc.. ~ -4f; CF-
SJ'TUCI\T)O,}
5-(Yt;+0.7)
lu =
Rfl
5 -(Vf+0.2)
lc = I. =
E.
Y.!:l
. v1'. =
••
. I
·"= I 8
.
+ lc
4.3 ··•· vf.
+ 4.8 ....
10
--·"
vf.
Chapter 4-15
4.46
lOVR + V« + IOV 11
""
4.3
+ 48
@l)
sv
(£)
Ve=2.49V
+ 0.2 = 2.69 V
Vc = 2.49
I•,kJl.
VB "" Vr.+ 0.7 "' 3.19 V
Check: lc
/ 11
= 5 - I2·69
t
'Lrr1A-
'V-a..= S-2.(''')®
·"' 2.31 mA
.:: l•tr'V
= 5 - 3· 19 = 0.181 mA
10
!.£· =
2 ·31 = 12.76 < 100
0.181
Hence, we are in satumtion a~ assumed!
(c) R6 I kU ·expect satumtion. use circuit in (b)
'fJ
=
18 = 5- (VE
+ 0.7}
RB
= 43- Vr
fc,. 5 -(Vt•+ 0.2)
= 4.8- VE
.
I
1
1
(b)
6V
1£=1 8 +/c=VE
43-V'r+4.8- VE = \'E
VE=3V
V8 ,3.7V
v':!. = s- •. ,:r"' ®
\l'c=3.2V
Check 111
=
Ic
=
4.8 - 3 = 1.8 rnA
!!;
=
.!.& =
u
18
.:: I· ~1--it.Y
4.3- 3 = 1.3 rnA
1.4 < 100
:. Saturation as assumed
J :~:'* -=
-o· :t- + 5
®
'2.· 2..
:=. l·~65mA
{, s;.~1
~
O·
'l3~"'A ®
'V't "'"I· '2-+o ·=1-
~
::: I·CfV
=
v, = -5-r 0·'15,(5.1)
~
-sv
= - ~·
'X-
2.0Cf \)
o ·"fact IV1 A ®
®
Chapter 4-16
sv
4.47
"'~ I·CI5SMf\ (£}
t---c
"l- . . 5 - ,.,
:o
~--o
f.= 1()0
sv
('i)
-lfltllcu-.m
IN{NA)
3t. Afl voll4,.tt
tit (V)
(t•q Sf::>)
I· t12~
®
v,_
Vrt; • o-o·"l-:: -o·f-V
,........._
V1 : - o· ot'ft(~l..)
"' z.,..A
-6\/
(e)
;::
-I· IB.'V
®
5\1
3·SkSL 15-V11
"lllc.Jl..
V
'-~5
~
.=
-5 + "l.·'l. .:re- @
.: :. -t· o~t
0 •
0
-o·~
I
c. ~T€
-:: :'
5- V,
3·2.
v,.,_ -;; - s . . ():. ~~~) s·t ,.. - o ·'2J.It:t v
Ie::.
1·1-?~NJA-
@ V· 3~1
V
\}
Chapter 4-17
@.)
sv
v,,_ =
.!.2.!2.. Ie
X
5· I -
5
:. -
101
0.
~ r=l-
v
- -
-sv'
~op
CD
5-a·-a:x:~ ~o·1--&sl-1·'2.:o
4.48
101
f :.
t.lOMI;._t
~
I
1'\0W\If\Al ol :.
VT-:.
-S+S·Ix~T:: -o·ii'"'Z..V
101
f\OWIIM.l
-~
'I: r;
noMII\A.l
I\OMI~\
(e) 'Uu
~~
~
'£
.../:l.A-
-::J~b/1
tt I
:::.
e£
101 :. 0 ·'\'\
,. \WI(\
J:c: • 0
.q '\Mt\
Vt:. sv
~....... Jo
~ ~.t ·.
Vee.== -s . . ~"''o
Raa
100 •
== 1·'2.2.Lf
v
5' .r,4 kJL
::
J4·"!>
~.. +- ~
101
RG =
(1)
14-• S
IOI
F\~
'F-
~ ~ 4[)0 1-o 1~0, ~
h. ~ ...tt.a. .....,. :_ L ;,...._ ~ ~ k>
..tl0°lo
'ik
-GV
0 ~. lwd\. ~ ~
~~
ilh I.- o,w~
•Bt> ( ~ Off'O~ ~ 'f' :o\Sb). To
C..,.
l
S-&.l.
~
t-
.........-.-it.J:
\00
~
~L
+o
so
~
~
~
~ ~ .l&.u.u.\#~
-ttv.. '--- ~
0...
~~ ... ~fr--.
(,OtJ-\ •
Chapter 4-18
4.49
--
:::. tt>·OLJ\l
--e::-V6 c.-?·3-I0·01f
:. -0·?-Lf
~~~·u Vsc.. <. -O·Lf.'J +kL -h'-AI\Sts~
SC\.+uro.ll..o ~
Chapter 4-19
4.50
p: = ·· = ra- 4. =J. -= r . s Ki-f.
'l
-sv
tf.z. •
+ 3V
V8•
Vt!
I-;('\";;
+'5v
Vuo.~
~
ve
·v~
= z..qv
"' :o -:SV
'.~tr..
@
-sv
Q.9'
- .:Z:&z) Jt 6. I
-=(~+ t) .::t.a2. 'II 4 .=1
+ o. ~
.:csz = o,o I 'iLl I.M.A
.I.ez.. = l. Y6 """A
V~=o.rv
Vt.t=o.gv
-
®
-:::;;;:=-
(I ·'l'-l -.I.s~) K ~
== o.-l + 1.3 )( ( ~+r )·.I.8~
.J: 8 3
• () • 0 8 8 MA
J:-£s = ~. 8-5'1MA
V5::. -4.3v
V~,=
-5v
=-
-o
3-= V£ )( to + 0.7 + V£
I l
~ v~ ==
.2.. .oqv
VB':I ~-~qv
Chapter 4-21
4.52
(a.)
to-vc
20
-S
: ({1.5-
tOle
'"-'fii:--t
~
ZOt<.
l(i
"------... •(t- <'•l.~ )TMA
O.Q$VC)WLA
1.40d~
Co.;- o.oo5Ve)+
5-o.:;
"I< ..
(tt -a, 1\fe) =-Uc:
Vc. = l. :3V
-~v
=
\lj ·
c~l
a~
~a.~~:~·~
td'"'Ve
..
= 3 .::: 4 .zq + 10 ::. /. 43k,.( A
rf
IO
0
ra e..d
It
4. 2 q ::.
I, 43
-3.
Chapter 4-22
4.53
(c.)
\JC. • I \1 1- b. \1
'-v•
1011;.
-I<.~
~
= - J!t:l. • -a~~
0'01.t'
-
Node etvo..bto~:
--
10-V +110-V •o.JV+O.IV+O.IV
~o
lO
3a -Jv + 10 -v = 9V
40
=13V
==> v =
Ckvs,
3 .()8 v
Ve3 t::: Vc.4
Iss - o.t v
o
ro
O~~f2.""'A
= 0. 6fZ. -0.308 •0.39'f...,A
(l !J toro•J = o. 38lf
.I.e'I
o8 v
o.sos \IVLA
.:r.e.s • l0-3.08 a:~3
:: 5.
•
o.sos
l o - 8. QB =- o . 2.3 I"""'"'
=
=
w'Jrtlt'IM'I~
~\T'
:: I · 2..5
30
0.1 v
o ·308 ~.4
:C::t>q .. o. 30& -o.2.31 o.o:!l~
(')'f fDt"Cl&d a () • 2. 51 = ~
o.o':l~
-
Ieq •
:o
\MA
/yM1 -~~
::: e>·O't.S '"' 1:1:_ .: 1·87-,.,IJ
'
.
::c~""""" ~ ~ ~~.._
~ ~~""""'"CAA..
..wa.&.. _:b...a. ~,..;_
Av ~ c.!.~
~ ~Mil M ~
~
r --
bc.r. -.or!"
4.
.-J.r"l: .,,~
~
~
,.-o·~
-!."0
"I•Cftt.,.-.V
Chapter 4-23
4.54
lOV
'U'e\io
VeE : : lO .... l.Orc... ~ ClfJIUdor
l>L
.,,,,.,.....
Av ::: - J:etec..
U) ft> cttlwla.k. tlut #..;.._ ~.
(?.),(}) totA.lwJ4k k
.u~-p.,t;/,M,h.
I'A ~ J..r_ lo :s.ku.t ;~ 11\.r.hMJ.. ..._.,k ~
ca... pa,..htu~Ar"
:!c.&. Ft)
Vr
~ ~
Av l''lv)
I
-40
= - .L..,r.t.O
2..
..
0•0'10
.5
-::::
-/k~ ~ ~ ~~ tio'O~V
.M ...,J;I..L ~ t.Tee 1-o ~
~rt~·
.•l> UC-= 0·5 =- ID- ie fIc..
=. L.b Xc..
r equation.
:::
-o·'!f. fo
Since we are a~suming linear operation we don't
~t!'i r
C v)
I
-1-· fo 2..
-l.f ·1- to 5
_,.
4.55
have to go to ir == Is e
.:C c:.. •
t
1
Chapter 4-24
4.57
sv
toV
=>
\J"'o
I pi
I pi
"=-
l.fo
\Jx.
Vee"' OJ+ IPI + fcRc
IAvf
~SV
::::..
:I: c. r?t..
'J1..
=-1oo .J/v
= (- )g .R "'" lcRc \'A
\w\(:lu&A~ '1ft..t.. ~~
tc.."'
"T;,. :.eJ/0.,. (
A.
....,..,.
rtl.s.ol~ k ~ e. -=-
M..~
k>;tf..
w.t.
n.<>~·ilJ:
.::: YAv,. t-rc.r(..
\J'*c.e/l)lt)
l t-
l!'r.t!!IJ-r
:I:t. e
.::
\) /1,
+ \)(IE
v-....-ttc!'k.
"'b(s-ar\t)
~·1 ~~~
tlt.. ~ vol~.t. ~ .
g4Nf;,:, ~.~ iW. uol~~ A.:t tt..A~
tN\L
~t
IV;#.. it... bu-~
-
~ .. :::
0
e>
o ~v =
J •
l~~
-~e.c./vT
-
\t>O
\ +
_;:G.::...;·S""~
IOb"'-l·S'*"
,
Vat \f"c
v~,
i ...,_.,.
···~
li~ ll~tt'llur
Chapter 4-26
4.60
+SV
(~ t:ct" ~
J ~L (}~ 1.>100/J.
~~.;IN.~~
A., 111 -rc ~/ \tT • 1\..\..o t.d...o ~
~
~~ ~ ~ ~.Jk+i
A11
ir.Alu,--.1-t.i::M.
:
-tlec..- Vc.na:t
Vr
.::...,5-O· 3
OV
o.ot..s::.-t
For l-'0 -= 2V, Rc = t kfi
rr "lv
5 - 2 =3mA
lc==
.
I
-l(·Rc·
Av"" - - · = -120V/V
Vr
AV0
= -l20X5= -600mV
AV 116
!1
"'
Vr lnfl 2 1ld
= EAtllJI!lt.'t = E.!i/2.~
II
(a) / 2 = 11 e~ns = 3 X 1.22 = 3.66 rnA
AV 0
= (11 -
/ 1 )Rc
= 0.66 X 1
= 0.660 V
Av = -66015 = -132 V/V
(b)/ 3
"" / 1 e-~n~
AV0 = {1 1
A~.=
-
= 3
X
Vu;;
Vr
lDD
Vc.tF"" ~
11 )Rc = 0544 V
{c) FeN'
AV 1111
AV 0 (exp)
AVn (linear)
-1·5mV
-·660mV
-600mV
-5mV
+544mV
+600mV
sv
-.:::. \}c.c. -
0.82 ""2.46 mA
-54415 = -l09V;V
4.61
f\ \f
--
01..
tic. ~ ~t o j tJ.f',..A-
~OS
~ w~.
vc.eN):
'!;·0
?.·~
(+) w
t.•O
\Je~
!:<
C>·<,;;
i.. -::
O·S
'-
:/J..;.s ~
5
.-
t
=Q·IMn
A
,_
p..RA-.;.
~1'..y.11cJ ollie:...
I
.
k. (,.Jt)·
o·"
o·S
(OV
.:Go~
~,f3
.
'
Lb
1&;._
= f> =
::. 5 kJL
= .Q..:..£
:::.
D·Ob5~t~~lr
o•q
/DO
Q...:.L
!00
";:::
0 ·001
>ttlt !
t
Chapter 4-28
4.62
4.63
(a) Using the exponential characteristic :
~
'c = lue
. .
ic
v~,...JVr
-lc
. vl,_ll'r
=e
giVIng -
lc
- 1
(b) Using small-signal approximation :
ic "" g.,
11
_£!.
Av =
-lcRc
= --·
ic
ButV,. = -icRc
See table below
For signals at ± 5 mV, the error introduced by the
small-signal approximation is 10%
.·. _icRc, _lcRc
vb,
Vr
The error increases to above 20% for signals at
± lOmV.
Now g,..= Output current ,. ic
Input voltage
flh,
.
.. g,.
vh
Thus, - = ---!
fc
Vr
Vr
t'h•
vb, """ ~. vn.
lcRc
c=v
R
iJic
Ex pan.
iJlc
small
+I
+0.041
+0.040
-2
-I
-0.03P
-0.040
+2
+2
+0.083
+0.080
-4
-2
-0.077
-0.080
+4
signal.
g.,= fc! Vr
g
=~=40~·s
25 mV
..~
"'
%Error
(mV)
1.1(1~
T
for lc = I rnA
4.63
+5
+0.221
+0.200
-9.5
lc mA) For~= lOO: 1,-={318
--5
-0.181
-0.200
+10.3
Load line for Rc = I kn
5 k - - r - - - - - - - - i · =50 pA
+8
+0.377
+0.320
-15.2
-~-----=~:::--------- ip = 40,.A
3+------""-..:----"7'...----i~""
.m ,.A
2+--------~~~-----~=20pA
+---------~~---i8 =lQpA
-8
-0.274
--0.320
+ 16.8
+10
+0.492
+0.400
-18.7
-10
-0.330
-·0.400
+21.3
+12
+0.616
+0.480
-22.1
-0.381
-0.480
+25.9
-~~-
2
0
I
Penk-to-peak. V, swing"" 4-1 "' 3 V
ForQ point at Vee! 2 ·"' 2.5 V
V0
, =
25 V :
I c = 25 rnA
! 8 = 25 fLA
~-12
·--~
--
-·-
Chapter 4-29
4.65
3W\ ~
X.e
=
VT
rrr = C>
lf Be• o.~ + lfbe
wker~ U6e • o. oo5v
(/~)
lc ;-I.e. (of+
er. (5.83)
VT
lC
Q:>
I...c. +- c.:c
t"c.
fe=
c
tfiM.
ill
where :
a .oos = o.t..""""
25""'
.1\MA +- O.Z."""A
Va. =Vee- L~ R.c.
*
e~.
c!>.1or J
~
sv-..z.v
qA.i\1\ ..
u
w/.Uie.
-z.v
=-4oov;v
V
tt.o
Ft
\2.1
F?r a.. 6ra..s curreu(; at l2C?JtA
J.e. 10 CiJM.e$ low~r:
J~
• l:J.!
~
4 .s "-«Aiv
la
frr • to,. z. .s = 2.5 K c..t
( t!! :. I 0 )( 2..0. l. = 2. o 6 c/1..
4.66
I.e::! ... Z.W\4
Q
fe
....;>
~IM.
=~
z-s""'v
W\. ..
0.005 v
... ~""'. R.c = -lw, .to/!!= -{foo
U
~~
~IV
48~
= 2.. M.\1
aoiMA IV
Vr 1 .::X:.E -= I.e (/l-1-1 )
=
.::r.e
~
.:te: = 2.lM4
f3 '
2J. • 2.. .0(( 1M A
.5o
f e = 25""" = 12 . 2.:5 c.ll..
Z.O{liAA.
frr-.
Jl =
0""'
5o
-::
'25<1l.
80111d 3
Ja.ivt : - ~"" )( R.c
For; \ ~c
aur/ D-te =~lc.\V
=SK cA.
(j ~ = - 8"0""""
=-.2.. v
lf
51< x 5"""-V
Chapter 4-30
.
+ .~"'·"JE>e tt)
Lct.t) .. .:I.e.
4.67
~ lM •
\)
=
60 lM.A =.I.e.
V
VT
~ ..I.e •
0"'"" \IT •
~Ow... lC 2.5"t
.. ;f.f-51N\.q
ftr· .2.1< =A=$:'(!.> =z:K.-A.50~
~" i9Q, .j~ ct = !£q = o.ii
101
IE = .:I..c = 11 · Z'Swvi = 1. 26~MA
-==
oL
~~q
0
::
tc
2.- o, G Slw we
=
lett> I()
.
L8t ~)
... 1
v
va.ri~s tro~:
t.z-.&o•jZWI~
lo o.g x Go= 4BJMA
p tlar{es tra&M. sa
fiH.{ b~.1~
=-
t.a Z.O.o
rrr = f5 10 """'
0
Lar ea {;. value : ( TT •
4_ w.c.a.:l! :: !:::!!£
$M Mi 4 9 M
\t
=
t.{.t..Kvt.
f) w.a/J.es t valve : ( rr ~ tM.h1 = 50
d--i2Z.k1
lA
wtiM,..,..
;::: 6tf 4 ()L
4.69
Vc~ ZV ==}xi= Vee -Vc
t<..c
.J:.c.= 5-t
3K
= -1tM.A
<(""' • .Ic :: IJ\M ~
(j
VT
25JV\
4 OIAA.~
V
v
t.t:J
+o.z . .s h-t. wl:
I
v
~~""
1100
=
t 0 + 2.
Volta.~e
4.68
R.c
Yeo::) -= ·5 -
\...)
ce)
~)
(11+
«
&.E..D.
vr
.
VT
• - .:t.c V
Vac- V8£
VT
-f
b
1.00 0.990
0
~
.. - '< ~. t<.e
Volta.se 0€M~
= -
a
tOO
""
lc
c
d
~
t
s
0.98
I
0.890
0.90
0.841
""
100
9
16
so
1.00 0.89
1.00
1.00
0.48
4.5
17.5
l.OO 1.00
1.02
1.00
0.25
5
18:6
0
0.002
0.5
!.10
(mA)
I< c
fe
~
(mA)
-V6e
0
0.010 0.020
40
39.6
40
40
0.01
180
700
t.({l)
25
25
24.5
25
100
5
!.34
r,(fl)
00
2.5k
1.255
00
!OJ k
50
227
In
(mA)
Vbe
g.
(mAl
V)
R..eit! = 5 ... o.:t -o.aos
t1 +-
=3.C,V
0-005
4.72
o.azs
JU fJS
Vc =
I
5 - 3. c
:t:.c ::: -1VV~.A , (:> = 1'2 a ot. =o. Vf2
3~ =- ~ = !._ = 40UA.~
;: -+ L 4 V
VT
A1M.pl4tode a~ ovb fu b ~et.tal ,.,
... ",q -( Vsc. +v&e)0
-=-
-
L
t.t - o.:t - o.oos
f e = Y.I
:I.e
Vo lt-\)e :!:Ja.itA. :: ... 0.695: -~J!_
v
().005
Volf;o..qe ~~
- J 0
=- (6o. -a.
7- o.aQ5)
oz~ a. cxr5
::::-
-1"
-(4~
ll1e. r;/,'frereuce
.
t'.! aavsed
6~
0
Otuerwis e:
Valbo.«._e ~~, -
v ()
-rlA.e.
IJb~
=- 3 K vt
40x10$
= ~ = o.
:3""'
'IZ :::- 2. 4 . 8 V2
Zj"();r o •.,
<:J
r.ovr ejuiva..leut aireu.l..t
w.oJ/e/3 a.Jc.e:
e,
0 c
...
V
Ctr
31C/\
5*11Rr
0 .O·Ll tfbe
t4
tl.
v
_,..
ftr
31<.1\
V/V
clec!ltM.tLI rav~tt~iue.-uf Of
f<.c Ia.
3~
v
0.695
elteel<
frr· /!J
'2.5
-= 12.0
c
c
.
oLLe.
•
O.f/92.. lt!
O.'=t lG
0~00_5
=-14 3
v;v
Chapter 4-32
4.73
(.> 'v~ l.v:5k v~
ot,.:::
I
4.75
:t:.. c.:.. .X.. E.= 0. 5MA
\Ia • -5 - -:r. 5 x 0-5 .:. + t. ~5V
:!.£. • o .5MA-
QW\ •
U
2.6~
Vi
1:'.
W~A
V
:4
:::::=
r-----.--o t!c
frr
~ ~ \Sle
+
N,;,
~- ~ IC. ell
R,.•
+
Nft
r,.
E
06&erve 6ka.1 (fbe:.- tft:
tke avbr,ut: volta.~e Uc. .H.J
fovrA.o/ f'a~AA.: U
{fc. • lfbe ~e. 1-. 5 K.
0""
~~ 1:. ke volt:~ e.
£.[:_ =
-10~0.1 ""3.1mA
If v0 = ±0.4 V
E= 30mV
. ~~ =
Vr,, =
lc = tlJ ft-+ a>- ..1:.!_ =
'
I"' "
I"'
0.023
ns
•
13 = 25:
0.32 X 30 m = 9.8 rnA
26 X 0.023 X 10" 3
r, = 41.8 n
4.80
c
g.=~"" 25126 "'' 23.!!!1
r,
41.8
V
4.82
hut; r
"
=VA=~
lc
Vr·g..,
~Va~_v ..
Vs
Vr
ifV, "" 25 V ~ V0
A
Vs
ifV"
=
250 V
"" -
~ Vo
v,
=
J(UV\
Y
""' V
~--10.. 000 Y
v
= 75 n
I _ 25 mV
1: 75 fi ""(0.33 mA)
RE "" -~.QI ,, ::!8 kfl
.
0.33
n
=:.
2.8
Rc = 14 k!l
Vo -· nRc
·V1
..... -;::-
14
oms
187 v! v
Chapter 4-35
4.83
~i l4 :: {?.. E [{ (e.
( e ~ l OOd kA.(.J£1A_
volts.
ca lle(!i:cr
Ualt~e
ZS .. 6
K.. t.t
Jalk ,. ~ =- 5 Rc =14 3 trv
Chapter 4-36
4.84
for la.r'(.e {Ji l:lte. oe 6QJe
eurrew.~ wltl
6 e """- a
valb~e a.l:
Tuos 6ue oc
l:ke /;a..s~ ea.~ bel" vwd
dlrea.l:l'1 va ~~~( 1;/Je
Bubsl:itul:lu.,j re • ~ =aut! R.E = 6 .. Btu
-
volt'0e tf.. c/ivlchir rule
ve,
=t
•t:\1 ae o
\00 ~
tOO+loa
o.~
VE. =- =J.s -o.-1
~IE=-
#
6-BV =
6
lfo1 t fl
=t. s \)
15 .
r/ tl ~ 4
t:l.lit
R.c.:::: q .SK~
~ t'f!e.S
~.g
a.ot5 + 6.8
= o. fJV6
6 .. 8 t-o ..oz-s
I tlte uoJe ltl.hele J tfat. ,·~
V COV\Ke.a.l:ed bo '<.rovw.d:
L..
lfuv\
-=
V
R..E.::. 0
lfat • -ct..R.c
l..e
llor ..
v7:
&.
~e+-fe
·e. D.
':=!
Given: R, = 100 Kfi Al = 2 kfi & R, "' l k!l
Plnd : R.,, Aw, R,
a) !Vi(e)j ~ 0.9j,~(r)j
1t;u 1 -
Also,
lt! .. U'b
if£.
-
S
~ 0.9jPsU>I
~~0.9
RIN + Rs
Rl
Rr + Ro
bi "oUl '"' ~.··-AI'o 11;(1)
-= - ot ll.c !ft..'
(e --r l!.i!
Ua2. =
IN f
,Rn; . R;l
1.
re+f,;~ ::=:> Gvo
R1 + R, 1t
= l!oitJ., 18
+
V;
R
= - - .'-.4t'o
R; -r R.n·~
Connecting the load
Consider the form
(R/e)ll R:
R,. "' R
RR
AI a+ R
I
II I' 1? r + (Hz II
1 '
Ll
R
+a
RtJ ]
+ g,.• IRz ·• Rt>
The circuit f{>r A,., is
~[£F1R,~
Chapter 4-38
4.88
=
=
Rc "" .to kO, V. 50 V.fl H)(),/,. "" 0;5 mA
I
0
. -3
g = _£ "' ·5 X IO
"" 20 X 10-.l S
.m
Vr
0.025
·
'
= tOO kfi
50 .
ro = V..t ""
0.5 X 10- 3
/11
r, "" J!. ""'
g.,.
JOO
2o.x to- 3
= 5 X 10J 0
Rcll ro = (10 X 103 0 100 X 103)
= 9.09Ul
Ro "'"
+
=
Riu = r,.
5 X 10~
The circuit is now (see figure above)
for values given
R = 99.90,Aw"' -9.9989, R, = 100
The dependence on Rl is
=
R·
100
m
1~)
If Rr decreases th~ gain becomes sensitive to R,
lfR1 --')oo R.. = JOO,A.,= -10
Av "" (
Gvo -
...-!!Jii...A .
R. + R
vo
1N
avg:
=
)<
S X 101
-95.2) "' -31.74
5 X 10~ + 10 X 10 3
max. Jno(t)! = 5 X 10-J =
IGvl
-99.9 ( -9.9989)
99.9 + 100
= -4..997 V/V
4.89
IGvl = f3
WithoutR1
Gvo "' (
·'
mall\ signal r~,18 is
witbR1
.~
R,)
= -(20 X 10-J' )(9.09 X 103 !1 10 X 10 3}
Av= -95.23
1100 Rr + 105
·
6
1100 Ri + 1.21 X 10
Avo "' -10 (;,1 ~·~
= -~
= -gm(Rol!
tJi
A,,
IOOJ~too)<- !0) =
31.74
Rei! RL
I
157.5 J.LV
ro
R,;g + rr.
-5
IF r0 --> x then Rc II R1.
Let
l?t'
II
r0
--')
Rc !I 1? 1.
= Rc Jl R,
R'
!Gvl = 13----L....
R.; 8 + r"
_!!_i__
~+c.:
13
13
But r,Jj3 =• l!g~
' .
!Gvj ·=
'
Rl,
.
~!it+!.!.
13
R/ '
0
g.,
10 k!l; R.,
I, "' 1 mA
··
IOkfl; f.1 ~" 100;
Chapter 4-39
a)
IGvl
.!.Q..4 + 0.025
ROV/V
4.90
10~3
100
!Gvl = 13 Rcil
b) If~ ranges from 50-+ 150
For~=50:
IG I ~.
v -
~
1·0
w•+o.o2s
50
10-J
R~. II ro ""'
+ r,
l3 (Rc!l
R, 18
"" 44.44
Ur 0
+ r,
R1)
R, 13
v IV
(Rc
!3
=
I!
RL)
7;
~+~
J3
13
Vr
109.09 V/V
~
II
I
~-~
g,
lc
c) What is~ range if G 4 ~ jG,j :5%
at!G,j"' 64:
~+
4
to
"' 64 => 13
104 + 0.025
13
to -l
= 76.19
13
Rc
II
RL =
v.,
lc
10 H. Rs;g = 10 kH. VA = 25 V,
and VT = 0.025 V
at!G,J = 96:
126.32
(104 )1! 25/lc
4
.!Q.. + 0.025
100
d) Suppose the nominal G,. is G, . ~"' .and lc is
lc
25 X 106 /c
variable
13 ""
0.8 G~' _ "''"'
50 => G 11
13· ,. ISO:::::> Cv =
1.2 Gv-nom
Then
H/__ =
_ __:..;::
OJ\ Gv ... """'
104 + ()J}25
50
Ic
to"
104 + !),()25
fc
150
Take ratio
104
150
+ 0.025
f,.
104 + 9:025
50
lc
0.8 => t,
J.2
() 125 mA
1, (ref)
IG.I
0.1
27.47
0.2
41.15
0.5
55.56
1.0
57.14
1.25
55.55
The values of I, that result in !G,I "' 50 are :
l X 0. 925 mA and 0.324 mA.
The 0.324 mA would he preferred since a lower
fX.lWt'r is required.
Chapter 4-40
4.91
4.92
= t11/ (r, + k,)
ih = i - ai "" (I ~
'
R; 0
i;,
a)i. = (l - a)_!L_
~+I
,
100 +I
I
1h
R; 0
""
r = VT
•
(jl
=
IE
VT
+ l )(r, + R,)
= a Vr
-101 R)c + (2 X 10') I,= (l01)(0.025)
(F.q A)
Equation becomes
lc
oms 3)
= 49.5
n
0.5 X 10
=
= 2.525
Our unknowns are lc & R, This is one equation.
i, "'u,!rn+g.,v,"" (1/r,.,+g.,)u,
(.! ])a .,
= p ' _
R,. = ( 100 + t )( 49.5 + t5o) "" 2m so n
Avo
l)R~lc + R;.lc = (13 + l)Vt
13 = 100; R.,. = 20 kfi; V1 =0.025 v
Given
lela
,. (0.99>(
rnultiplybothsides
- (Jl +
r, +R,
V;fib "'
l)(~~ + R,)
(fl +
by lc and rearrange :
111
(!J + I)
=
lc
R;n =
a = _L ""' _!QQ_ = 0.99
I )(r, + R,)
Vr
r, ~--
r, + R,
·
{!3 +
""
.l.
6m.,»'
-ni,Rc = ~nRc-1r, + R,
Avo "" -(0.99)( 10 x 103 )! ( 49.5 + 150)
now model becomes
=
R,(l13 + 1).Vr
fc "" + tt, + ~ h· v
13V "'
1
t!!'!ig
-
~~t>iR --
OJ - 0.005
R,lr
[ -1 +
13
=
·t]""R
lf' + R,;&"•
.. .JC
Vr
[-1- +
JOO
+ (_'\000 )(.'i X 10
f
Il[5 0.025
X 10 --ll.
j
.L
'
')I
( 100)(0.025)
('
0.005 = 0.202 RJ, + 10 1, (Eq B)
-~ __!_QHl_. (-496")
10 kn
''· -165R2
+ 10 kH
·
.20150!1
· · - 2015o n + 1oooo n
BVr
Chapter 4-41
4.94
tiquations A and B can be wived simultaneously
lc = 1.25 mA
R,lc:
0.00064
o:i,
=
=»R, = 0.22264/1.25 X
10" 3
l'•ig
lf~tg.
fllf
l 1si&
}(1. 250.025
X !0-J) =
= -{5 X 103
= (-250)(5 X IO-J)
Gt'
0.1
t•b,(t) = r,i,
-250
l'o(t) = ~ai,( Rc !I R1J
= -12.5
u (.t) = - r
"'
=
.
I3Rc
R,;g + {\3 + I )(r, + R,)
Suppose o. = I
lr
Jt•o{t)j"" ( 10
!Gvl ""
!3Rc
R,;g + (fl + l)(Vr! ft.·+ R,)
lOkU; !3 = 100;
Vr = O.Q25V:
I= lfnA
R.,
(X
_L
fl +
=
Gv
101
1.: = IJ<• = 1.01 X IO·•A
lfR,"'O
( 100)(10 X 103)
lOX 10' + (101)[0.025/(LOI
80
11Rc
h•s(tll = luoUll I Gl'
lr,,1g(t)j = 0.5/4.R8 =
ll'oUllm.,
= (0.5 V)
!i {t)l
"'
I '
)
at j3 = !50
G,. = 1.2 G, . .
-
w·' + Rc)
These two equations ~.·~m he solved simu!lnneously for 1?, & (;, "'""
1795 v
(} ,. """' .,.,
I11oUllnm
RI
•m•x
V
V ·
1
r, = -=>I~=
_!_
I r·
r,
rwm
(!50)( IO"L-~-10'1 + (IS l )( OJI25 / LOI X lO J + R,)
K · .· ·
0.1025 V
=
....J!:~~···
X I o·'
2
= 250j.l.A
r, = lub,(l)j,..., = 5 X 10 ·' = ZOf!
lic(l)j,,"
250 tJ.A
= O.SG,.. '"'"'
(50)( 10
Rcll R1
Rell Rr.
= a - - - · = o. -~:..._!:,_
R,;r + r,
R,,~ + Vrl IE
4.95
variable (note that a = 0.98):
JO'; + (Sl )(0.025! 10;X
= u0 (t) I t',;g!tl
X 10- 1)]
Suppose !G,I has a nominal value G.-.•um and
0.8 G,·num corresponds to 13 = 50. Let R, be a
1
rnA)( 10 X 10-l)
Gv = (to kO)!I (lO k!1) Since a= I
I k!l + 0.025! 10- 3
= 4.88V/V
lOO = 0.99
l
IGv[ ""
kUJ.~2~10 kfl)(0.25
luo(t)~ = 0.5 V
= IOkfl;R,."'
=
R1)
r,
r = l-'r
'
vo(t)
'o( Rcll
lt'o{l)! "" o( Rc II R!,)lv,,_(l)l
4.93
IGvl
E
+
l'o "" Vo ~
=
rtfi~
Rs~,
= 178.tl
Gl'
R,
B
lirU)j,,"
1.2G~, -nit'tn
,
(,,, ·=
1,,. -li,-(t)!.,,
f3
=
1.2mi\
20
= II''· 1ic(()jm,
lidtll,.,
Suppo&e
0.025
LS mA
I mA
100
(15 + I) R
.
...- ............1;••.•.•.. .,, .........
([1• l)R 1 +{f\+ IJr,,+R,i~
30.625
{ 101 )(:.!X lOll
(Wi);2 X
10 1')
+fl-fll)(2(i)"':
200~JOi
, ().49<)
Chapter 4-42
(
v6 ,.(t) I vo~ 8 U) = (r, I R1JG 11
v 0 (t)l v, 11(t),.. 1000 ~~
2 . 772 = 0.8178
This agrees with G ,,
=(24.752/1000){0.8178)"" 0.02024
v"(r)
= v0 (t) + v11,(1) =
vb(t)!tt, 18(t)"" Gv + (r,!Rt)GI' .,.. (1 + r,!R 1JOv
+ 24.752/ 1000)(0.8178)
v,(t}l "•;g(t) = {1
.; 0.838056
Gv"".
(~
b) ur,,(t) I v, 18{1) = 0.02024
,.lflo~s<
4.97
/,."" 0.25 mA; R.;8
V,= 0;025
= IOkfi:
(~+ I)RI.
+ l)R1• + l...,,
= !l!h,{t)l.,.•• t 0.02024
jv.;,{f)lm•• "" 10 X
10 3 /0.02024
= 0.494Vo!d
lu,,(t)lma• "·~ Gvh;~(!)!.,., .o~ (0.494)(0.8178)
"'0.404
v
c) lf R,_ is renl
=0
'.z.q
1<.62
v,(t) =
(g.,r.,r,..- RE)Rc
(r.,Rc + R~;r" + r,r,.. + g,.REror,.
vs~,(t)
+ RcR8
+ rQk8 + R;;Rc + RcR8 )
RE(g.,r,r.. + Rc + T 0 )
Rc + Rcr... + '"'" + g,Rgr r, + RcR8
+Rr.r, + r 0 R8 + Rr.Rc + Rr;R 8 )
+RE:ro
v.(t) ...
(r11
I'•ia{t)
0
=~
I
c
Part b) Nodal equations:
Re
S l(sz
o.&f ~s' .,. o.6 f ~61
4.3/ ~sz. -=o.6'i~81
Rs1 :.
=l>
R8
Solving:
ra
=
-i"
'"
a
+
c
+
@
s,·IAce
V&E ==
5 l Vss = o. 6~ ~ v
I
•
/;lu<_s Vae- raiA.~e" f'a~ 0.67811
l;o 0 . :r oztfJ
(!OIJ{.
Chapter 4-44
.:r: a .... .:r.
Fo f .Ia:
g
e
V&.4! 1 VT
jar Vbe-=- O;GfO-;.. Zc- .fwA
.-=:t
ror
.:I:..,
e
f.032.}(l0-f'S
VeE· 0.618
Vee -:
o. ':102. ~.I.e-:
..Ic (Q.""":t·'!J trotM
A, 62tM ~.
@ lt
-"~e=
o . .:l:_
4.101
tf. 'z'"
a. 618 H-t.A-
l:o
((c.::: 31( tJl
~ c.
Vee • 5-3 /t:. ._ o. Gl~ • .3. 14V
Va.e • 5 - .3 K" 1. 6ZtM..:: o.l4 v
th8 e/rCAM.i: Is l:oa seu..s/i:/vt!
l:.o fa.ro..IIIA. e. /::.e. r varla.t/a us
a&
3Uoww lte re fOro.. 1 ::. I.. c.
etwo.se
~
Rs=
If oo
==
r
ror tUest!!
19 +I 3
valve.s of R. Qk,j
f<.s :::
: I.e .. 5 - 0. =t )( 50
0-~ V
--
~ ·~$'C"
~~
r:J.. -= 0. ..:Cc = .2 ~ 8 o a-w A
:::
+ 3. 5 V
t~o: ..I.e.= t-5~A
Vee.-
~ 3. =r:J v
IS II 13 ::: 7. 55/Clll
:r:.e-"' 5.7=7 -o.? = 2.~3Wt.q
4SOIC
lff=
131<.1./L
\Jee = q x r3
-11< ....
I.e.. =0.,:50 kA.A
-::9 VeE ::: 5 - 3 K >< 0. 5 IAA
1g.~vt
----ltf
R,•R2.•/fK
5V- 3 K .A !MA = !LV
l ~ = 60
.scn'~a.61e. ~ ra res tstof.1'
f<2.=l2.3K.~
{)::: f'O:
Vcr=
R.2. ::: t2..SK
to
((, = /~.1-K
5-o.':J =- /1'11\A.
r<.s
4BoK..t.A
-7
lf
I?.E. is redveed
..Jt R.e - f tOc/l
~ ~e
I.e
6J ""'1-~~/t.
= .3 .09t.4-1-A
=3.051MA
·
Chapter 4-45
Is. wi tkiw ± 5~ Of w.oc.wic.co:l
4.102
0~6 :.
R, =
= 5 kO
vb = 0.1 + 3 =
R1
= 2..::_1Z
fe/2
Ie (('>::.~o) >/0.9J5ZE(~-=too)
RE
=
3.7
O(
I
\]
~- R1
IE
Suitable 5%. Resistors: R1 = 17.4 kfl
R2 = 12.1 kfl
li = 90:
= 12.3 kfl
RE .,_ R.e,
or,
IE =
3.7 - 0.7
5 K
+ 78.4
;>,., o. Y5 ( Re+ ~)
~I
RE ';?/ fR2 =
V
1'1M.f0.3e tL:.e COUS~fCUkf
Wt!! IM.V8t
Forji = oo,/8 = 0,/E = 0.6mA
~ ~ £J.T3
R.e
~vs, tl:= 1oo) aocu r~
a& btr.e low e.v..d Of (5 valves
(.(';=-5o) . Tf..tus, to Keef
::. Vee
-1+({&.,
3
Vse, =..L Vee. ( -1 + 5...:!2) +- o.=J
5
~
tor
VS 8
:: 0. S-5 Vee
+ o. ~
(c) Vee • IOV
Vs&~o.SS"IO.,..O.'t • 4.Z..V
~ f<'2- IC. ' 0 :r 4 . 2
/<.1 + R2.
R"
= o. 42.. 0
(<.,+It 't.
I.£. l<.e ... ...L Vee
3
C!OIVT.
Chapter 4--46
t
X
RE = ...L
-=-> R.t:.
I0
X
3
bul:.: .IE= .I c
= I. 6=1 KVL
Rs ,: 5. ~ 3
)C ' ·
6~
t::
y. 55
K
u
f{,.R.1 = 9.65
R..c+f:,
__, 6 aouz
6e~f'-'1 ~ G.t.Kt.t
=00 :
.:Le
e
0
.I.e.= I..£
Vs=o
UE.-=- -a.=#
Ic. • .3 -o.':/
4.104
(o.)
= 3-o.~ =a. S8t4otA
!<.~
Vt!.::
6'80
.3- g$8"""Y"'"-= -0.38V
'"For (1 ~ ~0:
.:r.e = ..-2..5
s.o~""A
-=
SSo+-6.fK
91
(.b)
{1 = (jO
-
VRE ::
V~e,
Ia
Is. t = 50
l
S
N J2. • 2. OQ I< r.ll.
tue 13Jr vsed kas
/;ke e~AA.iHer t!t~trt!!t.LI:
(t!!:s v l /:;/lA'{_ t:!.o.~ be f o" t;t.d f r;::>v...
£
s ."''f.l)
.Ie = Vee - Vse
R..c...,. Rs
1
f c
-
t.>+
= 6 - o . !l
6 . 6 .,.
a lA. tl .:r. s
=::
-
I
o. !.lltAAA
-51
Z-0()
o..:.!:!J_ a.- ~Jt A
51
/U.u...! fke collecta r w; U 6e
l,cr
~~ ke( 6Lta.-u bke. ba..se
U 3 "-O·Z. = {.6 V 1 a-lla~uJ
0- ± I.~ V tH5(.. vt.a..l swtu~
at bLte
colh!ai:o ( .
u
to;-
"':F""oc {)
==
150 :
:I.e ..
= o .sq &M.A
:5-o-~
G.b+~
I5 J
I
4.106
lb.!
=0
B
l:.ke
. -5 l(
'::. .!3 6.JtA
16t
collea.i::af" volt~e
<.»iU be iM;~ker t~ l:ko..'t'
af f:ke ba.ise £, lf
.
3·6 XQ.2. .:f}OM=!l.V
a..Uow,.111.j
.s
o.5
Fof
{!>-::
lOo ;
-
Is:::: .Is: -:: .
f-:' +-l
t"O IA..a...l
ra.r a 1Al'1
.s vv ,·0 . U
+
a. ~z.. v
Chapter 4-48
4.107
4.109
.:C.5 • J-c.tf> .SIMA/~o-=o.os!.t'4
Ve • R& . .:t.s + o.T
Ve • 1• 5 v ~ R. e • ,t4. 2..1<. Lt.
:r:..£ ~ ..:t..c • 3 .03\M4
1:
+5V
O.t mA
(+0.5V)
+t.OV
1>(.
::l:. • .I:.c - ..l: s a ..:r: E
:t: • 3 .03W'A
-sv
4.108
l1le conslraints imposed cannot be met
v.. < -0.7V forQ 1 active.
Change V.. to 1.5 V then
V.,= -3.5V
V.., =Vn + 2.5 = -t.OV
V"' =V.,. +l.S = +O.SV
Forp = oo
Ra • l.S VIO.t mA = 15 k!l
V11 = -3.5 + 0.7 = -2.8V
Then
v«•
= 2.8 =
VR2
22
!!!
R2
V,.=0(/11 =0)
Va = -0.7V
Vn = V11 + 1.5 = +O.BV
Vee
RC2 = 0.1 mA . 42 k
Thus,
ex •
I
_ ex/
0 -
_
E -
rI.R+R
y c:c:R2
t
- V ]
2
+ (Rl II
R
E
BE
R2)
P+l
n
For I.,. //fTr = 'f.64//t.S•/./5Kc.t
BlM. - ~"""" =- :P6 ·81MA =fo =VA :
v
-
v
Chapter 4-50
Vsg- Vse
R..e. + R.s
.:t.e :
(-;'H
3 -a.~
2..::
=> R.e =~K.J.t
~e. f l0/101
=II~ c.t
U:Je lle
Av;: era ::
Ch'.
Cft>
__.,....
__...
lJ$ \lt..'
, t;a.tA ( R.o llft.)
~
Us
R.i
-:;t.
f<.s+ lf
~ - I. 15
Tlt~ r~tAE:it.t.~ .Je wlli/:J!
(ft.'
fft.'
It
=? 6 . lJ X (2.. ll
f/2..)
l0+/.15
:::- -8.13
V/V
~
A t'
U.1 I ( R.J rl<.i)
-= cf(;) •
v
fo ..
At"-= fo :-: lfo. ~L
!{
J:E.. ~ 3 '>Jo.:t ::: .t.Of ~A
J +- lO/tat
::c.e. • ~~~. I..e. - o. qq tt 2. .a f = 2 ..o ~ \.c.4 A
_<{"" ,. .Xc. -.: :. 2 .. 0=1 • 8Z . '3 ~
VT O·OZ.S
v
f1f' :r f'.> ::: 100 ::. 1-ZI KCA.
f<.s t R. t'
iii
/tL.
~ - B. I 3 ~ { l 0 +- I. I~ )
2..
-= -45.3,.. AlA
J~
82 .. t 5K.
I:JAv~
v
awd' A6•I.s =0.6
Y! : :
( f1T --5K)
~
~
~
• f rr :::
5 2-!>0UL
tkeu. f<_s •
........___
'R_; ... 2.f .S8f13.S5
II:'
a.QtKI.Il
r"
Ra.,. rn
5. 85 K til.
fo-= tao =-l5/.5KCA
R.e,
~s
ae:>
R.;.., =5K.c.ll.
(#
O. OZ-.5
'I:.;
~s
IE""' 3 .t8 -().1 :: 0.66\M.A
\)
+5V
4.114
......___
161.51/6·8 nt
,. [.53~"'
t<_e
altoo.!le
10~ I<
r:
100 f(
vt
.
(kud :!.8 :r '1. 'T~JtA
::!'.E • {fo +I );!.8::: /01 ,c 4. ':f&.)lA
I.E= Q,lf81MA
x ::.I.e _., .:r: ~ o. s M.ot. 4
Jo
a.vo id .so.l;uro.JAoiA:
Vc.- vs ~ -o.c;
'Vc. :: sv -l
For l<.s =/OK , R.L -=
Cf!!.. -=- -5 w. (i.e II ~t.)
/OK
AIf .s
4.471<
Z.Z.3 V/v
-50\M.'tt
-
x -22.3
-:S·HO
(fC : Ub
-=- 5
__. . (/a
__...
lfa
trs {/b
:r:
zo t.«_i
v
2.<$M
=
roo =- 5 K ell.
lfa ::. tfrr x c.fo
(Js
c.Js urr
= -74 .s v;v
-=
-:: _
- ----
-
(Jb
'=
0.5 ~
~:w
...
Iii
~ -:::-
VT
= - 40'5 V/v
lfb
2..oa ~t.t
"=
~ Wl :
:
R.s =.2...5 K
,
fnfrr+ R.s
--
1r
-elMu
5
~ 2..0
5+2.S
((a II klll!.t)
(too II to II I a )
-= - 8 6 V!V
4.115
.Ie.• 0.6tMA
(a.)
IE
c
-o.~
15
l<.e
0.5 ::'
t-
B.L
f.:,+-1
/l/.3
10()
~
f<.e.=- Z8 .5=/t< 6.<8:: 8.Su
==
Chapter 4-53
m
(b)
(f)
1/bt
+
flt."'
~
l/o
___.....
lfs
K
lf6z
_..._..._
ll
lib•
.. O.SZ..'ll -68.!
w.
J
---lfs
=lf61
U'62.
~<.
-5lf.3
12.~2..-
"=
-
ifo
...,.......-..
. V/V
=-
~·1'1111
~st =
Rsz. ..
0w J
1 ::t
\M
R&::::: azKc~t
z. = o,. 1?6 =
as. 4 ""!.
O.OZ-5
ftrt = ftt!.
V
-
=
toa -:: 2~6
s s.q
Kef .: :. f<..e z. :::
K VL
fat= (. n. =< Gb
':::
V
tJbr ::
(J:
100
fe = Vr = o.oz.5 = J.6o«
-
::I.e 0 .I
(1.60+!.:50)
£iVI• fOl
• 50.-5 Kc.Jt
s' II r;.,
U6-
52.(/t..£=
R.;w1
J2 S t-
t.
R.i w = ( ~ + t) cr e + 2. 6o )
'I(
G. 8 K tit
(c) tliv. 1 =I<
4.119
':::;
/(.,;,.. I
Z-.4
5+Z-l..f
(d) l<.iYtt. = R.ez II ftrL
:::- 32..11 f...~
=t..
4 K VL
===-
Ub2. = - 0"-tt urn C f<.ct II f(lYL~)
=-5S.l.J cfb,
(6.811 /!..4.)
-
Lfbz. -::: -68.1 V/V
tfbr
== -:~\-W.Z
Vrr2. ( R.ez 11 RL)
'='
-38 ":it lf6z (6,8 lit.)
-==-
-51/.3
v;v
l5o • o:~z. ,t,.-/9.8 = -t4.z vlv
~
l.fs
"Fof 1f6e = 5Vv..V , 7.fe = 6'1M\I a.fsa
(s,·u.ce I.e= fe"" 2.5ovz.)
/tu..Ls,
lfb = 6 +-5
7.Jr. '""
Ce) Uo
lfo
tf6t.
cyltus,
?fa
:rt
#
IOIMV
lO\M = 1'3.
g 8 W\ V
__.
0 .'=ll.
/3 .88 X (l.(. Z = /<{:'/ , 2. 1MV
Chapter 4-54
4.120
+Sv
4.121
(;a..) l:e • o.f t.ra ::: - f o V/v
w
====-
o =Ve- o
=-
.aJ
K
t.o ( /l.C! II J..)
-
5 - (1.5 I.e.
t<.a.+-1
5 ~a. .,.. 5-o.!> I.e z... o.5 I.e. -0.2 ~c.= o
0.5/l..t: z. -l{ ."!JfOtA.
~
flurl ~ue volt~~~'·""
f5o/cJ9 jlr•t c-..o~~
Ve = R.,·
~Ri
y4
lfc
iTe.
Thvs.
fOr~ =Zoo, I.e:=~= ~'""
So ::::. o. 6
6a+ -5o
(Tc:ai:a.f rest'.r~e o.t:<:)
fe
~ -1 JC (lOO~tA. /1 ,f K\4.)
=-ol
Ve • + 6.5'(V
.f-
zat
6.tq v
(b) Ri•lOQIC.cJL//C{3tt){re~c•ur>)
= I oo II ( {>+-t) C re .. o.6J
IE • 2.. 41~A
Tor (lJ ='to ,
e,·
l:iAtJs
to. a~ Vt
fe •
too 11 '" x ((). Ol()~ t().f)
• I oo II 2..1
= (':~. 30 VI.
t:
J(
(9.9 V/V
~ ·1~.9 ~<0·5
(J.s
'foe
~
= UJO
~
/:)4 tJ~ ~,·
c
=
-
I.e.== f-
o. '=l
f -1- IOQ /(/S+I1
I. I!
•
5 .:5' 'l\M 4
fe = 4. 51
lOO
cJL
H 101.'(
60.3
A::~
--
= CJ6
•
£.S
-=
4.123
1
=too 1/z.at ( o.ooqs ro.s)
'::'
='?.Y VIV
(c) t!a
Cfs
(a..)
-
t.f-IQQ
~
==
60e~t
c
-
Vs •
4.122
Ike",
Ve:c.tl'l(.t.Cll• t..l.{IV
va • .t.l.( 1 f-o.!l· .3. uv
Rt.'
lfo
cf6
• (
1/t t)
i!S+Rt' C• llt)+ r~
For()== 'lo,
~ ...
l':l-.3
Us
I 0 + r:t . !
IC.
0.5
0. 5 .,. c. o '"' 1
= 0. 6Z I VIV
fo; (>
= zoo,
«fa • 5o. s
us
to ~o 6o.s
o.s
o. 5 + o.oo4~
= a.sz.:, vtv
Chapter 4-56
4.124
I.e: =
= ~ .oo \.11tA
5-o.~
a.~+~
.:I.e1 = 60Jt +..I..s.z.
::: 60 + ~~2. c 5CJ + 50IJO.
lOt
~'Z.+f
re "" 2 5 - t.5t.Jt
4#00
R..r.:· (fo ~~) [ re + ( s.st tJ]
• 130.0KtA
lfo : : Ub • Vo = Rl
VS iJ"S "iib
Tf..tv9 1
Lfo :
tfir
( .3. 3 II t)
lls+-tet te + cs.5hl)
~
)l
..l:& I= 0. _., "-tA
(60+1)
~ l. {/
e.
+
L
0' OZ.S
= o.=t9bl2.'*"')
e;+l
\.)
flt..+fet.
R.t.:: -1 K Lt..
-= 71-=t. 4 AlA
1/ [re
~L
=
tib2.
-1
Roul: ~ 3.s
£efer l::o Ft~ . P. 5. f4 1
tie
Uo /I<.L
ifs
o.s )( rs" = a. .-e.-zv
(_.3 •?, /1 I)
\
to
0.1""'A
Vat= 4. 580
tOI
X (
4 .(}05)
::::
IOL .5 Kc.t.
f.&:!
=
Lfba
--
fet-.. Vr
f!..t' bt..
~ t'6t. +- re.
= .Zbou
\OO)'A
-.:r lfer -: :
lOI. 5
-=-lJbl Lat . 5 + a. z s
4.125
(a.) -:I.Ez. = .?'M A
p,. . 6o,-- (!:,2.. =too
0.
9'17 V
=====--v
R..t..' = -t!tvz.lttJHt.t.l/(fl,+t)(fe,+R,:&t)
-: -1 //A /l5l){(o.zs+ ta,.s)K~
'=- 11/1 -1/l 5. 2. M l/2.
=.
0.4'!(WlL)V(1tt,t
=
200=>{WIL)V()v; 1
=
12;5
Condition 2:
l000=>(WIL}Vuv,2""' 2.S
lf condition l is met. condition 2 will be met
since the over-voltage can always be reduc:ed to
satisfy this requirement. For condition l, we
want to decrease W/L as much as possible (so
long as it is greater than or equal to 1). while
still meeting al.l of the other constraints.
Titis requires our using tbe largest possible
,,.,, voltage. V cs.• = 1.8 \bits, so V •'»~• = t .4
\blts that
W/l.. =
=
v.(W. 2 --
I
::::>
(J.LnCnr)(~)(Vov)
12.5 -
W /L -
(a) v.. i~doobled ..... r.,.ishalved. factor= 0.5
5.4
k. = 5 mA /v~
(c) Wand Late doubted -+
rmis unchanged.
factor ,., 1.0
{d) If oxide thickness t,.,_ is halved, and
= ~(IX~
2.5
- 028
12.5/1.4 - .
v,,u = 0.68Volts :=:> 0.68 ~ v,, s
(b) W is doubled-+ rD$ishnlved. factor= 0.5
c(IX
~ St 8.93
1.4
Condition 2 now can be used to lind va.u
With V,.. small. compared to V...
PS
=
Vof',t
5.2
r
ill
v, ""
0.5
1.8
v
Small tim
i0
= k.
( Vos- V,) Vns "" k. V,w VM
Hps = J_ "" k. V,w
l~:t
'vs
c,. is doubled. If Wand L are also halved, r1~
is hlllved, factor"" 0.5
tlos = 2.5V
--------------
5.3
I
The transistor size will be minimized ifW/1.
is minimi:r.ed. since WIL appears in the
equations that must be satisifyed, we. can
minimize (W/1.). Clearly we want to minimi7.c
L by using the smallest fcture siz.e.
L"' O.l8~tm
'Vcs= I.'SV
250pA
I
Vcs = I.OV
125pA
I
-~r::::;;;;
Vt;s ""0.5V
_ _ _ _ _..e:;.+-.,.
50mV
rns .,, _ _ _.:...-_
A;.tWi L)(Vas ·· V1)
I __
'os ""' _ _.;_
k:cw !L)tlav
tim
Chapter 5-2
5.7
p-Channel
(V)
(mS)
(fi)
Vov
0
g/>S
rns
0.5
1.0
1.5
2.0
2.5
5.0
7.5
10
(V)
Vas
0.5
1.0
1;5
2.0
2.5
0
v,v =
(a)
luovl
lias
= - 1.2
""
400
200
-0.7
V. = tJc;
P
· ·'
"' ( ···1.2) -· {-0.5)
133
100
'·'
t>S
= ·-0.7 v.
tJt>S = 11v S ·-0.7 V.
= l mA in saturation mode
:. kp =
VDssat=
= 0.5 V.
(b)fiJr Uav "" V, , Un• = ~(·•- V .
(c) ln
5.5
v.
v..
For
v,. =Vas-Vt =2.5 -I = 1.5 V
=:::> VDS sat "' 1.5 V
= -10 m V, ohmic mode
110
jD =
2iv
= 8 mA/Vl
( 11as- v,,) 2
kp(11aS- V,p ·-
~Vns)(vl>S)
In saturation:
.
lp
=
iv =
i0
!2 K'u
('f)V
L
2 ""
OV
! X~ X {1.5
2
yl
= 39.6
!2 KuV OVz
For ''n
V)1
= (1.125 rnA)
=
5.8
1
)l.A
···2 V, sat mode. i 0
'w
iD "' ikn[(V GS- v,)2
=
I mA
kn "' ll-nCox
for equal drain cun-ents :
2.5
0.4
5.9
·W
c
rorsmall V,.,ooi - k nJ..
-(V,..
·· V,) Vns
~s
(J
q for s~l t.tos; (tr,ob
,.~
:+ "
,.
v~S
tP
11 .=-o.4V
Vc,.5 ::: 'bv
:: --';.....,..-
K~i'lv011 "'
'-
f-fJi11 iD=. ~ f
'{,v. ~S
=IOOo
t:1..6.,-,..10':,.~ v
-..J----::--...,.--
-+{ = o.4+0."+= !.IV
o.lt Otf
.,_
50 X 10 6 X 20 X (5 - 0.8)
'us
=
2J8
n
Vns
r 0 s Xi, = 238 mV
for !he same performance of a p-channcl
device:
W,,
J.L.,
w.
J.lp
20 X 25
=>
1.5
:::'.> j-J?I'
L
~~
I.
50
w
-...!' X 2.5
L
Chapter 5-3
b)~~~¢~$~ N..,b,__; ..~Jr:'&s-VI:
'to= j: K"-1.( C. ~.s
cr !:q!L.f•IO xtOJ(I·,f:JJ.IIIA
-"f
c.) trltn:l• ~.D ;'&.: "o.s VG-$- \
il>., t~" 11~.1~1;6_,10)/(S-o.Bf~~t 9.5tMI\
5.11
fU69
I Ji
I 3
OA
0.4
Chapter 5-4
= 1.91.7 X I0-6 X 10[(5 -
iv
0.7) X 0.2 -
~(0.2) 2 ]
=1,6lmA
(d) satu.rntion region: V~~< > Var v,
i
1>
= !2 X 19L7 Xl0~ 6 X 10 X (5- 0.7)~
'"' 17.7 rnA
5.12
Sat mode, A '"" 0
(k.it.') ""' !2 v
2
C\'
Slope at Vov"" l V.
zvz
IV~
0 ...""'"---+--+--+---_.,... ttov
0
2V.
IV.
Ohmic mode, )1.
=
0
I
,
v., .. l/1.>~ - - v"
(~)
k,, . = " ' 2DS
5.16
htr {M.
rt.M-'.wot ~tS..-\~CAb :
cl\MM.\ U
'~t>:!.>~ "c,s-v\:. ...,.,..
'hs...o..or = ,.s_o.S': o.":JV
5.17
r," =
w
[k.' -.
V 0 v']''
[.
100 ,
50XS·(~·Gs-l)
Mn
I
·nor
-'!..
11
iillt>''
f'
·~' ttos '"' n
For prnos, change
l-'ns ·-~t
i·'ov-}'
V;;s ••· l
V~;s
Ll \!~ r 0 s
II V "">
l.~~o
tJs,c ----
~'os = -.-.--. kfi
IV tpl
~
10 Hl
=
r,,
IOOf!
won
~;;
r,. .. :£. 10 kH
Chapter 5-5
5.19
a) rn! a
~
so if Wls halved. r,. is doubled:
2000::s> rn3 ::s> 20 kO
b) r0 s a L so if Lis halved, r/)$ iulso halved:
50 n s rns s 5 kn
c) r0 s «
~
so ifootb Wand l~are halved,
~
stays unchanged and so does r,.,.
100 n s r 1u s to kfi
11»s
""
"Gs
= ~k11 ( Vos - V,)1
in
5.18
v,
or0
I
lk.
3k.
2kn
4k.
5.20
v00
i0
""
"'
0 =*saturation
v""=v,,-v,
.1'.
V rm ~ "s
I
ikn [(V 00
V,)- r~s]
-
1
0
0 s tis s O'tm- V,)
v~ 2 ( \! rm ·-- V,)
nS are possible.
h.( lie~- V,)2
2"'
+
,
-1.0 + 1.0
CU!-Off
r/i]
2k.
+1.0 triodt!,
··1.0 0
..:ul-nf[
+1.0 sat.
···- r--- r--··"· -·- .........-+2.0 sal .
....
()
---trindc.
--·~
with V"'negative, drain and source an: revers~~d
to show the st•n
v,_.
t' 6
"/
:tt:
A \1P.! \
::: .-.!..::!L
A.
1,.:2..
tp I}"G1 CoN:\- 2..
To CA.Itu.lcU'<... VA > Ct>"' s.>c:U r
VA
+ l.t • 1.-A. x r0
z""•)z]-t
. = 4r
1 ·4W(l
new r 0 = [ 2.Ak. 41_
• "-0 tt. J\.
Note that quadrupling Wand L bad no effect,
but decreasing the overdrive voltage by half
increased the outpUt resistance by a factor of 4.
-\-"'*- 1-.e.c. tn-(\u:
:l,. :to :: 40 IT ..,.,. VA, ?.6,'1/
c
)\ • ..l- ::: o.ol.'8 \f1
5.24
MC$
5.22
). "" 0.02 11- 1 ::=)
V~= 50V for
/, = I p.m
V 1 = t/At:
::=)
= 50V
V'A
v..
1
=
0
~
0.08
187~
,.
kfl
..
'l
'3
0.01
o.o)
o.\
0.00'5'
Vp., LV)
5\\
100
\0
~
rl) <.w\A>
5
3.33
o.t
0.2.
t;, (IU\)
\0
3o
\00
lOOG
ro-= .YA.
)\"' .L
Vp.
5.25
Vas"" -3V V,w;= 3V V,
llu.'
_ A V[)s _
- -- -
r,,
5
I
~-
., 7 _
1.,
:.
=
-4V Vsn = 4V ¥,4
). "" -0.02V
= 1. n I-LA
=
=
-I
;, =
I ·W
'
3 = 'J.kPL(-:\+ lf{l +002X4)
from SO ftA to 82.13 ftA .
Ain
.
2.7 'of change
. .
r,.
111
=
I)
Ai
In nrder to reduce ...,!! by a factor of 2. Ai,, has
ljs
A
to he hal vet!, or equivalently r, has to be
doubled. In order w double r,, V, has to he
dnubled and this can bt• done by dnuhling the
length. L
=
1 X 3
= 6
~-tm
.w
2.16ki'I
-IV
--!'iOV
for VI'S raised from IV to 5 V . i, increases
- 1-=:
4
).tv~
to
forL = 31-'-m: V4 = 50X3 "'· 150V
r0 '~
0
Chapter 5-7
5.26
v.
v.,
v"
a
+2
+2
0
ov.
b
+2
+l
0
c
+2
0
d
+2
e
r
IY,.1
\'w
Region of
Operation
2V.
cutoff
+IV.
ov.
ov.
2V.
cutoff/sat
{)
+2V.
IV.
2V.
Sat
(}
+I
+2V.
IV.
IV.
Sat/ohmic
+2
0
+1,5
+2V.
IV.
0.5V
ohmic
+2
0
+2
+2V.
lV.
ov.
ohmic
""'
pmosV.,=-IV.
5.28
5.27
+2.5V."'f',
dk~ t..T + oi0 1
oi1)i
~f~t,,,,
iJk~ ID dT
dVt t..T
iiV,Io dT
r 'w
-k.-(
l
u.,<- V,) ZJI
? "L. •···
~-
't)
pmos
v,11 = -o.sv.
Vs!l = 1.5V.
Vas~-
:. TJa
0.002 ·""'
c•
V,P => Cuwff
for \'1 "" + l V,
2.0V. =>Cutoff
2
(dk:)
tim> :S V,F ='> ohmic
:.V1;u :s
+ 0.5V =>ohmic
0.5
Cutoff
Sat.
Ohmic
0
"
1.0
1.5
2.0
2.5
.!. If/en ···(·.1_.)(M• 2mV)
k~ dT
4v
V,, = 5 V, If,,."' 4 V,
dT. -~ -0<)()3!C"(-0.3%C"J
k.~
c"
Chapters-a
v,. ... -IS ... )..•-"" ,vs • -?. +\Vtk\ •tlv
~-.... . =~ =')..c;'¥..1\...
:t"' order t.lt G:L\~ ~ t~~tu..
\.11\t.A.orl:' cd ;"' ""'•'d w:\io<. \\1. tl~, Vl:>s k.IIIA
\:o \Ha. «J{Qdl
to "Go:S-"~~:, So~~ ckV(u.
is o~ , ... "'t~ ~ .~ ~na.:ot'\..:
\'"osl• 'S,_\ •1v- . N;,te. ~ s.·. . ca. i 0 ,·$.fW~
Vt;..s SWA.'j5
0..)
~ s.c..-1. •
Rb ~ •
3 ;_l,.. 2.""A
\o) '~.2.•-:tV ->-
N6t& +htM-
v.,.
0
·5' ltJL
1
~
vl):-:l+:t.=o-.,.R0 ....:;,l=o.S
t's F'\o:&d tltt"''JII~k
c..) V'-$ =- ~ V -')' '{5
VG:S,.. S'i".
.. ~ rr 3 'I . Now
f:.o r
V'b$ bo
be -l.\t 1 Vp h,Q.S to be. \\1.
'\:.~
~~ .. 0· S'~ .n.
..
v....., __ '3 v
c& >
-
't. .. '{, •
2.". Aelc~.:rta -tW.. t"'t£,·s~
o..-.d dr~ N\t.CVI.S ~-*. VJ) \lAo bo
5 -1. .. -a v o.nA ~~ \u..~ , " vo\t:.c..ac. u-,p
\.c~ ~f.~ ~
""-
on -\Wl N!Ot ~I'D r •• Rt.fli'A~t
1:"
,., LlL
:= o.lS -.J\-
5.31
In= 1 mA,
V, = 0.5 V, VM = 1.8V.
To operate at the edge of saturation. Vns must
equatv"'
of"cl..r tc:. e»J.c.u).A.G.. -u- ~t puinor
'bJ· ~ ~~ n.ot:le. ~
oJ.Lad.
doa..s..l~
c:l.-w ~~
S,'>'\Gt..
~acJi
t.4.t.'"-~ ,t.h.c.. VoJue of! +~.
~$·~~... is. •• ~ ... ;C*.\·
('/ow we. c..e».l~ ~_,. ,(l<£Sc)~ ~ t"'-t.
ofo\tc.ae. d-~
\CQA.t
O..)
~.ss tW... w~ s.oo~f11l
,_~
v; ,.. '11 "¥
t\M.t'\..
v66 :
= ..!L-2-5' kA
SMc...,.r
1
Ft.
~
..,.-.,.
.,
\15 = 0a_..,=
,·s. ....t
'f)""
L8- o"' 1.s v
v,., = v,.,- v, = 1.8-0.5 = +1.3 v
with V.,. =· V,,. = 1.3 V,
v"'"" v,- v.• "'
Rv "' VIII>- VII "' 1.8- 1.3 "' 500 {l
//)
I mA
'
Chapter 5-9
5.32
5.36
R.;L.5:.. .. 3.64 ~
o.ns
2.
4
O.lls •..I...AOitlOlt)!L t-\·5-t-c.l))-')-W•lf.'I',UM.
c.t
lilA !l.
=
In circuit a: V2
10 - 4 X 2 = 2 V
assume saturation:
11» = 2
= ~X l X (Vr;s-2) 1
= Y(;s = 4V
::::> V,
5.33
= -4V
, VM = 6V
>
Vas·- V,
so our assumption was correct.
rn circuit b:
=\.2-S I<.A
R: 5-3.S
0.12.0
!
ID == I =
'J._}(WI
_.,.. W:2 '*'
X I X (Vas- 2) 2 =::i}
VGs = 3.41 V V; = 3.41 V
In circuit c:
10
=
2 mA =::i} V05 = -4 V
=::i}
V~
5.34
Vc;st = !.5 V
[20 11A =
!
2
w
V5 = -IOX2.5X2 = -5 V
!
X 120 X -1(1.5- 1) 1
In circuit d:
l
1/J = 2 mA ::::> V05
v(1s2
"'
2
v
=::i} v1
VGSJ
w,
=
-
4, 2
-4 V
=::i}
V6
6 V
v
If we replace the current source with a resistor
in each of those circuits:
In circuit a:
I
W2
120 11-A = - X _1 (2 - I)
2
I
W2 = 2
= V6
"'
~-tm
R = - 4 -(-lO)
= l.5V
2
s.,.m
=- 3.0!
kfl
(by looking at the table for I% resistors)
.. - Vr-) 2
now recalculate ID : I /J = !2 X I x ( Vt._)
Vc;s-
5.35
V1 • Yc;..s,. S V
, V6
'"t>s"" son." l!J).S.
to
t:t>.,.
= 0- ("
10 + 3Jll/11 ) - 2
= 8 - 3.01//l ::::}
..-Vt>.s"' o.oS V
=,..
v,
~.
o.oo\ "'""' 1""""'
So
R.c '(p()-"o ""7-o.oS ,.t.t.C\S I<.Jl..
~0
1.99 mA
I
'~~"os { Vc;.-s -"~;; .,..,. t.nod.~ re~:""'"
~ • K~ .!t£I-
((Ve;.s
I,.. loo.tci..!'. Y:L f
-V~;.) v,..yS
Ll
t'S"...!
-
v& ]'
"'-
)~o-o'> - o·~
~
:.
]-::rY:L"" So
L
~ v~ =
2.04
v
Chapter 5-10
3' r -k JlO·Lfx (.f-Ix t-1)2. _,.. :t .,,. !t"""A.
v, = -4,01 v
1:
v • 5-t. s~,.' ... ?,.,,.v
Itt circuit. b:
R
= lO -l3Al = 6.59 k::::. 6.65 k!l
= 10- 6;651
"'!X 1(10- 6.65!-2{~ I
= 0.99 mA
v1 "'
v
2
to- 6.65 xo.99
= 3.41
_.,.
3'- -t·lf v
" .. ::-S' +100)10•0
then
Vas
h) ; =7::,.. o· '-h ( ; - tc.o- :t ),z
:t~"""; o.o~,~
1\btc +hcla.t t'#O.o'IS""A ~~ c.i,~$ h. o-J·~
is ,.,,. c.c:upt'QI>fc 1 hc:c-.M ;f.' f'e,vlb ;......
,:s
fJ..u-
~.(~
~~tot t>~to'tP..U 1 p.ss: ble..
In circuit c:
R
= JO; 4 :::3.01
k!l,
Vag = -(10 +::tOll)
l
=!
2
X l X (- 10 + 3.011 + 2 )2
16 = L99 mA
V4 = 10- 3.ot x 1.99 = 4.01 v
V5
=-
10
+ 2.5
k X 1.99
=
-5.03 V
In circuit d:
R
= !2 =
I k so V 1 iss!.ill 2 V.
V!lw t.ss 11
:s
>'lOt-
er.~1'll\>\c bc.cJ;rMD4 H·
i"' V'Gs <.o ~ l.s Mt
-1>1
1.'1'1
02
po..,sill>kfof' ca.... NMt).s •.
va. .. -.l.s&v
'Tk..f"t!'H>roe
L ,.-i_
f'($()(1:-,
="r
Y.1.-t-S'~,.. .L,.l.(S"-'{-•l...,..
= ('1-V~
tJ
1~
""
""'> l.tN-tl.'1'"•
7 11,..2.lf4V
I
I
~
.. s.s,vx
ikt..s.ot.co~ CA.... s.....e.r i""d1l\ts. ,., ~ .. 5"-S".'S"& 'S-~-J -~-~ -J ... ,.. ~·2.'S, -~ 6)
V) ,€) •>' .l~-') ,..to-~ _..,.. v~ "'5v
®...;:- v, .. <'I-Vs->l.
-:>
~z_q~+-16 • .:>:::o>v,.... ,s.f.i\
V.,.,d:.ss
ruulb '"'- ~t> ,. 6·S"S"..,A
J
ov•d thi.sism•f"
7
Vs .2.Lfs-"
V..) • l.f.'ft;V
f>l-js~ pes$;brc. . .SC,
'vj: I0-2.qS' ... '1:5.S'V
\f•-2-ltsY'
Chapter 5-11
O.S ""
[<":m- I)(YsG- 3)- ~
2
v sG
=:
2
v
Vw
b)
=
IR = 10 X OJ
~
I V =>
saturation
V,w; "' 2 V ~ Vsu = 2 . t = I V
C! R ""
30 kfl
~
IR = :'10 X 0.1
"" 3 V =>triode region
100
=
8X
3.24 - 3 = 0.24 V
100"" 8 X 25 X
""VsG-IR
R "' 10 kfl
:=
X
= lO V ~ triode region
Vso
":w
100 kil~ IR = 100
or
Also,
YsG = 324V, -1.2 V X
25
0.1
2]
Chapter 5-12
5.41
Sine" V01 :;:; Vi>1then Q 1 is in satu~on. We
assume that Q 2 is also in saturation, then
beCause l,. lm, V,..., would be equal tO V«c·
=
V1;s1 = Ym't = ~
=
/1
2
I
V(;~ "" Yas1
I )1
"'
562.5 11-A
2.5 V. SinceQ3 andQ4bave
the same drain current, then
Y(m = VOS( = 2:5 V. This is based M the
assumption tha Q3 & (4 are saturated:
""'
VGsJ = YasJ ~ lz = las3 = lasl
'W(·(vaH- V1)Vos.c- :u~ut
')11
[*•7:
=v,D~ V»s.c + 1600'
[ (0.2.5
X
10-l )40((2.5- 0.5) -(Uti.....bi'.C:
The roots of this equation are 0.07720 &
4.04778
.
Clearly the v., .. 0.07720 is the choice because:
the other one is ~bove V""·
The curre.nt, i0 • corresponding to point C,
= 562.5 11-A
Vz = S-
regiont1
VIJS,c, +RD
= 2.5 V
! X 50 X 12{2.5 -
Parte
Find Yo;, .. corresponding to point c.~. v,_
=2:~ Volts and the transistor is in ihe triode
i, c. is
2.5 "' 2.5 V
Now if Q3 and {4 hl'l\!'e W = tOO 11-m then:
= ! X 50 X 100(2.5-
/2
2
l
l)l = 5.625 mA
or
lfl3 ""
/ql
If\
w,
=
= 2·5
~~720 =
l.514mA
An equivalent resistor value can now he
calculated at point C
~ ~ IQl
10
R
10 X 562.5 11-A "" 5.625 mA
VDs,t _
.
!iqllt\Uiem"' •
ltJs.c
-
0.07720
1.514 x tO-'
_ ..0 •98 ,..
-.,
••
This can be comtmed 10 the \t~lue of rns •
which Is really derived for t'll>' = 0.
5.42
Part a
TtJs ""--:,-....;_--
·w ~,. - v >
k.-r<
Find the R0 corresponding to point B, which is
the saturation-tri()(}c boundary with
1
I
Vvs,s=0.5 Volts
io. ff
5
R
[)
""
2
(0.2Sxl0-~(40)@
s
"'50
(0.25X10- 3){40){2.5- 0.5)
Also on the boundary
'W.
K -11
.L m;,6
5 1.25 mA
25 - 05 s t6oo n
-3.
The wlue is close to the equimlent re$istor '\0lue,
but they are not exactly equal.
Part d
V,;s ""0.8, so the transistl)r is in saturation.
.Find V,,.
l.25xW ·
Part b
Find "GS corresponding to point B.
""""= 05 ~ vi)SJI = Vm,s + v,,.= V; + Vm.e=
0.5 + 0.5 = LO Volts
= 2.5
\),s "< 1.78Vohs
The voltage gain is
A,=
·w
-k.T(V,.,-V1 )RfJ
= -(0.25x 10""3 )
(40){0.8 - 0.5)(1600)"' -4.8
~Mid]·
2 .
Chapter 5-13
5.44
5.43
Q)
Po;"'t- A :
""~=\", \1~ VJ>t> -:::.'5'1/
fc..,.. '{<'-';: , +~ frt~~ns;sf>t,, is nof 0"- >-~oi~tf A is wfvl,... ~ .. ve tMJ +W. fr6c,..ri'Sf-<~,.
"•=
.As
timtS on
~ irtcr-#14e.J 1 #....t ~ .-,..CJ(!6./d
~J ~ ~C$. ~ Jtve.C..S~S ~+INa. r-·Jf/tJ
RD • .2oJC.n. .~
v~o-2.v
='>'
ro,.o.t .... 4
}\,. ~ - 1. VAt;. -t-16 ._ - !:!.!, -> ".,:: t>• tt"
Vov
Vo11
VG>s••·'-V _,.If!. • 1.2-c:t.l.l• tJ.T\1'
.t
It>"' tJC~ ~ ~~~
_,.. o.f ·t-So"'t;3 .!f o.qz.
-
->JlL. ... .2.5'
\,:
t's t.&w 'i ~t~y 't vt~U=S. Af ~ p,,·.. t, 81~
HOS~~r enkt:J -#vi. -+-.-.·,~ "'#'.~"' ~ "~""~B-\1'1:;
it
or
lft>s:\lz.
-s; ,.t..
.. J.. ·"" (
.2.
v6-S- rl
=!>'
I· &l\T ,.,. Vr -::I. fP'
v65 c
e.:
Po;"'t-
v08 •
0
Vp'fi"'.d
J~\16;- '2.l~.s.,. '~
'{, :./." -1:0.61
0
v
II'
,,,\1' v,, =t.(:.t\1'
At B:
o.S"+I • ~
v()S
we. ~ve..~ A"' ...-2. v.. o/v().., ... -_at.v!>I>-~B)
Vov
All""'-:t.<.S-2) =-u.v~
t\.4.
'"fill;
is.:
ry-
-n. •
1 ._
..
gt.tUt
to X,Q-'{,6 :'2.-0.f>\
. I
.u ..•
.J
;...,plie-4 bj """" a-ptt't·vc.es. ;s:
100
~J..A,
'it.....:.,_
i !>
S.3j_ .U ~~ ~IV\.
(.c:..\c.u.l,d.d I.._ f>Ool'l.ttlo).
w \~e tr"~M--Sfu· e...~Jrtf«.
p~,..~~ li(\Uo..r.
W
[_
2 X 100 1J.. _ = SO
!00 1.1. X ((l2)"
(d) VI'D "- fvRn _,- 0.7
5- lOU 1J.. • R0 = 0.7 ==;. R0
fus
't.YuL i'"\~htf
d;~N.V\CL iS
d.WI- to t'\.-L ~ Ct-wJ: -tW.. s.~ ~ 0 f. t£.,.4..
I'\Ot-
c) If I 0 =
oP. ;v... qurpt.J;- vo\~ e s;~"4l tW
g......:...... ::: -.!:li.
,. ~~-'~ ~
o.u
V"
1-o.M_
AV j = ±0.5
o·= + I I 6 mV
43
-· .
In saturation: J0
C'c.Solh iS a.ppro~iMOli\~ 41u~l
!k;s
+ o.5 "" o.7 v
w~le t\,o. ~nsist-c:.r rctoM~i.._ lktb.!mkcn
1\,e. "-""f'lit"lldt.
V'
= 0.2
tha..t Ce:tM lo:>c:.a.flplic..d to
1.61-\.S:::o.\IV"
.,. t-~ •
= Vnv
= 0.2
~V;X43=8VO
,.vs·" tv~.,.,....,., v1 e.,.\·6t~~". "Tlw.A+h.t. \Qr8n,..
G\PApl;tvde. o~ .., .s:"" w<>.ve
= 50 VI V
-->!.A vi = (S -· 0 ·7 ) ""' 43 V/V
0.2/2
""' ot-t t";s pied pl')~n.t , Fro ..... e.t:tVC)h'o.... Y-41'
o. 5'
(0.212)
to allow for ± 0.5 V swing
t\Jaw t'o ce..ICM.lc::t.t'e ..,he. iA(..r.e.wle.ntc..l ~Q•'r-t
C) v1 Q.
"' - - -
Vvsl "" Voslu- v,
v /)$~8
1)'
10
5
J)JJ
-.-,_!ti;:__
( V0 v/ 2)
the gain is maximum when V,.,
is minimum ("" 0.2 V) and when the drop across
Rv ( == I,R") is largest possible. which occurs
when we operlle closest to point B
b) 1 • .1. .. 1 :11 o.Sl...ro.us..... A
Q 2.
~Q ... 5'- :J.'f,-o.tzs-: 2."'
V!Q ::: '16-.S • ""+\Ito
the maximum gain achievable is:
5.55
c.c,.,s,'dtNd ~f't.
Is
·"
43 kH
Chapter 5-14
5.46
for v,s v,s
v,
V., +
5.48
i,, and itn
equate
a)
(![) 1 [vDP-v~- v,t= (~l[v,- v,]l
[V
V
~
=
0
for
/ •
-
"J>o-f\ t~aiO-'b').E.
VI>.., 2..JV
•[V--V]
•
I
/(W!L>J
·~UV!Lh
rg~)
l(i)
~
=
JiO
0.5
-3.16
So w.:. e."""- w r, '1-«
"os- o .S' .f vG-s _,. o.l - "f:.
v05 ~ v65 .,. "·6-o.f .=-.,. v£>5 ~ ¥er -"'·2
(tfSo, F,......, +h.c. e+-N.r s•cl~ ; v05 + o.s~ V00
or
or v05
~
3;o.S" _
~
v0 J ( 2..-S'\1"
Chapter 5-15
5.50
We... ~~··t., ~ c..\~ ~,. lo.w-tst ()o~ra.c~
"l>s ~c..t' 3~ ~ .U.v:u.o~·o,..(ll
'S~v-~o ... ~
Vt>s. "Z>I> _
~ ~.
,,.. .. EQ.
ll'ov
vt>s.,. Yc;.s _o.'L.
Ro\ .,.>
ve7s-o.'l.s '3-S"o..,. :t0
M•'ni.M.UM
"bs
'!1.'1.-"crs
~
S'o
~\so' ~ro- ..., .
o.l.lt
Vt»-•·f
Jl.
_.,.Va.-s•l·4q\l"
-a ... R.o
Av ..
"1·"\-\' '"'" 2. rp
...
Vo-s _..,t:
,z.'l.-VQ:S
So
?
r
os
::: o.l
l F we
't/_ J)
0 co.o'l'( .... ~
-~
vDs
fu ~3' oF- s-Jvrc.Z.·o,..:
~ ~j-"'
Oy
tf>s-r:,._ "'~6$-IC-"f>
-1'\.1 ~.... v6.s ... v,: _v,
~
~
~ pto.."' All" w•l-1.. (i) :
'1.('11>1)-Vo_)
Vo~
VI> (I +
yl,...V 4- ~.
. oy
'
UVov ) ,.. V.ov ~ ~~ _... 2.~
~i
Vov
A.
....
vp .. Yov +""•' ~~~~tO ~IJ'Ifvovl
' + 2. ( 0"(1/Gv)
Yoo• ~v •
v. •
I>
A
:tF
17'
~· • :u> ""'v
""'•\= 1o- Yalt
V•
o.2.+ o.o1.+ 2.llll! to ..o.6ty'
I-+ 2.,. o.t
.,...2 l.o.6~\ • - !2.S.l. Vf"'
< O•l.
tc• too}\
,0.
~V
• o. aV
,.:o;,l.y_ _
z.
;o,4 "' I- '
"" ::. _ 2. ~o "tp
Vov
~po.1..5.l.tr..n..
.,.,,
2.'3. '1...,. 1,. Rp "o. I :r'>
o.-z..
Chapter 5-16
5.51
= 500c:m1 / Vs
= 2!10c:ml/ Vs Cox = 0.4~
1.1.m
Given u.
l.l.p
2
lr' • = u.
,...,. Cox = 201J.AIV
k·
p
2
= I01.1.A/V
Use equations
(5.55 )g., = k'
<5.56>gm =
![ v•.,
j2k'r'o
(5.57) g., = 21o
v...
case
lo
(mA)
IV... I
a(N)
0
@
b(N)
Q)
®
1.2
c:(N)
®
-
type
@)
@)
-
f(N)
0.1
g(P)
0
@
-
d(N)
e(N)
h(P)
I
j{P)
@')
(o)
lr(P)
0.05
/(P)
'@
i(P)
-
w
w
lr'!!
L
(Jlm)
L
(Jlm)
r
I
100
C)
100
2
2
0.7
®
®
0.125
400
8
4
-
Q)
250
Q)
250
5
10
@
-
-
200
4
2
@)
1.41
@
5
0.1
0.141
I
10
0.2
0.2
-
®
®
2
-
®
•
IV,I
10 10
IV.I
2
-
I@
(9
-
(2)
-
-
I
-
-
(4)
-
-
-
IJ)
I(:fa)
e)
-
®
-
-
Note- the eire: led ennies are the givens.
gm(ms)
(mAIV•)
.. See
conunent
I@
I
2000
20
20
125
1.25
5
10
0.1
so
0.8
I~
0.1
0.04
Chapter 5-17
5.55
NM6$;
.,/1 (!f.:r...
&.
ff
,..
"I)
=J2,9'•11 t0\:tt,.~i -0.~
.I
~ • !!t!. • h.!.. =16otu\.
r.,
. ""
o.a
-"• y
•
.2J2tt "'l".sd
0· S'
:::::: • o.::t
.1Jt.,co.3ff t
,....,..)..,,.. c o.1.,..o.&c2..• o.ofl.f w..A/\1'
5.53
, • • !t'.Q
_:I'
V.,v
v ...
~v
2..--·l •
·
o." t.
PMOs~ ''"' =b~t3011fi:s, ,_,..,
,.0 • !YAJ. • \2.:t; 1. == 2J.i 0 II'.J\..
T"
•·I
X. .o.l..
o.~ttv
•
o.2.tt
_.,. ,_.b =o. ,.,.().'2..\(= ~""A/u-
'(, ::: .t"t.J • o.B3V
" o. 2.14
5.56
.
w
1
V =tV k = - .. 2mAIV
'
• "
L
(a) de analysis V0 =
2.
IS
15 V,
= SV , assume
lp = 1 rnA
Vs ""' 3 V, Vcs = 2 V. V0v "' IV.
1() = !k'V~v
= 1 rnA {check)
2
5.54
VI>
=
(b) ro
g., ""
'~: ~::: _,.'ffwtA.fiT +.....-J
.:a..o.tf
'1.}...
~ .. _,,.. Cr;,JIR.t..> .. ..o.erJ(Io,._llt~).:::-B·31v•
Vm>- /pRJ> = 7.SV.
V 11
_
IOOV.,. JOOkfi
II> -
J;A
j2k,1;,
= 2 mS
=
-MAt
Chapter 5-18
(c)
Ru
3.33m!l
Where iN = Norttm's currentsouree ==
l
-l~j
where G..,
=
Ava
Ro
....
R!.
io
lOK
l·· fa·••
-==
~
and R., ·"" R, this is equivalent to Fig. P5.82
D
(d)Riu
AvvVI
R0 == 3.33 Mfl
If the output is shorted, i., = Gm\~ or
"" 0.97
= !.!ll
G.,
-8.2
viR~_=
0
I··
with a signal source and
o
load connected,
. .r~.~=r. m.:Q
R$ig
a .. ~·1
-=
"'="
by voltage division, Vi
.
S1UCC
Vo = GmV ( R o
f
,
~I-
"::" -=-=-
]I
v .. 5R1,
= R1• + R,;g
R. l- ) , substitution for v;
yields
Vo
82.6
9090
=
V,; 8 Ri" • G m.·(R o 11. R,),
sothat
.
Ri,.+R,ig
Chapter 5-19
5.59
note R1• = R1
R;.
ll
+
R~ "'Rl II R/,
1!t
RI
R1+Rf
R 1 +R{
I
= - - - + grnt·•,-,--
it
11
( l - - 1)
g,Rl
. Rf
{ =
11;-
'
=
R;.
n,
''o
J...
, 1
n'tn
I+
'';f I~ n;.(l-
=
RtL
g.,Rf)l
n;_+R1
~
n;
+R1
1 + gmRI
=
Rl
II n;)
=
R211
(R 1 Jl R · ) + R t
= J_ R[~ Ri.- R~ + g,.R 1R1
nf
R.r+
n;
R_,
100
=
n, R1.
stgu
I + gm(R I · Rsig)
Evaluate for
R1 = 100 k!l, R1 = I rn!l, g"'
~,
100 mAN
= 1 kU {R
. assumed)
· Slg
.
· ·
g.,R!.
n1 + n;_
A 1,0
R +R'
n:. "'' --~
__
1.
tn
Ro
,.n;
n'l + Ri
_
R2 (Rt-; oo)
~ =
=
1+ 8
n;,
+~
l
=
.r
R1 +
l + om
o R'
I.
I
+ g..,(R~
R1,=R1 l!R,.,""R ll
l
RJ
"
I
Rt
II
R,r,l
Rr + R. !I R1
.
'.
+ g"'(R 1
'"'
-!OJ)
R11 = 100
1i
R1)
n
/(,is out in half by R1
Given R,;g ·"' 100 k!l, R 1 -+ "''• and
R 1 = 1 mfl
Output resistance assuming R"' = 0
Ro ~· R: fti Rr""'R,
C'r - -".'.o 1'-is
Output rc$iSI~tnce including R,.,
r~
"'"
g.~R.· ~
;ig
1+ ~
Rt
Gv
·-455(R 1 -;·:r.),G~.=
(1? 1 = l m!l)
l
R;..
. n·.I..,.~
1.
+ R ( -- gm
= .50 ktlll6 kfl
= S.36 kO
5.63
WITH R,=tO kO and assuming losse!l due to
g.,= 5 mS
source ilnpl:dance are negligible
G11 = A'I) = -g.,(r.,URoliRt>
= -0.4m11A(S,36 kfii!IO kfi)
= -1.40V IV
F6r a o.2Vpealt output, tbe input must be
ot:
g,.
:. n
= OJ43V peak
-
I
I+ glif R• -
S
.i
= soon
g..,
=
J
.m
5.61
5.64
= I kfi
Rs
-g,
l
R'
L
-g,.R'L
. g
··
b) If Gv =-lOYIV,
R£ "' 15 kfi,Rn""" 2RL
= 30 kfi.
what is gm?
Gt•
= Au =
-g,.(R{)!loo~R 1)
-IOV /Y = -·g.,{30KJI l5K) "" -g..,· !OK
gm=~
v
therefore:
/1) ""
V.v."
2 .,., "" 0.2~_Y. 2lmA/V ,:.• 0.125 mA.
-
m-
for A
2.lc> VA- 2VA- 2·12.5- 100
V,, • ID - V,.v 0.25 -
""
+g,,R3
t.t
=
·-15
-30
I
R""'
•
I ms
"" -lO,Iet R =
.f
l..
Sm
=
2 kfi
)
Chapter 5-21
5.65
R- =
"'
+ I KRout ;a: L6K 2 + L6KR!)U(
2
2K
..l
g,.
400 <:: 0.6Rout
""250fi
ROUI
~~~~
Rli,
n
Gu ""' =
g,(RDII R1)
t'_.;11
R, 1!t + Ritt
g 111 = J2k.fD, so for
..l = Rsig•Km
g,.
+3.3
must
s 667fi
for R4
111_.
_ Rl.. mu
R1.. mlli + Rout
:$ (
LlO)
Rt..nom
Rl., nom +
Rout
3K
:s;
2.4K
3K +Rout 2K +
decrease to 112, and I tJ must decrease to 114
R,.,
Rw 1 s2 kfi
Therefore RL. min is the ruling case and
Rout :S667fi
Rv
5k
g
- 1., = J2k!v2:
•
6670
k.
= l6mA/V2
:. Io a:: 70 p.A
R.;g
I
v
= lip
gm
8V
= 0.093V.
5.67
Source Follower
!u85! :5 50mV
jv,!
5.66
::5
O.SV
RL = 2kf!
R
~
SOOn1V I
''" = g.,"~' '·:::?g..,~ 50m V 2k0
5mS
For low distortion. keep
< 0.2Vot' ~ V
lu.l
,.,,f!
11t.'
...
IK < RL< 3K
I'
1n~-r~ JJ1 t1712
0.6 5mA
in. peak ,~ 500mVpK
2kfl
Rl.,nom = 2K
= 0.25V.
g
=
"~'iO·~Apk
~-
'
ill, rna' "" 0.625 rnA+ 250 ~LA = OJH5 mA
if>. rnis = 0.625 rnA ··· 250 ~LA ~o (}.375 mA
11,;& =
l'p
+ ,., =- 550mVpk
Chapter 5-22
""
VDS ,...5'
a+ +'t-M =- LA- 1t<::>-:U." (3.'-1-'9".2.1:
2.
• - 2...4)
! )...........
:t~ ..,o.c"""""~
5.69
. '
.
I.
:r.... - 2,..,.. A "' L lc.., ~ V.:0 v
v -
~,. .. t-26
'Ins ,..y0 1/
2.
t..
..., >'
2.,. .L"
2.
-3
So.. wx ~
2
.L.f V:"01/
v
e.(::i" of.
tro"od.t!-
fii'Oifo)~~of:. c...t.toH t"os=~n>~ b~-~"'-}"f.
(.'IDS= V~HI) is ..,) \u.f\ Vb$"' ?,o ... t. :ll.
tr:ode o~ .. ,...
Ve>s.: 1'3. 6311
l.
-.r-,.. "s" -~-~-.v •'>''\.• -2,-32.-t\5
1.
"Crs= :l.-3:t
R1 ... 6.3Lf~e:..n.
v.s + "o.s"' -2..32.+ \S. 63., t3. 3\ v... ~ R • 1'.5'-1~-~\
0
2..
Rn ::.o. as KJ\..
V0 •
5.71
~
c"J,.'M
~.r
.t!) =l;: .. 1. t ~.t _
r 0 ,., _
'-)"
v.,_
1:
l.v
» ~..
z
~
rl>"' "• :to-1..\t.o ""'>'I.I>=l""ll-,/.\,_
:t 0 .. 4 ...,,.. re.!>vlh '"' v..~~c.h
\'"' - \" ....~~\ole ,""'-a..-e
~o..- ~ 'et.
rl') "'\.,. 1 t
f-ort t:.D '== \ ""'~ •
So/_ ~t'r , (.c.. II!-'~=~""" "'JVL
4-fb -l.) l..->rt> =- ,_ \3 .... "'
:t 0 i>~'<.NC...S.~S \o'f
1 ~/••
IIU\.
Chapter 5-23
5.72
vG:s,
s-~-~v
,
r.,. ~.&..=2.-~
R,s
t.
I
:S:oc2.• ~,.1,.(.1-') _.,..!1·"'11 .. ~
f. ... .., ~·u ~\-\..
!!! - 2Rs f!RKaT,
iJio =
aK
K
V"
=->'{•l.5'ct
"\.,.. l·'i'\-•·'5"• t.o'\", -
:r0 • ~ "' 2. "'t.s- ~·' -•· o"t )2. """ .r0 • 2. Si ~•
v.s. 2. •.3"\f
4K
ilk
ill"(l
+ 2A Rs) = !_Kt> ~
sfo-
alt>fS. =
aK
K-
iJK/0
I
I
+ 2A Rs
JOO~~oA/V 1 , A:= :!::10~.
b) K =
V, = I V,lt> • 100 t~-A
5.73
--··""""''-..r .,..:.. ,..,.
TO
de.\~" ~ ..
'41-t. 1-.-t p.ss;~o~~
c...s.·..t-.t ~-1
w.·~ cuto~o~.·d ' VI' p-p ,.~-1 J. • '
"o
'Swi~.
vb~ •
Vb--.._ • VJ) -1
V•- vt.
.. o- 2.
-••"
Yp-1 • -1 •> VJ> c-IV
=>l ltp c
19-tj? clltc.A
....
rD=t..r l.[o-t-JotJ,.,_,) -'-] •' .,. IJr(f-~)z.
=---r==~~====~
I + 2 J100 X 10- 3 X 100 X 10-lRs
~Rs
= 45 kO
=
R,c'PKJl.
=
Now find V"' and Vss when 10
100 t~-A and
K = 100 ~~oAIV• 100 IOO(Vas- I)'
~Vcs=2V
Also V.,. • Vs.f -1,/l,
2 = V.,- 100 X 10 • X 45 X lOS
5.74
= 6.5 V
~vss
k = h·~
2 L
C.ForV.u=5V
_
-Vas+ Vss
lo
Rs -
s
- 2+
= 30k0
100 X 10- 6
=0.14
• ForAK = •tO% Alo = ±1.4%
..
/" = K(O + V58 - R51"- V,)
illo
ilK
= (Vss- Rsln- v/ +
1
K
-
' 10
Chapter 5-24
5.75
e>ot\o.
vl>tr
c.a.4U c:..t"&.
5.77
r... ~m.t.:...... ~o"' , b~
t.
0 .l.:~ 'C.0·'-'"-(;V6-$; -l)
Vo-lo-5' .. \c&; v-
o..) , .. ~ .. o.'.i'
vt>S :::
\>)
to=o· 2.-.J. •
f..sr
>. . "t: .
' v~> '"""Go ..
t., • '-~ • 3'Stc.J'I..
"G-S, 'J.V
'l.
.. t''c,.s-') •'>"6-5. .. '!.\1 '"s•-~V
rrv
""
t =d.- , \.l.S" •
1
tvG-s- ,_)t .-> v.e;s "'s.'!V, \Is"'_'?>.?>
vo.s•1"
o.~
Q ...V-
, ... vo .. o.2.( 'i>-')1.->
36
"os=t.:sv
L
-> ~r-i:_~<•·&.ftvD-t)
SAe\c..& Rc,•36 KJl.
v., .. :t.v- , t:D""o.:U -A.
5.78
5.76
Vm; = 0
5.79
a)
10- 1'0
a)--10-
I
- X 0.5 X ( F o - !)
2( to- 2.5)
I
2
2
=-15 V/V
Vr; = 2.7 \l
b .) .I f
1.1
•
·o~
I.~. halved ( Vov "' 0.5) then In is divided
=
h.}. 4 •..
1. e/ [)
=>
v/)
=
:1os v
= 0.125mA
0.5
4
Since VDis kept unchanged at 2.5 V then:
= 10 - 2.5 = 60 kfi.
R
0.25
f)
2lv -
~
Vov
v
~" - - · - 0.5 m
R
"'
I'
'·'·
o
A
V
10
_2 :=!> ,.
=
V(>
°
_
""
15 x 2
l."i_
4 X r 01 "' 4 X 0.5
=
"'' 600 kfl
-30V/V(withoutr;)
Chapter 5-25
c)lf we take ,., futo account :
Avo = -gm~<"t) n R,)} =
-o.s(600* H ook)
= -21.3VIV
R.,.,t "'R,>n r0
cl) R1~
= 600*H ro*
=
54.s ~rn
Ra = 4.7 Mfi
""
R0 == 54.5 kfi
G
v
""'
= R.
4.7
R.
m
RL
A-
+Rsg.
·1. VORL+Rv..
m
4·7
X27.3 X
IS
+ 0.1
IS+ 54.5
Gv = 5.77 V/V
e) As we can see by reducing Vov to baJf of its
value or equivalently multiplying drain current by
4, Avo is almost doubled, while RIM is multiplied
by4.
As a result G,, which is proportional to both
Avo and
R!., is only shightly reduced.
(G, was -7 VIV before and it is 5.8 VIV now)
Chapter 5-26
5.80
+5V.
5K
c~,,
1~
r
2K
vgd
+
•
;.o - o..5
u,+m
:=
Von
s V, = 0.7V .
,;ce max =
...
1.07 V pk
__
max _
v~P
tl_,;g,rnax ·- - 4-.1-
v,= 0.1 v.
Vr; "'
,
IV.
gm
--l'~g1 +g,.R.,
then u~,
V~+
+ 2V
V 0 s = +IV.
ll 0 r = 0.3V
0.5 rnA=
h
2
V 2 =}k
a·
t''"
"'
··" 11 lmA
•
VD = 5- (5 K)(0.5 mAl =
Vun = -0.5V
<
V1
:.
Gr ~ 5!_ =
i!-s~g
( 5 K H ,., :! 5 K )
J ..BmS
~
G; =
v
l;;
=
)00 k!l
.. 4 . !
g,(5K!i5Ki\ lOOK)
+ 2.5V.
Saturation
b) R., = 2(X) K J! 300 K
,.,
v?.
8.! 2
120 ld1
g,.Rl.
l
+ g.,Rs
II'-t.:.·•I =
=
tto + ....::_0.5 s
4.06
~
261 rnvpk
!_ = 300 !1,
d) Add R 1• =
\!A= 50 V.
a) with 10 ~ 0.5 rnA
-
;,,,,max
=
4.06
0.7 V.
0.96 V
Chapter 5-27
5.81
5.82
~)
+2.5V.
':rl) • o.l ~
3:,. o.S,. V zv
0
v
.,.'(, 11 -.oS
~vGs= o.s-.. '"' ,.sv
Dol
-'·~"
x--t~
v(O..o -vs•
~ ... -1·'5 -l-~) ., 35\C.J\,
0·1
$
"t,s • 5'- R0 wO·I
"D!~
..
-S"
~e.$~ p,p;\J\e.
Ro
is o..~&ve.cl f-or Vo,sh\on.
R0
v,P =
-0.7V,
.., --
.,..~ 0 • ~."SSIO.A
~
ftf:.,
•Vov .,.vl>.s-I•Vov
.,.vos~
V•s-Vtr
• ..,.vi)••I+O·'S•I·SV
-2.5V.
tt..,
-
tOM .s\. ·
v_. __..""
a)for 10 = 0.3 mA,jV0 ~ = 0.3 V ·
V.m = l.OV.,Va = 0
Vs = 2.5 -l~s = 1.0 V.
:. Rs = 5.0 kO
o.. c--.......o.,.,-'8.c:IIINL C'cw\.h'1u,...&·o0\ .~• vd~f
2/
b) g., = _!!. = 2 mS
Vov
~
Gv = .;!. = -g.,Ro = -10
vJ,,
:. Ro = 5.0 kO
c)
vsd
+ Vao 2: v,P
= -0.7
-I;+~~+
!V.2: -0.7
• 10
;0
s
R....
c;..,.
v•• max- 01SSV
.
pk
v, = 50 mV, changed Ro
-li•o + g.,Ro
vo I+ (2.S -foRo) 2: -0.7
d) for
11
for g., = 2 mS,/0 = 0.3 mA
-~~~g,.R 0 v, 11 + 2.5 -10 Ro2: -0.7
g.,Ro
(t•si•• = 50 mV)
- 7.88 kO
Ro"'
Gv = -gmRo = - 15.8
~~
1o
£.,.
'1. 1"2.:
tt.s.·a
,...
"o II'\)
,.o.'t .. (o
•
,·s ~nllel..,le.cl ~ ~ ut"c:W:I.:: bc.eo-.s,
a.
C.) I 'fo
i5 grov..d.td, '"ht."
t~ c.i~ 1.-c..co-.es
!>oorc:a.. f-ol\o~er '-"~;aor~o"'.
e.,. 't.lo3
: ,
~.
.. ...r.e.__
'"•+..L.
l(,....t: .,..~ \\ "o '"~lllfao s-
R....t •
e:)
ll'
............. ~
l.too
&f40of'h
o.
.2..&fB~J\..
J:~ "
eo.....---.-
\s ~..d·cl
"'.,;a,
, ~ c..i~'t:lt.u::........... ,
c.o"' t-l~uro.L:o".
~·~- .1..
\\
,....
Rs. "511.\\....L..
.. .2-331A -tV /V
g.,
,,.,R
I'IUI
.l.
=
g.,
!5
=
= 0.2 kH
is a oonunon • gate configuration:
b)
5 = 0·-?
I = I
kfi
R
=
A~
=g.,(Rvl!Rr) "" 5(5Kli2K) = 7.1 V/V
iii
g..
The test for region of operation for a depfe..
tit>n mode MOSFBT is the same as for a ellhancement mode MOSFET. The threshold voltage is
negative; however.
V1 = -3 Volts, n0 =0, t~,t=O Pva.= 0
a) n,. 0. t Volts Pv,. = O.l and tim- 111 "' 3 IU111 11, so transi."'tor is in the triode region
b) tlo"' 1Volts 1'1•..,= t and t1""11'=3 Pv/>11< ll,;rt.'to
so transistor is in the triode region.
C) t>\):=3 Volt.<; F'vy,=3 and ll.,.-fl1=3 l'lu:= Thf-111'
so transistor is at triodtHaturation boundary.
d) TJu =5Volts f'vn, = 51 alld th;-t~ = 3 Pvm > u,,- 't\so transistor is in the saturation region.
c) lfwe connect both stages together, then: for the
R
first stage: Av = A voRL +~
·~·
where Rt is fact R;9 of the second stage.
Therefore: Av
I
=
I
X
~
0.2
+ 0.2
""
5.87
0.5V IV
For the second stage: A tt = 7. i VI V
overall gain
A
v
= A
A
'l-'t
7.1 X 0.5 = 3.55V I
1.~2
D
v
5.84
V
s
Vw;
= v/JS
Vns
<
=
v v,
VGs - V, (always)
First. when V = Vas >
•
is negative so
v,
From TABLE 5.1, this is triode region
in= k~(~)£r "'0.6 v.
0< V,.< 4V.
V,
V,., +
r!J2d·;·-:;:·l\ 8
-
J?o1 ]
for 0 < j/,. < 4V.,
\-',., < V, < F,, + 0.685V.
so 1V. < V, < 1.68 V.
Since t .,.
(i:ji\i.~€~
.
. .
r·-:·:-·-.
an w.:re;~><> nr 4~ tn1
,I
'(
make.~
k:(zX1 v- - v. l'J
(.
~; t
C,, 4x lower. ;md V, !x:.::omc$
IV V1 Vws• = -V
(VD'S' is therefore positive)
Since VGD' =V < V, this is saturation region
(see Table 5.1)
S{}
V(Vl
v;a: Y,
vl
Chapter6-1
6.1
Forl::dO~A:
g
r
=
"'
Vr
=
1T'
,.
1•
o
!\
• 0
6.2
!_ = IO ~tA = 0.4 mA/V
.! ""
Vor
100
T
= 250 ld1
g.,
OA mA/V
Vd ,
I
10 V = I M!l
lO !LA
,.. g
lo
g.,= -,so
25 mV
ID = g.,Vov = 2 mA/V(0.25 V) = 0.2S rnA
2
2
r = V~ = __!Q_y_ = 400 V/V
vr
.m 0
0.025
v
Fromcbapt.
~t.Cox
5, k; ""
sioo: g 111 = ./2t.L.Cox( WIL)Jfn,
For I= 100 Ji.A:
gm
r
=
=
1T'
r
a
100 JLA "' 4 mAN
__!QQ_
4 mA1V
~tA/V 1 )( W/L)(250 f.LA)
2 mAIV = J2(200
25 mV
yielding
= 25 kn
WIL = 40
so that
= ~ = 100 k!l
!00 !LA
·'\, == 4mA/V(l00 k!l)
W = 40(0.5 J,tm}
=
20 11m
= 400
6.3
Assuming that the MOSFET is opemting
above Jl',,
_ V"' j2( 11.c.,}{WL)
Aorr
-.Jlo
If I0 is t]e(;reased !() 25 JLA,
A
Fori"' I rnA:
g
r
r
15 mY
=
Tr
o
100
40 mAfV
is increased by
=
J1i4 =
2.5 kn
1/2
If ! 0 is increa<;ed to 400 tJ.A,
""' IO V ·"" 10 kH
I mA
A
0
is t]e(;reased bv J_ =
•
g,. increases by
~LA
"" 2
so, g., is t]e(;reased by
A 0 = 40 mAN (10 K) ,, MJO
10
~
"'1/4
g,., = j2(p..C.,){ WfL) · Jl~,
I mA = 40 mAN
=
"'
()
0.4 mNV 250kD
I 00 !LA 4.0 mNV !25 k!l
1 M!l
400
I 00 kD
400
~~--jf-----1------l----+---1
I mA
~0 mA!V
k!1 10 k!l 1 400 __
j2.5
•
J4
J4
=
2
!2
Chapter 6-2
Vm) = 1.8
v.L
=
r
v
5V/JJ.m (0.36!:!-m) = 18 kO
100 JJ.A
_!I_ .,.
o
10
A 0 = g,.r0 = 253 JJ.AIV(I8 kO) = 4.56 V/V
c) Now, with a new Wand
V.:w
= 0.25 V,
10 = tOO 11A,
g
"'
__!_g_ =
=
I OO f!-A = 800 J.1.AIV
0.25 V12
V,:w/2
r = ~·L .., 5 V/JJ.m(0.36 JJ.ID) = JS kO
The edge ofthe Saturation region is defined as
when IV0~ = V(;J - !V,I = IVo1i
I
:. The highest instantaneous output voliage is
100 11-A
10
0
= 2 ~·L ""' (2)(5V/J+m){0.36 11m)
V0 r
0.25 V
A
0
Vrw - Wo~ = 1.8 - 0.3 = 1.5 V
= 14.4 V/V
d) 10 is now 10 p..A, first. find k11 :
k = 2lo .., 2( 100 !!-A) ,. 3 mA/Vl
n
2
1
V0 r
(0.25 V)
6.4
L
a) g..,
= 2 (0.18 11m) = 0.36 J.lffi
"" .!.JL "' 10 JJ.A = 80 J.lA
v()v
0.25
2
2
so, now with I 0
v
_~To_
-k -
VOl'-
V,
V/~-tm
(2) (0.18 Jl.ID)
10 ~-tA
2 (5 V/f!-m) {0.36 f!-ID)
Ao ""_V_o_v =
0.25 v
"'14.4V/V
=
=
k
"
Zln
\l1~V
=
tso kn
10 !LA
2 {1° p..A)
(0.25 V) 1
2 0° f!-A)
3.2 mA/V
lO !!-A
0.07912V
= 0.079 V
253 p.,AIV
-r; -
= 2 ~·t = 2(5V/JJ.m)(0.36
o
o.079
v(IV
180 kO
e-m)
v
"' 45.6 V/V
e} The lowest Ao is 4.56 VIV
320 JLV/A 2
Solving for V0 vwith In '~· 100 JJ.A:
when V0 v
=
0.79 V, /D
100 J.A.A,
L = 0.36 f!-ID
The highest A0 is 45.6 VIV
with /ll = 10 ~-tA. V0 v "' 0.079 V
vOf'
If W/L is held cnnstant, and 1- is increased
IOtime.s,
2 ( tOO J.I.A} "" 0. 79 Volts
310 JL"./V 1
,
10 f!-A.
_ ~·L _ (5V/JJ.m}(0.36 f!-ID)
ro 10 ~-tA
A
2~. L
b) with In
lo
.
g,. - V01' 12
so.
5
'
_
5 V/11m
!\
=
_
In
g,. - ~
OF ,_.
100 f!-A
0.79 V/2
=
since A
(J
253 JJ,AIV
+
2V-L
(or since l;m remain~
\10\'
constant., and r0 is increased by L J
Each gain is increased by a fuetor of I0:
Low A0 = 45.6 VJV
High A0
=456 VIV
Chapter 6-3
6.5
6.7
lv = ~k:(~v:~.
W ""' . 211> .
L
, .2
Aj, vov
=
2(100/'A) . z
2oo p,AIV (0.25 V}
=
16
so, W = 16(0.4 p,m) ,... 6.4 p,m
ez; v)
It; _ 100 tJ:A = 800 IJ.A/V
!tm. = v.wt2 -
rtJ
~ L "'" 20 V/Jl,m(().4 !LID) = 80 kll
lo 100 !LA
""'
l.f L "' 0.8 !LID,
W ,:. 0.8 J.LID(l6) "' I2J! J.Lm
g,. =
Va = VpJJ- Vsm
"'Vuv ·-IV'Pl- IVmJ
1.00 :.tA , 800 uA/V
....
(o.zg v)
=2.5 ·-05- 0.3 =1.7V
160 k!l
_
6.6
2(100 JLA)
I LI
- (200 f.l.A/Vz)(0.3V) 2
1=50~tA
2 ( 100 !LA)
=
_
22 .2
( 100 fiA n;)(0.3 ) 2
v'An "" 1!v'.1 p.I
Since
r
•.s·mce A (f
01
L = AoVt!r- = 100(0.2 V) ,
.
2(20 V/fim)
0.5
~m
,~
r
02
o
=•
r
= 100 kH
lt.<.>.t
l'ov
0.3 I 2 V
2/n
2 (50 J-~.Al
(200 f.u\IV 1 )(0.2 V)'
12.5
20 V/p.m (0.5 f.I.Ol)
100 J.!.A
J..l. A!V
.
"
so.
'"~g.,
r0 •c -3(667 J.LA/v)(lOOkf!)
= --33.3 V/V
w ...
l
")
A1.
r - - YovJ = 5 - 0.53 = 4.47 V
Now to Find the linear equation for segment Ill,
we can write i01 = i02 :
Ak:Pn<,,-,,.>2(1 + ~:.)
= !k:(i)}v-lv,rl> (1 + Vo~= "o)
~200(V1 - 0.6) (1 + ~)
= 65X0.53 x(l + Voo1 ~ Vo)
6.11
+1.8V
2
2
2
(V,6
-IV,PI>tl
+Vat~ "o)
200
(V,- 0. 6 )2
65 X 0.53 2
=
1.5- ''oliO
I + !1!
20
7.J(u,- 0. 6)2 = I- tt0 1l5
I+ !1!
20
I- 0.067 v0
I + 0.05 "o -I - 0.117 "o
~
Vo
-1.5 v
(a) If G and D are open, and no current Hows to
either gate,
V0 = VG and 101 = 102
I ·
2
lm = ik.(WIL)j!Vc;-V.r-V,)
= 102
VIA =0.86V
To detennine coordinates of B, note that
v18 - V,. =v08 or V18 - 0.6 =V 08
Substitute in I:
Vc;
-IV,!) 2
or, (VG- (- 1.5V) -0.5V)2 =
(Vc;+ 1)2
=
(1- Vci~vc;
=o
SO,
/02
= lo, =~(I
mA/V 2 )(0
+
1)2
= 0.5 rnA
(b) For r0 = "'.the small-signal model
becomes:
+
v,,
V08 = 8.57- 62.57tr~ 8 ~ V08 = 0.36 V
=
-
(l.SV- VG -0.2 V)l
= 8.57- 62.57(v1 - 0.6) 2
If we substitute for v0 A = 4.47 V, then
I ·
= ikr( WI Lh(V00
=
V18
0.6 + 0.36 0.96 v
Therefore the linear region is :
G
0.86 V s V1 s 0.96 V or
+
0.36 V s V0 s 4.47 V
VI
R
*
lim V,,
*
D
+
Vo
+
v,,
-
lim V,,
*
V0 = V1 - 2(g,. v8,)R
v,,
"::'
= V;,SO
Vo = V;-2gmRv 1
Atr = Vo = I - 2 g R
VI
.,
":'
*
Chapter 6-7
substituting values,
Av = I - 2(1 mA/V)(1 MO) = -1999 VN
Substituting in numerical values,
=
R
(c)
~
r0 =
=
llol
r01
Adding
I Mfi
'" t-[ I - 2(1 mA/V)(I Mfi) J
= 40 kfi
20 V
0.5 rnA
= 24.9 K
I +(2)(1 mA/V)(40kfl)
and r 02 to the model, we get
(d) If tbe gate is driven as shown:
i;
a-
D
s
s
KCL at D yields,
v-v
= 2 g ..
0
- 1- -
R
v,,
V0 =
v
v,, + r___!!_
R1•
V, 11
and since
100 k.O
012
v,.
=
!JR -
2g
Av
=-
IOO k.O
v.
m
Vo
v1
= Vo
R
I
+ 2 Vo so that
R
+ 2S k.O · ( -39.2 VN)
= -7.84 VN
must be i!: IVot~
with V6 = 0, Vest = 1.5 V, Vscz = 1.5 V
.!-2
.!.+.1
l"osl
(e)
ro
R
g.,
= ---
+ R1•• A,,
25 k.O
:. IVot.-1
I - 2g R
= --"'I+2R
r0
= 1.5 - 0.5 = 1.0 V
Given the :t I .5 V supplies,
r0
-0.5VSt~0 :S0.5V
Substituting numbers, we get:
Au = I - 2(1 rnA/ V)(IOOO kfl)
I+ 2000 k.O
40 kfi
-39.2 V/V
6.12
To find R10 , note that
lanF = 3
..
R· =
_ 3- VBE3
23 k.O
a) /REF = In -
!J,,.
/RHF
v0
=
v{' -2g,.R)
2R
1+ro
~30.7
= 0.1 rnA
=In= 5/o
lc 2 =I= 0.5 rnA=/= 0.5 rnA
'
3V
3V
Q,
so that,
R,.
=
~
11
= --.,.-...!R.!.__..,.
I_(I - 2g.,R)
I+ 2R
ro
b)
IVAI
= SOV=rot =
r 02 = 50 = 100 kfi
o.s
50
~
= 0.5
I
= 100 kfi
Chapter6-8
Total resistance at the collector of Q1 is
equal fu
II
rOl
r 02 • thus:
'"'• = 100 k!l ~ 100 k:H = SO kH
r 101
= SO kn
g.,. •.!EL =50 p.A(2) ..,_, 0.5 mA/V
V0 v12
0.2 V
-10 VI V
'o = -(1/2) (0.5
R0 = rm
Av
=
ll
r02
Q 1 _ _ 40 k!l(0.05 rnA)
or J• •t 5 V/p.m
I!
100 kfi
-g.,. 1R0 = -20
0.4 p.m
for Q2 and Q3.
2.5 kfi
""
llol so.
~
]. --so - 2..5 kfi
r,.,-- g.,
20
""'
= 40 kll
lv~jl.
c) g 1 = 10 = O.S = 20 rnA/ V
·"'
Vr
0.025
d) R1" == r, 1
mA/V)
X
100 kfl "" 50 kfi
=
50
-1000 V IV
I.
=
= L,
2
40 k!l(O.OS J.LA) = O.JJ J.tm
6VIJ.LID
'
Since we want L 1 = l-1 = L3 and L he an integer
multiple of 0.18 J.Lm, we choose
L
=
3(0.18 J.Lm)"' 0.54 J.l.ID
(Note: Choosing 0.36 J.tm results in slightly less
6.13
than-IOV/V.)
checking,
V L
= 2L,
5V/~m (0.541.1.m)
01
10
0.05 mA
' 02
rm
r
54 kfi
= 6V I p.m (0.54 JLill}
0.05 rnA
= 64.8 kH
Ali = -g,.0 (rm
II
rm)
·-0.5 mA/V(54 Kll64.8 K)
-14.7V/V
If the gain is to be doubled. and the
For an outpu.t of L6 V,
VS/)2min
!V(
v!. = L8 .... 1.6
• 1
0.2
=
0.2 V,
v
(Lw)
1 .,
~(p.pC<>x)(WIL)V 1:\.
(w)
L
the san1e, r 01 II r01 must dooble.
If r01 and r02 had been equal. this would have
meant doubling L and W, making the area 4 times
L 1 = 0.8 J.tm
L2 ·= 0.67 p.m
-~-2/vL_
J
(p.pC 0
The closest integer multiple that satisfies om
,)(Vvvl
2(50p.A)
requirement is (0.18 p.m)(5) ""0.9 JLill.
so. with L 1 "'L~ = L.1.
29.1
! 86 !LA i V1 )(0.2 V) 2
5 VI p.m (0.9 fl.lll)
0.05 mA
ForQ 1•
(I.w) , '"" ps7 ~-'A2(50;vp.A)
)(02 VJ
1
2
A1• must he at least • 10 VI V,
and
111'
= -~ g,,( rO!!! toz)
If we wam to make r01 and rm equal.
Ar ·
!\(\,
1,;
... l,
ratios kept
greater.
Foragainuf ·-20 V /V,
Since Im "" lm = lm = 50 p.A,
and Ill=
I
r
2'~.-~~ "
A;-
·::jJI'Re,t
6.46
90 k!l
Q~j f!.lll
(0.9 fl.ln)
0.05 rnA
This result.• in a gain of
Ar"' ··(05 mA!\/)(90 H1
AF
'"
-
Ln kfl
!I
1.n k!l)
26.8 vI v
This rcprci'ents an increase in area of
JL'Z-.) 1
(054
'''
2.78
(instead of 4)
.
,
Chapter 5-9
2WL = 2(0.18 J.Lm)(O.l8 J.Lm)n =' 0.065 n
AS$umi.ng that the driving NMOS tmnsistors bilVe·
similar g. and R0 ,
6.14
K=40=g2r
.·
Si)
"' ·tn
.,._!L
J'()y,z
Y.t
= -~g..,Ro
Ar
diat
= .K'Ylw
=
2 .
40(0';iv) = 4 v
2
Av = -4(0.1 mAtV)(810 K)"' -40.5 VIV
rr r'= s v tJ.Lm;
"
L = Y.,
y~
=
For /, ,.. 0.36 p.m:
2
l
= 0.8 p..m
4V
IR .,; 2(5 V /p.m) (0.36 J.Lm) .,. 32 .4
S V/J.Lm
0
R
v
(0.2 V}
= ..B:1_Y.. = 3.240 kll
0.01 m.A
0
Hm .remains unchanged
Av = -i(tU mA/V)(3.240 K) =-162 V!V
6.15
Area= 2 LW = (0.36 1J.In)2 n(2)
= 0.26 np.m2
For L = 0.54 Jl.rn:
2
IR = 2(5 VI p,m) (0.54 J:A.ID)
0
R
0
2
(0.2 V)
=
A.=
= 72 .9 y
J!l!...::!_ = 7.290 kO
0.01 rnA
-~(0.1
mA/V)(7.290k0)
= -364.5 VIV
V0 v3
= Vov4
Area = 2(0.54 n){0.54)
Now, use I = 0.1 rnA:
= Yew
.
,44
A
J~·\L
=
g
g,...trol
=Am ... !"
·
r orrZI
(0.2 v)
Ro = fRo =
I
A,. =
so that
2jvj}L .lv"iL
Ro::: !Vorl
f-Inally. fRo ""
Now, ~~i1
""
z!v. r£}
R
~-tm, Wov\ "" 0.2 V
For L ,. 0.18 11-m:
.
IRa .,.
2!5 v 1
e-m>
'
(0.18 !J:mr
(0.2 V)
·- IR, ,. ..J.J...r_
Rn ~ -T
0.01 mA
·-
g.. -
rnA I v)(81 K) "' -40.5 VI V
=
= 8.1 v
0
=
810 k!l
0.1 mA!V
2(10 n)(O.I8 JLrn) 2
!lf,l-111 2
= 0.36 ,._m:
32.4 V ~ 324 k!l
0.1 mA
A. =
!
Area
= 2 WL
2
(I mA/V)(324 K)
= 2.59
""
IL ., JU!.L.!!!~- "'
(0.2 v r2)
w(;r112
-4(1
= 0.65
For L
lVl
, Ot1
~
8.1 V , 81 k!l
O.l rnA
Area = 2 WI.
I
5 VI
2
(O.l mA)(2.) = I mA IV
"'
lv,.iL
I
Vo4 =
1 •
Smce 111 = - kr (WI L)V1) •
2
"
WI L will be ten times larger (1 On)
V.=V,:V,
Al
0.58n !J:In2
L = 0.18 J:Lrn
=L
L4 = L 3
=
-162 V /V
= 2( 10 n)(0.36 J,trn)Z
O).L012
For I. ·""'- 0.54 1-llll:
Ro ,., 72.9 V ..... 729 k!l
0.1 mA
=
Chapter 6-1 0
L
=
L..1n
=
0.18JJ.m
IR0 = 8:1
g,..
Ro
mAlV kfl
L= 2Lmio
v
IR 0 =
= 0.36 J.Lm
32.4 v
L= 3Lmin ""'0.54 p.m
IR 0
= 72.9 v
A~?
2WL
g,.
Ro
A,,.
2WL
g.,
Ro
A."
2WL
V/V
J.l.ltl2
mAN
kfl
V/V
J.l.m1
mAIV
k.fl
V/V
J.l.llll
-162 0.26 n
0.1
7,290
-364,5 0.58n
0;1
810
-40.5 0.06.5 n
0.1
3,240
1=0.01 mA
WIL= lOn
1.0
81
-40.5
0.65n
1.0
324
-.162
2,6n
l.O
729
-364,5 5.8n
l=O:OlmA
10.0
8.1
-40.5
6.5n
10.0
32.4
-162
26n
10.0
12-9
-364,5
/=O.OtmA
WIL.=n
58n
W/L=lOO.n
A. = -!(1 mA!V)(729 K) = -364.5 V /V
"
2
Area = 2 WL
=
{2)(10 n)(0.54 JJ.m)2
"' 5.8 n ~-tm2
=
Now, for I
For L
=
1.0 rnA.
0.18 J.l.ln:
I mA(Z) = IOmAiV
(0.2 V)
Ro "' S.l V = 8. I kfl
I rnA
Av = -!(tom:\ !V)(8.l k)
2
""-40.5V!V
Area "" 2 WL = 2(100 n)(O.l& 11m) 2
= 6.5
n f.l.ffi 2
For L = 0.36 11m:
=
32.4 V
=
A •.• =
-lno
mA/V)(32.4 K) = -162 V /V
R
0
'
l rnA
2
2 WL = 2( 100 n)(0.36 J.l.m) 1
Fnr L = 0.54 p,m:
R 0 = 72. 9 V = 72..9 kfl
I rnA
-~(10 mA/V)(72.9 K)
= -364.5 V /V
L
=
=
= 26 n p,m 2
Av =
~ = lOOn
Rur
Area
'P4k!1
·-·
Area = 2 WL
=58 n iJ.ffi~
The table
= 2( 100 n)(0.5 f.l,m)2
summari1..es the calculations.
Comment~:
(a) R0 and A v are increased hut at the cost of
larger device area. As 1- increases by a factor of X,
A 1• and R0 increase by a facwr of X2. The
device area increa~s at this same rate.
(b) gm increases with III. but R 0 decreases with
l
m·
The device area increases with ill.
(c) Smallest area= OJJ65 n ~-trn2
Largest area= 58 n p.m 1 Gain and g., have been
increased, l'>ut at the expense of increased device
area.
Chapter 6-11
6.17
6.16
2/p
g.,, ""' .,--.
•ov
!10,
Io = g...,Vvv = t mA/V(0.2 V)
2
100 v.A
2
Ra = (g,.zrm}rm
However, if we make g., 1 =
T
Since all devices have the same VA and ID•
'm = 'm = 'o3 = 'o"
Rml = g.,
=
and r01 = r 02 = r0 , we can say
that 400 kfl = I mA/V ·
r~ =
rt
~
ln
= ....!.Y_ = 40kfl
0.1 mA
Ro" = g.., r~ = (0.8 mA I V)(40 kfl)2
·"" 1.28 Mfl
400 k? :::::> r 0 = 20 kfl
l mA v
.
10
100 A
Vo:
= (0.2$ i)/2 = 800 JLA/V
=
R"P = g,.
VJ.L
t:,
= L28 Mfl
= 640 k(l
Ro = Ron H R0 p
smce r0 == / ; '
Av
=
100 JLA(20 kJl) "' 0.4 !LID
5 V !JLm
-g.,,Ro
-800 v.A/V (640kfl)
-512V/V
I
Vo
Va2
o--t
6.18
Since Av=-gm 1 Ro
Qz
Ro
o--t
Q,
=~ =
-200
-g,.,
-2 mA/V "" IOO k(l
If all have the same I 0 and l'~. and
+ V·I
since R 0 = R.,. R Ror• and
g,.
= J2
JL. C,, {lV l L) ·
JJ;,
so that
gm,
~ Xm2 ~ RmJ = 1:,4 ~ g~,,
solving for r0 , we get
( IOOO !LA/ V)1
= 12.5
2(400 JLA I V 2 )(100 !LA)
For maximum negative el(cursion at the output.
we want the MOSFETs to be biased so that each
transistor can mach Vus "" V 0 v
v,. + \''0 v + V,w
0.5 + 0.2 + 0.2 = 0.9 v
:. Set Vm =
=
=
minimum output ''oltage will be
2 V 0 v = 0.4 V
2( 100 kif) 2 mA/V -
1=
g,!Vm·l
2 mA! V(0.2 V}
=
2
0.2 V.
=
10 kf}
2
0.2 mA ," 200 !LA
S.mce r0
"'
lv:iL
-"-.
In
]0 kH(0.2 mA)
5
v / J.l-01
0.4
~m
Chapter 6-12
In general. circuit (c) has a higher output resis-tance, and for the same V0 v of transistors it has
Since g, = ./(2 p.A C0~)(W I L) · ./f;
(~I "" (~1 ""
lower output swing. The output swing is limited
to 2 Vov on the low side for circuits (b) and (c).
but limiled to only V0 v in circuit (a)
8
2 ~t. ~,,,10
(2 mA/V)2
2 (400 f,LAIV 2}(200 ~A)
25
6.20
Similarly,
ForQ 1,
Vov "" VI - V, 0 = 0.8 - 0.5 "" 0.3 V
(2 mAIVt
100
Since all transistors are identical, and
k.l "' k.: = kp3 = kp4
with Im "" 1112 = lm
!Vo~
= 104 •
= 0.3 V (since 10
"'
!2 k!Vol·)
with V02 and V03 fixed,
6.19
Vs2 """ Vm- VGs2
1.2 - 0.5 - 0.3
0.4
v
2.1
v
V:n "" Vm + V(;s)
Forsame /:
For same /, if
I
Vvvh
V:lva
1.3
v,2 + V0 11'2
=
0.4
+ 0.3
0.7
is divided by 4,
V:w is multiplied by 4, or equivalently
v is doubled g., = p..c.)[ V Thus g,. for
V,53
-
V0 n
= 2.1
- 0.3
=
1.8 V
so the output range is 0.7 V to 1.8 V
01•
\ 1on
= 2.5 V
circuit (b) is half of the one for circuit( a).
VA
2V~L
.
X - = -,-· . Thus. 1f
fv
\'ov
Lis multiplied by 4, and V0 v is halved, then Ao is
A 0 = g,.r 11
21
0
= -,-
'-ov
Vm = 1.3 V
o---1
doubled for drcuit{b}.
In summary, for circuit (b), Vo~· is doubled, gm is
Q1
halved, A0 is doubled.
(b) Each transistor is circuit (c) has the same V0 v
as the one in circuit (a).
Vm
A.vo =
G,.
2
-tlo "" ----(R,.ro)
= g., 1 '=
V:w
as the one in circuit {a).
8mi .," gm (satne as circuit. (a})
Note that for the transistor in {c), the f?m nnd r,
are the sarnc as those is circuit (a).
Thus. the intrinsic gain for circuit (c), A,.,= ... A~
where A0 is the intrinsic !Win for circuit (a).
V;
,.0.8V
g., (same as circuit (a))
doubled.
(b) Each transistor in circuit (C) has the same
~,
o---1
1
Note that for the transistor in (c), the gm and r0
are the same as those .in circuit (a). In summary.
for drcuit(h), V,,,. is doubled, J?m is halved A 0 is
G"'
v
The highest V 0 is
then
V0
+ 0.5 + 0.3
The lowest V 0 is
Ql
Chapter 6-13
6.21
+
Vo
,------+---D
5 V !V
--<:2--__,l...._..._o
i··
+
Vo
v
F L
II' dis shorted to ground, the current Howing
-·1~~-
lo
through the :;hnrt is
-~£iA-~--R,;~.
"'
+ R;,
r0
L?~Y..L!!:!!!li9-36 i!'m 1
200 JJ.A
••
9 ldl
Chapter 6-14
This raises 1<0 p to
I '(W)···· ~
Zk•
T V ;VI
In =
R.,P = {g.,3r03)(g,4ro4)ros
F>
Vovt =
2 <200 p.A)
400 p.A I
g,;l
=
. = 0.26
v
vzf 5 ·4 )
Setting R,.(%.9 ).
R1. + (96.9)
~0.36
= 200
e-A
0.26/2
lm
Vov
= (0.769 rnA f V) 2{9 kfi)J "" 431 k(l
We can now find an R1• that will allow a gain of
-100 V /V: Since
431 kfi ~ 12.'5 .kfi = 96.9 kfi
=
!.54 mA/V
2
100
0.52
Rr. = 197 kfi
To find the gain of the CS amplifier, we calculate Rm:
Rm = [(431 kOft 197 kfi)(l.5 mA/V)
(9 kfi)JD (9 kfi)
~tA 1 v 1(1A)
0.36
ROI = 9 kfi
v
so, An = - g .. 1R01
lOOp.A
-769 1L A"
.,.,3 =~><>m4 = (0.5
I v
2 V)/ 2 "
"' -13.9 V /V
Ron :::: (g.,Jri1J)r0l
= (1.54
mA/V)(9 kfi) 2
= 125
k!l
6.24
Rop = (g.,Jrm)ro4
= (0.769 mA/V)(9 k!l) 2 ""62.3 H!
R.,
=
R,r II Ru•
= 62.3
w
The value of R0
R f)
II
R l --
.
65 kfi, we get
Rm "' (Rop !I R~_)(g.,Jro3) R rm
2(200 p.A)
=
=
n 125 kn "" 41.6
II Rr needed is
t1v -g,. 1
-- -
w
-100
1.54 mA/V
""
64.9 kO
This is greater than R 0 !
This ciln't be done with the present design.
One thing we ~'Ould do is double cascodi: the
current source to raise R,)(>:
b) / 1
=
V;g,.
By current division,
+Vno
also,
i3=.~
"o + RJ
= ( 1.54
rnA I V)(9 kfi)
Chapter 6-15
c)
V1 ""' - V1g..,(r0
II R3 )
'-Vig,..(ro)(i.. + ro)
-=---,.-~~--
'o +-+'a
g..
-V;g,.r0
I +.....!5!_
l
-+ro
g..
""
-V1g,.r0
l
-50mV
+-··~~-:::
I+ _l_.
g.,ro
V2 "' v,g.,[(g.,r~) U (g., 11 r~)]
i
.2
= 2(g.,ro)
::: - 1I v,g.,r 0
d) V1(f)::: -
1l2 (I)
50 mV
~ V,,i1.,r0
with V1P••< = 5 mV,
Vtpea<
VJt)
= -~{5
mV)(20) =-50 mVpeat
c~ -~ V,(gmro)l
Vzvcuk ""· -!(5 mV)(20) 2 =
·I V pe>k
6.25
Since I 0 =
w
~(~t"'C,)(r)1Vovl 2 •
2!0
L
2(100 t:J.A)
Iw ,~ so
(for all transi smrs)
I_V1jL
'I
r"
c .•.
f0
=
=
{6 VI fLm)(0.18_J,i;Ol)
100 flA
!0.8 kfl
To permit the maximum swing, each Vos .
tliiJl
should equ;1! IV 0 .,J . Sn.
Chapter 6-16
Vat = Vnp -IV,PI-IVo~
o.s - 0.2 "' 1.1 v
= 1.8 -
=(1.8 -
0.2) - 0.5 - 0.2 = 0.9
v
Rinl
=
R,,..
1.. + r,
g.,
= Vntma•- W~,,I-IVo~~
Ym
= (1.8Ro
d) If R, =
Vm mtx -jV,pi-Wo~
r(/2 ""
0~2-
02)- 0.5- 0.2 = 0.7
v
=rm(g,.lroz)(g,.,Jrm)
6.26
a) Assuming that all transistors have the
samegm and
3( 10 kil) + __
2_
r~.
ZmA/V
-129 V /V
= '•
Rm
H.,,""',..
H.,,""' (11m3 '•3 ).
Rm
ll
R02
= g.,roGro) = ~g,rz1
6.27
~"' 50,
VA "" 5 V, I= 0.5 m.-\
If the base currents are ignored. we can use the
same r 0 and 8m for each transistor.
=
I
z
J.g.,ro
c) If Vo is shorted to ground.
R-~
'"·
=
_!_+_Q_ = _!_
lim
gmrt~
8m
Using current division.
Rm
=
!J.
= 0.5 mA
\'I
25 mV
5
v ,
20 mA/V
10 kO
0.5 mA
R,,
( 20 ~ 1 A)oo kflH 10 kll !I 25 k!1)
R"
400 ld1
i,
Chapter 6-17
6.28
6.29
If the transistors are identical,
= roz
ro1
8.,1
'•I
=
=
= ro =
R =
o
c
= g,. = !fPT
8m2
'•2
= '• =
Ro
R =
•
~~AI
~
r
0
=
~~~[
=(g,.zr02)(rolll
(l!£1
~)(~ II
Vr'IId lid
[
PVr
lTd
lTd
with/c=l,
•
fRo
50
rnA.
rnA)
.5rnA
=
If T 1.0
= 400 kfi
rnA.
Vr
II
.1!. =
g.,
25 rnV
__!QQ__ = 25 kfi
4
rnA) =
I rnA
rnA IV
R 0 p)
100(1 Mfi) = 100 Mfi
R00 = IOOMfi
= 200,000 VN
g
"'
"
=
~
= 0 ·5 rnA = 0.02 A IV
.1!.
=
Vr
=
g.,
25 mV
when R. = 0,
100
0.02 A/V
= 5 kfi
R 0 = r0
a) For R. = 5 · r ••
R = 2 Mn(.l
o
=
200 kfi
5 =
[I + g.,(R,II r,.) )
+ 0.02 AN (R, II 5 kfi)
5 = I
= I Mfi
=1!£1=~
r,. =
-g,.,( Ron
0.1 rnA
so.Av = -4mAIV(IOOMII lOOM)
r
R0 = 2 Mn(· 1
=
Av
.!QQ..Y.
=
R 0 "'r.ll +g.,(R,ii r .. )J
. _ I _ = 2 Mfi
~+..!.. O.lmA
If I= 0.5
"'
T
6.30
5V
5v
=g
~
R•• = (g.,zr.z)(r.lll r .. z)
For IVAI = 5 V,l3 =50
lf/=0.1 rnA,
"
"' 2
= r0 =
= (4rnAIV·I Mfi)(l MOll 25kfi)
=~~ =~.
I
Vr ' 1 + 13Vr
Vr _1_ +...!...
IVAI
13Vr IVAI
IVAI
- + 0113)
R =
=g
2
Rop = 13r.
lid
IR = ~[IVAI· PVr]
o
Vr IVAI+ PV1'
IR
"' 1
0
= 4 mAIV
r. 1 = r,. 2 =
PVr)
~ Wcf·wr ]
Vr·~+
g
r,.z)
= r
1
Chapter 6-18
RR 5k!l=-4= 0.2k!l
•II
0.02JVV
""'' .
R, · 5. kfi _ O., k"
.,..,Vlng, R. + 5 kfi - .~ u
R
s k!l(0.2 t!l)
=
4.8 k!l
t
b) FOr R<~ ""' lO •
""
since g,;,r"
t-.:&.
loa n
g. "' rI) • --R!3ro
1 +...!
r.,
There are several ways to derive the equation
for Gm.
Method t:
Take the basic small-signal model:
A VfJ --
r.,,
10 = I+ (0.02 A IV)· (R,U 5k)
9So durt R, . 5 Hl .,. - = 450 !l
R. + s k.n
0.02 JVV
Solving, R, = 495 n
c) For R11 "" 50· r;,.
R,
=~
-
c..,.._;,
B=;t
so ,. 1 + 0.02 JVV(R, n s k!l)
R,·Skfi
R, + 5 kfl
= ~•
+
""2.45kH
V;
0.02 JVV
v'"
r1r
= 4.8 kfi
E
+
6.31
With the output unloaded, the small-signal
model can be drawn as follows:
c
B
Note that V, = V1 - V,
+
V;
v1F
+
ro
r'"
Vn
io =
'o'
'::"
+
E
v,
R,
l
~
0ro
v, + 8m v"
- -v, + l?.,n ("'v; ro
Assuming that 10 >> i 0
V, :: i 0 R, Then.
-ioR,
in = - l'o
. Rtllm
+ gm Vt + lo
Since no current flows out the collector.
V0 = - g,. V~r0 + V, By voltage division •
V,
..!.l!.L
r.+R,,
and V _ "' V;r ~
.,
r,.+R,
substituting. we get
V" = -· g,r0 r" + R,
Avo=
Avo
•!
~,
r ~ +R r-
=
dividing by r" .
I,
-gmro· ···---! t
!!!
~'e
v)
,.
since U,.
- << l usuaII y.
ro
Chapter 6-19
Method2:
Con&ider the model
_.,&_
-gr~
1110
R
t
l +...!
r,.
1,.. -(S mAJV)(JOO kO).
R0
ra
"'
t+--
-
12.5ldl
Avo=
+ (Re II r,.) + (g..r 0 )(R, H r,,)
or
-784 V/V
G ""
g,.,
.. - 1 + g,.R,
=
r0 [1 + g.,(ReU r.,.)]
R0 -
0.25ldl
lOO(lOO.kO)
0;25 kJl
8mAN
l
+ 8 mA/V(J2.5 kfi)
2.67 mAIV
ShOrting the output removes Rr. from the CKT.
"r t -R
.
-A,,() = ""' 0 "
' (from part. I above)
r.,+ R,
G ..,
g,.,r0 r.,- R.
+ Rt) + g,.r0 R,rfl'
Dividing by r0 r, , we get
R,
~
r 0 (r#
g,.--;-;a.. = .......,,...-.........,~o-'"::..,_
r0(r,
+ R,) +.
r0 r,.
g,~
R/!"
. r +R
R
smce _!!.___.! :::.1 and-' <<
r-,
'o'"
G
g:,.
-
"' - l
~
with
g,,
+ g.,R,
= 100, r0 = 100 kfl,
lr "' 0.2 rnA, and R, = 250 11,
11
"'
= l.&j =
Vr
r:,. = ~ ""
g.,
0.2 mA "" 8 mA/V
2'i mV
100 = 12.5 kfl
8 mA
v
R0
.::
r0 + (Re II r ,}{I + g.,)
.::r11 +r0 g.,(Rei! r~)
R0
::::.
100
Ji
.l
l'ov;
2
Rin =
~' () 'l "/'l
-~
r, 1
'·"
"
lm/\JV
~
25 kfl
Cl?m;r01 )rm '" (! nv\/V)(.'iO K) 2
2.5 M!l
4 m/\JV, •·, 2 ·= 25 k!l
gml
Rje
=
Y;;
Ro
=
(g,.2rrnHrm
II
r wll
R 0 = (4 mAJV)(50 k!l)(SO kfl II 25 kfll
·""' 3.33 Mil
..... GnJ?o
Avo
Avo'" . g., 1R 0
.... 3.:13
X
10' VIV
-I mA/V(3.33 MO)
Chapter 6--21
ConuneniS:
(l) A MOSJo"'IT for Q1
6.35
~~'~»='·•v
miikes R;. -4 ""·
(2) The output re.~stance when Q 2 is a BJT is limited by r:r1. lit cases (a) and (d). R, was higher
due to the value of ra and 8m2·
"os, v., _vt:.
"o--... •Vos ••"''-'•="t;.~-Vt*7
5
(3}111 these four cases, Avo wa~ highest wilh two
IJJTsA,,
lowest with two MOSF&'Ts.
These results could be changed with different
biasing.
was
->-'{;, '= \ ·8'- o. 'l::::: t"
R..
• L.:.o.s_p... '""'"" S'J'"" , vt. = o. '1' 1/
.,.,to..U...
Rt:
"* 1. 8' _o..,. = o.oso
,. "<>Q-"vs
~
~
-;;..'1
r
N.
R = ..22\C.ll..
Gl,ow..d Q~ ""'~At. -t-M. s-..-.4 Vc,.s .11i.t.lo~sl- voJIAL
oF \{,
or Von.
"*' "" o .'1 ....o .'5 .. o.2.1/
~....it "' o.l.v
Jtertt41
'0,.
,·s wMn. lll>S:::: "-s _
YA. .. VA\..
:&C>
.,. 2.0)<0. c;
..
~00 IL.l"l..
o .OS
I:e>
/>. f"c:oct AIJ'.., -::: ..L_.,. 5 LIA -> A!0 : S'j(A
r..
uot(
; .
6.34
?~~~,. .. ~5o)''Yv"'
1
~.,_!lt:>~"'
~"' 5/..,..:>~"b-:::.S.J'A
to
.
,.-0
::::_!!::!e. ::
ro
= '41... =>-
,._to
1. SS"
<;;P
L.•
t:o
=
31 o
h.r
,
'\=o.6v
.t>~•I·'B-o.'JS::!·s/
"-5\.
I:JY' .!f- '"'0'.1 ,..~:d·'SS'}!...,
2.c .==-=-•
VI\
y
-"t:. .,. o.ac; =>v6S,.. o. 'l.5+0.~::o.!';
V0 _.,"'-.,.. v65
R= Vet>- V~S-..s
.,. 1.
ED
r-o .as- ... q,.,.-._n..
o.1
_.,.w.!"t 't't-
r,;:> .. .L ;., C..)' 't:lL\VG.s-vet
::z.
.:="7W:::
.._
!ll!: t.S'!>w;IOO .
:
:L'5'o (O.'tS'-o.£)'1.
2.
1<(. I>:S-Vt)
\9.a-4f'-
=..!.,__ :
:2.
~,.1.'3'0)i~l
2
.
v
5 0 ,. .l.-,.2.S'o~<5:..(VGS-o.r;)..,.>VGs•o:1V, o.'l
:1.
o.")
f.o
Y.!!:
I 'l. .'!S tt..Jl.
r 0 o.oJO
l:o ..:..1-J;~ (.VcA -vt:.f YL
6.33
v~:o.~v :
. .. ·:t-·
t;...C.V(;s•"~::>
....,
Chapter 6-22
6. 3 7
I
·
Vest
= "vc;sa
(WILL
So 3 df'l!kteJ' PJ-p~ ~ ""re p•S.Si/1/c.
Jept-J,·, ,n w""'h h.uo +r-s/r:lr>rs A~
T,;; - --'"and
(W/L)t
50 that En -
(W/Lh
loz "" [Rfli'(W/l,)!
cJ,'#t:k _ tp,h4-cnJ •
ln2 "' lm
(WIL), ""'" lv 3
Vos3 "" Vc;s4 50 that ~ - J,14
10
"" /p4 "'
6.38
11= tkL
f:a."
!3
too ~C &!a
e fOO,!C
!tQ.
1ft
to
If
.L;c. lo;t. J!L( "S&- o. '1)
.z
I
.,..,.
Ct;t,(-,3" ~·,, .
W: II> i$ dt'oJ&-t:.o>U'I~chJ_,
#?IJOfiA
tF- tvt trDt...,d:.o-o.,. witJ.
11'1• 11.0
,·s
d.~-c-.....uhd
'I2.."' 100,. !.1!.. ,.. So~ A
tkl."' :
~Gr:
z
:l.oo)IA
:o
tO
we CA.JtuJoJA..
Vs&o"' I
0-11 "'h"-c -tiM s-. ".sc.. f.,,. e-1 aiV'It...
(W/l,h (WR.)4
lR!il'(W/l,),·("Wii:h
~'.ffo.r w,'tk.
tNM.:
f~o r::.
lm
In4
_
-
H#r.>
6 •39
. tllc small-signal
model can be drawn as tollows:
.2.0
~.,.too ,. ~ -.2..ooft/lo.
,26
t~ -tt.... b--_,;stor wo~ W-::.'"1o
~:
t.. "'
too'!(
....
,·s
di6tlt:-c...,.,.uJ~J,
a'"'\0 . . :J..!';r/>.
X:.l:.'\OC> IC ~ ,..
Lfo
f!;;ojt,..
+
$.,. for CA.s~s #vJ:oH'1 ~
t.ol'\,..u.h:c/, 1
+rc<+tS"t"Sior
,·$ d.i>c:lc
+
v,
~ Jd/;~rt.....f: o...fpr;..t ~ a.re
possi,/e..(dcf1tNui'"J """ rk (j;.,.,(:.,/~"'"' we.
(l) V0
dtoo>e..).
V.,z =
Um -
(2) Vx.vt
=
IF
1
fraASi.Sfors Gl"«
.f,L
of!.
fM vJ/df-t.s of! ~e>J:J...
f-N.,.
J:, ... too Y fl!l. =J3~,)4A
w/Jtk.. is 'ft.. -'""""
r~
tF-
=to+ 1P
Wt,tJ:, ,.
~I+
int r.tn - Rm~
\l,t.1·l
rn~~
-v, .. 1 .~,.. 1 -~
substituting (3) into (2). we get
\l.\1"·12
= - VRrl
-fn,......Ji> lor:
wdf-
R/.)V~,~
g.,2 vg,l) rm
rol
f'f,..ay c..c.t- cu
:rr
II
-gm3 (r01
....
2o f Iff>
1-t()t-(0
3D
t~ 1:"11 •IPOJ( ~ r
t:~
r;,
f6.7,JtA
.,.fOOl(!£_"' #o~t>-
>o
(4) v<.o} ~·
.
suh~tittHing
y~f'
Kml
-V
"} rrp
rot "". ~.:: -
·!'1
rm
J1
ml
r
m
(t + roz
+ g.,! r,J)
rm
.·
(4) into (I). we get
,
gml \,J{.ll
.
rol
J.,'1.~3o
.,. .l(lltf
Chapter 6-23
v0 =
-g (r
mJ
VJ
II
R )[
/_
-v,,, g.,, 'o2
(
I + ' 02 +11m2 'm
'o1
since
V,, 1 =
~v, = g
)
l
v,
v,.2 =
since
ix -
=
Riu2
(r
mJ
m
'rn
]
/. (
'm
)
I+-+ R"'2 roz
R )[
11
llml
'o1
=
02
ii; -
V1,
=
I
=
Vx
'o2
+ g,H2 Vx
_!_ + g.,2
'm
~
=
'x
'mil _J_
8 ..2
1
-g,., ( g-11
'mil
"12
lim I 1/.,)(roJ II R,)
2)
( ..!. +..!. + g
'o2 'ol
m
Assuming all r0 values arc>> I,
~ = g.,, g.,3
V;
6.40
R,_
8m2
•
1
Smce
/ 0 = -k
·(w)
- V
and V
ov
"'
2
2 ,, /..,
1.,. 2
Vov2 = VovJ , Also, /1., =
__!_g_
making g., o / 0 which is a
.!:!:'
VG.'-1·
V0 y12
Tlv,.,h.r~
I
o,
R,.
Here, L.z = L3, so we could also express the gain as
Vo :: g.,,
\1.
(W3)
RL W,
•
Now, to find the resistance looking into the
diode-connected drain of Q 2, we apply a test
voltage Vx:
~IOM.4 -~O.'f>~Vef:(o.71tiV
is rtuy lwj"-~ .r6
~ _. CREF
.----ov,
,
10)4A ( rREF!
L
,;8;11G.f
I
Vx, ix
The CS gain is
vvd,
divide out r0 :
~
;K = vVx + g,.z v,,2
,'$ ""'-' t•!Jlhle.
to;tAt ~ fo ( ~~ ""A
li:CJ lwt.u
Chapter 6--24
6.43
6.41
Is,_"' Is, ~...,
' rc:., .. t;;
.rQff. ,. Ic. +' ~ +' ~ (i)
va,,• '~~!3Ez ->v: t-.&.. 'f. e.. .h.
rs,
r
~ .le...
Ee
.!.sa.
ru... "'""'
-:.ore. .. ro;....
~s,
"r s ""$,..(hth'"3 f.o (' ~c. ,.,. (j) :
~Re-F- • h. ._ b_ + ~ =:;> 1sz_. c
,...
.f,(>~f
JS
,rr,IJ
_k.. = __.;IY';.;.;...._ _
~E-F
I -t J:t.!!!
fr +- ;3-
-"'....,--,..-
I fo
iJ'
1Nj re.Htll: ,·s +N s_.e
()1..
1 E' · 6 · .2.2
·
reto = Cc., =I. I 6"" A
6.42
"C-9 ::: VC:.ID .. Vat"
= S' -"· '1=4· 3 V
tc.t, =- rc.ro ""1. 86 ... A
11c.n
For identical transistors, the transfer ratio
is the same as eq. (7.69):
.!JL = __I _ = [REF
I
+ 21 f3
1I + .1
20
= 0.91
= t. ,., "1
"'1. 1"6v
Chapter 6--25
6.44
6.46
0.)
/l.::IDI<.Jl..
v, =-0-~V'
1 _ 5 - VBE _ 5 - 0.7
E 4.3 k!l - 4.3 k!l = I rnA
J:o)
R,. took.I\V,= _o."f V c:.> 1c, ::. -0·'1-of'/D.':I ,. ()./..,A
' (J
q
;z
____,.
I,. .2 "C.t~
r : O.'l""A
v3 ,.o.1-v
VI~. -'O·,
-.
, v~, 5.1_o.7.
+l"' /00 .. _(). '1- v
s-"
Ro""ro+g.,ro(R,II rwl
In this case,
_ lc _ I rnA
lim -
v
/0
since 13 >>1, I= lc-IE= I rnA
To find the output resistance, we
canuseeq. (7.50)orsince g., r 0 >>I,
Yr-
25 mY = 0.04 A/V
rw = ~ = IOO = 2.5 kO
liS' :. - I o · 'f +- o. I J1 JJ2Q. • - S' •'1
.04
g.,
2-
ro =
~
lc
= 100 V = 100 k!l
R0 = 100 K
I rnA
+ (.04
A I V)(IOO kO)
(4.3 kOII 2.5 k!ll
6.45
T::~,.lMA
R
-.,. It .. T.
'w:.n.
R0 = 6.42 MO
If the collector voltage changes by I0 V,
6.1 = A V = __!Q_Y_ =
Ro
6.42 M!l
1.56 ~A-A
Chapter 6-26
6.47
t\11 i\u.. ~S:$tc.r.S
1"1 ~s ptbbt~ ~n.
(!.\::a. btAS ~of t>-S~~>tA. ~ ~
!50.tL
f"e •
3m.....:lo """AJv ,
'
M
S.·,. at C;.u ,..
a.} c,... _.., - Ew.l H'
J -.I
~ - c ; > . . - . q "-01!14! t
I
,..
~ (f;;t-t rf;;.>%.
~ ... ~~'
1
_..L_ :z
2n
ZH
Zr1
)I
9.,. 1 1'1
"'"r- f>"le :
!p, ..
·
r
.
1
"" (;.lfMffl:
.2rr 5'-2. ~
I
lrJ;l+(tt
tn l< i!JOJ(.
::.5HHt
HH1:
'lv
·
p
(c.., + 2 $t.1) -2-,.,c._tt:J_f_l1"-)l<-."-,..,-,>
f.Pl-""
l
,.,., c.
r:
nz. t.t
..
I
i>
=53oKW~
211 1<6 ,.c.os
~~h~k.l
=
.£n $tt !1- Re
J!.n -,.2. "!'- I 0
J.p.3-:::. :f. 'f 611 ffZ
i'kM..s: J.lf~
2-n ( t !'(flak..) (3>.,. 2.,/
QJ::pt>J:- p.. lc. ; Pp,2.... I
.., J.
,., -=f. '}6
1o +S
4.~':1-HH-a,
ft+M..\.clJie,.;
so '7.;
=I.' Hlt2-
=- too,co.'l'fJCto~_, 6 v
l-n (~~II,.,)
Jpofe:
2.nc;, 1 r~
., IOO)IIO
~~1-2.rn
~;
Lt<>llpF
.At- ~pu.J:" : fp~... I
CA$c-..d~
C) CC- Cf>
.n.
IDI
o: IOIC..J\..
~) li1eJ~J
?? .. .J!~,_~
( .. ,!.rl,.2~'.clo""
I
..,1-.Cf6
HH?:
_._~,_
1
Stl11 .. tltJ.t
~~ A_x, ~ ";;, ~111 + ~ !{, *" :;,,_ ~~"):: 2,. 4f.BI-r6)1fo·'""+lfoaxa.ILt'1 ..,..2.)(10
J.lt
Jbr ( Hs;; 1/'itt) (Cm-t '"$u)
fi>~,.
j-1-j ::
'Y"
/p,• - ' - - - - - - -
ou.l;p,J: paM·;
'
=- 'S "II
t;:
A-<2 & =-
:J.,p:d pole
.._'f.tf/ICr
='5/J .Jh:.Eb-=
flflfJ'l..
'+:l.i>JitS'
o1e.s-t rnz.
1'1-,_, ,.,.,~
ntr 't ~~
lrtr't+ r,a)
~htrf! c..r""' c..,.,,_.,. ~ 1 (.t+,,..,z.. Rc)=~-tZ(I+UJo)
:::: II'Tit.f-f?.
~th.:
A"""'-
'tJ "II ( ,8 ,. t)
Ptl
~1 * ~
b)
411
~)IP• lt/4 //1~>111 [.o.b;+,.J
\
Fa ..
c:
;. rq "' iiiC)
... _ lt1f)l(OP:~f~ .-tlflf~
,,., 'S'+(fllht~
y
f!s.-,.,.. r,, +(,A+ t)r;,l.
~l·'f.,·t:...-f
A,..""'_ rrt . ., 1£. ... _.§.._
~tsg- 'Tr ,.,
to+$
f,
d) cc.-.cs. c.......st:Atl.-:
AM,. - ( ~1-f: I);J.tt.~:.
e
frt10
~, ... lok..J\..
l.oAvc.:
~"tt"" SI'A
. "' pi!
J.ttu:l'1
.:J.p J! ...:::> <;., ,. 6p F
,.
Of't.mfl(J
Jf::;;;:1. of" J·1~
c:
=->
"-t·l H 1-t&
•
Chapter 6-27
F> C<=-Cf!> CAsc...le:
A • ~+ tJI'f,.~ ·. ....
l~>l~to.'f,~t• .c'58 ~tr
~1-;
pote : ~., I
24-f ~ ju
·
.
.,
• 6.'tt-tfU:
::::".f.'t'
I
H In
:;!Jf 1< < >t'tt>
"::5HHi!
J~+;.,,. .
St~wtllltA/
",_.
6.49
rt5tJ{fJ:
C,nP•;urai:.:...,
0tH2)
f). (HH't:)
Al-f l~)
G.t3.
25 J.tA,
/REF ""
25 11A = 11 W1
/4 =
wl = wl ""'
0.)
CE-
b)
~c..c>cl<•
e.\
c.c _c.s
CAD~
ell c..cce.~Cc.clt
(.) lbld.t.d ~~
tJ c:.e.L..B ~<.
-66.1-
o.ll":f
-n.s
-6'
4-!
2."H
-r 5"0
5.0
'2.$"()
-lql.f
1-6
3!0
-'-'
-tS'o
lf .t
27-/
5.0
:2.50
40 11 rn
J _ I k, W 1 f2
I -
:)
-L \ OVI =:> 25
n
I
~
1X
= -2
200 X
~I
v;,·J·v1 =:> V,JV
· t
= ~'ovo ~ !J, "'
Vmq
•
(WI L)l
(!!:)
J,
f •.
~ /2
= 25 X
7 ""
354
V
= 0····
I
500 ttA
12 = 0.5 mA = ! 3
10
xll6r "' ?Pft" =
rlf =~
.. r:z. • r3
vGJ
,...
,..., }
11}
Y-
ov
v"ALU. ..s •
/
VGs ...
Vc;. 1 ::
'r
~a
z.
ov - .. o.08':-i:-,.4><"v-"'bv=o.2..V
"ov- + ~~"t; ::::o .a..o. S" .. o. Y.V
~
.=o.:J y .c v54 """ VG4(::: 0.'1 + V6 $lf"" t.lf V
-=> VG3 = f,Lfv ,.,. ~'::. l.~v-~::.. o.'lv
....... va"' vo,.
'"""
As
...
=~~ . . ~v
=o.qv
',.?
+ V,
=
0.354 + 0.6
=
0.954 V
= vGst + v<>s4·
Since / 1 = / 4 and W 1
l'o.n =
...,.:r~
VOVl
Vcn = 0.954 V
A fl +ra-s;s fo,- s ~V"£ +Ju. s-A: q
L
0.5 mA
=
llv.\'1
6.48
r
=
~pi
= roJ + U + (g.,3 + K,.v,>rmlr,_n
+ [1 + 2.82 kO X 40 kO X 40
= 4.6MO
Ro = 40 k:O
X 40 kO
QI"C
.kO)
Q.th Q~2.r-·~
~Ef
all ~~~~~.aJ;dued t" Q 1 :
to, "" ~.a,.. . • • =I.,., .. ! 0
Q
.f-: 1AAt tt.n~lffu .:~I 'b s,ppJ~.s I ~
j ~,
~ ~~ 1-orcdl
::
&-l'\ .,..t -th.A. ba.bt.. , f! Q.3 Y,o.tliJ ~
rllt:F "'r_+ (n+ot., , 11w.t.J .-.L._...,_
. .,..~.J~~
J9>1 f!..v..., v~ •f.
,8'1..
(f!SS flt--.Ml:
ill.::::. I
"'"'7 n+l.: I
6.51
I~
·-=>,... ... e_, =>11.:: s>
:l.
,_
I+ ~
~'&.
11'1
fl:l-
ipil(
6.53
Q" R"' 96t<.IL
=
lr!l = lo2 "" lr13 ··· = lon lo
The emitter of Q3 supplies the base currents for
all transistor so
1. "" (n + 1)/o
1·3
fl
for deviation of . I% from unity;
99.9
1
:=)n9E9
100
I + (n +I)
100(101)
~~ is ol!hM<11f".d wk«. Q3
ls.scotur~te.,l,
.l;u,:l.r--=rV.. .. 'S... o."i_o.?..'t.Jv Jl
t'.-"\)1'
=
~
Chapter 6-29
6.56
..) See lhe analysis on the circuit.
2
I .. = I+ .JLt....LI = II! + 213 + 2
13<1!+1)
Rbf
lor = loz =
/01
loz
_
/REF -
IHI3+1)
!
1!..:!:.11
2J!+l
= ! 13(1! + 2)
/RF.I'
2 J! 2 + 213 + 2
=!x
t
2
1+2!(J! 2 t213)
..&t_ =! __I _
2 I + 2 113 2
/REF
Observe that !he deviation factor - -1is
I + 2/13 2
independent of the number of outputs or the value
of each output, i.e.:
The current /REF can be split into any number of
outputs through l!!! appropriate combinations of
parallel-connected transistors. (Q 3 and Q 4 in !his
case) The reason the error factor remains
unchanged at - -1- - is !hat lhe base current
I + 2/132
that need to be supplied by
/RP.F
(subtract from
/Rrw ) remains unchanged.
b) The I rnA reference current can be used to generate lhree output currents of I. 2, 4 rnA by using
3 transistors in parallel having relative area ratios
of I, 2, 4 as shown:
lo1 _ I
I
- - - ---=>lo,
/REF
7 I + 2113 2
= 0.998 rnA (I
I
--21+ fj2
rnA ideally)
Chapter 6-30
(e) If a small-signal model is added
to account for Q.. the cin:uit is changed to
G)
s.
~
~ --'--, => loo
=
7 I + 2113-
/REf
= ( 1.996) mA (2 mA ideally)
~
~ - -1- - =>1m
=
'••r
7 I + 2113'
= 3.992 mA (4 mA ideally)
Since Y0 ,, = Yes• = -g.4 Y,, 4 r,..,
(no current into gate 3)
=
Vcs• = V054
0 so that Y 01 = Ym and
there is no effect.
Ro-=- (fnrJ'oJ)rnl
(a) Fi ... t. we need an esrimate for V.,.,
and V.,. Since the currents are all approximately
g
I
'
the same, nnd //J = i(JJ..C.,,)(
W /L)Y0 ·v.
R.,,. (I mAIV)(200 kfi)' = 40 Mfi
6.57
v.,. =
OV
= ....!JL
v.,,./2
=
~
0.2 v/2
= I mA IV
2 1,
11-.C,.,( WI L)
JJ.A)
= J(4002(100
jLA/V)(I2.5)
y
.,
= 0.2 V
since no value is given for v,•. we have to estimate
this with
11-.c., = 400 JJ.A I V 2 ,!
+
this fabrication process is similar to
the 0.18 11-m technology. We will therefore
approximate v.as approximately 0.5 V.
Yes = Y,. + V.,,. ~ 0.5 V + 0.2 V ~. 0.7 V
(b) V052
= Ve 51 + VcSl = 1.4 V. which is
,
=12 · Y.,,>
r =~=
"
!J.J
I
20V =200kfi
0.1 mA
= AV,,. = 1.4-0.7
'·
= 0.35 ILA
200 kfi
'· .. ,REF- M
so that,
/,. = 100 - 3.5 = 96.5 11-A
(C) /,."' /RCF
(d)
....
v.. . = v,. + 2 Y.,.
=0.5
+ 2(0.2
V)
=
0.9 V
~:''l
..__-'--..-...::..
.
'"1),,
:
::
i..:&- =: daot.2- 'I:J:SL @
v.1s 2. .... v_,sJ .. v" .
s,.,., c.c.
Q:l a-d Q_, he~ow t-~
.s_......
~h:rs
a-J ~ ~, tN.rc for~ v.f$.t•~s.l
Vx. • 2. vj.S.J.. ='7 ~s2. = ~
S<~PSI"it-tl~ for ~.s~ •'"' (f):
i..)t. ,. , ...l. ,.
tr-
R. -~=.!:-
,..,_
c..
'"' !L
Chapter 6--31
b)
k_ = o. 1
11r.EF
VR6 = Vr
6.59
.......,.
e.... to::
Io .. to j4 f>..
57.!>~ ,..11
A~~ D· S"(,... 5. '"161U\..
'"
'0 = ~
"'...1oHA
10,'1
, ..... , ....... A;.,.
1\, • (I.,J,.. l~e) '0,.. 33H.!l Q,~ to ~"o•/OMn..
Vost• 2-Vc:;..s
V.Ds:t.• V~s
~~~·zt Q
~I
T
Vers
I
~- r:R,( t~~ ~(1/h.s.Vt:J&(I+~loo .. ..L.,.boo (1/&,s -o-6f
z.
:l
I ""(Vc;s -o. 6) (lo-t Vc;.s)
v6-.s ~ o.qr v
1• •
Zi>z .. t
::
!2..o
( b;t it-~r,J:·.,"')
Cl k _ . o.ol ...,.t0
r#ISF
VRE .. VT
~< :J.t1~ c"c:.-.s -o-6/ o.'f -:r ~ .qO.J'A
e......1, :l.63MV
"·1
~.v ~.63"'" ..
'fC1,.UA
'-1-:SJ\..
ro " 'iA- "" 1.11 f-1 .1\..
!0
''":: 3.t;
"'1-'v-
'\, ... ( ,..., ... ~.eJ-o •I.23MA..
e.._ too .,. ltSMV
~·lf2 =IISf'.~
I
'"' ., o. o "1 ,., A/.,~
,..5'6oH.t\
.. U-t,,.. '\>'0
6.62
:CR.Ef
IJ~
r 0 : ~ .,(ooH.J\..
~ #lOO.S17/'fA.24
Tlw.o -there ··:. ~ or o.'Sf erll'o~.HoJdJi"'"l
.
1"0
"'
(T
t-M. "·..-""'"" .:u F,·,. 6. 61 c. e.nsv..-v -thM- Q,
-.1 ql '-.IIY. f~ .s.....- vos tMJ +'-.1 dt~~
v_ .. V..r
....,.
-
C...,:.~ t"~· 1'ffl-
Chapter 6-32
6.63
tAJ R,.3 ,_,11J.l..=
,..H.=o.cr?s~
u.
R,.P .,.o.ct' ~ 1,.,. :J.lf.ltA. =,.fH c.- S"J'. 3''1H2:
b) f?:s.h.:fotc.Jl.;
J!.u, ct Jl KJ!. ,
\f\1"'
R,.=I2:~J2. -,. lj,_ .,. 1. 2.'?-t-ffta
C) Rs.-,t~t./tlo#C.A:
1\H..,.o.q
AH• o.ltct''ilf(v-
ll;c•SO.,KA 7 /},..,6.27-4-:> f!14
.c
r.3~.HH'il.
Chapter~33
~· ~ 1-, + c;., IJ,, + c;. Rr -+ c.}a + 'i.) ~z
tH .o..2,c t• + I.Jf~·.15'+11of. ~-q.S#f2. +(O.Z.,.I):rttooo
6.64
-
• j + o•'r'l + .:l?'f.tS+ Uoo AS
~
'1lws .( 'i .. ~,.)A.)!. a. f$ ......... J.,.,-4,J,.:t~t;
~
+h.·
-.Ck. •p;
.........,.s/tl.r s ;s ope,J;.: •; illob&
•~s ~4J I# 4(1(*i~ t•9A .~:
' ..,_..!.!!e. :.25'Kn.
,,.. .. ~-If""'"/,.v
~~tr
~
r'e.t.2'io..n..
!}
CL_
I'll
Cy,+7f"' u:::-::
.«t.·
.ltr.;. 111 'lfllfOo"'
·iG •
tfX> •
t>•l
,,...n..
:::I•S'~' _, C .. I.:Jt,.F
n
~A) f{;,... • (~+I) [ret+ (rnJ.II"bt1]
R,~ • lot (L5'o~tti~.2.s~Cu
. ..
/IH • • ~.·n
rMA1• .t.s HA.
f'ru. U~'e•
r4 +(r,1 /lfiJ,)
~·... +tls,'l
111
3 r,._
,..,.
/4 ,. _. g ..;" )t ,J.'S"Ill/11-t
. ,.lot JCIH
H
.t.~o.ol o-15'-t( 2J"IC/I l'i)
~To tAttA.tltd£
1!11
,
.
(} lt~Cf'U>.S•'"d +ht b:e..s ~b hf 4 hitt,.,y
of. lo:
,., ... ifq.,....
'i'tr
'7
'i t: .t."1' J'L
'n*'
; ,..,
,. o.
.2 .'5f'A.
ftJ, 14/L
#
I+ o· 5i ..-10 a6. "f,F
J., 2't'f IC/L
" z.s"-nteoc.
by
<;..=
(}.(12$"'+ (2.~/ltJel')
6. ?-t:
<1·2.{1+ l.foxroo) ..
Rr= ~ .'5'-Jt to"If .1·S'tlO
•
IO I
lf1.1JL
RT t:S. a.f..,.o~f" redtJud hy a (Q..c/or of! 3.
R;n, 2. .. r-n1li tOt::"- .lfltiVl..
R.n, .., lo" :r '-· .lf6f
,. 13-o .tL
~
•412.$'
Rnt is «IMt~tsf" Jec:.~
~'1."" '02.=-tootc.IL
ilt1
~.yr.,
c..,. s 4r.z+SNz.l t+p..tJ 'Oa.)
'r • 1-31tP·.t.ltt~
1l..rlf}f:J..
r.,J..Jir. fl
''
'D;t :r
pv,i h.i
""
f()CC"JK
tO)
!1. 1?-. ct 2
n5
T1..uc -Ht.t... do~ ef:.l=ec-1:, +hJ ff>l- +lu.
DJp~ pole.) is
rtteiVCAd hy
4
fQ.c.lru· pf:- (D.
~r>~~c ('i_ .... ~2.) r~-..u.s
c-f$~ e.t~Ai..lc 1'02. dttc#'t!ASt!S It>)' ~ f,:...eh:JrDf
l">k>OO)..,B'or.6pf'-
P+t
lilt. fr>..c.O.rof
li'ol> .,,.o.ffg + f.2.)cloo
l; ... 1. ~ 2 r o. 8'" + q S'". 2.+ 1.2.0 ...
"Jh.,:.s occ.&J.,.~
'"'z.vm..
rm"f'l\
,.
~u..t
10.3
.,.
t'(J'L
r~:a. ""J,...._
h+l
lr...., +
~'!l
}
j),t•
~-.(-"' '-·'5'+....L-_(;So-t- ~js.IAf.fJ'\.
IDI
t<•l
N~l«-,,_,, 'ii :
IOm:lR,
A~:rJqo"~v-
nn
6.67
I
= _!g_ "' 10m "" 99 JLA
p+l
101
~~~
In "" 1112 = 99 p,A
In "' a/ f. I
= :~(99
In = alt:! "'
:~(10
p,)
~
m} "'
98 J.LA
9.9 rnA
Neglecting Early effect
using resistance reflection rule:
R;. = r. 1 + r,ztll 1 +I)+ I K((3 2 + l){P +I)
R;.
=
J!Vr + (3V1 (101) + lk(IOI) 2
In
lcz
., 100(25 m) + 100(25 m)( IOI) + 10_2 M
98119.9m
·
R;. = 25.5 K + 255 K + 10.2 M
~
R
I)
-
R ,
o
~'w?
PJ
IJ
100(25 m) + [25.5 K + 100 K]( I )
<101 J(9.') m)
lO'l 101 WI
1_-) "'
R 0 = 2.5 + (253 + 9')0)( ..
·
JOI
t\,.0
A,=
= 10.252 M
+ [ r vi . !()() K l( I )
+I . p, +I .. p, +
P: + i
I, 000 VN
I X I,OOO
14.8 + I, 000
"" 0.985VN
14.8 f!
, Alf" ..
,,,o~:~t:>
,.o.IJB"f~
l"f.lt~t"""
rv
Chapter 6-35
0.5 "'"'
NMOS PMOS
I
v,,.(V)
v.,
O.I3J.1m
0.18 """'
PMOS
NMOS
PMOS
NMOS
PMOS
0.32
-0.54
0.27
-0.46
0.23
-0.48
0.2
-0.42
1.02
-1.34
0.7
-1.08
0.71
-0.93
0.6
-0.82
0.5 ""'"
NMOS PMOS
I
0.62
gM (mA/V)
6 ·69
0.25 """'
NMOS
0.25 """'
0.18 """'
0.13 """'
NMOS
PMOS
NMOS
PMOS
NMOS
PMOS
0.73
0.43
0.88
0.41
1.02
0.48
0.37
If the area of the emitter-base junction is
changed by a factor of 10, then 15 is changed by
the save factor. If V8 , is kept constant, then I c is
2 X IOO = 11.98 - 12
267 X 0.25 2
For PMOS:
(IV'J
also changed by the same factor:
=> L
V8£1VT
P
lc = 15 e
15 a A, lc"' 15 => lc n A
=
2 X 100
93
X
= '14 4 - 34
0.25 2
• .
-
A,= 10 A, =>lc 2 = 10 leo
If I c is kept constant, then V"" changes:
VBF.tl'T
1.12 = l01 50 =>1se
--.-.,-
VHL"I-VBE.2
r
=Vr lniO
6.72
· - '"''. := 2J.L"
I C (IV)
I v
= 0.058 V or 58 mV
'"" -
=
Ill
~IJ.pc.,,(T) V0~P
It
1
OVIt
(I)
p
we also have g,.,,. = g 111 ,
6 . 70
1, =
I
= 10, If) = 100 fl. A,
! ( .!!' v'
2
L
gm
21n
===
Vov => Vav,.
(Z')
()V
(j), ~ => ...:::.:..£
(~)
I. "
6.71
IVo.l
1
f)
= 0.25 V,l" = 100 IJ.A
=!k'~v'=>~=2!t?_
2
L t)V
L
k' v :'
J/
"()y
For NMOS:
- ~67 ~
•
v'
=
= Vov,,
460
~ = JLp
160
(2)
= 2.88
Chapter 6-36
6.73
6.76
V0 v"" 0.2$ V
for an npn transistor:
g
"'
L = 0.3 p.m. In
!.£.
"".J!l.
= 4 mAIV
Vr
0.025
=
f-orM NMOS with the $."line g .... i.e.
"
nm
"'210 .,. 2X100X w··l"" 1 mA/V
V0 v
0.2
g
'"
g., "' 4 mA/V
we will have :
10 = 0.5 mA
"' VII X L = ~
= V,.
r
""· Vov
21o ~ Jl.• "~ ~~
" X Voy "' 0 5 mA
2
-
•
~
/0
c•• zVor=:>
8,. X L "'
I X 0.3
p..C.,Vov
3R7 X 10-J X 0.2
W "' 3.88
Assumiugl~e
6.77
g.,.
L
10~
-;::===::====:::::;
1.58 k!l
PVr
()V
<11 +
,. = 250
,! X
nt,""' I
2
0.025
OT
~tm
C W y2
2 ......
,L
ll -
+!3I )gm
Vr
W = 6
JUII,
_ I
I
For case (b) we have
(fj
0.3
Vov = 0.2 V
X 200 X 10 X 0.1 X 101
r!l',. g: -
).I.IU
r. For both transistors, for
case (a) we have r = 1
r =
= 15 k!l
g,. "" .....
6.74
Jz
~I
A.= g,.r."' I X 15 ., 15V/V
W ,
r =
100 p.A,
= 0.2 V
Vr>v
387
.
X
J!_
0.3
X
0.2J
155 11-A
= 0.155mA
10
0.25 k!l
210 = 1.55 mA/V
n
Vov
~~
3
L C m + C(>"~ "' ~3 WLC,
~~+ WL 0 vC.,,
6.75
2
X
100 X 10
0.5
C,.. = ~3 X (, X 0.3 X 8.6 + 6 X 0.37
0.4 rnA /V
=
12.54 fF
2.22 fF
25 X I = 250 kfi
0.1
fy =
,.-.,-::::"'g"'-'"-:-;:-:-
21T(C., ·I C,.,)
--~''-.5""'~~
2-11( 12.54
g,. XL
0.4 X I
If we
IJSC
1'-'0'--'--::; = 16.7 GHz
X 10 1 ~
+ 2.22)
the approximarion formula:
J..l,,C,,,lVov
, 1.5 / 450
.f,::
~.,,..,
X
10
,
4
X
2 X -.r X ().:\' X 10
The g.,
0.35 X 10-• X 0.4 X Io-'
:
=
lr =
C1,
C.
30 X 10-9 X 40 mNV
""-fr =
21T(C. + CIL)
g.
21T(c.
fr = 5.3 MHz
i
0.4 X 10
_, "' 26.1 MHz \
21T(2140 + 300 X 10 )
C4,
= .,,g., = 0.35 ns x 40 mA!V
g,. = 10 X 0.4 = 4 mAN,
C1,=2Xl=2pF
c4,
C,. = ;::O.JpF
= 1400 IF
c.
c.=34001F=
fr
lc;
g..
4 x to·'
= 21T(3400 + 300) X 10- 15
= IO ILA. Low-voltage process
= 10 X 10- 3
0.02S
:
c•• = 10 X 10-ll X 0.4 X J0-3
c1, = 2 X 5 IF = 10 IF
c. = c._+ cJ• = 14 IF
c 11 :CILo = S IF
21T(C~
=41F
Ao = g,. 'o
lr
=
g.,
21r(C_. +
2nawLc...
fr:
In Summary:
Standard
low· Voltage
npn
lc=
IOp.A
26.1
MHz
172.1
MHz
OHz
3.35
c,. + ca.)
I
X
2
3WLC.,
.. 1_ 11 Vov
41T "L2
As we can seefr can be detem1ined after knowing
Vov and L, it is not dependent on either I0 or W.
lc=
I0011A
I 1.6
MHz
~WLC.,,:
k~WILV0 v
2n
lc=
100 I1A
+
c,, =
r.t.I
·w v;..)
'
(replace 10 wilh 2
= S IF
= 11.60Hz
lc=
1011A
-y;;-
c••>
lentlyc,,>>C,,and
C~
I
2VA _ 2V.AL
= Vov -
If we assume that Cov is very liJJJall or equiva-
_ JOOX 10-' = 4 mAN
0.025
g .. -
Standard
High-Vollage
npn
210
V,..
= Vov
X J;,
2/ufVov
lc = IOO ji.A, Low-voltage procellS
4x1o·'
fr = 21T(SO + S) X 10- 15
= 391 MHz
Therefore A0 is only detennined by setting values
for Land V0 v
• 3.350Hz
fr
2 .,(~~~)pF
6.80
_
c4, = IOX4 = 4011'
c. = 40 + tO = 50 IF.
14pF •
= 14 + 2 = 16 pF
=lr =
0.4 X 10-J
+C.) - 21T(5 + 14) X 10- 15
g.,
fr =
172.1 MHz
0.4 mAN
40mAN
211'( 1200.6 + I) pF
+ c~)
Elu:JuuJ:
= 100 ILA, High-voltage process:
= 10 X 140
1200 pF
= 2C1,. = 2 X 0.3 = 0.6 pF
= 1200.6 pF
C~;:: I pF
= 0.3 pF = 300 IF
-)
=
lc;
c 110
.,,.g,. =
C4, =
= 2000 IF
=
c 11
40 mA IV
fl!um.p:
= 140 x 10-" F = 140 IF
CJ• = 2C.Jr• = 2 x I = 2pF
C.., = c1, + c1,
2140 IF
= !.£.
v,.
J
Chapter 6-38
6.81
Vov "" 0.2v, t.
"" 0.2wn. 0.3 f.llll, 0.4 ~tm
= SOLVN
fr
:::L5~'!.~1W ,, ,15 X450 X
2nC
"'~.1
1?" X ~~2
t: X II) •
4
2 X 3.14 X
GHz
Ll
L(j.lm)
0.2
0.3
0.4
A0 (V!V)
10
IS
10
fr (GHz)
53.75
23.9
13:.4
6.82
L "' 0.5 JLffi. V0 v
fr '~ lOOMHz
= 0.3 V, C1. ,. I pF,
= 21fC,,
~=41?,_ m -- "r.Cf
,.';,. * l r
fT
., 2rr
g
I pF
X
100 MHz "' 628 p.AN
X
= llo =41
V01..'
i'Jf
fj
- g
-
X
tff
<
V'(HP,.., ::::. 6 - 28 X ().3
2
1, '' 94.2 1J.A
I
_ 1,. W 1_,1
D -
;;A~tr•·(.~\>'~
'""'
2LI,
»·, _- -.-,-.
/c,~ v;.W
2X0.5X942 = 5.51 flnl
"' -· 190 X 0.3~-
IV···'· S.Sl tJ·l!l
2; X
'"15 Mlh
I
l pF
X
1(!6.2 ld1
Chapter 7-1
-
Le) Ve.111tx. =- V 1:: +- Voo - ~ Ro
2..
= o.~ +2..5- o.t..c 2.. s
7.1
Vr>o = vss
K\11'~
vz
L
:r. =
o.2.'-'1AA ,-
Vt111"' o.t-v
l<..o
V~s = Vov + Ve
==
-=
j
O.Z6 +o.~
-\.
7.2
(a) \Jov
z.q- Q. rt6
'='-
-- -'2..2 \1
J.::t..lt- :'!.:. x Ro
= + 2. 5 - o. I v 2 . .; 2.. = i.. . 2.5 V
(_ c ) I f u-C/WI :: ..... J! v
Vst :::: Vs z. = -+ I - 0 . q6 :: o .0
Ior-=-.l.oz-- 0-IINIA
VDt :::. VDZ =- -2, 2 ~ V
Vtt = V~z = -I -0-96:::
J:..r,. ,.. .:I.o 2. • 0 .I w A
Vor "' Vo2. ... 2
-:zsv
4 V
-==-
tb) For Q' OM.d Q2. f::o
l\1\ sa.tu ra.hia v. :
Vo s {:. V~s - VI::.
--t;. Ve.~--t .>,.. (~ Ro -
retM.a..iu
Voo) r
-z.-5
VcMWAiY~ ... O_:l,e2.
2..
VI:
-0~8
-:;:: -2,b v
To allow .5vt;../eJeuf
u(.,M. - -1 v
~
-0.45 -o.g
-: 0 + I • 2.5 • + I. 2.S V
VoL ~ ~ )( R.o - Voo
@ Vsr ';. V_gz. =
.I.ot :;'; ..Loz = ~
-= - o. li5V
= Vov.,.. VI:::=
u,,-=- Usz.
0-2.\M.A
Ld) tJ:
'tJCJ1tM.i\4 - V&~
'1M i\4::
w IL •
o.L ; ; .., o~Zf v
""' O.Cf6V
@
~
--
c
v.'
':!.
\J S
= 6 K.vt
J I.. / K..
(a.) Vov
+2.95v
(J) 1JeHIN. - Vss + Vcs i- v~ + Vov
I
.. -2.5 +o.?> +o.':l +0-ZG
-== -L Ll.l V
=2..5V
3w~ ;
•
c
-I.
~6 v
===--
for
vo!~e
4;ue. e-uHeUi: c!3ovrce. l::a V
Ofera.l:e r,rf/fer/'1:
Vef-t ~ V.!S - Vcs + ( Uv 1:: +- VQv)
-t>
VcM
!M.A.)( ...
z.. 5
-:: o.
-o. 5 -1 . 2. 5
:t5 v /
Chapter7-2
7.3
7.4
V<>~ "' 1114 i/)1 "" OJ I mA
Vtn "" 0
/0
im "" 0.09 mA
I ' W
- k -(V., 5
2 n l.
._,
2
V,)
-
ForQ1:
O.llm""
-t ll(;q
iI 5 m(V(;$ 1 -05)-~
"" lt71 v
ForQ2:
i.I
0.09 111 "'
-25V
v, = -v652
ll'cr: ~ 't1id
2
"'·
-0.69 V
+ V GSI ·= - 0.69 + 0.71
= Vs
,, 0.02 v
V 1J
Wht-n all the current.is on Q 1 :
m( l'<;s 1 - 0.5)
Vest = 0.69 V
-1
V 02 = 0
5
Vm- V 01 = 10 k!l (i01
i01 )
-
··"" 10 kV (0.11 2 0.09) m
=0.2V
thus
Vm-Vot=~=IO
= V,+ /iVov
and Vc,, is reduced to V, thus Vs
iben v1J
"'"
when i 1, 1
-v,.
t!c;s• + Vs
"'
"' V, +
().02
= 0.09 mA and
im = 0.11 mA
Ji Vov- V, = Ji. V
is the reverse condition from the case we just
studied, thus v14 = -0.02 V
0 ,
In a similar manoer as for the NMOS Differemial
Amplifier, as v; reaches -
Ji V0 ,,
Ql
turns off
and Q1 on. Thus the steerine range is
Ji Vnvs V,::s; -J2 Vov
7.5
For this particular case
v()~~
0.25 rnA = 0 _25
4 mA/V1
=
v
V,. + V 0 v = 0.5 V
V fH
V G4 = - V SS + V <'S = - 1.2 V
-0.5
J2 X -0.::!5 ::<£ V;J ::<£ Ji X 0.25
0
4 ld1 X 0.5 mA - 2.5
Vm
0-2.5V
im
VS
J>
(~)
- V,z ·= ·;· OJ! V
1/111
when
1}
l'tl ·:::.
-0.5 v
. (.
=
+0.:i5 V,
~ l'GSI ::::;.
Viii- Vtl
!.US V + 0.8 V "' !15 V
··0.5
v
=
1.2V ::_0.2 V
0.2 mA
5kfl.
(!!::.) = 0.4 ,mA[k ·V 2]·
~
L.!
.
ll
OV
2()
(11'
)
. l.
0.4 m;\[ 0.0 I mA J
1
40
l
~)
..
\ L ·•
f,
Vm "" -25 V
I!,;
V
vi).
~~w- Vm
0.4 mA i 2
!
+ 0. 7
mA
= -25V
0 : i 02 ~· 0.5 rnA
l 1td·
=
R
im ~- 0.5 mA, i 02
Vs
(U
V = 0.7 V
v
l',o-
R
-0.35 :5 ''. ·....
ForK= (U)1
AI = 1./JO,...,:o,.,..,-l(:--1-"7"0.0~1)
=O•.J98 X 1
v,.,;..•• = 2V0 .,J[Oi
FQI'K ""o.t
AI
= 0.2Vo~·
= u J-o.~-u"'"""-o.~.> ,, o.st
Vu,.a. = 'i.VovJiS!i.
=0.894. V(JV
7.8
I.o =L
w (Vs-s- Ve) 2
z..".l..4.\.( eox 7:
.t.aa
2
=..L x 'Ia
z.
==t>
lf&-s =
2
)( lao ( Vtr.s -o .. ss)
TI
I • It:[ V
Q~-w • .t..I.o =- .f..'l( loa= l.o6~
Vs~- Vb
U
tf;d
(I ·I '1 - l )
If~fv.!!t:r::q
{ 1:-:: R
\.l
(VGS- Vt;.)
== 0~2..1-V
~o double ~Lu·s va.lve , llc;s- VI:
7.7
( ViJm"'/2)2 ,. ,. K
.
Vav
Q.E.D.
...• ,·f)l ..····_. 2
I ~
·• I ,.,jl\
r;;K·(--1··~::;···-.;;··)·
)\
"·-~
V
£~LU.Ls {; be doul:,le.d wl.u·a.Lt tM..elUcs
fkal:
tJIA.ovl d 6e f.va.dru/.e . :I el.Lo.,tA.01e d f:o:
f'ed.
:r..p
~ODftA
Chapter7-4
ivz
im
Vo.\<
VosJ
VG-
(p.A)
(ItA}
(V)
(V)
(V)
50
50
1.5
1.5
0
0.5
33.3
66.7
1.4()8
L577
-0.17
0.8
47.4
52.6
1.487
1.513
-0.026
1.4886
1.5012
-0.013
im tim
7.9
-
-Vov
o.ZwtA-
::Z:O w j_)tCA (tOIL
'2
.~
2.
V QV
L
too :. ..L y. qo" ~ Jllo.z.) z.
2
47.75
0.9'>
50.25
For lfJJ f ioz ·"' 20 ""'> i112
Vi:;J
= 4.76 JlA
im "" 95.24 p.A
v.
v()$2 = 1.154 VGSI "' 1.690
Thus Vm- Vm"" 0.536 V
L
= 55.6
~
7.11
(a)V,J = Vl.>!-VVI =
{Voo- imRv)- (Von- i,.R 0
~
50 =
~
For
X 400( V vS -
= vc 1
"'
+ (-/
Yo>·
0, v5
(im- im)Ro
!)~
VGs"' 1.5 V
TIGI
""
)
"'
)(VId)
I_
2
(V,dl2)'i]Rl,
Vov
-1.5 V
tl 01 = 11 Gl =2 v; t• s = +05 V
The dmin currents are equal in both cases.
For
ForV02 =0:
(b) see plot
slope of linear portion
Tnreducei02 by 10%,
im ,, 0.9 X 50 "" 45 J.LA
=
d (IRn. V 4) -_ IR,I Y ,
JV:1
0
dV 1J V 0 v
im "" 55 JLA
~+I=
I
"i
Thus. V c 1 "" Vcs, - "os~
iu~
i 01
plot
when the bias current is tloubled, V0 v so
IA7V
"' L5'">
400 ~+I
.
.~
To increase ioz by 10%
'
(~)see
=
v
(UJ5 Y
v , v.fd =
~,JJ
1)
2JR
Ji
Vov
f _(Jiv,4 ')1
iolcrease$ by a factor of
12
---
Vov
Ji
the slope of the linear
part hus increased by a factor of
Ji
55 11.A
"'
(d) sec pl01
45 fl.A
''un.
~ 152 V
''(;$)
= 1.47
co.}~·(;)
lfWfL is doubled, l''ov reduces hy a fa~!
-
as high.
'= o.t
7.16
HAlF-CIRCUIT
-
.:t..
-v
Vov
For a differentia! amplifier A.,
g,.R 0 with I= 210
So the differential pair requires twice the bias current as the CS ampli.ficr.
The power dissipation at the diffamp is also tWice
0.$1'
.
also:
Vov -~
:r.
I
/(VI' W!L
~
!:!:!..
--t:>o
tt
V
(o. at6J :z. ,..
t...
R..o
rs IMA
O .. t ~ )( (Yt)
o.
v~
r: g5
~ 51<-t.€
=t>
Adr:(J-w.f<.o •
3 ~ x S l V,then Q3 will operate in triode.
r,"'l =
[ k.. Iw v,.J
]-!
(See solution to 8.21)
l
~ ------~g~~~··~z_R~o~----I + l!mt,l W
[
(~).'
·•
(r ). ·2 ·g.. ,.~
.. 4
I+
l
~ OVl
Kml
(d) rps;t = - , - . (i)RI ""
- L :<.•
_.!_ .·. Vo~J o• V,,
.~llfl
From(b)V that
.\:!:' "'-
2 //} '
t
f-1.
c.,_,v 0,.
ForQ7,
( .\:!:') "'
L
1
f.I.Al .
2( 100
24.7
90 J-lA t V!(0.3 V} 2
For Q4 and Qs.
(~t
=
2!50 ~M
(~)~
12.3
9() J.I-A I Vl(0.3 )!
For Q 1and Q2·
(~),
=
{y)l
2(50 ~tA)
~;.'_
30 JLA J V~(0.25l
53.3
For Q6 and Q3.
For l RfF
"'
100
15- {-1.5)
0.1
V.,:n "' V 0 s4
"'
Vt-..v 6
:;:;::.
"'
30 k!l
m
V,,, ·-- V,..,
! .5 - 2.5 -~.., - I
I - 0.7 ~ ti.:l V
v
V 0 v_1 = \'r.s- V~!~ = - I - (-0.7} ·.: : : -0.3 V
Ftnm section 8.23. we know that
Q2
Q)
Q4
Qs
Q6
Q1
J.L(-:.,_(
30
30
30
90
90
30
9()
JLAN2
ln
50
50
100
50
50
.100
100
I-LA
....{).3
0.3
0.3
-0.3
0.3
v
74.1
12.3
12.3
74.1
24.7
·I
I
1
-1
I
Vo~·
w
-{1.25 -0.25
53.3
53.3
L
A., '' g,.,(r,d \1 r, 4 )
Since Q1 and Qz cirruits arc symmetica1
VGs
.·
-0.95 -0.95
With I '" I REf --~ I 00 j.l.t\
I,_J = !2 = 50 ttA
20
v
50 ;u\
-tOO kH
7.20
So,
Hll V
1
V = g,.,,(400 K Ji •1011 K)
ami
g"'' o=
.tm ttA i
V
Since V1;.1 - I·, is equal for hoth transistors :
.cC)
JIH
I0,
but I
2(50~
400 ;•.A /V
025
v
~Ov
1' 1,; 1
Vun •• Vm
-0.95
v
V, 1,
••
·-
14.1
I. '
Alt
II){\
~'cs~ '" - 1.5- (-15)"' 1 V
Von "' V,w, "" V,.,.,
v "-'" "" v (;.\I
2(100 I!:A)
30 f.I.A I V 1 (0.3 V)~
In summary. the results are as i'ollmvs:
~lA.
R
(~)
{y).
-25V
0.15 ·• 0.7
}_]OJ
ln1 +· ln2
J/IH
v
Chapter7-8
1m = 113
= 2113
lm
(b) VQV
""
IVovl = 0.2 V
For A 4 = JOOO VI V and
V GS- V,
2n
!Vo.-1
woo=
Vov• "' Von ""· Vov
2
1
VA
<::>
J5fij · 0, 2 V "' 4.47 V
trlv~l
=
I. "'
4.47 v
JOV/p.A
= 0.447 j.LIIl
10 V l}l.M
For high g,., the bias current should be high, but
with :!:.0.9 V Supplies the bias current must not
exceed ~ = 0.556 mA to keep power dissi·
1.8 v
pation at I mW
-g.,,
tl(\t "'
T. RI>
X ";.t
"' _!_I_ · R, · ll·J
3Vov
'
7.22
t!·d
Vov
"oz = + 8'•• 2 x ~ · RJ)
'i
K!v1 1
~ "ol - "m , (~ + ~)· _I_ .
fl14
.3 3 Vov
R,,
1-·R
,zxVov
I>
.~\M.
U
= f7iT = o. t.6 v
W/L I} 3
J... ,
= .;:!:_ = o.Z\MA -:.
o,uv
Vov
(a.) 8iu~e-
0.~7~
V
eudul oul::fvt.~
7.21
/1 4 = g,. 1(R.,. [I R"P)
"' c,.,[(/l,..Jrt·
mcm) is sutlicient.
6.1 1o-4
)t
7.25
Cli I?.R = 5. I z.
-= {_f 6 80
6. ":f- 'lf to-4
-TI'
?l.:_l.l
-riB
If A, '"· 100 (40 = 0.25 mA
t ....
for ie
= 0.5 rnA .V Bl
= 0.683V
Vn -
-s • 4.6 x
-0.4 V
ifV 81 "'0.5Vand VH 2 = OV,V;, = 0.5V
1
0
1+
-5V
7.30
0.5 rnA
1 +eo~'o2s
- O.S rnA "" I X 10-" A
- 4.85 X 108
V..: = 690 mV at ic = 1 rnA P = 50
VcE(SAT)
= 0.3V
Rc = 82 kfl Vcc = - V F.£
I = 20
= 1.2 V
~tA
(a)
10005
WI
. rnA= 0495
.
::-:1\)
rnA
V8 ,
Vco = 2.5 V- (0.495 mA)(8 k!l)
v
-sv
e-o.~v;o2~v
= 0.5 rnA = 0.5 rnA
I + e-lO
= -1.46
=
lkfi
I kO
0.5 rnA
I
· 'c•
-0.3 + 0.1
= +fi.4V
-O.JV
v,
vn
= 690 mV + 25 mV In ( 1
= V 8 -V 8
=
v0
= 575 mV
•= -575mV
= 1.2 V - ( 10
~tl\)(82
kfl) = 0 ..18 V
Chapter7-11
(b).
VcMMAlC
., 1.2V -
~ 10 p.A ' 82 kfl + M V
"'OJ!V
[r))
VeMML\1"" Vu+ Vo+VIIf
"' - 1.2
v + 0.3 v + 0.575 v ~~
-0.325
v
Zl'L.
.(,
;JtA
~+\
So -0.325 V < Vnt < 0.&0 V
-=~ :r.. ~ 4 t(->+t)jtA
/uvs, :Z::.. • l.i )( IO~A"" 03iCJ'ljtA
!3eleel: :r. =- 0.4 tMA
(c}
-viil ~·1'
(t82 = e
- V.tln(0:82) = V;,1
-
5 mV
+ SmV
ifVez == O.Vnr"'
lOKvt.
: t.
7.31
Willt only common-mode at the inputs
Vn = Vc 1 = Vee- a{Rc t-V,
7.33
therefore the ripple voltage directly appc;rrs at the
single-ended output V c1 and V n
However, because the differen!lal output
.
I +I?
V.,J = ¥'c- V n does not include the
G,1
.,.,
lf(!. I
1
t. .,...
2-
(mV)
nccnn.;.
lf·
i.::l,te
c.vrre.u.t- is :3teered
& , , l::ueu
tf e 1 = Ve_ (! - :r... I< c. , a.. e Lt. a.M...je
ob:
vcz..
~((!!
:=
Vee.
4- ~
2
I
R.c
a..
e~ol!.
_ 05
5
10
20
30
40
9.97
9.87
9.50
8.95
8.30
Observe that the gain stays relatively nmstant
upto V.tnearly 20 mV. Then it decreases ~ignifi·
cnntly with the increase in signal level Whenever
gain depend~ on -signal level. nonlinear. disto11inn
Vae. - :r.. . f<..e
(b)
I:; o
= /1n
= !!.it'll
G,.
\,<\A o::u::_
T
I
I
Define 11ormali:led Gain
tXI
(!/-(
lJJII- 1 '82
I
7.32
u
=
-v-
lie 1 == in-112
common·mOde output, the ripple voltage doe~
not appear on the differential output.
This is. an advantage of using the differential
output compared to using the single ended output
(_ Q_)
I
1£1 - - - - ''t!J
VJ
QiJ
Chapter 7-12
7.34
7.35
Wlftt:
VtH-
i~ t
"'
i£1 "'
vsa.. ro t-1Av
-=*:.--...,......-- : :
t + e- tott5
0. 5 f 8 :r:.
{ec +{e. c.. "'
with
:r..
io ,:: in ""
For a.. eolleal:-ar res ia baM..ee l!.t:.
(f o ... lf~,-
Uez.-= (U{!(!; ... t'e, Re.)
(..Vee.. - iez.l!.e.)
(tl:..z. -L~t) ~c.
-= - ot ( t'E.t- t.:et) /(e
~ - 0.
-ruvs
1
I '16.. ::t:.· a. c.
Now
I.~ ZI.M.A ,
=l
5 mV. and n
= t.
-- 0551
•
~mVJ1lmV ··•
1 ,
= 0.45 1
I+ imVI,looY
=~
"'
VM
,
+ 0.55. IRe '"
OJ IRe
(0.1) IRe _ .,.,
0.005 V - (~0 JRc)V IV
(b) Each collector is biased at V cc
- ~ Rc
If we want to mainmin the same differential input.,
below its bias value.
so,
Vn.,;•> "" Vre- 0.5 IR,- 0.05 IRe
lflhis is permitted until l'cn = 0.
V /CM(m"') = VC(mon)
VC('
-
0.55 I Rc
lfthc gain is 20 IRe
-
De Cb itU) vo U::a...~e at
e0-au
ea llee.E:-or
\J
-= Vee - :J... R.c = l o- l ~
-
!V
l Rc =
A
so that
.....!'
20
vtcM'"'"'l
C.
··
vcc- 0.5:i
----w-A,.
=
lice- fHl275 A,.
2..5
-= --=f. '5V
For a.. -l V oul::ful: .swi~,
t:-1.-t.e vutz...titMUUA. vo lba..J e. b.f
ea..eU coli e e. bar ,·g :
-z_
l
I +e
"'
1
= Vee- 0.55 Ike
l::.uu!J
1-. -5 - 0~ 5 = 7 ,OV
T!Av s , Vx elM tM.~..
tl8 ~
/•a'"r
each collector should be allowed to fall by
R.c= 2. .5KVl.
"C."'
-
= - 0.45/Rc
A
~2-
7Ja =-tv / 0.1<16 I.l?..e
::I: R.c. :. 5 . l 0 2..
1+
r
Vn- Vn "'CVn:-iczRr)-Wcc-ict Rc)
0.1/k('
for
. ""' ---.:::,......,.,..
I
and 'n
,.,
1.1
= 11111
V1o1
in:: i~~:• ~'
o. 402 .L
iez. ...
. .I_,
1+ e
so, for a given VCC• A ,, reduces the maximum
allowed V1cM·
A.,(VN)
100
200
300
400
11cc-
2.75
lice •c
5.5
~·n·-
(V)
8.25
Vee-·
It
IRe (V)
5
10
15
20
Rc (k!l)
5
10
15
2{1
VJC.Mtm:;,'tl
fur example, if Vcc '" I 0 V, a gain of 2!)!) can tx~
a<: hived by incre:L~ing Rc to 10 k !), the muximum
commfm-modc input voltage would be Vee -· 5.5
4.5 V 1fa gain of300 is rcqmrc(~ if c;;~ be
ad1ievcd by changing RC to 15 kO. l!owcrver this
means thnt V1cl!imaxl = Vee ·- 8.25 '~ f. 75 V
Chapter 7-13
7.36
:::C:
= b k.t A
(c)
lue etJrrw.l: wt'U (Avicle 6el:wettJA.
bke. ?:.wo f::rtJ.usl:Jta/9 ,·"' f~'of'''t:a,.
ttJ l:lt.eN
tvltlt.
etAA Hrer ar~s. 1ti.v,
t'~..tfVI::
(.(0
For
Vnf.J
('~ 8)
690 mV + 25 mV ln
~"
690mV + 25 mV In
Va,; 1 "'' V"n
-~- { V 111. 1
=tet+Z.Ez.-== z..s I.Ez = 61M.4
Iet =2. .q t.wA
=
""
"" 640 mV
(-~2 )
""' 620 mV
~· 20 mV
200 mV =' VM ""· Vno- V,n
1
Iet• 1-~.:C.r:z.
'IEt
V Ml
+ 138 1tA
-(l!neJ
Rt + V 1,)
+ 62 p.A Rt: + Vd
200mV == V 111
-
V 82
+ ( 13ll1J.A- 62
~tA)R£
180 mA "" 16 JLA • Rc
c3.6&-UA
Rc "' 2.37 kf!
CoL:: {
Iet "" .5.6 u-tA
I Ct.:: Z.
1../r.M.A.......,.,._
~
I o etva.kte tlte eotJeei;ar
eurreut!J
we tlff,l_~ a c/,fJ.e!'euce st:~~.~ta..l
tftJ /; 1 V"Bz.- (/6 yl wlt ose va.Vve c.a.~.-t
=Iset
e(e ((
with
V6t -VE)Ivr)
w ke f e ::x :JE t 1 I :1t::Z. =- I . 5
AJ ow , (Gt -= (Gz w !te~-t.
(va, ,vl!tz.)tvr1 ~ 1_5
?JcJ ;- Ue,t -/}'61 c IJT lv... L5 ""10.1
~A
mV causes a differen-
LO mV
V~t- iE~vr)
LEt. .. ']:. SGl
J~d of20
tial current of76
G,., = .~6 ~A = :U mA/V ~ (263 H)- 1
be c!e&er&M.tM.e.r/ asc£otlow:r ·.
(El
(d) Without Rf:, a
RE
= 2.37 kf!, a <'M of 200 mV causes a
ditli:rential current of 76 f,lA
76 ~
G'm -- 200
1)18
m·V -- ..
,_,
e
t.uV
'IV
rlli'\
-· (76'
-~., kf!'
-o
_..,...
)
·I
So G,. has been reduced by a !actor of 10. This is
Jhc same fa;:tor by whkh V;,1 increased. So we
have traded differential gain for a wider usable
input range.
7.38
7.37
(,a)
v810 ~
Rc
690 onV
0. Vid ,, 0
ib) Eqn SJJ
tcr
+ 25
mV ln
CL!J
2)
•o·
6}2 mV
Ea.e. k de vle. e ; s af.'e rQ..l::,.&A.~ u...f:
a.. CUHeuf:: ar- 150.JtA = 0-I'S&.uA.
i!Avs 1
o'M ~ o.ts 'MA ;: b wA
IVa ..
"::
t{·::
-v
2.$ W\V
2.(~+1) (e
-y
2. )( 150
::::
Z frr
'?'5 K Ut
~
Chapter 7--14
7.41
7. 39
R. i J :>/
I 0 I( ell.
i
Ad
:=:
2.. 00 vI
vi
(}; i"/ lOO i VC!e :: I OV
RuJ .. 10'4 ... z.rrr = 2.." ~
cS~
((,.""'= 2-0tMit/v
=I>
ilws eUfA. dev/~e. fs 0 f.era.b,·w.J
o.~ ().51MA
a.tA.d I • ~A
fe •VT
I£
:: ZS""
="""'
'Z.~cll
VoLE:~ e. Oo..iVI .. ,~....... f<.c
\.) z.oo , tb R.c.
=l> f<..c.. ~ ( 0 ~Ill.
:
O.f V
2. ( 2 ~ + I oo)
(b)
t£,:: {+ o. 4
(t;z.:: 1 -
7.40
Ill
~
L:!i \-\14
o. 4 = 0.61Nf..4
5""'v
fe. : Vr -: 2.S~V
:X../2..
Ha..f r..-
,
'=
6oo vt
SQ)'tA
el rav,·t ((a.i"'-=- o<.f,zoo
f<.lvt ,./ 8oK
Chapter 7-15
(b)
VC'M .....
~ ,.111. = l ( ()+ t ) ( re -4- R. c )
o;.
=1>
2
re
'It
ZOI IC t.O (e.=
gooo
Thv5 ea.e.~ deviee. i& ore,.ab~f
a): a.. e.vrre&A.I: o~ Z.~ = t..SM.4~
IOcA.
~
.:C::
( v,. + Tli'ld)
= 5V
+ 0.4 V- S mV- 100 (30 mV)
= 2.39SV
li',J = A 4· "'d = 100 • 10 mV = I V
IRe = 21rRe
Eqn 8.80 g.,
= !.£
v,
1,. = g,.V,. IRe= 2(g., V 1 ) Rc Eqn8.93
A4
IRe
5~MA
Ad
= 5 v + 0.4 v - 10 ;'v - 100 ( 2S mV + IO ;'v)
80kfll
to (.4
C" 80000 :.
li',. = vec + 0.4 v - T
= g,. Rr:
= 2V1 A4
= (2)(25 mV)(IOO) = 5 V
1 = quiescent power = S mW
Vee- (-V££)
10 V
Rc = S V
I
= O.S mA
= 10 kO
(c)ForVc.,_ = OV
= SV
OV
7.43
A
(a) V sc s 0.4 V
V8
-
4
(for
+ 0.4 V-
= 5.4 V -
V;•- A
4
(25 mV +
V~•)
S mV - A 4 (30 mV)
= 5·395 V = 180V tV
30mV
V, 4 = 10 mV)
Ves0.4 V
CVc., + v,.l2)- CVec- inR,.) s 0.4 V
7.44
with IRe = 4V, andBSl>1lmingthata =I,
V eo = V n
.. = !.£
v,
g
=S-2=3V
Tji>.d
-[~g,.V1 RcJ+(g.. li'tRc)]
VeM-• = Vee+ 0.4V-
= Vre + 0.4V-
li';•- [A
4
I
= V ce- 2.
Re
Vee= +5
V1
+Ad~~]
= Vce+0.4V-~4 -A.[v,+li';•]
v.,
v,
Chapter7-16
Vel (t)
(a) .V!r1
= l + 0.005 sin(...,,}.
llirJ "'
I - O.OOS sin(~»t)
we see
,.~--
3.0~--
ll.S
that since VM = 10.0 mV .,. 1M,
Vr
25 mV
the output will he fairly linear. With the information given,
since ic "' l,
in ;:
l+e
,!vc
It! I
•J 1
and ic,::
l+e
(~i!l:tr)
-
-----------
----·
,.BE_
•
VoW
3.0~-
0.5
---------
---------- •
.. ·t
or
V.,;~
7.45
=
withlRc"" 4 Vaod
jV14 =
10 mV.
1
=sv( l+e~wt~
-
t'
•><~mat
•
1
)
l+eiOJl~.
"'989mV
,,:E
~
IOmV
3.0
2.1\04
79.2
Diffenmlinl
1/alf- Circuit
--------------- ------------
IVA!
lc
(b}
t~8 ~
URI
= J
+ 0.1
-· \I d
\7,'
Sin(t.J/)
II, -
=
LO V
_'s_.
25 mV
= 1 - 0.1 sin(wt)
flere,
200
v,J "" 200 mV = 8
Vr
25 mV
sec that this will dearly n::prcscnt large-signal
operation with significant distortion.
Using the same equation,
V,J '""'
·"'
4 V
1 ~..·- (-25
1 + t"
;(11},-
t -+
l
e~f~J/~~
7.46
)
;:4.0\'
\
-
~~ d -
v~ul
t~·~.J
5
0.2
v '"
v
rn+(fl+I)R,
25
waveform is distorted; uppo::r extursions arc lim-
ited to 5\'
and
I,.
So,
Chapter 7-17
(a)
+Vee
Differelllial
2
1/al[-CiiY:uit istM
same as rhe c:ireriit of
Part (a)
v,.,. + 2V.> r.,
7.48
1:~: ~ ",.~c Ifa= I,
t.Jt
tb.
-===::
l£!!. ~
/((!
v.:r:(!M z.~t
I
V.!!:l - Rc =
= Z.o. IC t.t
- 2r,
v~
na
(b) A,.=
... o . a, v;v
.1.!..
2(50)
= 20 vrv
2R:~c r,
-4oal8
Of
--===
7.49
tla
Vl
( Z. f e + z.. 1t. J..oo )
Wllt.re re ... VT
-
= o.os v- taocA.
0.5
0.'3f2.,
--
R.i
=
-=
2.0 aoo
ifo tft'
l..ll
\t-\,4
3'!1. ~v tv
6ot;}
(.{!>tt) (Zfet--ZJC1.tX1)
rat x 2. x 301 ~ 6o 1<.. c..t
-=-
£ a..a lt tra tA.s is tor l s af e f(A/:.,·t.A j
a.f:- :r..E • tfwA
bkv'
re • 25vz. t:tiAJ rrr * fat)( 2.5
1
'='-
-1/t'
Vo -::. ()(..
x
?
.s
I<.Vl.
(tfe. + 2.0Q)
~ l ,., ( ~ + I) ( f e.
t.(
t Z..OO 't (' e
with V, .• "= 0. V, "· -1}.7 V
~'A.:::::..~..t! ' :::...QL::::.~. ::. ~ "
,. " !':r
I,
Rr
25 mV ,=
~~
4J K
so n
t5 z~ lll
.-
7.51
'
A -
;:! K
- 023
··- 2(4.3 K) +50 - .
(c) CMRR (dB) = 20 log,.
IVoiVIJj
A,,.
I mA
) ~ l. 5 K til.
"" 20 log 10
l.l!l ""
0.23
38.8 dB
{d)
V 111
"'
0.1 sin 211 X 601 + 0.005siu 21T X 1000 1
V 81 = 0.1 sin 211 X 60t - 0.005sin 211" X 10001
V.., = 0.01 sin 211" X 10001
v•• = 0.1 sin 21TX 60 1
so that
vo
7.50
(a) 1 "·'
common-'mode half circuit
If u "" I,
=
Iv,4v.,. v,.
+At,..
v,,.,
V.{l) = 20 fO.ot sin 2:n- X JOOO tJ + 0.23
£0.1 sin 211' X 601]
\IJl) = 0.2 sin 211' x 10001 + 0.023 sin 211' x 601
Chapter7-19
~
AJ c f!o
.........
ct. .[
.
V'i
Rc.: ll(Rt..IZ.Jl
fe + RE/2..
.
~
i0//5
'= U ·T V/\1
a.ozs -ro.too -===::llr
R.id • 2.. ( ll& ll tf..+r) ( fe. + R£/z.)
1
[.son 101 (o.cns .,. 0-10&>) 1
lt.
'B K.JA.
._
":. 1..
'=
Tke
C!OtM.Wt.ou ..
Aetu. •
-
tfo
-:::
U:uz:w.
WA.ode ko..lr ~ireuil
0
-.300
f
7.54
J""' IOOp.A, fl "'SO, V-1 = 20V
ForQ,,
Ret"',.
'
J
o
=
v,.
I
= .lQ..Y_ = 200kfl
(UmA
20V
V
r "" r8 • = r , = ~ "" - - = 400 kfi
"
'
•·
112
0.05 mA
Using Eq. (8.103).
I+ Rc
R
"R
lcnr::. 1-' U.
1
R
+ c
J3r..,
+ 2R .
n
'o
I+
R
- 50(200 k) • •
""' -
andR('<< r.,,
7.53
(_a)
R;,.,.:: 50 (200 K) (.5) = 5 MH
Ad { s r"'s'e~eu.JQJ
-:::
e>t.. (~e.
z.r e.
oubpvl:
fe.
WUefe
0-025'11 -:::
c
11 r o)
lOQu
0· Z5u 10- 51- 0.51
I= 1.8mA
Vn = Vn =IV
=
A= 20 kH whercr
2r,. •
c
1.8 sin
~ol.
V
r.:~:_I_V
I
rc~Q._'!_+IV
rAQ.:
; 40 dB
R;
7.58
25 = 27.80
0.9
Thus. A,= 360VN
~
ARc
cm,R;
flRcl Rc = 2%
0
=
A
.,
ARc
.. 1
2IR c- /12R5
V, c 2''"w
v<.uvn~o
th- tb
Acm flRc
. • Rc
ARc
Thus, 20 log - 2-
-t
J
=
R;
IR c- /12R c 5.
Vn = VC(.- 2
2SIOWI
"n = Vee-
Rc
CMRR = ~=~
e:7
~+
~ aRc. !lRc
Chapter 7-22
£1o.J'
1
-=J_ ll o. U:~ K o .o~
Vott
2..
-= t .t.s 1/Vt.v
-::===.=
For :L =
Vov"' Vs-3· VI:.
Vo., = .L lC
= o. aI' v
o . 0 t • 8 . t 6 UA.V
--=Tkos bo~IA Ad a.M.-d Vos r'uerQa..!t!
~~ tlte ~c.:L~M.e. f'o!ra "tw.ee.. 6oH..t.
V art! fns1rarb,.ow.a.l l::a [Z'
.2.
0. 3l' JC
7.59
fO;
:I:. ~
z.oa fl-4- :
•
J
f:ywo. ... J2. K~~t' WfL :r.~ ... Z~t 4 ~~.a ·• •
-=- 0- 8-~ WtA /v
12o::. lOKc.ll.
itu.u.> A J u
9.. \M ~D
I
'#-
.. ( 0 ~ 0~
9 f ;: SR ~
=./1
7.60
Wofs b eases: 4V c • IO\MV
Af.o =o.o4; A CwtL} = o.otl
R.o
(_wtt.J
Vos 1 (!ve l:::o ARo) ..
\ftN A~D •
rro
'=
6tMV/
o.s
T
'A:
o.o~
Chapter 7-23
---
Vas,_ (due i:tl Alllll-) • \bv 41J.JIL. r: o.,J{(J .01[
2..
W/{..
2
::: 'I,4A. v//
7.61
Vov
c
Voss CrJve./;of:.JJI:):: AVE.:== lO'IM.V
-==--
Oiwee tke,e Off3ets are H.Ob
aor re l1'"r'[;;=w~_..;_ll_-=------.
Vo:s :: JVos, z .f Va~ c.'l. ,... Vas;"·"
Vo 5
-:::
J 6 t+ 6 z -~o l o ~ ""' 1.5. ll u. v
::::::=o===r
~
c
,-,;;;;-'
J~L J~Zo
=
o. t2.!V
we obtaiu.
Vas '·'du~1 to A..t.o fRo a..s:
Vo$• Vov Al.o =O.t..Z5)f..O.OS
T l<.o
?l
-::
f"r () lM
6.~~1- I c "" 2 11lus, the
7.64
v,. 1
300 v
lO
Vo3 = 25 ( 100 -
= L7mV
10)
300
Chapter 7--25
:r.. .
--
AR~
1.. ((';+I)
-2.. re
-AI. . R$ - .6J:. .
.....-
.! (~H)
.2_
= I.05JMA
Icz = o. YS"'"A
Otfs e 1: /A..v lltl.t.l{ t'3 a.c.IM.e v e.J
(b) .Ia'
~ :..:!_ .6."! + A I R. s t- A:t. f e
.!c:~+-~ l
2 z¢.-'>
6I. [ (e +...&_]-;:
!1 wlteu x. v /s .!Jua.I.A i:::lw..t
"·05 (X +5) :or- 0.95 ((1-x)~~)
T
_!:_. AfH
A\J~
-==
7.69
-e.:z:.. Rc.
::: -.::r.R.c 6R..s
_A-:-_
fe + !;;> f
re
lvo~l ~ ..b.(/1Rs)
z~
.I.s\4.o.cu:
=
'
7.70
&.E..D ·
7.68
.Ie-r • kJ:__
.3
+{Jl , 1 ~
::: /!.cz. + ({-X) Jt. IK01.
-5. Z5 +-X =- 4. :ps + ;f- X
.-t> X .,
o. Z.5
I..ez"' ..L:t.
3
tw,-u flu a.reo.. Of
AIJ c ~ U"c.z.- Vet
adlu'eve..fJ' wUe.IA. ~ Y'-3 -'tJc.'-< i:W
~
aMd
6. Z.tb K..vt
R.e 1.. "" -5 X o . 95 = 4 . 1- 51< lll.
Peft.e.e6 oftsel: f/ltJlltu~ will be
R.c. t
~ ~ :. ~A
I/Z.
{31M•'" + t So+ 1
Is \M.,· . . • :I: I z.. ... ~ I.;Jjt A
f?:.KA-•11 +I 2.co + I
Ios -= .I:sM4A¥ -.:z: SM.t•'"' .. ~A-
(_(y,
Rc.t "' 5 )( I ·05 :
c.
ret-~
r, +I
r;-rt
(a.)
a.zz5
e:
Qz.)
1.. I ~c.
3
--===-
NowJt.talLI.f ,
ArJ c: vl?.c. = :x.. R. c.
-(e
LVr
Vo!l =- .6-lfe "" z.. vr = /6 . ".? wt. v
'Ttw~ 1
AJ3--
S IM-ail-
8r'q_ua..l a.ua..~'f s t' 5
prediats tl, ~ . . . r OA • _ __:__
2R.,;$ ] + Km:. Ytr.~
l'i(ttt
A,,. '
-v,._, ·=
Ia) /"' c~· I "" 200 FA
-1
v
-i:~~~~. l + 11
mA...:..IIV,.,_}-{l-!JI_lk_)
"" -1).()1
CMRR!dB) "''
" 2() I<> I!,.,
(J_Q_ ). "'
' . 001
~
. m!A...:!ul
20 lt•g
74 dll
Chapter 7--28
7.74
7.75
CMRR = (g.r0 )Cg.R..)
(a) For a simple current mirror
R.ss = r 05 =>(for/ll = 1!2)
CMRR = (g. r ")(g. r ml
_(21"
v,)
(21o
v,)
V
1
V v 21
0 ,.
1J
0
0
=2·~·~
Vov
=
Vm,
z(~)'
Vov
Q.E.D.
(b) for the modified Wilson current source of
RSS =. g., 7 • r07 · r 0 ~
·c.• , 'u7 ro,>
=> CMRR = (gmr 0 }(g,.
for """
For Q5, 6, 1.8:
Vovs =
J.w
while for
0 1• 1• 3, 4 :
k
11.
~I
Vov~Jk:!L
=> Vov.< = .fi Vov'
Thus, (for I= 2/0 )
VPO t- Vss
21
vA vA
jiv 0 • . T. T
7.76
4 VA'
VA''
v.,.
vov
.fi - , = 2 . .fi --.2
Fork. W!L = 10 mA/V 2
/=I mA
IJ.AI = IOV
I mA
10 mA/V 2
= 0316V
.
=> For the simple current mirror case:
CMRR = 2(_!Q_)'
0.316
= 2000
-+ 66dB
For the Wilson source:
CMRR =
-+
=:2. (r.5)+Z(1.35)
=
CMRR = _!_. ~ . _!_
V0 v (/ /2) V0 v.
=
=VGs
- \1 ss-+ Z VGss-• t 2. VGst-t-1 -= Voo
21
2.fi~
(0.316) 1
R9442
99dB
::; g,.rj,
ROf,
~(g..,.
rf'f>) r0 •
= g ... r~~
A, = s~
I
1!
= g,. · igm r"
5.~v
=
Chapter 7-29
thus. g.r., = 2V,I v.,
~A 4 =
Q.E.D.
Note, through, that this is inconsistent with KVL.
lfiu"' o. Pro= o~ but Ym = Ym = -11;,14.
2(VA!Vov)z
v14 but .Ib'IS C(}hfl'ICIS .Wit. h ··''J)l
4,
=0, V t>J
If 110
f'or V,1 , ·""· 0.25 V & V, "' 20 V
A, "" 2 (20 I 0.25)' "' 12300 V/V
L-·
vctng
v,d
-4.
It appears that the appro:dmatioos for Vir.• and "•
prevwt a clean 110lutron. If these were more cuct.
7.77
all current and voltage relutionships should be
consistent.
7.78
I
.
I
_ I _ 100 f.LA
m = I m ""' I m = t.N - 2 - · 2
g
. ml
.ml
0.5 mAIV
""
v... --
ro,\
2_I Rm r o
%
6
tl.(J- i!il
T~4
=
In
20 v
-~·= 400 kH
0.05 mA
!l.~; ~A
= 240 kH
R,. = r,.l II r,~ "' 400 k lr 240 k '-"' 150 k!l
However. if we usc KVL.
i
=
50 1>-A
=·~=
50~~oA
v.,12
02
v /2 -O'l
- ., mt\IV
=g
G., ""g,. 1
~"
A4 = G,.R. = {0.5 mA IV)( 150 k) = 75 VI V
Gain will be reduced by a factor of 2. if
RL = R,, = 150k!l
f'
-id
::::::;:·
rH
g,.l
v--i"d
4
7.79
R.;r)- ({j+ 1) 2. fe.
(which is the same a< i3}
-
j f e = zs""v • -sect~.
~OJ' A
__:p.
R.o =
0
gml'_\{.>·.:t
'
4I $,,
R. ,.rl ~ I or Y. t aoa
Vd
0
• I 0 II< «A.
(ar.at faz. ;:: fo i fo== VA
L
Ie
Chapter 7-30
Ad=
c;~f<.o
W;biA.
.. Z1<
(600:::
szaa V/V
svh~e~ve.ut s~a.~e l..o..v,'tcj
tt
a.. 100 /(.c..l. 11Acvb re& t5totAI!e.'
Ad =G~AA (to {{ taatu)
=
188.2 V/V
==="
7.80
~IM. •
:r.tz • 5"""
7.81
v
Vr
.X. • 26oM
R•
•
Gl,
6-(:-~-v,. ..~----co
IlL
-'P
fo:::
VT •
i:/2.
wU.tU. ,
! ~ \M. V
~IAA4
R.id = t5l X !.X 0-2.
'l!. :
lOa
:r
•
11
Z.OO c./l.
~0. C( /( "t
SooKvt.
::I.e o.tzs
& ,. [!! .:: 4.ool<.c.t
'Z..
-=---
Ad-= OIM R.o = 6~ 4oo ... tooo v lv
.:1:8.
=.!.!:..-. ~ .. 0-85}t4
(.lf-1
/51
j'-" ;>6~/clv
.:t: • ~ 4
( rrow. Prol>lttM. ~.68
above.)
-=Joo.
o.z.s
f e. •
S~M is s~rll ~(~J
l:o;
.!.:! :: 3~. z I< c.t -5"
R. irl c (~ +1 ) 2.. re
Rt'tJ" ({h• ).t (re+Rl) a~.,........_,.
ll-ud
fe::
Y!:-=
-VEe
J..ooc/L
1
(o•
80QJCIA.
'S-It.
lk lid • IOOKvz. ~
U lOOK • 151" 2. 11
--t;t
(zoo+ lle)
flE. • 13I \A.
lo ob Go.ht Ad:
Ad= GIM. .l..o (E~a.t· 1.16~)
A& iu l:.U.e. d~riva1iou ok t.oz. £M.
£ i u . (-=?. I 6 Z), & z. ca."- be ff~"~
bvt ~lt,.w
tiiMe.v w,dl,·w.( ll<.a.6 I.e aJ 6ke.
uiiu'{ €'f"'-.
(6.151},
etMA:!ta ollJQz
{) (el
.,u,:
+ i.ll£
Chapter 7-31
~~ • flit. [ I+ 5....f{ lao ~Zii:or)P fn~
Ro't •
9001< [ \
l.ot ::
+ 5""" lt2M+ tJc t&r) (~o.zt< ~
(~+l)fe
J..,' tO K Cit
AeiNl :
- fo'l
15o,c Z'lOI(
= -
aJMJ
t!H.Rt.
1
Js.s._vlv
=lt.l.(oo.r~.
-J B.StA4
,;o, QSl
f,e.. 105 dB
~ 0 • ~ot.ll fol(
-=b>
• - (Zle ZllOK)
{3a~Ec
- (2.'2() UBOO) 1<. •
AcJ• ~~ )( ~IS a: =
6!3l
sumtiluting in for if.
A<"nl
:S:./z.
1. 5 I< '-t.
tor a. ~ twp le.
Z.ttao~
V;.,
[A., i 1 - G"""" V1..,)R0 ,.
v,.)
i .
t;
I - .r:~} ~ I - ~
r(,J''
= A. it so~
A,. i; .
1-
Si!l(:e 8m4
g.,4foJ 11
,
g,.,,,
I
1+-·
Rm'J .rl~Ji
C~mtinuing,
IAion
{"ffJ
we can substilute this i1110 the equa-
o~p: 8{-·--·-' -lJ
(a):
-
III("HI
tnl:'
1
1+-·-g>i<(\ ro:t
fl
Chapter 7-32
7.84
for a wilson current mirror,
A"" '""
I -('l + _ t )]
[ , . g.. t~l'
(~-)·
~R,.
lo _
/I!U-J+
J
1+-'-
g,., r.,
I
.2
. lHP + 2)
·.
As no active load,. thi!l menM that one collector
-
current will be n/, while the other is
2
+
(1/(1
2
:r·~
)
2
al(1
T r-13Hl +-2) - t)
l!.lr'I
=
nl(~]
13<1~ + 2l
a.J
-::-
':"
'm
r,,
tib
+ib
f·
T
a/
g., "' Vr = 2Vt
G.,
=
IV
.! _
()Sj
_
Ai _ . al
Gm - 13(@ + 2) a/
-
2\lT
for (3p
"=·
50.
2(25 mV) IV m I -- 50(50""'
2) -
A
'"
'"'(_!!_).{3 + 2 -
2Vr
13<13 + 2)
19 2 V
· ~t
I
1 + 2/13
7.85
Now, suhstiW.ting into the resulting <'> 2.
(!'!-@- 2)
~
+2
..
=
-r,,. (--,L')
,.J3 +
2Ru:"
(;MA.Ro
2~
wl
Vhit..umtll
=:- Ran.ge of
is:
VJJia!
-~(L4
V
VA
G
m
Ro :: UO/ o.tu.) II ( 12..0/0-l&M.) ='46e.lil
c;JAA. : :. KA-1 ""' J .:t:. J<. I< p' w I L'
r""' Jo.L/t.wJr~ 6-4111AA/vz.
::.=t> Gtu =l. 61M 4
=
5!l, f1~
150 &
Ad: GJM. ~ Ro
... 1-6UA.~ x. 46 ~ lA.
12()
= g r
~-'I
<3 V
---v
c- n..t << vbi"·' "'' 4.11 v
For: l " 0.4 mA. 13"
I/2,
d
V cr- 0.2 -· 0.7 = VM"' ""''·
And: V 0 ~ t' 1,~(Jsmm
:I/2..
= ~:.?.!!0
'g!:O.~
15 mV
V
~V>"
A-rl:
v
7 3.6 V/v
==--
Chapter 7-34
7.88
1m
=1
=1
= / 01
08
116
n• = !_
= I ·2
= 1m
1,, = lo2
= I
jVd
= luF = 225 11-A
225 tJ.A
2
g.,, ·· •
= 112.5 !-LA
From Eq. (8.180}, systemic balance will occur in
this circuit when
r
(I),
(w) .
2
-
(I\ (w) (2}(~)
4
=
112
~
0.225 rnA
= 80k0
9V
= 40k0
0.1125mA
'o•>
-(O.l
=-
V SS + V CSJ = - 1.5
with
jV nsl
= IV ovl·
+ 1 = 0.5 V
this would be when
V 51 = - 0.5 + 0.25 = -0.25 V
= 0.25 v
/OJ
=
~
1.8 mAIV
A 0 = A 1 X A, = (-36)(-36) = 1296 VN
60 11-A
IV o•I'·.
=
=
0.25 V
A, = -gm6(r0611 ro,)
W6 = 20
!Vovl,,,
=~
112
=
=
20
To find
0 1- •
= 0.9 mAIV
= 2(225 11-Al
A 1 = -g~,(rolll
·(.!Q)
( 60) 0.5
0.5
-().5
SO,
I
= 225
11-A
0.25 V
I V0 ¥JN
'"•-•
T. • - (![), T.
= _f!1_
I V ovll2
8 "'5
(I).= 2(I),
(I).
= IV.! + jVovl , so all are
= 0.75 + 0.25 = 1.0 v
IV G~
~
;:-c
r-n
Vi•mi• = V 51
2( 112.5 11-AJ
Ox
180 IL Atv'(.!Q)
0.5
V,~mu
= 0.25 v
V 50 = -0.25- I = -1.25 V
= Voo -!Vovl
v,• .,., =
21
-
Tile upper limit is when a~ leaves saturation:
2(225 !LA)
=
1.5-0.25 = 1.25 V
Vsmu- Vsr. = 1.25-1.0 = +0.25V
so, range is (-1.25 V to +0.25 V)
For the output range, V omu is
60 11-Atv( 60 )
0.5
= V oo - IV ovl = 1.5 - 0.25 = 1.25 V
Yamin = - Vss + !Vovl = - 1.5 + 0.25
= -1.25 v
= 0.25 v
V omu
so the output mnge is (-1.25 V +1.25 V.)
Q,
Q2
Q3
Q,
Qs
Q6
Q,
Os
I,(!LA}
112.5
112.5
112.5
112.5
225
225
225
225
IVovi1
= 111...,-""
200 J.tA
= 400 flA
No requirements are given for Q6 and Q1, so
choose
* Jm,
(d)rm = r04
= ~"'
lm
r06 = rcn "'
~
1 ;
= 2..Y_
0.4 mA
111
""' lm "' 2/ flU
400 j.i.A
""
g
2..Y_
0.2mA
= 25k0
= 12.5 kO
·2 mA_2mAN
.. , -~
- ,,..~ -R- vovn - 0o::rVn
-
Q,
Ql
Q)
Q4
Q,
Q6
Q1
Qg
g ·~ = JIJJ6! ., 0.4 gtA = 4 mAIV
25
25
100
too
50
100
50
25
A1
(!
"'
Vc"n
0.2n
= g;. 1(rozll
r 04 )
""25VN
A2 = -g.,;t(r06tl
= (2 mAIV)(25 kQ
-4 mAN(I2.5
r 07 ) =
25 k)
kll
= -2SVN
(!)u.s
"'
A0 = A 1 ·A 2 = 25(-25) = ··625 VN
2f11n·
k~(V 0\')1
25
(t)H
21Ru
2(200
k 1,(V,;;)!
400 ~LAIVJ!0.2
=
7.90
~tA)
I =
v}'
= J(j()
(!L
(a} V0 ,.
by tn
g., = 21/Vov = k ·
"' 200
Ideally. 11 0 (dc)
.
4(200 j.t.A)
if k increases by 4
gm increases by x 2
(hlA 1 =gmROl
100 flAiV 1 (t.U VJ'
o
·me
!Vosl
= - V 5s +
-
I V + 0.2 V
= -0.8V
-0.2 v
0.4 + 0.2 - 0.8
The upper inpntlirnil is when Q 1 and Q21eave the
.sn:turatif)n region:
l - (0.4
Vm ·= Voll- VsM
Vas! =
o·
V 01
IVm·l "'
,.
v,.
0.2 V.
=
+ 0.:!)
so
o..t v ""
"'· the range of input vohug;:
! --0.2 V to +0.8 Vl
0 '
~·> A 1 increases X 2 as does r\ 0
(c) Offsets due to V, mismatch are unaffected.
Others reduced x
OJ!
v
~X
90
4()
OAV
~
-O.R V
><)range i;; 1-0.1\ V to +fUl V)
"A
,...
= 112.5
~
90
~.
22.5 flA
=> V. "" 22.5 J.l (r,.6 ll r.,7)
i~
=
since .4 0 increases X 2
7.91
/' 1
"·
o,q
W,,l
i
= 112.5 J.tA
Output off.\ct ~urrenl "" ID7 - ID6
(c) The mu.ximum output voltage is
- Vss +
V(>V
~
(h) For the common-mode inpot range:
lower limit is when Q5 is leaving saturation.
Vm
J¥
If K Increases by 4 ~ V ov de.:reases
(,(V,w)'
4/RM
kr(V(wl'
L c
=
2(2/RtF)
"' 50
('!:\::)
h. V~v
2
- I
+- 0.2
=
_I_O_ = 88.9 k.fl
112.5 j.t
v" ""
22.5 11 ( 111 k 11 88.9 k)
~I.IIV
I mV
12.5 k)
Chapter 7-36
7.92
Oftsel:. euaeut .. .I oz.- ~ott
....
~
Io3"'
z.
.I..o~
( Vu-s- Vl::)
-
fer:. reL.: 2~ .. Zt50t.ll
(b)
0-l"'-A~
~O't
~
(es "'..t5Joo\A.II1\4.1-4
f~ • ~v""'
I.vq "" K ( VGs - (Vt -+AVt))'
z
.:r: o
:: r
I
#
=
o3 -
"For
I 04
l(V(Js- Vc- Vcs! t
VI:- tAVt:) ><
)1
z
2.
.:I:o =
f_eaa..lf
=J>
l2..5ut
2..\MA
( V6s- Vt + Vo-s- 'Jc -AVe
A \It.~ ( V6.$ -2v1: -AVe)
~
Z..5lA.
GW~r
=
.2.'1/.A.V
0· 31M
flo
la.) .:I:.e• .. Le2. "' o.t 'IM.A
::!::
.:rr.s , .:te'4
IE6:: Z.IMA
zso
V
z
lo~ I V/V
Far tk e. eoWt.IM..O w- e IM.i tt e.t :
A 5 ;:. - 0\M.S . I< cs
-(J.f!.L -=- -too,. tOJ(
res
z.s
-= -4o, ooo v;v
For fl.te e!M.i lte..t J.oUowe.A;
A~6
~·
"For AVt "' 2w..
v
Vas = 0 · .s~ )(. 2\,V\ =
aetlve. loo..ded
rqkc ; Reeo.ll trOIM. q~;t.
~_!._=_I ;
K ( VGS- vt.) • A \1 E::
J'-"3 .A\It
6ue
t{ a.€
(7.1~1)
t:.1< e dow.A.tA.o.r.A-E: lcwt~~iueu0 role.. ts .!Je/;
Ce 9<
(C..) .S1·"'-ce
fp ..
,...
=t>
c :::.
-=
zrr.r<..o1
60
= LOOH~
_.,
(A6.,.1)cc
....___.__.._
e/1
~'I lUll eA.
ed
"/ ( zrr J<.Z5'i5 l( 'IOK ~ 100)
15. '=i-6r~
Chapter 7-37
7.94
r6
225 p;A
""
V OYIJ
,_,, C0 x = lllO p.AJ'fl
=
..
then
/2 X 225li ,., 0.25
,J60f,J.X:I20
Vou
v
_21o_2.x225~=t
c.... 9
IV ••~ -
-
0.25 + 0.7
= 0.95 V
+ 0.95
= Vc;•o = 0.4 V
Vas "" Vov- Vs~:,11 = 1.5 + (- 0.25- 0.7)
= +O.SSV
Finally from the rcsuhs above:
(Wl L)10 = 20 I 0.5
(Wi L) 11 = 20/0.5
(W ILJ12 = 8010.5
(IY U)u "" 20 I 0.5
"'~
I ""
IV o•l."
IV ov•l = 0.25 V
Vt;u = - 0.:55
p.1 Cmr
60 p.A I V1
For Q1 & Q9 : WIL = 6010.5
=>jV••f
""
~ V,;su '"'
0.25 V
,. I.SmAIV
Since g.. of Qu1, Q11 &. Q,3 are identical to g., of
Qa & QA) then v,.n = 0.25 v
ThusforQn
·
2 X 225 11
- 180 p. X (W IL)n
,1 _
(0. 25 '
7.95
--. ( W IL)jJ = 40 i.e, (20 l 0.5)
Since Q12 i., 4 times liS wide M Q13 , then
(W/L) 1, = ~
.•
R,"'
= 80/0.5
2
Jz k~ (W !L),ll,
J2
-t
0.5
X
·(
ISO.:~ Xl25. ['-'--.;-f4~--,-]
R11 = 555.6
A'-:.
.tt<..c. If f<.,·dz
Z (i.Et +ret)
f.,·rJt•(('.>+l) (2.(e.2.)%
=P At = tz.s v1v
n
The voltage drop on Rfl is :
555.6 X 22511- = 0.125 V.
To Qbtain the gale voltages:( assume IVrnl
0.7V)
2 X 225
"' !V,rl =
Ai.:: /;;;::
p. .., 0.125 V
80
180 !A- X ().5
VOVIZ
=
VG$12-
_, v"·"~
-=
+ 0.7 "' OJUS V
VGI~.n = V., 511
0.825
+ 0.125
==-
"' -0.55 v
+ IIJR 11 - Vu
- 1.5
7.96
- ...!!.L+r
~~ t- I
'"
Ro -
= R
c
Thus R5 aiTccls R0 We want R1~ il 3 k "" 76
"~
R0
'" 7~
""}R~ =·'
= 7.34 kfl
f>l.i
f.t'dz. + l.t..c.
4
L4 x 1.0 AlA
thus,
%>
2..fAn = 8513 X "iOO"+i5:! = 3378
7.98
{)-=- 100 ~
0(-=-
..:Z:EI = 1.03\M.A
Loa ~ I a\ .. J.oz ""'"A
101
~ lfooc
=
Thus,
(3+-1
-·
Vc1 ~ Jov- r.oz""'A x s.1 KVl. :: 4 .g v
= 1\\UA
4.5 K\..l
_. :X:.c z. ... o. ~ • too :: 2.~1<\.ll
~w.·
~
--~z /I too II Z..5)K =l~S~I<~
.,.
,.,
wltere.
/l,·IA
=-=1>'
,
t.P'. " -- t.rrx s-szp
r,-,
l~
/1 [ fe3 _.,
Root.:: IOK
-=
l1. 81+-t)(fe!>+ lot<.)
-= to. 6 KIf rot x ( .t5 + ra K )
-::::: l0.5fs '" V ov
So,
Vllmi••"" Vcs•+Vov+Vss= V,+2Vov-Vm,
Tho maximum limit occurs when Q1 or Q2 hegins
to leave the saturated regiQn: For example, when
Vs, = Voo-Vs(n-Vovt
( l _ _1_)1
-
Since
i
=
"
g,.,
.1J.. 1
rmJJ
g,.4 amlrm
= r0~.
v·w"'[( I - _I )(I - _1 )]
Rss
'
Cm_,roJ
g,.~rM
V 0 = i 11 (r,6 ll r 08 ) and Since - 1-
>>I
gltf3ro)
V,(mnl =
Vn•••"
=
V, -1· Vot
+ Vsl
V,+V 0 ,+V00 -V,+2V 0 v
V,(max)"" Vm,- V<>«J
So the range is
(d) lfthe current souree is fabricated as a simple
current mirror, R 5 ,
(- V 0 /l
+ V, + 2V£w S V 11:M S
v,
I
7.100
(a}
+5V
-5 v
\1 00
-
~' 0 .,)
Chapter 7-42
DC All!lly$is
1M! -1-4.3! ""79kll
R·""'
~;t
Node v(iltaps:
v:. .,. -4.3V
!ii.,.l
+ "'
.._
~
~-
IX
!x
l.65 X to' = 3~
2
v
100 !LA
v,"'
Vc"" +0.7V
VE ..
Vel
-().7V
'l'bu5; !!/.,. 3300 VN (Polarity corrett)
"" = 0 v
v,
vt... + 4:3V
+ 3.6V
.(d) R1•
Vc·"" +4.3V
..
2 "••
""2X. IOO
:50\.0
4
(b)
Transistor
ic;(mA)
em( mAN)
r,(mfij
a,
'0.1
4
2
Ql
0.1
4
2
Ql
0.1
4
z
Q4
1.0
40
0.2
Qs
0
0
QA
0.1
QB
0.2
Qc
0.1
Qo
1.0
Q£
0.1
Qp
(U
Q(i
0.2
""
R. =
""'
···················
~-···
2
0.2
n
n· T
DO • 04H
[r
. ••
+ rozd
f! +~'t»]
I
"' 0.2ft 0.21 [ 25.10- 6 + l J
:~X 2]
""16.4 kn
(ch•u.'M1miol = - 4.3 - 0.4 + 0.1
= -4V
VICM!iau1 = VG
+ 0.4"' + 4.7V
(f) The vohage at the base of Q4 can rise to Jln.
( V E)
········~~-~"······
T
+ 0.4
""
+4 V
b!lfore Q) saturating. Thus "o can go up to + 3.3 V
Thll voltngc at the output Cllll go down to v"- of
Q0 + 0.4 "' VA - 0.4 = - 4.3- 0.4 .. -.4.7 V
Thus the linctlr range at the output is - 4. 7 V to
+3.3V
(g) AI the positive limit of v0
······· .......
(c) Total resistnnce at '-"<>llllC1or Q3 is
"'l:l.•'o.•llrodl
(!34
+ l)(ro.l!
ro/))
= IOOX2U2ft t01(0.2lj 0.2)
"' 1.65 Mil
I..arge
.?
(ignore.)
f-tmA
+3.6V
R,.
+
v,
i.e. "t) "' + 3.3 V and Q2 just cut off
R, ,, . 3.3 V
·
9.1 mA
= J63 n
(this is the minimum allowed R, for+ 3.3 V outrut)
Chapter 7--43
-33 V and
At the nq1,ativc limit of ''o Le. v 0
l
2
Q1 has cutll roc= ((28.3 X 5)(2.5 II 2)) II s
= 4.9 Mn
+
v mA
-:f
+2Ve-i
= 28.3 X 4.9 =
and vo =
138.7 V /V
2kfl
(rooll r06)
Crooft 'o6) + ...L
v.,,
8 .. 6
(I
II
I)
lo =
... I
(IQ l)+_L
200
0
•
-
V,)
2
=.) V0 ""0.17V
For the lowest posible output, the circuit becomes
~=138.7V/V
R;. =
I
W
2"•CoxL(2V
oo
R.,.,=roollro611 llg.,6
= Ill •II 11200 Mn
-skn
(e)
VtC"Mimul
= VG
+ V,
= +4.3V
VICMimiol
=
Where:
Q6 cuts off and Q7 conducts
V GSI + V Rmin
= Vest+ VA- V,
= 1.7 - 3.3 - I = -2.6 V
(t) Vo.,.. = Vc.,..- VGS6
Io =
=
= Vt + IV,I- Vcs6
=+I +1-I.S=+O.SV
Vomin
= v.i- v, =
- 3.3- I = -4.3
v
Vo
T
-
0.05 mA
~IL,.Cox('~)<- tio + 4.3 -
=.) 9o = 1.45 V
That is, the range of v0 is
- 1.45 V to + 0.17 V
1)2
Chapter 8-1
8.4
8.1
CTOr = Cs+ Cn + Cc~"" 3 ~tF
rl)"' ,_ .. • 'l-"" '""'"/..,
U$o"' ..., • 't·fCf
Rc = 10 MH, R,, "' 100 kH, g,.
W« ....0!.\J'lt ~
}.. . = _, ... Q.p
:: -~
"H•I·'i~~;,..R.s
,.,,..,
= 2 ;A.
R0 = Rt"" IOkfl
-~ x 1o···J . ....!_
"
(,., = -"-"'-~C$ = · •
211'· Cs
·
2n
ln
4
_ 3.18 X 10
- - fn _. __
I
/,., =
211' · Cn · (RG. + R_.~~it )
8.2
~Cr
=:)
.,.
'"l
. -7
~
~(1\.+ ~ollro)
);
l. /
="7
~to""
I
I
/,
IC1l(l.I111(1C"-t' 1/tl)t>lf.)
Cc 1 .::.0.1-)A?
=> t~
f,,)
-;>C. '), o. 6 7)4F
Cz. 7
.. Cf.62Ha.
! f r 0 i; cluvbtecl t.V,·t-h.. botH. r0
~ RI> hAivd:
f I
::::
~- ~rt,.o.;r(lo~<+~ll~)
!)...
)..;
Cc, = L57 X 1{)-s
.r,.,
= ;:;---:;:,---!---2n-. Cr1· (R 11 + Rt.l
=> Cu "' 7.95 X 10' 6
fp,
If we choose· J, - ~ r
•
• 1'2 --'L>ll'l = fti25,fl'3 =frf5
(ror"" 3 ~-tF = c5 + c0 -t Cn
3 ~-tF "' ~ + 1.57 X 10-a
ft.
.h/25 .
13.5HI:
+ 7.95
X JO-~
l,15
=> J, = 120 H1. and C, ~ 2.65 1~f;
Cn "' 33 nF.
0.33 ~·F
(:{'2 =
If wcehoose: _r,, = f/
~ -·
-
ClOT = 3 ~-tF
= C5·
,-l!.
=
f.J..
I
_ LJ:
') .
I'J '" :!5
+ ('Cl ' Cn
~
3 ~tF "" l.!.!.~_I_Q_~ + I_ 57 x 10 s
f,
f,_i'i
+ 7.95 ><
w·"
---.1;125
"=> f,_
= 172.3 Jb·.
c\ , LX 1~r.
Cn = 455 pf~ Cc "' 1.15 ltF
8.5
7.44 kH,
If c" ,~, 50
R,_ '" 72 n.
l.l Ul
R,, ,
,._._r. Cc,
=
Cc~
= 2 !-'-"
f/', -- ------'-----
211. Ccl Rc,
' ··-.-~+ ·--.. -
J..p_, .. __,.1_ _ __
2.nC'L (.IL
+~)
-'D
U
.rrX2<10
lt.,
= 2 rr . (:;~~--:
I
R'
X7.4-txro'
10.7 Hz
Chapter 8-2
8.7
8.6
Zrr· 50
X 10~~
xn
=
100, lc = 100 p.A
13 =
'21rCc~ R0
X Hf.~X
6J Hz
13 X !OJ
1:.,
= lc ,.. 1.00 p.A = 4 ~
Jl
h "" fr,+ f,,l + f!'J
= 40mA/V,
r 0 = 25 Hl and
Thl'Jl,
n
II
Rn "' (Rn
If I c is reduced by half, since
_ Ic
, 20 rnA
,lim - V, 4g,.
V
J!. 4
rw)
g.,
4
+ R,; 1
25 K) + 20 K
Rn = 42.22 kU
RF: "' ,.,
R,,,
Rsll
+ fl + I
°
r, = 50 n
"' 247 5 + <200 Kft 2 K) "' 427.52 0
.
101.
Then:
Rn
II
r rr = 5 kfl and
gm
r, "' .£.
w--·.•
> "' 247.5 n
g:t:t "' 4Xo.9'10-J
"" (200 K
r~ =
V
_!!!L_ ""· 25 kfl
4 x
g,.
"'61 Ht
r, ·""
Km
25mV
Vr
FmmEq9..19
r, = 25
20 kO, Rc "' 20 kfi,
""'
Rn .., 200 kfl, Rc "" 10 kO,
I
.... 2rrX 2
R,;1
44.2 Hz
= (R 8 tl
Rc2 = Rc
r ~}
= (100 K !I
+
R,.,
5 K)
R = r + Rn ft R,,.
£
<
13+1
+ RL
= 20 K
+ 10 K
= 30 K
lfwechoose ft"' 100Hz and
+5
K =
9.76 kfi
SO+ !_Q9_K
!i
5 K
101
fr2"'0.9Xft
cf..
·
=
21T(0.9 X
'
.fd
X
R£
= 97fl
Rc7 "'· R.- + R,_ = 13 ld!
Fort;. to contribute SO% of ji_
21f
X
2rr
X
0.1 X 2rr X 10() = _ __,;..,_ _
Cn · 9.76
90
wo
=· - -•..
X
427.52
= 4.2 tJ.f
X
JO'
5% off L each we have:
C
-
cc
"'
2'11'(0.05
. Cl -
X
100)
X
(42.22 K)
I
--.,...-.,...---,~---:-:-::-::-:-
2'1T(0.05 X 100) X (30 K)
The resulting
0,! X 27T
I
Selecting Cc 1 and Cc such as they contribute
100 ~ _I_-~ C£ = 20.51 J.l.f
.
Ct: · 97
For Cn and Cn !() crmuilmte W% o(h each
0.8
X
'"' 0.8 tJ.F
I f.LF
f ,_ is:
L_. ···~
Cc; · J:'\ >: l(ll
fl '" 2'JT{4.2 fl. /(427.52)
··> Cn = !.23 t-tF
Tb verify Ihe value of fi that results,
f,
:}; (elf X
1
20.::> I J.l.· + 9.76 K :;
+
L(~tf~
99.X9 Hz
+
p.~JOK}
J, = 98.65 Jh.
The 1<1tal capacitance is:
cI
li
0.8 fl. X 4:1.22 K
0.1" J.l + I J.l
+ 4.2
ll
= 6 J,l.f
Chapter 8-3
8.9
8.8
Tu select eli: so that it contribute~ 90% of the
value offi_
l r 11 ~"- ll cr~ + r11 )
t..vl.ere ((,-= .53ktll ' Rz. = Zl K.C4
l(,·va
s
I
.
.
0.9 X 100 R.: = 110.8 fi
(Fmm problem 9.1.1)
f X =50 a.Vt.d,
frr-=- {)o '= lZ..O
=}
--0\M
-= lZ.O = lOK...t
O. 3X L(Q
R. iVP·
·~
•
,
211" Ct: · R1;
!> 3// 2.2 II I o. 05
""
15.9 11F
Cr.~
so that it contrihutes 5% ofJi..;
Ret"" 10.7kfl
\2
~
5.9 /(Ill
'Co
C k'
To select
c, =
-~-....!--~-
21T· 10.7 X IOJ X 0.05 X 100
" .2.97 J.tF
AM ,. - Ri\11
ftr . . . 3~ (~! u~ c.Ura)
Riw + Rs frr + f>~..
-=- ~ •_JE._ . 12 (4.)-f16.~l!:Jao)
f
.
To seleet Ca so that it contributes 5% of ft: R ..,
= 10.3 kfl
(_
~Cc2:::::::: -~~.....:,-----
211" · 10.3
"" 3.1 fJ.F
5.~-r-5 10+-0.05
'"""
-JG. Ll
V/V
RSt'j .. ( rr If [(X
1
+ ( R.t(l ~zit rz,,~
)]
= 2..69Kut
R.c 1 "" t< s +- LR. e, 11 ( r" +- rrr J1
= ro+[1olf
R. e ' • R. ~ (/
2.~3rF
""' (2. . =t 3
=
Z 1T ¥.':/00 .rol.
I
1 ).
-
l.=l~pF
-:::
= s3rF
fH
!J/ [ Tf e j \1\ • f!. S i ~
-::. "/2. n 33 .t0 .2.Y 6q' I o'!>
~
1
1zll
1(
-;.
l . -=1 )> 11 Ht:.
--~-*''"''''''
v--'"'*
-----·-······------·-
+ o.r +- (_ 1attta)
100+)
-::: 5'=1Vt
fof Ce
e} vall
Of'
err +Sf<-(I!+.J\M.R.'l...)
t . ":l 3 p + It f' ( If+ 12.x z. 6 3 )
Civt =
~ ~II t
av...d
J
pF
.f rr .r r x + ( Rsll 1!. ~ )
po+-1
= fo II Rc 1/t C e
&,
1
Ce R.e'"' Cct. R_ct
:::
Rc 1
Rt.'
-::
ro. <:( 9
= 1'13
o,o51 ~
Chapter 8--4
8.11
Thus 4> = tan
t,. = f:l· '· = 13. v,
Rs + r,..
V0 = -lc(Rc
I
R,.)
= -13
(Rc U Rtl.
Rs+r.
wp£
CJlpc
=·-(tan- 1L
v,
/,E
t tan·tLJ
he
=-[tan -I L +tan -I.LJ
_ V 0 _ _ 13 (Rcll R,)
A.,-v,Rs+r.
100
10
Thusatf= lOOHz cf> = -(tan- 1 1 + tan- 1 10)
= ( + ~)
c,. r, 13 +I
b)PoleduetoCE: "'n
-tan -1("')
-1("')
.:: 129.3°
I
PoleduetoCc: "''c '~ Cc(Rc + R.)
zeros are both at <
40dB~
20 loa (100)
=0
26d8~
__.e:..--+- 20 log(20)
I
c) A(s) = A.,. (s
[s
+
(
I
C, R, +
d) AM -
IR,xc,l
~)]X [s
13 +
=
~c•.
c,
I
--
I
31.8
2s +
w,
I
.fpc= 10Hz~ 10Hz = 21T Cc (RL
I
21T
X
10(10
:I..e
== .:r. e
Vo • -R..c I-c.
- R.e
fe+-Re+
A(S ) = ~
+ 10) · 10
+ Rc)
, = 0.8
fe+~~ + _1_
s
.,.r
+- _,.,,;.___
ce (fe+~a)
fe+ ~e
lJJ L •
32dB
40d8/Dec
OdB
-L-'------1---~lOHz
lOOHz
Unity-gain frequency must be an octave lower
than 10Hz i.e. at 5Hz
g)
A(jw)
...
2
~ -A.,.
(w, ••
+ jw)(wpc; + jO>)
I 4{).
(w,.
+ jO>)(w,.,. + jw)
f
_-1-'---
Ce;.Ue-tl!..e.)
. V..s
SC£
= -..:..R..:::c::..___
- t?..c
12d8 -----,
1..
SCE.
20 dB/Dec
f)
31.8 Hz
La.) I e
V.s
= 21r. 100. (0.025 + 0.100) X IOJ
=
~
tS9.1
= 12.1 .,.r
~ Cc
It 1$9.1
s-=-s-
I
I
-40 VN
e) Since the resistance that fonns the pole "'P£ is
very small, we choose to make "'P£ the dominant
pole, thus:
I
/p£.=ft=IOO=
(
IOK)
2"
---20db
8.12
+ Cc(R: + R,)]
I
- 100(10 II 10)
100
10 + 40
-
201or
-f;l (Rc II Rtl
Rs + r •
=
A(s)
0
+ Wp£)(s + w,c)
ChapterB-5
(b) A11 is reJve.e.d 6(} l:ue.
(e
fa..el;o I'
qx
= 4.3
lo "" 100 fl.A ""
~X
vov
v
=;:>
V05
"'
wldau
/s l:lt.e sru«l! a.-1 6ue
-~ctt'&-t. re dvebtou 1- a.e&ot. {l-w.J 1
U tke vo.ive of £e cau 6e o.1ed
as l::b.e ra.ra 1M ete r ~or
exeret'srue the Ja.!w- 6a.udwld!lt
r0
=.J....=
A/0
J::: ~-= lO,aoo.,.
re
'C
ZtrCe
Po lower fL
5 use:
Of 5
z.s
re
4ao V!v
-
-1_~-
'::< _ _
t..n~~.roth.to~"J(z..s
~'1 a..fael::ar Of
V
lootA. . ilte<~o...i1.1.
towf!re.d
to
Vlw
X
=
0.05
J
~Saturation
1
X
0.1 "' ~ 00k'!
.,
7
_0 ·5
~ 0.19
2J0.65+1
r
2~
g,.b "' Xfl,. "' 167.2 JJ.NV
Cov = WL 0 vC.,,, = 20
?
- 63.71- Ht:.
,·s atso
1.5 V >
v,;.
X~
X
0.05
4.3
X
"" 4.3 ff
f..e:::o:
R.e = 4re ""'
= 0.23
193.5
g,. "' = 880 JJ.NV,
Vov
trade of!V
f L.
~
21o
(e.
lAM
10' 3 F/m~ ,~, 4.3 fF! 11m~
,, 193.5 1J,A/V2
is redve!e. d 6_'1 tlt.e
aetar ( 1 +- €!E_ )
V
t_d)
X
k: "' fl.C., "' 450 X 10- 4 X 4.3 X 10'~
(c.) Wt..
t
"' t~:., = 3.45 X JO' II
'••
8 X 10 9
C
+- R.e
re
If + l(e
re
-
8.13
80 v;v
6\1
a.. Ft-or
=
C11"
j
X
"'
c -
20
X
C0 v
I
X
43 + 4.3 =
(i 16
fF
= 4.3 IF
c.....
,, -f-f- ~,:; -·- J ' -~ "
15
9 6 fF
7 fF
Chapter 8-6
8.17
8.15
The intrinsic gain A 0 is
Ir
j2 ·"'· ·c,):!:' · 1,
fl.,. =
Also
"'!1 ·ro =
'"
V."·L
[,
c,,:: ~WL ·c.,, if C1 ,
>>
c,. then w<
then we have :
Ir
Vnv
lo
= 2 V• ancl
Vov
= 5VIJ.LmXI.
2 X 5(V/!!;m). L =50
0.2[VJ
Linmm.
• From problem 9.19
c,d· lfwereplaceforg.,and c,,
can ignore
(21o)·(~)
X
l. VN with
3 x 45o[vc~s'] · 0.21 v 1
Ir -
3!!;. Vov - _ _......,:...:_...:::.,::,___ _
-· 4rr ·1. 2 4111. 2
= ./21J.., • C,,.(W/L)/ 0
2 · ,.,.. ~W
3 · L·C , ..
2.15 X 10- 1
L.'
/.nun
Therefore we can see that the higher the current I 0
then the higher isir- Also the frequency is inversely
proportional to the size of the device, i.e. higher
frequencies are achievable for smaller devices.
8.16
I ,. --
fl. ••
2'1T(C,, + c,dJ
= 0.18 X 10- 6 m
IL_
2L_
Ao(VN)
9
18
27
36
45
f;.-[GHz)
66.35
16.59
7.37
4.14
2.65
3L,_ 41.....
5L..,
For C1, >> C8d and the over lap capacitance of CB·'
,
negligibly small:
Also g~ =
c,,:: jWLC.,,
·w
v2/oov = k.,zVov
If we substitude g. a~
formulas:
Ir
=
8.18
c, . in {)from the above
k):!:' Vov
1
21r x
L
~WI.C.,,
4rrL"
Therefore, fm a given devicefr ;, proportional to
"Vov
~~
~TT ( CTT+ C.)t)
-=
=>I.,= 3"'·v~,.
Vov·Ir
fT•
t!G>< to·!
f. rr (to+ 1) J( to-n
:: 4.~4GH1:.
For L = I IJ.IO, V0 v = 0.25:
Ir
= 3 X 450 X to-• X 0.25
4 X 1T X I X 10- 12
For Vov = 0.5 V
:ITJ
In
= V o•'
v,»,
""'In = 2.7 x oois
fr = 5.4 GHz
2 .7 GHl.
fr 1 (3o = (4.Zl.ljtso)x
~
.2..9.2.6MHe
101
ChapterB-7
AI:: I.e '"' if .OwA 1 lkte.l = II. G
tJ.t f .. oao 1-t Ha 1 l:b.vs :
o
r, ,, 100
'il
~- lc
r
=
·"'
v,.
n
C.JL = ~ -11' C tr .:- $ ~ -C)t
z_ rr fT
2. rrt-r
C11 (:r.e• a.t\.144) =
81t to ·.3_ a.o5.to 12
Ctr+
Cn (.:tc.
"
zn w ,.ts.tall
0. '1686 pF
4ono·$ -a.o&,no''
tt{.Q\M4)....
!1. n
~
15.7 fJ'
C,:: 2C~,"' 2
X
20
-:
40 fF
Ct. ,, -r.- g.., = 30 :< 10· 1 ~
=
X
20
X
10·J
600 IF
!0 '
211'(0.64 + (}.0!6)
1.04~6
---;
X
J0
pF
f.{O,t0 1
ca0e
21r(C~ +C.,)
20 X
--=
6.8wiQ"
't:f .. I. 04=1-6, •6
SolvilA.{ Ef"'"· (.a' a.w.d (2..)
tke't ,~leLJs,
gm
r =
l(
S,·wee C1T "''fj·e t- 'tf,~'" ,
Cje + 8..- ro"tr ::: a.t/6'1/t> X lo-•1..
CJ'e +
C • = C1, + L~,,, = 0.640 pF
1,
5. S c!:-Ha
2..11 ( err + C:.Jt >
= 0.5 mA = "'0 mAN
25 mV
~
VA = . 50 V ~' 100 kfi
J,
05 rnA
"
c
fT '"' 11.6 ~ 500 •
fr ==- 0w. ·
=
Cje ::. 'b.'i5pF
4.&5 GHz
1
L
12
lVT = j~ I Ccn t- C.Jt)
l "' te \ = tT I r
oAt :Ic=0.2.1M.A, ll1te\-=-Z..5
a& f = soo JtHt: , tuv.J:
ZTT
xSx IO ~ = Z.Oxto-3
C
_
fT+o.t
(CTr+O.l)XIO-t~
-
zo
=- o.¢4oF
!OTT
f
Clf• 0.64 fF
Lr::: t.-5 x !!Jaa ::: t. t5 &-He
l
- -
[3\M = 2a-w.Atv
ftr:o
t2;s~ = 15oft-a ~
fft:: fr
!(>; ..
(2.)
LF =,f-!flps
8.21
8.20
(I)
::;, s /(t.l.
Chapter8-8
8.24
8.22
le
.,..,
g,.
(fl}
(mAN)
"•
Po
(rnA)
{kO)
I
25
40
2.5
t.(l()
(a)
{b)
I
25
40
3.13
125;3
(c)
0.99
25.3
39.6
2.525
1.00.
{d)
10
2.5
400
0.25
100
(e)
0.1
250
4
25
(f)
1.0
25
40
0.25
100
10
(g)
1.25
20
50
0.20
10
8.23
i! •
f)(+
:.
J.. +J·""~'II"
fTr
.,. . r,..
I t-j W.--
110
c ..
c..
(pF)
(pF)
(MHz)
(a)
400
2
13.9
4
{b)
501.3
2
10.7
4
(c)
400
2
13.8
4
(d)
400
2
157
4
(e)
100
2
4.4
1
(f)
400
2
13.9
40
(g)
800
I
9
80
Wt1
-frr
10()
=? too
( affro)f()
'•
8.25
+ ~~~)1
+(~)z
\
lr
{MH:t)
CONT.
I
-~-
C;. =
c,, + c,4
"' c.,+ C8d
(I
+ g,.R~)
= 0.5 + 0.1( I + 29) "" 3.5 pF
Neglecting R c; :
. Cl . R
fu "' .,
..,.'1'1
f.>
IOMlJz
""> R,; 8
~...
~
~i-
i.e Rc; is very large if
l
>R.
21T·3.5xto·•~x106
.,,
< 4.55 Hl
Chapter 8--9
8.27
8.26
g.= I mVA: C1, = I pF; C84 = 0.4 pF;
Since Rc; is very large:
I
Cm "'4.26pF;AM=
v ;f11
-7V
=382KHz
~GB =" 7 X 3&2 ·IOJ = 2.67 MJh
We also Know that:
Cg, ·"'
C;. ·.,;; 2 tr X 10 X 106 X I X 10'
C,.,;; 15.91 l'F
-·.~(··
~,._,.,
~It-~
+c
5 x 10 ~z
+ IX
10· '\I + 5 X 10
' · Rt).
,;;; 15.91 pF ~ RL,;; 1982 fl
Since Ra is very large: t\" = - g,. · RL
A.,;,: -5
x w·l.
1981 ~ .4,
z -9.91 vrv
Gain. bandwidth product: GB ..
GB;;;,: t}.l x !OX 10" tm
If .f 11
2:
;;'>
IA\1i · BW
91 MHz
W ·Cox+ Wl. 0 v · C 0 x
WCo.t(~ + t 01)
"'
(I+. grn R~l,;; 15.91 pF
~yJ-
~
C8a"' WL0 v C,._..
=l>
if W is reduced by half so are Cs• and
~ c,, 2 = o.s pF
c,,11 = 0.2 pf
CR.~
In saturation:
1 ·w
~
.
. .
In "" -k -(V00.) =0 For ln to remant
2 11L
·
unchanged while w is halved
=l> ( V ov) 1
is doubled
!f MHt
I
--g
.fi"'
n
then : R~ s 8349
A,;;. -47.75VfV
GB;? 139.2 MHz
·"" 0.707 mA IV
we can now calculate the new values of A~h Cu
fHandGB
-.l.·V/V
.fi
"' -4.9 V/V
C, 1 ~
(J ·I
0.707 t 7.14) · 0.2
X
10··!!
1.21 pF
.1.71 pF
2-rr. 1.71 )( 10
950KHz
GEl;!
The
4}) X 950 · lO~ '" 4.65 MHz
r~tins
of new vs old vahws are:
1,
Chapter 8-1 0
8.28
R,,~ • roof)
;!&.
t';,j
L
R~~$+1-tf{~s
.!:L .. -" '{
1- s
I+S'<~+~)R~,.
c - 111
~)
IF-
""a...
'"
'J
8.32
= 2 mA, p = 100, f r = 800 MHz
= 4k0, r, = 500
v. = 100, C~ =I pF.R.,1 = 5k0
~) =~(1-t..l-- +5..!.)
(1-
=875KHz
R1 = 50kfi,Rr
Vs~-<,,..~s+~+"'> w _, ..!&.-
.:t:
~~
"'C,J
R1.
llm
r
•
= 5 kfi
=
2 mA
25 mV
= 80 rnA
V
= g.,
~ = ...!QQ.
80 m
= 1250 n
r = ~ = 100 V = 50 kfi
"
fc
2 rnA
c.+ C~
'
= !!!
=
80 X 10-l
211' X 800 X 106
U>7·
-j,., t(
I+ S(cc.+Sd)RL
C~
=I
pF~C.
16 pF
= ISpF
.,----·.
_
A
- R•
--;-''•:....X:-g7;.,;-RT,t'-;;--:-;
R8 + R,11 (r • + r, + (R 8 II R, 1,))
= '• II
> R,ir.• 'x << R,.,,
=87 pF
R•i• >> r 11 , g,R~ >>I.
The resistance seen by C r is R r.
g., R~C~ >> C1r
R,
1
fu =
21T X 10- 12 X 100 X 2.5( 103 )
= 636KHz
/ 11
= r.ll
I~~:l / 11 = 6.36 · 106 Hz
when A.,
HR, IIR3
II 21 lito = 3.2 w
=572KHz
R't=25k0
A.,,:
40
R,
=6.25 II 68
Thus
I
fu= 2-CR
1T 1" T
I
2'1T(6.36 X 10 6 )C~ X
f:l
~
g.,
= 2 GH7.,
= 120
CJ.L
= 1 pF,
R,;,= 0
= 80 mA
v
= 120 = 1.5k0
0.08
L
~x10•
2tr(C, + CJ.L) => •
=
0.08 X JO'l => C = 5.4 F
21T(C. + 1)
•
p
= -IOV/V
a)lfA.,
C• = 1 X 10-"
f:l
= 100
R~ = 250
n
-10
8.34
II
where r. =
R,
r
•
R1•
=
"'
0~0~ 5 ~
= 17 pF
Thus /
32 mA/V
2tr ·
A.,= R ;·R.
s
••
2n · 17 X IO·I. X 133.3
= 3.2
- 4 ·72
kfi
,
X
-g,.R,
X 32 X
10 + 4.72
=-32.8 V /V
5.1 -0.8
=
1.5 pF
1.6 GHz
u
= ,.. wII'"f ,. =
~
n
=> R;_ = t3.3
c,. =
f 11
32
10-.1
27T X 10
'"
= O.S mA = 20 mA IV
3.2)
25 mV
=~=~=5kfi
20 mAtV
g,
v. -
8 . '3 E5' T; -
c,. + c..,_
4.3 pF
X
r
X
------.
=5.1 pF
~
b) A.., = (-I VI V)
''"
C 1 = 4.3 + 0.8 (I + 32
MHz
3.2
C,. = Ctr+CI'-(1 +11.,R,l
where C1r + CJ.L
c,. (R~)
-----'--,,:---- = 70.2
6811 2711 6.25 = 4.72 kfl
R-
C1r
11
200 = 6.25 kfi
32
R~ = Rcll RL = 4.711 10
=
=
1.;~-~. · 0.08 X R~= 133.3 0.
= 5.4 + 1( 1 + 0.08 X 133.3)
Vr
lim
g,. =
c,.
II '•
J! and 11~ = !.s:_
R,. = R,
1
c. t c..,.o +g. R~>
=
=
100
v
0.5 mA = 200 k!l
!G!
w,
=
211'
20 X 10->
~ 4 pF
X 800 X 106
Chapter 8-13
z, ""' ___;,.-.__
c,... ""
Since
l pF -t C1r "' J pF
The midband voltage gain is:
(g.. +~)+ sC'lT
-R~~
R8 + R,;,~
r#
;::--t r, + (R
il /?,!I R;.
:(2oo!l 8 n s) kfl.
+ sC11'r,
.!. +sC11'
1! R,.g). g.,. R~,
11
re
,.,. __1__ ,
,
re
wilh !?,_ ""· r 0
Since C 1t contains a component that is proportional to the bias current, it ft>llows that at high
,,,3 kn
Then ~:.,R;;='20 X Jo•60 V !V
-100 - _ ___i___-·
100+5 5+!l05+(100i! 5)
= -29.1 V! V or 29.3 dB
A.., ""
currents C'lf >> C11 and
X
60
T() detem1ine [ 11 :
!r, + (R 8 H R,,,)J
· 5 n 10,05
Ttm~.
/
11
"noo !I
(at high currents)
The phase angle will be -45" at ,.,
0
r~i!
Thus,
l+s/w1-
4 + l · (I + 60) '- 65 pF
R;,.
g,.
l
::::2,;C;:::11r · C'lf r,
z, = __r_l'_
C.,.+ C~J. (I + g,.,R~)
C'"
f1
5)J
f = J1
2.45kfi
=
wr, or
= 400 MHz
__,_·,_
For a lower bias current so that C 11
fr = - -1--- and
4nCnr,
z.'
1 +....!...
2w-,-
-45. angle is obtained at w
GB = 29.1
X
999
X
!OJ
=
29.1 MHz
In example 9A: A., = -39 V /V,
754 KH1.
--~
GB
29A MHz
Nn1ice how operation at lower supply voltage,
thus I r n::duL'ed tho: mid-hand gain. im:r"nsed the
( 11
while keeping the gain-bund width pr(lduct
consrnnt.
8.37
=
2wr or
J =2fT= 800 Mlh.
(Assuming { 7 remains constant which is nol
necessarily true)
Chapter 8-14
This figure is for 8.38
20log/A($)
40db-+---....
20db
I
'
I
I
I
I
I
!
IM
fH (MHz)
(log)
-20db
--------~-~--------------
8.38
40 db = 201og A0 ~ A0
+ IOO
2rr~ 10
1 )( 1
(l + _!.--:)(I
r-
(b)
20log A
2tr~ 10;,)
X
10'• ·
+ --w._.v6 )
2n
X
10
60
-r---- Slope- 20d0/decodc
~
40
{)r..s ·-.p
IK
;~·=·~
·)~
•J'--:!:r:XH)"'.
f11 "' 0.995 MHz
8.39
+
(t+.~)
+ 100-
2r.
Wn
100 VI V
(I+ sllOO X}()' X 2tr)
(• +
A(s) "'
""
··t 6 t . ;f =-L~ y)
. (= .
2-:;y_·J(I/
\2.'l'f·<'.;Hf'...-
~:~
-
---
I() K 100 K I M
••
-
-- w
••
1
(c) Gain-bandwidr.h product •"' HlOO
)<:
f(J·l )·.
z
log scale
10 K
IOMHz
{d) Frnm the gain plot, unity gain
frequency= 10 MHz
(e) From the plot unity gain frequency = 1 MHz
Chapter 8-15
8.42
201ogA
SlC "" - 2f) dB/de<'ade
60+----"""
40
2U
f~z)
0 -1---.--+--..---'~-..l K 101<
lOM
8.43
8.40
fu(s) ., ---..,:'-:----
wH!!:'
(t + :J(t + ~:J
tn,., < Wr~
I
~str;.l
w 11 t::
I
fir'?J.s•SJ, .. IPF
using the root sum of squares fotlllula:
"-'II:::
wrl
fl +
I
~ ...~, w~1
~ J;~·:-··w,l
1
L,J
o'l11
is:
(
f'ro,.. EJ,c-.plt 6.6)
,1{ .,3.33k.J'L.
•j.,.etrMA/11"
=--~--~~~--~~~------~
to 1 );.~-tli'2 ( tl~.33;1.
411'1 "':2-nJ,,oJ(
If
""6'f.o~ =>R•o2.6tc..ft.
0
"'r!: ~"'"
II
w,.i
f1lrt
8.44
for dw"
where: r_,, = C,_, · Rr< = C,_,('R,.. IJ R,;~)
JO%z0.l""'>l! '' 2.07
(I),~,
for n«)n
I%= 001-"'"
r,,1 = C,J · Rsu:>Mto.>:
to
ro'rndfs ~
.
R,, · N,, 1
Ri
-t
R,!l
ll
RL
I;., R I + ~il
r.
!l
'.IF) ..!
Chapter 8-16
b) r ""' C1•
•
R;,, where R~ 11 ""
c,. "" c.,+ c••
where
RoO R,11 and
c." U.the result of
applying miller's thenrem to rellect C,a to the
gare-groond nodes.
Frumf..q9.76
_llsCt"
I - ( -g,.R~)
z''q "'~
I- K
_ , !_ _
sC,u( I + g,.R;)
~ C,, ~ C,i I + g,.R:)
r =
Thus
(c., + c,a(l + g.,R~)) . Ra ~ R,,!!
Rc; .
R,,~
Evalvating for: R0 = 420 k.O
c.,
C3;~ = I pF R~ "" 3.33 kH
=
R,,1 =- lOO k.O g,. •"" 4 mAN
( 1) For the complete expression found in part a)
'Tn "" 1.230 11S -t '"n ""' 813 K rad/s
(2) For the npproximate expres.~ion found in part
b) .,. = !.228 fl.S
-t w 11 = 814 K rnd!S
('Til-T) X 100/'T = 0.163%
8.47
8.45
a)
If a capacitor CL is connected Jn parallel
R~ then
with
w 11 ""
I
.
Vo-= -g.,Rt.Vs,Ol
V.~-~ =- \l.'lix-RsXgm\l,gA
v_,J I
+ g.,Rs) =
v,,,
....... + ..,J + RL. CL
the values of r 8 , aml
T, 4
remain unaffected since
each is derived by selling the other capadtors to
lCfO. When mosidering the re,;istances seen by c~..
C~.,"'
CnJ = 0 a.nd Vc, = 0
~ the open-circuit
time constant of C1 is: R ;, · C L
~""II
'"'
t~fJ.~ .. tlf.M))
y_
w- 9
H-'-----·
~ 13~x 103 2nx w--l~
"'" ·.•, 765 K radl~
/ 11 ""
<>• 11 !21r
=
121.7 MHz
Rl..\ = ~::~
i\
R
Chapter 8-17
= 100 !l:
Rs
~
- 4 X.5
v,,,
I
R~,
= J+ 4
= -14.3VIV
+ 4 X 0.1
100 + 0.1
X O.l "' 11.5 kO
RtJ ,., 5 + IOO + 4 X 5 X 100
I
+4
1533.6 kH
X 0.1
I
IO·I! X 11.5 X IOJ
!Oainl
Vc,. -Ri_v
-
10 ·ll X 1533.6 X 10'
623 K rad!s
to calculate R~
V5
+
V,,}
Riv-
X
Bandwidth
= 14.3 X 623 K "' 8.91 M radls
2so n:
R,"'
V 5 ""'Rs X g.,V,,
-4
X
5
-10 V/V
I + 4 X 0.25
100 + 0.25 """ 50.1 kfi
1 + 4 X 0.25
5
+ 100 + 4
X
S X IOO
+ 4 X 0.25
I
1105 kfl
"'n = -...,.,------,--'1---:-::----......-..,
10
I!
X 50.1 X 103
+
10 " X 1105
X
10~
= 865.7 K rad/s
lgainl X Bandwidth
10
X
865.7 K
"' 8.66 M radls
Summnry mble:
Rsoll.)
tl a,
c.c..ltA.Lt...h.
~::_;_;A_ _
1000
""
I
r::
30 ~ 31o d"-t 2.'1 3.75'"
.2"t6."1~
:1
v.....:.'( i.....,:.., 1-re'fu.t.M.'7J :
l~fz:l
v;-
f1.5
~.S -t c;..t -t ~
H
r...
1+): ~~s
-1000
Chapter 8-19
8.52
Using Miller's 'Ibeorem,. i" each c•tbe
capacit;mce nt.the input is C(l.-A) and the
capacitan«\ at the oolplll is
c( I - k}
C,·,...
'flillll:
IIOpFa:nd C""" II pF
~, ~s +~ ~"' +1.1\.t.
t"tt •
Ss Rs.J +~ f ~,.~ <~+;... ~> +R:J + 1.'{
7;,~: .2~ '"I)"+ ol[~«c,.,. ;,J.DJ '+ .t:.J -t ,,,. .2o"'
lH •
= -R.
{Miller's theorem)
v
'r--tf "'.26 'f 11S
b) IN = .:...!!!
R,;g
R,,,ll
R,.
f.11
~ ...1- • 6o3 ICHa.
i',..,
•-9,., R~
/LTI~
~,.s
:
~J
:
:(F
If R.,."" Rthen:
-~=~
,U({!
Nfpt-"''f
Flf
to
'"''1'' J1
ffv,.
t>lf.A.
olttcu..~J
nc:> f'(a t~
Prqblc,...
dJAA te> lt.Affke.l-r),/
+Jw. fl..e. c.onslb..n.fsof'i A--d ).s .
KCLat /\:
Vo_ V
2
C,.,tr/i>,J;'4,. oF ~4ek. fr'i+w:.~
'~-·' 1
6'.6/i ~ ~
R
c)
N~t~,
IS.1Z
;,. p,-,,/eW1 6. (; 6 1 tJJ e
R,v = Rl! (-R)"' '",.'•·'" l"
2---~ + :!11
.. -'S',..2o.:-llloVfr
n.3~.
rL ;
"" R,;~
J.olt.
8.55
8.53
R., =
(E.,.6.,.;}
{. ..., . 6 .'544}
1n~ett.t~ R:
•(2+o·l (1-t'S.,:lo)j:).o +Cf.,.o.t),.lO
• :2.6'f n S ,. 26Li,.lo"l ~e.c.
~Pf>t'OK•'.........G., 11'-..L..t.of. fp 1
is :
~_;
I
oH~4
I
.tn[(C,HC.,cl (lt-3., R:J)~g + (C.Lt1 ~ Wp:a. .....~ ~....re.:
""'Pr .. '3.81.1"''., rod(5
Wp;t "'-7"3 .\2.
f'"""'/!1
~~- f;, vs,"; &1. .&·C6 ,·, I·Ji lowrr
+"-. tft.. cx.cu.t 1/tJw.. _,,.;1...'/c #/.&.ut-I~ oi- ltz
is .,f,rJ....t
fkj/.4,. fl.-lfs ~~.
esfo/M...G-
1.3z
... ..,. f-r .. ~ .. 6l\.1"§"-ll~
"
'1''0
fpJ. = 4S.67 hH•
1h~
1,
!> =CI
->
I
'-1'1
F-p,_ = ~ : 'f5.o6Ht+~
::tn
c>bJ-c.c.i.wl us~
=
[cc..."c;J.J)s +-tz. c;,] R~ Rs,)
2"
-t'fl.o
J
fCtH c,., ,.~
=[l.S..+~.a)~s -r~c..,.d ~at~...
(L
is f-p1 :
+kL tr-...skr-
fv,c)-ton oF e.rr....,.:C.·~~~~(6.6o),wko.se. ~-.\inQ.~or
is
9:
J.p, ,__d ;.~ << '-: ,
~S~,",.,Q):~ {.,,. }H
8.57
R.' = ~ICJ'L
:
AM"-,,., ( = -'SJCS' .,.-.2-~ ~V"
f .Pl ...
~P."'
1
I
1rt~C.,'t'~JCI1",1.,.(~J?\') -r-(li.+)c~IR~]
I
.ttt[( :t+O.I .. (I+ ')~S)).2.o
+(ftD·1)15]
tp1 ::/.63 HH-a
~PJ" [Cp-t5f~C1-t$=(J]~·,.,. '5,.-c;p\
<
f<~+~l'Js-t-£C,c~] Rs.~ I'.ZIT
IJ.z..,.
C.t. + o.1( 't')•5')).to + (t.,.O:!J5
(<•+11·1) .. :1_ + /•9·1)
f.PJ"'
s~
..CP,
<< t.p~
S",.J.o
K2n
675' MHi!
g_
::: .R..:::_
C,J
"?
o-cJ lp, {<
-,..t .,,
n
q__
S"'
~ ::: -L!!:!.. ..
~2
J"C,c1
-..
-5>,
2rr11.1
iJ tilL
~Hl
J;;:~
Chapter 8-21
t, ~ ~p,• I. 63 HHl:
~~ f•
AH.-
e-Jw.'t/H,.,. .1.5,. f.'-' ..
teD. U H fla.
r11 .,.1}_ .. 12.£ .,5'~ frM =-
~t. .,.f1411ffl;., =' lt0.1!FJ.IH._
S,·,Le.
~«
?.r
,,..
(1,./{>
I
I;;
( .1c:h< 5')
+ o. :J. + '5
"l'v:
c.,. 6'~t:
AM= So.65
/
c,." ::: c11 -+ ~ r ,....; .. Rt. >·•" + o. 5'£1+ .1ol( ~)= &o,;~
AH• -5",ctO::r-So"{11
1 ·
):.P:J.""
rn
~a,r,+~
Us,-,; M•llu".s T/..t11rtwt ~J
/(::.co 11\.Jl...:
.2n
J.o
~I ~/ << f.i ,I().. Jt>....c.'..-.t
pal• e.xrsts.
~Pt..
8.58
G-'f· 6. '"lf, A;
.
. . . ,.
.
ftz. +et.l (H f"~IO)}JIJ +(I-tO .I)JCtDj
:a • r11 ff (~Srj .,. rx) = S~l(tl<+o.2)
R;,,3 "D.11f(JL
1.o.li HWe
('l.+D·ICI+~I'to)).!tJ t (1+!411!11 :. 52'/~MH~
(}t-f<<'•IJ.2. 1" (J·U ID•lf>)<2.Ji
f 1 ,.._a__ :
7.ti6GH"J.
2rhtN
<< f!(>L
fp,
t
f.p,
«Fa.
-7
J1.t
pole ct.• J +1-.Lrr {-.,f'Y.
!A,_, I·
tS t-1.. do~'it-J'
f.p1 ,.
/.
o4f H HJ.
Fi-t "' 5o>< 1 ot.; ... s:l MI-l ,
,
f.r"""
Pr,Jcft,..6.6.1..vtht«K.·
= 603.16 K rl :~,
-the.
IA,..I.f.u
S·i'IC.C:
tA
<.< F-<1 , rJvrt.fo,..~
;~
tkffroY-,-...ItbJ
?p,
!AMI· f:~
ihe rese~lt~
,t~rrFcY"e. f.r
,·s SJN..II11'r~
ar<. 5vwt"''ar:ee:d ,.,. "'this ~-q bfe.:
s-/(1\..
IO
t<.J!..
td'l.
!l.O
~..15
-50
~loo
1.63
1.04
o.6o
5:uo
6c.32
~1>.15
R~j
+ 0,3
kHJ=764n
p[{ I + 40 X 5) · 764
+5KJ+3pX5K
r,, = 7J>411$+ 47.57 ns + 15 ns
!,,
,...60.32. MH~
<. V"M/•f,..
+ C 1,(( I + g,.R;) · R.. , +
7n = (10 p X 764)
·c:::·
f.11 !.Cf.h::: 6<.3.16Kfl'l
fp:~.
Tn = C, · R;,,
R;,g = r,ll (R,,, + r,)
= z.sxnn o kO+o.t
J.ol"''...irt~ pp/l! a.nd ?11 ~ b!.
by?,.
.A.,= -139VN
Using the method of open-circuit time constants,
fmm equation 9.100:
+ C,·R;
f.P.:t = 4~.61 Hf/1<
J:~ ,. "f.'f6 Gl-/2;.
AJ.U"7't << ~~ A4AJ
8.59
70.21 ns
~
._-rr-ru
:: 2.27 MHz
The .(.l:> ('Ontributions to T11 1[Jr.cach capacitance .are:
CJT: IO.ll%. Ctt : 67.8%. c~,: 21.4%
f, is 10.6~'f· higher than the,/;,
nhtaincd in this pmhlem
Chapter 8--22
8.61
8.60
~'1c-8o.6S"l'V'
,
R~3"o.l'f7'-.ll..
Ji. • ..1m_
= 2o ..,
.tn7-t .2rbo.~P
F. "'
• 6. 3 '1 (r/-1 ~
I
p,- 2n [ (~ + C,UCI•!..R(JJ~., -t-Ci+ SAJ!t~]
f.. ,.
,
p,- Z.rt [' '" 1- o.'>(t+ 2#,.s-)Jo.n-r 2.s;. s]
Fp,: 2.)/.tMH~
'
'
_ ( c, 1- ~,..a.,.,.,.IIJ)Its.;
1 .,. ( C... +$11Jil~,.
.,.tz. 7<] R$.3 ~~
~
I
PJ.- 111 ( c.ttcc;_.,.<;..)
fp_
"3
C' (
fCI + 11S(I + 2#,t!)p.f?
+ l·S"J
c • + c•
40 m
211' X 400 M = 15.9 pF
c..
= 15.9-2 = 13.9pF
Pole frequency:
2Tr(22:
IP =
+c.>
J,. = 213.74 MHz
=
g.,
90 X IOO X 100
.
1.6
= 1060 fLA/tt
=
rlll = VA, = 12.8
40 m
-
X
gM = 1.06 mA/v
Zero frequency:
fz
j2
jz.,..c 0 ,~1 1,
g.. =
w-'
40 x
27T(2 X 13.9 + 2)pf
21TC• - 211' X 2 p
OT
loo
3.18 GH1.
'o~ =
~ =
l1 1n!
19.2 = 192 k!l
0.1
DC·gain = -gm (roo
= -1.06
(128
X
128 kO
II
II
ro,l
192)
=
-81.4 V /V
Total capacitance between output node and
ground
8.62
"M"
f: ,.
H
-3.., ~I.=- 5~<.2.0: -roo V/V'
I
2n(tj_ 1' 'y.1> p~
•
I
2n( I +~.fhJ.o
=
c,., + cd., + c,,.,
o.o1s + o.o2o + o.OJ6
=
:a 1.l'JHHi.
CL = 0.071 pF
Write a KCL at output:
<"'~tc t/..,d' ,·.,_
-c:1
S:_,H
Wf
"'~H
IUUf
+lv,
W,.
• = 200 J.1. = 8 mA and
""
0.025
v
21)
8"'2
~t.CoxUH V••
~t.Cox(T.),v••
-A.
1 + Lo
"'T
+ A.J
=
.-!QQ...
= 0.5 Mfl
200 J.l.
Thus the OC-gain = -g•. r 0
= __-....;g;;.o'":,:l_l.::g;::.m·"-'_
V;
=
Forlr = 200 J.LA, V, = IOOV:
r0
Thus:
~
_ _
.r [
I - sC e I g., ]
- g.., 0 I +s(Ct+C.)r 0
(g,., + gm/oad)
w,
w!
= -8 X 0.5 X 10-' = -4000 V/V
For C,. = I pF, C. = 0.2 pF
I
(I + 0.2))> X 05 M
0.5 .
Chapter 8-25
""L67Mradls
8.67
f .... "'f~"' 265A KHz
fz ""
~
21r · c~
8111
2r. x 0.2 p
""
fr ""' [At~ ' IH "'
Bode plot for !AI:
4()0()
.~o
1 .• where
2TrC;nR,1,
lu::: fp; ""
6.4 OHz
R·. "' ~ = 20 kfl
"'
2
2
X265A "" 1.06 GHz
R;, "'
IAidS
·~
n
Ill kO
+ C#4 (l + g,R;)
C;. '" Cg,
4000 V N e,, 720 dB
=
r 0 U r 0 ""
~""
c.. = 0.1 p + 0.1
where
10k!1
p( l + 2 X 10) "" 2.2 pF
~ 1·II -- 211"··X 2.2 Ip X 10 K "" 7.~..
"3 MH z
-20dB/d!Xade
i) If the bias current I is reduced by 11 fi!Cior of 4:
vi,v "*
For Into reduce by lf4, Vov
Since 11, o
is reduced by 112
2111 ). = !.1
!(2VM
AIV
2~m.
- 2(1,/4)
g,.; - (
vovm
I rnA
v
4X20ktl=80k!l
8.68
8.66
tt- =- ~..,...
.11'1 (
-~
ll_ ~ C,.l
c:;_ + c:...cl =::1
To
1\R\/'C-
~ b""' "'\f-lO
"3
:2.1"\ ~ l.l' tO"!
~n.:: u:; 1'\~,
I MJ../IT '
~
then:
"';2.6\'1'2:.
C;"
T
= +q.61
u:
we ....~,
R;,,
~·
R~
=
"' !:22
= 80 kfl '' 40 kt1
2
0.1 p + 0.1 p( I + 1 x 4tl) ,, 4.1 pf
"*ln
1
··
211'
;..; 4.2 p
=
09'>
.•. !\>1u
uz
40 K
X
ii) if the bias current is increased by x 4:
-'> V 0 v is increased by a factor of 2.
'!>
C.l-"" C. .l ::. I Z:ID
"' I 5"'\ · 2.3 fr;:'
:1
2 !\ ,.,..,. \11'\
Thv11 »>"- n.. e.cl a- a.d..U. t.\o"""-\ C~cilc""-"C£-of-
_gmil
=
2(4
X.1 0) _ 7(2/D)
~ V0
{2 X V 0 y) -
2X2mA,o
v
lhcn
1?;,.
4!1~
v
R;, " r 0 }2
=
= 2.5 k!l
c,o= 0.1 p+O.I p(l
+ 4
fu = 2n x L21p
2.5 K
·.~
X
2.5) '"-· 1.2 pF
53 MHz
+ CL. Rcr
C"". 1(,, + C,,1 ·!!?;.·, •. (I r , .,. l + R' l
'
- ~"'" J.
t
·ru = c,it'•R,>P +
+
X
,,
CL
(·,'f:.!Rr.:d
·R;
Where
R;,,
R,
20 Hi
··~ Tn
·.~
100 ldl
i/
12 !d!
0.2 p
7.5 Hi
100 K
HUp[ HWJ f.;( l + 1.5 x 7.51 + 7.5 K I
c,, · 1<; ·•
20 ns ·I 246 ns
266
c, . 17..'\
flS j
K)
+
c,. (75 K)
Chapter 8-26
8.70
a)lfCL = 0=>T11 = 266ns-+f 11 = - 1-
2-rr-r,
b)CL
= 598MHz
= 10pf=)T11
B
r,.•
+
= 341ns-+/11
oo
=467 MHz
SO pF =o-r11 = 641ns-+ f 11
c) CL =
v,.=- -v.
=248 MHz
Using the Miller approximalion : Eq 9.80, 9.82
C;. = C,, + C,J! I + g,.R~)
= 0.2 p + 0.2 p
(I+ 1.5
fu
X
7.5) = 2.65 pF
= 2.r · C110 • R,.11
W~ observe that v,. vollage 111 the emitter. is equal
to - V • . We can write a node equation at the
emitter:
2.r X 2.65 p X 100 K
= 600MHz
Notice how 1his resull is clo!ie 10 case a) whete CL = 0
and diverges further wilh increasing value of CL•
showing !he importanCe of C1• in delennining.f11
I = - V
~
.(l .
r•
=
+ sC ) - g,. • V .,
v.(f. + N. + .• c.)
Thus, the input admittance looking into the
emitter is:
!.:.
l + g + sC• = r,! + sC #
v"' r, IJf
Therefore we can replace lhe lransistor at the
input of the circuit by this admittance as shown
below
=
+
fro,.... ~·
$; .,
PI
,.,,.o;
...!'------
..oc. '1\A\I't. ~ •
1"1 ln~,t~,. \\...1-)
I
• 5S7tiH'I
.2rt,.Zrf(IIC"....1--)
Fro~ ~·
v,"' - v.
'3 ',:~
s.,o:us
6·10'
o.>Ja h\lfe:
Sin u. ~ .(,(. f-p 1 1 -t~ S:pa..
1•5
+hi,
c.h,..,.;n-l: po\(.~1
ftt ~ S.pa.. '3 ,,,1"114 ..
a) As we can see above, !he circuit can be
separaled into two pans, each wilh its own pole:
fr,
= 2.r· c (~. n ) (inputsidc)
11
f r'
2 .,.(C
P
"•
r,
~ C L )RL
(outputside)
Chapter 8--27
8.72
If we compare the poles
for MOSFETS. we observe that these equal ions
are their bipolar counteparts:
1
fr1 =
c,.(R,;,ll t)
211 ·
R0 = 2r0 + (g.r 0 )r0 = 2 X 50 k!l
+ (I
9.108
X
50) X 50 kO = 2.6 MO
Av = -g.,(R 0
Rd
11
-1 m(2.6M U 2M)
=-1130 ~
v
=?
C
•
~1T( .cJ
{po
.
I C
+ ,)R,
9.109
Av =
b) For C1r = 14 pF, C • = 2 pF.
C, = I pF. 1,. = I mA
v
+ R4 , T 11 = - 1-
271/11
Ras ,. R.s, ~ " :zo "'.n.
c,,R,, +
T 11
fu = - 1-
p)
8.71
11
= 22.5 kO
MHz
f, >> f,. and(, »
If /
41 kO
+(Rtil R,.)(C, + C•• + C, 4 )
P( I K II
14
X
I X 50
= 'o n R;., = 50 kO II 41 kO
•u = R,; 1 (C,, + C,.(l + g,.R 4 ,!1
As.•uming 13 = 100
/"' =
= 50 kO + 2 MO =
RJI
= 40 rnA
2'11
v
R;., = ro + R,
g.,r 0
0.~25
R,; 1 = I kO, R, = 10 k!l =>g.,. =
-1130~
530.5 p = 20 f X 1.38 K + (5 p + C~) X 18.7 K
Thus c~ = 21.9 fF
Since the original CL in Eq. 12 was 15 fF
~ We must add (21.9- 15) = 6.9 fF at
the output to reduce/,. from 396 MHz to300MHz
c
"" 1s ~s
""C,.. ~" -r1. ~
I
); .. 2.P,.Jl.pi<;1.0 • IOOV/V'
•
01
1 1"'
"'v
Ro!Ak • '0 1 + All'ol. r01
R~.
A....r.Av:
•-11\~too"
~ '1' ~..X
0
.:: ').(JK ~ 02.1
'-to ...
";~oo") "1.411 M.l\.
••'lB."'VJ'
!l.o
1o-t:l.'l'40
---I.!J
u.,.,3 E,.~.13l,
~" .. Rs. 3(c~., "'l"·t"'·a ... ~,> 1.. f\i,Lc-,.~,-+ 'a-.i'" ).J
-t tR'-11 Rout )lCs,_ + ~o~,l..)
Rei,= ro,
11(-1- - ..,. .!J,._
:J...'J.+'l .. ..a
""'.. 1.
Rd,,.l)..(,"'lll
+ .:l.o )
t:r :toICr
5''1' o.t...'5
,
I = I rnA,
?',..
, 11
A., =
r
r\S
J..-
f\ = 100. r0
= 100 k!l
p
~.'-l•l1 ) (.I+ o . .l)
1;. • t5. I
. R,;1 = 4 kO, RL = 2.4 kO,
.. o.32'li<.SL
I :Z.I
,
8.75
!l.to.2(.1tS"0.31.'t)J+o.:n:J(o·2.t0·2+ l)-t
(.2.0KI(
~H .,
l
c 2..1.2 Mt4~
g
=
•
m
'•
'• + r, + R.i,
g,.(flroll RL)
..1!
= ___!.QQ._ = 2 5 kfi
!s_
V,
= 40 rnA I V
g,.
=
X
A.,=
I /0.025
2.5
2·5
+ 0.05 + 4
.
X40X(IOOX 100
II2.4K)
A., = -36.6 V /V
.,
+ R,; 1 ) =
...
=
R: = r .II (r,
R : = 1.5Sk!l
8.74
R.,
A" • '6cl~ • 1"lqS V fv--
.A.:t•
"~.t R\,
R._..,
c.~Md
-i. 0 ,..t
~~..-R...t-':>A..,.A
AL
~ ::1.
a. a. o
A~ 0 =ll+,,.,r;, 1 ),..,,ro 1 ~ ,.,.'0 '"(~:~ •¥c;-)·(~:ja.
..1...~ ( .2.t 10
..,.> 1'\1'5 •
~
t -..,. .
5>nct'\;a
'lH
'6
~ ( 'i.
s ..... c...H ,
~....t)
.. c.1 4 J(Rt.. II
£1)
CJ•/
,.
A~.:~. r;, 1 -t
~u..t
(fM
fOz.
c
(
1-+
~...t .. 6SIO IU'\. ""6.5 MSI-
1;;
S:11
~
( IPF+ o./fF)
,.ftl.f,S ~<.Ha
I
R/,. Rc>~
"
(~.): 35Fo.5 ns
.:t
I A"'J·FI+ . . . I 'f't 5 )C ~ tt. 5 C' 8 11. 8 M H 2
If:-th.£ (D.SeoJe frAnSo'$t'or is r"(OVIO\I'e.c) 1 ~
J.'"
!:
we ~ve. "'- c..,...,,..,Oif\-5ocJrCe..
C-dT!
~•3vr~h'ot1..
+4
R, 1
...
+ RL )
R,, - 'n• H r,, { 'o +'oRL/f\
+I
~ 100 KII
100
101
K( 100 + 2.4)
100 + 2.4
101
=
R, 1
I kfl
R. 1 = 1.55(1 + 40
X
I)+ I = 64.55 kfl
= fl · r,. = 10 M!l
Tu =
c.,R., + c.,R., + (Ccs• + c,,)R,,
+ (CL + Ccn + c.,l(RLU R•• ,)
V.,tf
j,,,r;,) r; "" ~
(0.05
-
R•• ,
~ ..,,YA. , !E.. ::-loo~.Jt , q .,l!.Q .. o.631 "'""!.,
II
= R: = (I + g.,R, 1) + R, 1
V011 cO. 3t'f 'I
~~
~
a.
->-I: ,...L...;_:t£..1/' .-L .. .2.0o,.\O ,.,o,.o.3tl :.o,t.._A
·-o .2.
\... ov
2.
2.5 K
T 11
=
X
T 11
+ 2 X 6455 + (0 + 14)
+ 2)(2.4 K II 10M)
14 X 1.55
I + (0
= 169.6 ns ! 11 = 939 KH1.
K)
Chapter 8-29
8.76
I~
Q)
.1-
C,.,S
e.~f>loy f"'lil\oer't. ~~- to
IIJ(.
Or l ? ~ptat'S
c...... ~
.25-s
, , ,
t.D
R;'§ (C,., + .2~) e->"-'Ch
rdl)
fb
R,,,
r
,..,
0
-'rt/?t':1(c;.,T~)
rt.fu f.,. ~~.6.4L 1 .,.:Jt/fl SUikCI.t'fi...R.
ou.tp...t"
pol~ is : ~.t :::---..:..'----~
-2tT('i+c:($,_+_5ue.) RL
Rs,~ .. JI<..n.
b)
->
1{3.,. 'W 1/~j"'
',i:.tJts- II'
....,. R's,lJ "'o. '11'1 k.n.
~
. "'
f>1
h ..
P,.
I
.
..
::
31.85 HHA
2tt"o.?t't1s+.:t>< 12.4) + 12.4 kH
R.,
645 kH
c.c:;>·r 11
= C, 1 R" 1
+ C" 1R,. 1 +
+C,,,, (R 1
1J
C,~Rn
R0 J
where
R,,
= 13 1 - r 0 ;_,
7 11 z,
0.54 p
X
=
x 12.4 K + (U
r,
· 5.08
100
x I MH = 100 Mil
12.5 K + 0 I p X 645 K + 0.54 p
r { I00
tvf
I~'
~ tr'Tu
' 3U Klh
'I 100 M I
Chapter 8-30
8.78
where C,, = C,, (I - K)
and K
=
R~
g,.
I+ g., R~
c,,[l - c..R~ .J
c,,L +;mRJ
~ c,. =
=
I+ g,.RL
1- . ]
c,. = c,, n c,, = c,. + c,.[-I+ g,.RL
At>:= -L- .. o.83Vl
Ro •
..
a~·~a b
•
1+)1.
lt~•-lo
d) w:n R\..= lli.S\.
II.
, ...
I+
.,..,
I+ g,..RL
Notice that this is the same result as obtained in
problem 9.86. This estimate is higher than that
obtained from the method of open-time constants
reduces the contribution of c~ from:
I+,...Fl~
1\. . '\. nro K..L=1""n 10"-\1 .J..!
,..,..
1-6
~'"'~:r"'
~)
since it neglects the contribution of C,. to -r 11 and
,.. "' ,..., R;.
~
21r · R, 1,( C, 4 +
'"
1\r"o = ...L....! \\10... .. to 3.S\.
-'}
'!!?30Z'IQ
v-.,.a. S"'l:)
:O·"'HiVt,
'(
&
3f<>A.
C,, · R,,, h
C•' · (R.,, + R~)
. to
.• t us
I + 1/mR<
I + g,.Rt.
effectively reducing the value of 'Tn, and
therefore increasing
fn
g,. =_VIc'= - 1-
O.D25
8.79
Using the Miller approximation, the
resulting input equivalent circuit is:
8.80
fr =
r
'
r•
= (13
___!!_ = 25 n
+ l)g..,
g.,
21rcc. +c.>
~C~+C•
c~ =
0.1
= .[
g.,
= 40mA/V,
= 2 GHz
= 3.18pF
pF~c. =
J.os
pF
= 2.5 kfl
'o = ~ = ~ = 20 kfi
I
lr
r. is in effect parallel toR,_, so R~
= RL
II
r0
R~ = I kfiil 20 kfi = 0.95 kfi.
_R_,,'---+ r X + R.
A., =
Rsia + r •
13 + I
'
0 ·95
I + 2.5 + 0.1 + 0 _95
101
R~ =
R;,,ll
(r,
= 0.964 VI V
+ (fl + I)R;)
R;,, = R.,, + r, = I + 0.1
R. = 1.1 K II (2.5 K + 101
= 1.08 kfi
= 1.1 kfl
X
0.95)
Chapter 8-31
8.81
l:.luv>.
"tH ::. 3 0 f }(.
2..0 1<.
5~ [2..0K( lt-0.8)(19.18)+l$Ja~
t ~o,.. ... sf)~< IS.ts~
-4-
"' +
L.H ::: 0- G'-\s
.... l.64vt'3
+ I. t2.\lt.S
(b) Ad:: _q_tM. ( R.o tJ fa)
whu e. Uro .. I!A -:r 2..0 ,. z.oa K.tt
-o.z..;z
(H
==-¢o>
"" 8.
A-d ""
0. 8 wt
K
(Z.OK lt W K)
-= 14.54 V;v
....-::::='
tc.) Tor a- cs ~t.c;tfer wue~-1.
R~l~ i9 low:
-1
{
H-
_
_;__ __
8.82
t~
I
::o
t. TT C, Rs s t
sfJ ~L1
fz
a~ArJ
eL'
8i~.<£!e tar a... ~rou~;~.c/~d .sovree
t.M.
fo.ft.o.llel
wl6lt f:l..e laa.d.
~ cc.· = ro +5== Y?rr
=
fl =
~
= ..1Q.Y.. = 300 kll
10
100 J.O.A
I
21r · 100 f · 300 K
= 5.3 MHz
If V.,. is reduced from 0.5 V to 0.2 V while I is
unchanged.
For the current-source transistor: when V,,. = 0.5 V
0
~k~(~)V~v __. 100 !LA
=
=[~k;,~11 ] X (0.5) 2
-"1
2rr ( i~+s)ro'$"
"" 6':f. S4 ft H ~
~
I= I
tb,
f l-(
Tf ( o..tp )(root<)
= 21r · Css1 · Rss
Rss = r0 =
.., 1'8. t8K.t./Z.
-= Cc. +- C.d6
C.dW t'5
1.
w,
w,
= 2500 ...
400
w,•
= 6.25
w,
The width of the current-source transistor is
made 6.25 times larger to operate at V"''
In= IOOftA
If Css is directly proponional toW:
Cs.l
r;,J·IC... +c,.n +g,.RcH
I
:lir((JO.ft + 100)1 5 It l·() ::
2//) _ 0.2 m .. I mA
02- V
\1 0 , . -
The low-frequency differential gain is:
I mX 100k!1 =
f,,,
fr,
8mRo( l+
jt~CIRJ
At unity gain: !AJ( '"' ll
"'
Ro
2'tr :X 100 f X I!Ml K
..!G._
21rCm
I
25 fF
v
response of Ais) passed the unity gain
Thus:
c,
ro = 100 ld1
1
15.9 MHz
(d? GHz.
•J
,-) I ~
./1
fz = ~:..Ji!!.t~· .~:;. 'lXfrL = l2.74GHz..
::::K::"':::R:':"==
:!rrC 111
+ (w,{;N~~~
a ~!!
2 >> l
,.. ~f:'
lfM)V/V
NPticc in the fh>dc plot the ln.;-;•tion nf the
unity-gain frequenc'Y
-~
--? W 1 ....--- . .. >>!
Cs.
Chapter 8-35
8.90
8.89
~l A,.. .. -Ar>
~!..
~j..+l;i')
"to
A~ c R~.rl R0 ~
R,,. "'~. ,
(I+
..,._ 4 .-:
C...,~)
(J
riT
..,...,..
(b)JfjAMj '~ 20V/V
..:lbF-.1\.
wht-N> &"""' ''"'
and
t-H "'
i'f; •
f
'=
II
R85 ., R5 ~ .... ib~
20
"'
100 VIV, Rt = 20 kfl
Aa = g,.·r0 =*ra"'
...
RL ""
~~"'
20kfl
r 0 = 20 kfi
Wtl Cllll
)s R3.s ~ )d f)~ + c.~.. 1-?Cz.
AM = -G..(R0
~.o:tH+ I l< ,1e~K,. 2bl.t\S
'lit,.. ...,.1o o- >t 60 'l .If "'E. a· 'l'S ,.. \01Ro
R ·!- 20
= 80kfi
0
Since
Rs c
b)
R0
Soo.n.
1\,u,l:: •
fo [I
AMe-~
I
... l~..-'}...J
'0
R..b
~'t
...
R,..l:;
Rs,J ,.'-\o[l-+(..5-+l}o.S') .,lbo\~e"to,.__.!:!.2..- =-...,OVh,.
"\0+\~
\1
1\_ := Rl..ll fl0 ..;,. "''1¢r. \\ 1&0~" 32.~JL
{+ G.., R~)
'o [• . . .la ..... ,... J~)
~'"' ~~a
&.,.,
1
im"e
G-,.," 2x ~0
.,.l.")..t;..-.A.{IT
Lto t• + L<:> -t l)o.s)
~~.t "' .:to IC- () + 1.:ts'"'x 32..'~') "' g 2..o "'-J\.
l)s..
R.s.;, + H:s
1-tC,,.. T 'J., 0 )
f?J s ,.
:lo" ... o s
-
ft I'! c...
tr:.
,_
8 . 2 J0l..
I+( ,.,_, )o.S ...!::Ul.1-fe~+lfo
R'-t..'" fl~.-11 R......t -~~"" 3:1 tc.Jt
?;t
""C_,1 ~.r+£!,.,
7'-11
*' I'Jq. Lf MS
(.H "...,!__"'
.2n 't-fl
/Atff·f-t+
=>
1.; tl]_ ~.::: .2x '1.2+o.t.,f'2o.,.32.>< 1
I· .22. HH 2;
41!.8 HHlr
"'
ra(l + g,. Rsl "'20 K [J + 5 m X Rsl
= 80 K=}Rs"" 600 H
Chapter 8-36
K
f- nn•thod:
RP XC,,, + R,, x Cd + llt. :.<; C~_
>u
0 we can see thni
Setting CA'-'
R, _-: :.
Tr~
C;.;-d
~'n
obtain
ll~,J
set C
the fi',Ho\-\:ing drcuit:
0 we can
S('t:
thal
IA,I· fn
= 20 X 56.8 M = 1.14 0117.
(b) for the low-frequency analysL~ of the CD-CS
consider the following circuit
Chapter 8-38
ortheCS:
Vo "" Viol(-g..;roi) (Eq I)
For a CD amplifier:
A _