Fundamentals Of Microelectronics Solution Manual

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Solutions Manual to Accompany
Fundamentals of Microelectronics, 1st Edition
Book ISBN: 978‐0‐471‐47846‐1

Razavi

All materials copyrighted and published by John Wiley & Sons, Inc.
Not for duplication or distribution

2.1 (a)
k = 8.617 × 10−5 eV/K
15

3/2

ni (T = 300 K) = 1.66 × 10 (300 K)


exp −

0.66 eV
2 (8.617 × 10−5 eV/K) (300 K)



cm−3


exp −

0.66 eV
2 (8.617 × 10−5 eV/K) (600 K)



cm−3

= 2.465 × 1013 cm−3
15

3/2

ni (T = 600 K) = 1.66 × 10 (600 K)

= 4.124 × 1016 cm−3
Compared to the values obtained in Example 2.1, we can see that the intrinsic carrier concentration
13
in Ge at T = 300 K is 2.465×10
1.08×1010 = 2282 times higher than the intrinsic carrier concentration in
Si at T = 300 K. Similarly, at T = 600 K, the intrinsic carrier concentration in Ge is
26.8 times higher than that in Si.

4.124×1016
1.54×1015

=

(b) Since phosphorus is a Group V element, it is a donor, meaning ND = 5 × 1016 cm−3 . For an
n-type material, we have:
n = ND = 5 × 1016 cm−3
2

[ni (T = 300 K)]
= 1.215 × 1010 cm−3
n
[ni (T = 600 K)]2
= 3.401 × 1016 cm−3
p(T = 600 K) =
n
p(T = 300 K) =

2.3 (a) Since the doping is uniform, we have no diffusion current. Thus, the total current is due only to
the drift component.
Itot = Idrif t
= q(nµn + pµp )AE
n = 1017 cm−3
p = n2i /n = (1.08 × 1010 )2 /1017 = 1.17 × 103 cm−3
µn = 1350 cm2 /V · s
µp = 480 cm2 /V · s
1V
E = V /d =
0.1 µm
= 105 V/cm
A = 0.05 µm × 0.05 µm
= 2.5 × 10−11 cm2
Since nµn ≫ pµp , we can write
Itot ≈ qnµn AE
= 54.1 µA
(b) All of the parameters are the same except ni , which means we must re-calculate p.
ni (T = 400 K) = 3.657 × 1012 cm−3
p = n2i /n = 1.337 × 108 cm−3
Since nµn ≫ pµp still holds (note that n is 9 orders of magnitude larger than p), the hole
concentration once again drops out of the equation and we have
Itot ≈ qnµn AE
= 54.1 µA

2.4 (a) From Problem 1, we can calculate ni for Ge.
ni (T = 300 K) = 2.465 × 1013 cm−3
Itot = q(nµn + pµp )AE
n = 1017 cm−3
p = n2i /n = 6.076 × 109 cm−3
µn = 3900 cm2 /V · s
µp = 1900 cm2 /V · s
1V
E = V /d =
0.1 µm
= 105 V/cm
A = 0.05 µm × 0.05 µm
= 2.5 × 10−11 cm2
Since nµn ≫ pµp , we can write
Itot ≈ qnµn AE
= 156 µA
(b) All of the parameters are the same except ni , which means we must re-calculate p.
ni (T = 400 K) = 9.230 × 1014 cm−3
p = n2i /n = 8.520 × 1012 cm−3
Since nµn ≫ pµp still holds (note that n is 5 orders of magnitude larger than p), the hole
concentration once again drops out of the equation and we have
Itot ≈ qnµn AE
= 156 µA

2.5 Since there’s no electric field, the current is due entirely to diffusion. If we define the current as positive
when flowing in the positive x direction, we can write


dn
dp
Itot = Idif f = AJdif f = Aq Dn
− Dp
dx
dx
A = 1 µm × 1 µm = 10−8 cm2

Dn = 34 cm2 /s
Dp = 12 cm2 /s
5 × 1016 cm−3
dn
=−
= −2.5 × 1020 cm−4
dx
2 × 10−4 cm
dp
2 × 1016 cm−3
=
= 1020 cm−4
dx
2 × 10−4 cm












Itot = 10−8 cm2 1.602 × 10−19 C 34 cm2 /s −2.5 × 1020 cm−4 − 12 cm2 /s 1020 cm−4
= −15.54 µA

2.8 Assume the diffusion lengths Ln and Lp are associated with the electrons and holes, respectively, in this
material and that Ln , Lp ≪ 2 µm. We can express the electron and hole concentrations as functions
of x as follows:
n(x) = N e−x/Ln
p(x) = P e(x−2)/Lp
Z 2
an(x)dx
# of electrons =
0

=

Z

2

aN e−x/Ln dx

0

2

= −aN Ln e−x/Ln
0


−2/Ln
−1
= −aN Ln e
Z 2
# of holes =
ap(x)dx
0

=

Z

2

aP e(x−2)/Lp dx

0

2

= aP Lp e(x−2)/Lp
0

−2/Lp
= aP Lp 1 − e

Due to our assumption that Ln , Lp ≪ 2 µm, we can write
e−2/Ln ≈ 0
e−2/Lp ≈ 0

# of electrons ≈ aN Ln
# of holes ≈ aP Lp

2.10 (a)
nn = ND = 5 × 1017 cm−3
pn = n2i /nn = 233 cm−3
pp = NA = 4 × 1016 cm−3
np = n2i /pp = 2916 cm−3
(b) We can express the formula for V0 in its full form, showing its temperature dependence:
"
#
NA ND
kT
ln
V0 (T ) =
q
(5.2 × 1015 )2 T 3 e−Eg /kT
V0 (T = 250 K) = 906 mV
V0 (T = 300 K) = 849 mV
V0 (T = 350 K) = 789 mV
Looking at the expression for V0 (T ), we can expand it as follows:
V0 (T ) =




kT 
ln(NA ) + ln(ND ) − 2 ln 5.2 × 1015 − 3 ln(T ) + Eg /kT
q

Let’s take the derivative of this expression to get a better idea of how V0 varies with temperature.



k
dV0 (T )
ln(NA ) + ln(ND ) − 2 ln 5.2 × 1015 − 3 ln(T ) − 3
=
dT
q


15
From this expression, we can see
that
if
ln(N
)
+
ln(N
)
<
2
ln
5.2
×
10
+ 3 ln(T ) + 3, or
A
D
h
i


15 2 3
equivalently, if ln(NA ND ) < ln 5.2 × 10
T − 3, then V0 will decrease with temperature,
which we observe in this case. In order for this not to be true (i.e., in order for V0 to increase with
temperature), we must have either very high doping concentrations or very low temperatures.

2.11 Since the p-type side of the junction is undoped, its electron and hole concentrations are equal to the
intrinsic carrier concentration.
nn = ND = 3 × 1016 cm−3
pp = ni = 1.08 × 1010 cm−3


N D ni
V0 = VT ln
n2i


ND
= (26 mV) ln
ni
= 386 mV

2.12 (a)
r

qǫSi NA ND 1
2 NA + ND V0
Cj0
Cj = p
1 − VR /V0

Cj0 =

NA = 2 × 1015 cm−3
ND = 3 × 1016 cm−3

VR = −1.6 V


NA ND
= 701 mV
V0 = VT ln
n2i

Cj0 = 14.9 nF/cm2
Cj = 8.22 nF/cm2

= 0.082 fF/cm2
(b) Let’s write an equation for Cj′ in terms of Cj assuming that Cj′ has an acceptor doping of NA′ .
Cj′ = 2Cj
s

qǫSi NA′ ND
1
= 2Cj
2 NA′ + ND VT ln(NA′ ND /n2i ) − VR
1
qǫSi NA′ ND
= 4Cj2
′
′
2 NA + ND VT ln(NA ND /n2i ) − VR

NA′



qǫSi NA′ ND = 8Cj2 (NA′ + ND )(VT ln(NA′ ND /n2i ) − VR )

qǫSi ND − 8Cj2 (VT ln(NA′ ND /n2i ) − VR ) = 8Cj2 ND (VT ln(NA′ ND /n2i ) − VR )
NA′ =

8Cj2 ND (VT ln(NA′ ND /n2i ) − VR )
qǫSi ND − 8Cj2 (VT ln(NA′ ND /n2i ) − VR )

We can solve this by iteration (you could use a numerical solver if you have one available). Starting
with an initial guess of NA′ = 2 × 1015 cm−3 , we plug this into the right hand side and solve to
find a new value of NA′ = 9.9976 × 1015 cm−3 . Iterating twice more, the solution converges to
NA′ = 1.025 × 1016 cm−3 . Thus, we must increase the NA by a factor of NA′ /NA = 5.125 ≈ 5 .

2.16 (a) The following figure shows the series diodes.
ID
+

D1

VD
D2

−

Let VD1 be the voltage drop across D1 and VD2 be the voltage drop across D2 . Let IS1 = IS2 = IS ,
since the diodes are identical.
VD = VD1 + VD2
 
 
ID
ID
+ VT ln
= VT ln
IS
IS
 
ID
= 2VT ln
IS
ID = IS eVD /2VT

Thus, the diodes in series act like a single device with an exponential characteristic described by
ID = IS eVD /2VT .
(b) Let VD be the amount of voltage required to get a current ID and VD′ the amount of voltage
required to get a current 10ID .
 
ID
VD = 2VT ln
IS


10ID
′
VD = 2VT ln
IS

 
 
ID
10ID
′
− ln
VD − VD = 2VT ln
IS
IS
= 2VT ln (10)

= 120 mV

2.19
VX = IX R1 + VD1


IX


IX
= IX R1 + VT ln
IS
 
VT
IX
VX
−
ln
=
R1
R1
IS

For each value of VX , we can solve this equation for IX by iteration. Doing so, we find
IX (VX = 0.5 V) = 0.435 µA
IX (VX = 0.8 V) = 82.3 µA
IX (VX = 1 V) = 173 µA
IX (VX = 1.2 V) = 267 µA
Once we have IX , we can compute VD via the equation VD = VT ln(IX /IS ). Doing so, we find
VD (VX = 0.5 V) = 499 mV
VD (VX = 0.8 V) = 635 mV
VD (VX = 1 V) = 655 mV
VD (VX = 1.2 V) = 666 mV
As expected, VD varies very little despite rather large changes in ID (in particular, as ID experiences
an increase by a factor of over 3, VD changes by about 5 %). This is due to the exponential behavior
of the diode. As a result, a diode can allow very large currents to flow once it turns on, up until it
begins to overheat.

2.22
VX /2 = IX R1 = VD1 = VT ln(IX /IS )
VT
IX =
ln(IX /IS )
R1
IX = 367 µA (using iteration)
VX = 2IX R1
= 1.47 V

3.1 (a)
IX =

(

VX
R1

0

VX < 0
VX > 0

IX
VX (V)

Slope = 1/R1

3.2
IX =

(

VX
R1

0

VX < 0
VX > 0

Plotting IX (t), we have

0

0

−V0 /R1

−π/ω

0
t

π/ω

−V0

VX (t) (Dotted)

IX (t) for VB = 1 V (Solid)

V0

3.3
IX =

(

0
VX −VB
R1

VX < VB
VX > VB

Plotting IX vs. VX for VB = −1 V and VB = 1 V, we get:

IX
VB = −1 V
VB = 1 V

Slope = 1/R1

−1

Slope = 1/R1

1
VX (V)

3.4
IX =

(

0
VX −VB
R1

VX < VB
VX > VB

Let’s assume V0 > 1 V. Plotting IX (t) for VB = −1 V, we get

(V0 − VB )/R1

0

0

VB

−π/ω

Plotting IX (t) for VB = 1 V, we get

0
t

π/ω

−V0

VX (t) (Dotted)

IX (t) for VB = −1 V (Solid)

V0

IX (t) for VB = 1 V (Solid)
(V0 − VB )/R1

0
0

−π/ω
0
t
π/ω
−V0

VX (t) (Dotted)

V0

VB

3.5
IX =

(

VX −VB
R1

∞

VX < 0
VX > 0

Plotting IX vs. VX for VB = −1 V and VB = 1 V, we get:

IX
IX for VB = −1 V
IX for VB = 1 V
1/R1
Slope = 1/R1

−1
VX (V)
−1/R1
Slope = 1/R1

3.6 First, note that ID1 = 0 always, since D1 is reverse biased by VB (due to the assumption that VB > 0).
We can write IX as
IX = (VX − VB )/R1
Plotting this, we get:

IX

VB
VX (V)

Slope = 1/R1

3.7
IX =
IR1 =

(

VX −VB
R1
VX −VB
R1 kR2

VX < VB
VX > VB

VX − VB
R1

Plotting IX and IR1 for VB = −1 V, we get:

IX
IX for VB = −1 V
IR1 for VB = −1 V

Slope = 1/R1 + 1/R2

−1
Slope = 1/R1

Plotting IX and IR1 for VB = 1 V, we get:

VX (V)

IX
IX for VB = 1 V
IR1 for VB = 1 V

Slope = 1/R1 + 1/R2

1
VX (V)

Slope = 1/R1

3.8
IX =

(
0

IR1 =

(

VX
R1

+

VB
R1 +R2
VX
R1

VX −VB
R2

VX <
VX >

VX <
VX >

VB
R1 +R2 R1
VB
R1 +R2 R1

VB
R1 +R2 R1
VB
R1 +R2 R1

Plotting IX and IR1 for VB = −1 V, we get:

IX for VB = −1 V
IR1 for VB = −1 V

Slope = 1/R1 + 1/R2

−VB /R2

Slope = 1/R1

VB
R
R1 +R2 1
VB
R1 +R2

Plotting IX and IR1 for VB = 1 V, we get:

VX (V)

IX for VB = 1 V
IR1 for VB = 1 V

Slope = 1/R1 + 1/R2

Slope = 1/R1

VB
R1 +R2
VB
R
R1 +R2 1

VX (V)

3.9 (a)

Vout (V)

Vout =

(

VB
Vin

Vin < VB
Vin > VB

5
Slope = 1

4

3

2

1

0
−5

−4

−3

−2

−1

0

1

(b)
Vout

(
Vin − VB
=
0

Vin < VB
Vin > VB

2

3

4

5
Vin (V)

Vout (V)

2
1
0

−5

−4

−3

−2

−1

0

1

2

3

4

−1

5
Vin (V)

−2
−3
Slope = 1

−4
−5
−6
−7

(c)

Vout (V)

Vout = Vin − VB

3
Slope = 1

2
1
0

−5

−4

−3

−2

−1
−1
−2
−3
−4
−5
−6
−7

0

1

2

3

4

5
Vin (V)

Vout =

(

Vout (V)

(d)

2

Vin
VB

Vin < VB
Vin > VB

1
0
−5

−4

−3

−2

−1

0

1

−1
Slope = 1

−2
−3
−4
−5

(e)
Vout

(
0
=
Vin − VB

Vin < VB
Vin > VB

2

3

4

5
Vin (V)

Vout (V)

3

Slope = 1

2

1

0
−5

−4

−3

−2

−1

0

1

2

3

4

5
Vin (V)

3.11 For each part, the dotted line indicates Vin (t), while the solid line indicates Vout (t). Assume V0 > VB .
(a)

Vout (t) (V)

Vout =

(

VB
Vin

Vin < VB
Vin > VB

V0
VB

π/ω

−π/ω

t

−V0
(b)
Vout

(
Vin − VB
=
0

Vin < VB
Vin > VB

Vout (t) (V)

V0
VB

π/ω

−π/ω

t

−V0

−V0 − VB

(c)

Vout (t) (V)

Vout = Vin − VB

V0
VB
V0 − VB
π/ω

−π/ω

t

−V0

−V0 − VB

(d)

Vout (t) (V)

Vout =

(

Vin
VB

Vin < VB
Vin > VB

V0
VB

π/ω

−π/ω

t

−V0
(e)
Vout

(
0
=
Vin − VB

Vin < VB
Vin > VB

Vout (t) (V)

V0
VB
V0 − VB

π/ω

−π/ω

t

−V0

3.12 For each part, the dotted line indicates Vin (t), while the solid line indicates Vout (t). Assume V0 > VB .
(a)

Vout (t) (V)

Vout

(
Vin − VB
=
0

Vin < VB
Vin > VB

V0
VB

π/ω

−π/ω

t

−V0

−V0 − VB
(b)
Vout =

(

Vin
VB

Vin < VB
Vin > VB

Vout (t) (V)

V0
VB

π/ω

−π/ω

t

−V0

Vout

(
0
=
Vin − VB

Vout (t) (V)

(c)

V0

Vin < VB
Vin > VB

VB
V0 − VB

π/ω

−π/ω

t

−V0

(d)

Vout (t) (V)

Vout = Vin − VB

V0
VB
V0 − VB
π/ω

−π/ω

t

−V0

−V0 − VB
(e)
Vout =

(

VB
Vin

Vin < VB
Vin > VB

Vout (t) (V)

V0
VB

π/ω

−π/ω

t

−V0

3.16 (a)
IR1 =

(

Iin
VD,on
R1

Iin <
Iin >

VD,on
R1
VD,on
R1

IR1
VD,on /R1

VD,on /R1
Iin

Slope = 1

(b)
IR1 =

(

Iin
VD,on +VB
R1

Iin <
Iin >

VD,on +VB
R1
VD,on +VB
R1

IR1
(VD,on + VB ) /R1

(VD,on + VB ) /R1
Iin

Slope = 1

(c)
IR1 =

(

Iin
VD,on −VB
R1

Iin <
Iin >

VD,on −VB
R1
VD,on −VB
R1

IR1

(VD,on − VB ) /R1
Iin
(VD,on − VB ) /R1

Slope = 1

(d)
IR1 =

(

Iin
VD,on
R1

Iin <
Iin >

VD,on
R1
VD,on
R1

IR1
VD,on /R1

VD,on /R1
Iin

Slope = 1

3.17 (a)
Vout =

(

Iin R1
VD,on

Iin <
Iin >

VD,on
R1
VD,on
R1

VD,on /R1
VD,on
0

0

−I0 R1

0
t

−π/ω

π/ω

(b)
Vout =

(

Iin R1
VD,on + VB

Iin <
Iin >

VD,on +VB
R1
VD,on +VB
R1

−I0

Iin (t) (Dotted)

Vout (t) (Solid)

I0

I0
(VD,on+VB )/R1

0

0

Iin (t) (Dotted)

Vout (t) (Solid)

VD,on + VB

−I0 R1

0
t

−π/ω

−I0

π/ω

(c)
Vout =

(

Iin R1 + VB
VD,on

Iin <
Iin >

VD,on −VB
R1
VD,on −VB
R1

VD,on
0

0

−I0 R1 + VB
(VD,on − VB ) /R1

−π/ω

0
t

π/ω

−I0

Iin (t) (Dotted)

Vout (t) (Solid)

I0

(d)
Vout =

(

Iin R1 + VB
VD,on + VB

Iin <
Iin >

VD,on
R1
VD,on
R1

I0

VD,on /R1
0

0

−I0 R1 + VB

−π/ω

0
t

π/ω

−I0

Iin (t) (Dotted)

Vout (t) (Solid)

VD,on + VB

3.20 (a)
Vout =

(

Iin R1
VB − VD,on

Iin >
Iin <

VB −VD,on
R1
VB −VD,on
R1

I0

I0 R1
VB − VD,on
0

0

0
t

−π/ω

π/ω

(b)
Vout =

(

Iin R1 + VB
−VD,on

V

+V

B
Iin > − D,on
R1
VD,on +VB
Iin < −
R1

−I0

Iin (t) (Dotted)

Vout (t) (Solid)

(VB − VD,on ) /R1

I0

0

Iin (t) (Dotted)

Vout (t) (Solid)

I0 R1 + VB

0

−VD,on

−(VD,on +VB )/R1

0
t

−π/ω

π/ω

−I0

(c)
Vout =

(

Iin R1 + VB
VB − VD,on

V

Iin > − D,on
R1
V
Iin < − D,on
R1

I0

VB − VD,on
0

0
−VD,on /R1

−π/ω

0
t

π/ω

−I0

Iin (t) (Dotted)

Vout (t) (Solid)

I0 R1 + VB

3.23 (a)
R2
R1 +R2 Vin

VD,on

Vout (V)

Vout =

(

Vin <
Vin >

R1 +R2
R2 VD,on
R1 +R2
R2 VD,on

VD,on

R1 +R2
VD,on
R2

Vin (V)
Slope = R2 / (R1 + R2 )

(b)
Vout =

(

R2
R1 +R2 Vin

Vin − VD,on

Vin <
Vin >

R1 +R2
R1 VD,on
R1 +R2
R1 VD,on

Vout (V)

Slope = 1

R2
V
R1 D,on

Slope = R2 / (R1 + R2 )

R1 +R2
VD,on
R1

Vin (V)

3.24 (a)
IR1 =

(

Vin
R1 +R2
Vin −VD,on
R1

ID1 =

(

0
Vin −VD,on
R1

Vin <
Vin >

−

VD,on
R2

R1 +R2
R2 VD,on
R1 +R2
R2 VD,on

Vin <
Vin >

R1 +R2
R2 VD,on
R1 +R2
R2 VD,on

IR1
ID1

Slope = 1/R1

Slope = 1/R1

VD,on /R2
R1 +R2
VD,on
R2

Vin (V)

Slope = 1/ (R1 + R2 )

(b)
IR1 =

(

Vin
R1 +R2
VD,on
R1

ID1 =

(

0

Vin <
Vin >

Vin −VD,on
R2

−

R1 +R2
R1 VD,on
R1 +R2
R1 VD,on

VD,on
R1

Vin <
Vin >

R1 +R2
R1 VD,on
R1 +R2
R1 VD,on

VD,on /R1
IR1
ID1
Slope = 1/R2
R1 +R2
VD,on
R1

Vin (V)

Slope = 1/ (R1 + R2 )

3.25 (a)
2
VB + R1R+R
(Vin − VB )
2
Vin − VD,on

Vin < VB +
Vin > VB +

R1 +R2
R1 VD,on
R1 +R2
R1 VD,on

Vout (V)

Vout =

(

Slope = 1

VB +

R2
V
R1 D,on

VB +
Slope = R2 / (R1 + R2 )

R1 +R2
VD,on
R1

Vin (V)

(b)
Vout =

(

R2
R1 +R2 Vin

Vin − VD,on − VB

Vin <
Vin >

R1 +R2
R1
R1 +R2
R1

(VD,on + VB )
(VD,on + VB )

Vout (V)

Slope = 1

R2
R1

(VD,on + VB )

VB +

Slope = R2 / (R1 + R2 )

R1 +R2
R1

(VD,on + VB )
Vin (V)

(c)
R2
R1 +R2

(Vin − VB ) Vin > VB +
Vin + VD,on − VB Vin < VB +

Vout (V)

Vout =

(

R1 +R2
R1 VD,on
R1 +R2
R1 VD,on

Slope = R2 / (R1 + R2 )

R2
V
R1 D,on

VB +

R1 +R2
VD,on
R1

Vin (V)
Slope = 1

(d)
R2
R1 +R2

(Vin − VB ) Vin < VB +
Vin − VD,on
Vin > VB +

R1 +R2
R1
R1 +R2
R1

Vout (V)

Vout =

(

(VD,on − VB )
(VD,on − VB )

Slope = 1

R2
V
R1 D,on

VB +
Slope = R2 / (R1 + R2 )

R1 +R2
R1

(VD,on − VB )
Vin (V)

3.26 (a)
IR1 =

(

Vin −VB
R1 +R2
VD,on
R1

ID1 =

(

0

Vin < VB +
Vin > VB +

Vin −VD,on −VB
R2

IR1
ID1

−

R1 +R2
R1 VD,on
R1 +R2
R1 VD,on

Vin < VB +
Vin > VB +

VD,on
R1

R1 +R2
R1 VD,on
R1 +R2
R1 VD,on

VD,on /R1

Slope = 1/R2
VB +

R1 +R2
VD,on
R1

Vin (V)
Slope = 1/ (R1 + R2 )

(b)
IR1 =

(

Vin
R1 +R2
VD,on +VB
R1

ID1 =

(

0

Vin <
Vin >

Vin −VD,on −VB
R2

−

R1 +R2
R1
R1 +R2
R1

(VD,on + VB )
(VD,on + VB )

VD,on +VB
R1

Vin <
Vin >

R1 +R2
R1
R1 +R2
R1

(VD,on + VB )
(VD,on + VB )

(VD,on + VB ) /R1
IR1
ID1

Slope = 1/R2
R1 +R2
R1

(VD,on + VB )
Vin (V)

Slope = 1/ (R1 + R2 )

(c)
IR1 =

(

Vin −VB
R1 +R2
V
− D,on
R1

ID1 =

(

0
V +VD,on +VB
− in R
−
2

Vin > VB −
Vin < VB −

R1 +R2
R1 VD,on
R1 +R2
R1 VD,on

VD,on
R1

Vin > VB −
Vin < VB −

R1 +R2
R1 VD,on
R1 +R2
R1 VD,on

IR1
ID1
Slope = −1/R2
Slope = 1/ (R1 + R2 )

VB +

R1 +R2
VD,on
R1

Vin (V)
−VD,on /R1
(d)
IR1 =

(

Vin −VB
R1 +R2
VD,on −VB
R1

ID1 =

(

0
Vin −VD,on
R2

Vin < VB +
Vin > VB +

−

VD,on −VB
R1

R1 +R2
R1
R1 +R2
R1

(VD,on − VB )
(VD,on − VB )

Vin < VB +
Vin > VB +

R1 +R2
R1
R1 +R2
R1

(VD,on − VB )
(VD,on − VB )

IR1
ID1

VB +

Slope = 1/R2

R1 +R2
R1

(VD,on − VB )

(VD,on − VB ) /R1
Slope = 1/ (R1 + R2 )

Vin (V)

3.27 (a)
0
R2
R1 +R2

(Vin − VD,on )

Vin < VD,on
Vin > VD,on

Vout (V)

Vout =

(

Slope = R2 / (R1 + R2 )

VD,on
Vin (V)
(b)

Vout



−VD,on
2
= R1R+R
Vin
2


Vin − VD,on

2
VD,on
Vin < − R1R+R
2
R1 +R2
− R2 VD,on < Vin <
2
Vin > R1R+R
VD,on
1

R1 +R2
R1 VD,on

Vout (V)

Slope = 1

R2
V
R1 D,on

Slope = R2 / (R1 + R2 )

2
VD,on
− R1R+R
2

R1 +R2
VD,on
R1

Vin (V)
−VD,on
(c)
R2
R1 +R2

(Vin + VD,on ) − VD,on

Vin

Vin < −VD,on
Vin > −VD,on

Vout (V)

Vout =

(

Slope = 1

−VD,on
−VD,on
Slope = R2 / (R1 + R2 )

Vin (V)

(d)
0
R2
R1 +R2

(Vin − VD,on )

Vin < VD,on
Vin > VD,on

Vout (V)

Vout =

(

Slope = R2 / (R1 + R2 )

VD,on

VD,on
Vin (V)
(e)
Vout =

(

R2
R1 +R2

0

(Vin + VD,on ) Vin < −VD,on
Vin > −VD,on

Vout (V)
−VD,on
Vin (V)

Slope = R2 / (R1 + R2 )

3.28 (a)
IR1 =

(

0

ID1 =

(

0

Vin −VD,on
R1 +R2

Vin −VD,on
R1 +R2

Vin < VD,on
Vin > VD,on
Vin < VD,on
Vin > VD,on

IR1
ID1
Slope = 1/ (R1 + R2 )

VD,on
Vin (V)
(b)

IR1 =

ID1

 V +V
in
D,on


R1
Vin
R1 +R2

 VD,on
R1



0
= 0

 Vin −VD,on
R2

2
Vin < − R1R+R
VD,on
2
R1 +R2
− R2 VD,on < Vin <
2
Vin > R1R+R
VD,on
1

−

VD,on
R1

R1 +R2
R1 VD,on

2
VD,on
Vin < − R1R+R
2
R1 +R2
− R2 VD,on < Vin <
2
Vin > R1R+R
VD,on
1

R1 +R2
R1 VD,on

IR1
ID1

VD,on /R1
Slope = 1/ (R1 + R2 )
Slope = 1
R1 +R2
VD,on
R1

2
VD,on
− R1R+R
2
−VD,on /R2

Vin (V)

Slope = 1/R1

(c)
IR1 =

(

0

ID1 =

(

0 Vin < −VD,on
0 Vin > −VD,on

Vin +VD,on
R1 +R2

Vin < −VD,on
Vin > −VD,on

IR1
ID1

−VD,on
Vin (V)

Slope = 1/ (R1 + R2 )

(d)
IR1 =

(

0

ID1 =

(

0

Vin −VD,on
R1 +R2

Vin −VD,on
R1 +R2

Vin < VD,on
Vin > VD,on
Vin < VD,on
Vin > VD,on

IR1
ID1
Slope = 1/ (R1 + R2 )

VD,on
Vin (V)
(e)
IR1 =

(

0

ID1 =

(

0 Vin < −VD,on
0 Vin > −VD,on

Vin +VD,on
R1 +R2

Vin < −VD,on
Vin > −VD,on

IR1
ID1

−VD,on
Vin (V)

Slope = 1/ (R1 + R2 )

3.29 (a)

Vout (V)

Vout



Vin < VD,on
Vin
2
2
= VD,on + R1R+R
(V
−
V
)
VD,on < Vin < VD,on + R1R+R
(VD,on + VB )
in
D,on
2
1


R1 +R2
Vin − VD,on − VB
Vin > VD,on + R1 (VD,on + VB )

Slope = 1

VD,on +

R2
R1

(VD,on + VB )

Slope = R2 / (R1 + R2 )

VD,on
Slope = 1

VD,on

VD,on +

R1 +R2
R1

(VD,on + VB )
Vin (V)

(b)
Vout =

(

Vin + VD,on − VB
R2
R1 +R2 (Vin − VD,on )

Vin < VD,on +
Vin > VD,on +

R1 +R2
R1
R1 +R2
R1

(VB − 2VD,on )
(VB − 2VD,on )

Vout (V)

Slope = R2 / (R1 + R2 )

R2
R1

(VB − 2VD,on )
VD,on +

R1 +R2
R1

(VB − 2VD,on )
Vin (V)

Slope = 1

(c)

Vout (V)

Vout =

(

Vin
VD,on + VB

Vin < VD,on + VB
Vin > VD,on + VB

VD,on + VB

VD,on + VB
Vin (V)

Slope = 1

(d)

Vout (V)

Vout =



0

Vin < VD,on
2
(Vin − VD,on ) VD,on < Vin < VD,on + R1R+R
(VB + VD,on )
2

R1 +R2
VD,on + VB
Vin > VD,on + R2 (VB + VD,on )
R2
 R1 +R2

VD,on + VB

Slope = R2 / (R1 + R2 )

VD,on

VD,on +

R1 +R2
R2

(VB + VD,on )
Vin (V)

3.30 (a)

IR1 =



0

ID1 =

(

Vin −VD,on
R1 +R2

 VD,on
+VB
R1

Vin < VD,on
2
VD,on < Vin < VD,on + R1R+R
(VD,on + VB )
1
R1 +R2
Vin > VD,on + R1 (VD,on + VB )

0

Vin −2VD,on −VB
R2

−

VD,on +VB
R1

Vin < VD,on +
Vin > VD,on +

R1 +R2
R1
R1 +R2
R1

(VD,on + VB )
(VD,on + VB )

IR1
ID1
VD,on + VB

Slope = 1/ (R1 + R2 )

Slope = 1/R2

VD,on

VD,on +

R1 +R2
R1

(VD,on + VB )
Vin (V)

(b) If VB < 2VD,on :
IR1 = ID1 =

(

0
Vin −VD,on
R1 +R2

Vin < VD,on
Vin > VD,on

IR1
ID1
Slope = 1/ (R1 + R2 )

VD,on
Vin (V)
If VB > 2VD,on :
IR1 = ID1 =

(

VB −2VD,on
R1
Vin −VD,on
R1 +R2

Vin < VD,on +
Vin > VD,on +

R1 +R2
R1
R1 +R2
R1

(VB − 2VD,on )
(VB − 2VD,on )

IR1
ID1
Slope = 1/ (R1 + R2 )

VB −2VD,on
R1

VD,on +

R1 +R2
R1

(VB − 2VD,on )
Vin (V)

(c)
IR1 =

ID1

(

0
Vin −VD,on −VB
R1



0
= 0

 Vin −2VD,on −VB
R2

Vin < VD,on + VB
Vin > VD,on + VB
Vin < VD,on + VB
VD,on + VB < Vin < 2VD,on + VB
Vin > 2VD,on + VB

IR1
ID1

Slope = 1/R1

Slope = 1/R2

VD,on + VB 2VD,on + VB
Vin (V)
(d)

IR1 =



0

Vin < VD,on
2
VD,on < Vin < VD,on + R1R+R
(VB + VD,on )
2
R1 +R2
Vin > VD,on + R2 (VB + VD,on )

ID1 =



0

Vin < VD,on
2
VD,on < Vin < VD,on + R1R+R
(VB + VD,on )
2
R1 +R2
Vin > VD,on + R2 (VB + VD,on )

Vin −VD,on
R1 +R2

 Vin −2V
D,on −VB
R1
Vin −VD,on
R1 +R2

 Vin −2V
D,on −VB
R1

IR1
ID1

Slope = 1/R2

VB +VD,on
R2

Slope = 1/ (R1 + R2 )

VD,on

VD,on +

R1 +R2
R2

(VB + VD,on )
Vin (V)

3.31 (a)
Vin − VD,on
= 1.6 mA
R1
VT
=
= 16.25 Ω
ID1
R1
=
∆Vin = 98.40 mV
rd + R1

ID1 =
rd1
∆Vout
(b)

Vin − 2VD,on
= 0.8 mA
R1
VT
= rd2 =
= 32.5 Ω
ID1
R1 + rd2
=
∆Vin = 96.95 mV
R1 + rd1 + rd2

ID1 = ID2 =
rd1
∆Vout
(c)

Vin − 2VD,on
= 0.8 mA
R1
VT
= rd2 =
= 32.5 Ω
ID1
rd2
∆Vin = 3.05 mV
=
rd1 + R1 + rd2

ID1 = ID2 =
rd1
∆Vout
(d)

Vin − VD,on
VD,on
−
= 1.2 mA
R1
R2
VT
=
= 21.67 Ω
ID2
R2 k rd2
=
∆Vin = 2.10 mV
R1 + R2 k rd2

ID2 =
rd2
∆Vout

3.32 (a)
∆Vout = ∆Iin R1 = 100 mV
(b)
ID1 = ID2 = Iin = 3 mA
VT
rd1 = rd2 =
= 8.67 Ω
ID1
∆Vout = ∆Iin (R1 + rd2 ) = 100.867 mV
(c)
ID1 = ID2 = Iin = 3 mA
VT
rd1 = rd2 =
= 8.67 Ω
ID1
∆Vout = ∆Iin rd2 = 0.867 mV
(d)
ID2 = Iin −
rd2 =

VD,on
= 2.6 mA
R2

VT
= 10 Ω
ID2

∆Vout = ∆Iin (R2 k rd2 ) = 0.995 mV

3.34

Vin (t)
Vout (t)
Vp
Vp − VD,on

VD,on + 0.5 V
0.5 V
π/ω

2π/ω
t

−Vp

3.35

Vin (t)
Vout (t)
Vp

0.5 V
−VD,on + 0.5 V

π/ω

2π/ω
t

−Vp + VD,on
−Vp

3.36
Vp − VD,on
RL C1 fin
Vp = 3.5 V

VR ≈

RL = 100 Ω
C1 = 1000 µF
fin = 60 Hz
VR = 0.45 V

3.37
IL
≤ 300 mV
C1 fin
= 60 Hz

VR =
fin

IL = 0.5 A
C1 ≥

IL
= 27.78 mF
(300 mV) fin

3.38 Shorting the input and output grounds of a full-wave rectifier shorts out the diode D4 from Fig. 3.38(b).
Redrawing the modified circuit, we have:
D3
+
Vin

+
D2

RL

Vout

D1

−

−

On the positive half-cycle, D3 turns on and forms a half-wave rectifier along with RL (and CL , if
included). On the negative half-cycle, D2 shorts the input (which could cause a dangerously large
current to flow) and the output remains at zero. Thus, the circuit behaves like a half-wave recifier.
The plots of Vout (t) are shown below.

Vin (t) = V0 sin(ωt)
Vout (t) (without a load capacitor)
Vout (t) (with a load capacitor)
V0
V0 − VD,on

VD,on
π/ω

2π/ω
t

−V0

3.39 Note that the waveforms for VD1 and VD2 are identical, as are the waveforms for VD3 and VD4 .

Vin (t) = V0 sin(ωt)
Vout (t)
VD1 (t), VD2 (t)
VD3 (t), VD4 (t)
V0

V0 − 2VD,on

2VD,on
VD,on
π/ω
−VD,on
−2VD,on

−V0 + 2VD,on
−V0 + VD,on
−V0

2π/ω
t

3.40 During the positive half-cycle, D2 and D3 will remain reverse-biased, causing Vout to be zero as
no current will flow through RL . During the negative half-cycle, D1 and D3 will short the input
(potentially causing damage to the devices), and once again, no current will flow through RL (even
though D2 will turn on, there will be no voltage drop across RL ). Thus, Vout always remains at zero,
and the circuit fails to act as a rectifier.

3.42 Shorting the negative terminals of Vin and Vout of a full-wave rectifier shorts out the diode D4 from
Fig. 3.38(b). Redrawing the modified circuit, we have:
D3
+
D2

Vin

RL

+
Vout

D1

−

−

On the positive half-cycle, D3 turns on and forms a half-wave rectifier along with RL (and CL , if
included). On the negative half-cycle, D2 shorts the input (which could cause a dangerously large
current to flow) and the output remains at zero. Thus, the circuit behaves like a half-wave recifier.
The plots of Vout (t) are shown below.

Vin (t) = V0 sin(ωt)
Vout (t) (without a load capacitor)
Vout (t) (with a load capacitor)
V0

π/ω

2π/ω
t

−V0

3.44 (a) We know that when a capacitor is discharged by a constant current at a certain frequency, the
ripple voltage is given by CfIin , where I is the constant current. In this case, we can calculate the
V −5V

current as approximately p R1D,on (since Vp − 5VD,on is the voltage drop across R1 , assuming
R1 carries a constant current). This gives us the following:
1 Vp − 5VD,on
2 RL C1 fin
Vp = 5 V

VR ≈

RL = 1 kΩ
C1 = 100 µF
fin = 60 Hz
VR = 166.67 mV
(b) The bias current through the diodes is the same as the bias current through R1 , which is
Vp −5VD,on
= 1 mA. Thus, we have:
R1
VT
= 26 Ω
ID
3rd
=
VR = 12.06 mV
R1 + 3rd

rd =
VR,load

3.45
ID1 =

(

0

ID2 =

(

Vin +VD,on +VB2
R1

Vin −VD,on −VB1
R1

0

Vin < VD,on + VB1
Vin > VD,on + VB1
Vin < −VD,on − VB2
Vin > −VD,on − VB2

Vin (t)
ID1 (t)
ID2 (t)
V0

V0 −VB1 −VD,on
R1

0

0

−V0 +VB1 +VD,on
R1

−VD,on − VB2

−V0
−π/ω

0
t

π/ω

ID1 (t) and ID2 (t)

Vin (t)

VD,on + VB1

4.4 According to Equation (4.8), we have

AE qDn n2i  VBE /VT
−1
e
NB WB
1
∝
WB

IC =

We can see that if WB increases by a factor of two, then IC decreases by a factor of two .

4.11
VBE = 1.5 V − IE (1 kΩ)
≈ 1.5 V − IC (1 kΩ) (assuming β ≫ 1)
 
IC
= VT ln
IS
IC = 775 µA
VX ≈ IC (1 kΩ)
= 775 mV

4.12 Since we have only integer multiples of a unit transistor, we need to find the largest number that
divides both I1 and I2 evenly (i.e., we need to find the largest x such that I1 /x and I2 /x are integers).
This will ensure that we use the fewest transistors possible. In this case, it’s easy to see that we should
pick x = 0.5 mA, meaning each transistor should have 0.5 mA flowing through it. Therefore, I1 should
be made up of 1 mA/0.5 mA = 2 parallel transistors, and I2 should be made up of 1.5 mA/0.5 mA = 3
parallel transistors. This is shown in the following circuit diagram.
I1

VB

I2

+
−

Now we have to pick VB so that IC = 0.5 mA for each transistor.
 
IC
VB = VT ln
IS


5 × 10−4 A
= (26 mV) ln
3 × 10−16 A
= 732 mV

4.15
VB − VBE
= IB
R1
IC
=
β
β
[VB − VT ln(IC /IS )]
IC =
R1
IC = 786 µA

4.17 First, note that VBE1 = VBE2 = VBE .
VB = (IB1 + IB2 )R1 + VBE
R1
=
(IX + IY ) + VT ln(IX /IS1 )
β
5
IS2 = IS1
3
5
⇒ IY = IX
3
8R1
VB =
IX + VT ln(IX /IS1 )
3β
IX = 509 µA
IY = 848 µA

4.21 (a)
VBE = 0.8 V
IC = IS eVBE /VT
= 18.5 mA
VCE = VCC − IC RC
= 1.58 V
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 710 mS
rπ = β/gm = 141 Ω
ro = ∞
The small-signal model is shown below.
B

C
+
rπ

gm vπ

vπ
−

E

(b)
IB = 10 µA
IC = βIB = 1 mA
VBE = VT ln(IC /IS ) = 724 mV
VCE = VCC − IC RC
= 1.5 V
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 38.5 mS
rπ = β/gm = 2.6 kΩ
ro = ∞
The small-signal model is shown below.
B

C
+
rπ

gm vπ

vπ
−

E

(c)
VCC − VBE
1+β
=
IC
RC
β
β VCC − VT ln(IC /IS )
IC =
1+β
RC

IE =

IC = 1.74 mA
VBE = VT ln(IC /IS ) = 739 mV
VCE = VBE = 739 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 38.5 mS
rπ = β/gm = 2.6 kΩ
ro = ∞
The small-signal model is shown below.
B

C
+
rπ

gm vπ

vπ
−

E

4.22 (a)
IB = 10 µA
IC = βIB = 1 mA
VBE = VT ln(IC /IS ) = 739 mV
VCE = VCC − IE (1 kΩ)
1+β
(1 kΩ)
= VCC −
β
= 0.99 V
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 38.5 mS
rπ = β/gm = 2.6 kΩ
ro = ∞
The small-signal model is shown below.
B

C
+
rπ

gm vπ

vπ
−

E

(b)
1+β
VCC − VBE
=
IC
1 kΩ
β
β VCC − VT ln(IC /IS )
IC =
1+β
1 kΩ

IE =

IC = 1.26 mA
VBE = VT ln(IC /IS ) = 730 mV
VCE = VBE = 730 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 48.3 mS
rπ = β/gm = 2.07 kΩ
ro = ∞
The small-signal model is shown below.

B

C
+
rπ

gm vπ

vπ
−

E

(c)
IE = 1 mA
β
IC =
IE = 0.99 mA
1+β
VBE = VT ln(IC /IS ) = 724 mV
VCE = VBE = 724 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 38.1 mS
rπ = β/gm = 2.63 kΩ
ro = ∞
The small-signal model is shown below.
B

C
+
rπ

gm vπ

vπ
−

E

(d)
IE = 1 mA
β
IC =
IE = 0.99 mA
1+β
VBE = VT ln(IC /IS ) = 724 mV
VCE = VBE = 724 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 38.1 mS
rπ = β/gm = 2.63 kΩ
ro = ∞
The small-signal model is shown below.

B

C
+
rπ

gm vπ

vπ
−

E

4.31
IC = IS eVBE /VT
IC,T otal = nIC



VCE
1+
VA

= nIS eVBE /VT



1+

∂IC
∂VBE
IS
= n eVBE /VT
VT
IC
≈n
VT
= ngm

VCE
VA



gm,T otal =

IB,T otal
rπ,T otal

ro,T otal

= n × 0.4435 S
1
= IC,T otal
β
−1

∂IB,T otal
=
∂VBE

−1
IC,T otal
≈
βVT
−1

nIC
=
βVT
rπ
=
n
225.5 Ω
=
(assuming β = 100)
n

−1
∂IC,T otal
=
∂VCE
−1

IC,T otal
≈
VA
VA
=
nIC
ro
=
n
693.8 Ω
=
n

The small-signal model is shown below.
B

C
+

rπ,T otal

gm,T otal vπ

vπ
−

E

ro,T otal

4.32 (a)
VBE = VCE (for Q1 to operate at the edge of saturation)
VT ln(IC /IS ) = VCC − IC RC
IC = 885.7 µA
VB = VBE = 728.5 mV
′
′
(b) Let IC′ , VB′ , VBE
, and VCE
correspond to the values where the collector-base junction is forward
biased by 200 mV.
′
′
VBE
= VCE
+ 200 mV

VT ln(IC′ /IS ) = VCC − IC′ RC + 200 mV
IC′ = 984.4 µA
VB′ = 731.3 mV
Thus, VB can increase by VB′ − VB = 2.8 mV if we allow soft saturation.

4.34
VBE = VCC − IB RB
VT ln(IC /IS ) = VCC − IC RB /β
IC = 1.67 mA
VBC = VCC − IB RB − (VCC − IC RC )
< 200 mV
IC RC − IB RB < 200 mV
200 mV + IB RB
RC <
IC
200 mV + IC RB /β
=
IC
RC < 1.12 kΩ

4.41

VCC

VEB = VEC (for Q1 to operate at the edge of saturation)
− IB RB = VCC − IC RC
IC RB /β = IC RC
RB /β = RC
β = RB /RC
= 100

4.44 (a)
IB = 2 µA
IC = βIB
= 200 µA
VEB = VT ln(IC /IS )
= 768 mV
VEC = VCC − IE (2 kΩ)
1+β
= VCC −
IC (2 kΩ)
β
= 2.1 V
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 7.69 mS
rπ = β/gm = 13 kΩ
ro = ∞
The small-signal model is shown below.
B

C
+
rπ

gm vπ

vπ
−

E

(b)
IE =

VCC − VEB
5 kΩ

VCC − VT ln(IC /IS )
1+β
IC =
β
5 kΩ
IC = 340 µA
VEB = 782 mV
VEC = VEB = 782 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 13.1 mS
rπ = β/gm = 7.64 kΩ
ro = ∞
The small-signal model is shown below.

B

C
+
rπ

gm vπ

vπ
−

E

(c)
IE =

1+β
IC = 0.5 mA
β

IC = 495 µA
VEB = 971 mV
VEC = VEB = 971 mV
Q1 is operating in forward active. Its small-signal parameters are
gm = IC /VT = 19.0 mS
rπ = β/gm = 5.25 kΩ
ro = ∞
The small-signal model is shown below.
B

C
+
rπ

gm vπ

vπ
−

E

4.49 The direction of current flow in the large-signal model (Fig. 4.40) indicates the direction of positive
current flow when the transistor is properly biased.
The direction of current flow in the small-signal model (Fig. 4.43) indicates the direction of positive
change in current flow when the base-emitter voltage vbe increases. For example, when vbe increases,
the current flowing into the collector increases, which is why ic is shown flowing into the collector in
Fig. 4.43. Similar reasoning can be applied to the direction of flow of ib and ie in Fig. 4.43.

4.53 (a)
VCB2 < 200 mV
IC2 RC < 200 mV
IC2 < 400 µA
VEB2 = VE2
= VT ln(IC2 /IS2 )
< 741 mV
β2
IE2 RC
1 + β2
β2 1 + β1
IC1 RC
1 + β2 β1
IC1
VBE1
Vin

< 200 mV
< 200 mV
< 396 µA
= VT ln(IC1 /IS1 )
< 712 mV
= VBE1 + VEB2
< 1.453 V

(b)
IC1 = 396 µA
IC2 = 400 µA
gm1 = 15.2 mS
rπ1 = 6.56 kΩ
ro1 = ∞
gm2 = 15.4 mS
rπ2 = 3.25 kΩ
ro2 = ∞
The small-signal model is shown below.
+
vin

rπ1
−

B1
+
vπ1

C1
gm1 vπ1

−
E1 /E2

rπ2

−
vπ2
+
B2

gm2 vπ2
vout

C2
RC

4.55 (a)

VBE2 − (VCC

VBC2 < 200 mV
− IC2 RC ) < 200 mV

VT ln(IC2 /IS2 ) + IC2 RC − VCC < 200 mV
IC2 < 3.80 mA
VBE2 < 799.7 mV
1 + β1
IC1 = IB2 = IC2 /β2
IE1 =
β1
IC1 < 75.3 µA
VBE1 < 669.2 mV
Vin = VBE1 + VBE2
< 1.469 V
(b)
IC1 = 75.3 µA
IC2 = 3.80 mA
gm1 = 2.90 mS
rπ1 = 34.5 kΩ
ro1 = ∞
gm2 = 146.2 mS
rπ2 = 342 Ω
ro2 = ∞
The small-signal model is shown below.
B1
+

+
vin

rπ1
−

C1
gm1 vπ1

vπ1
−
E1 /B2

C2

vout

+
rπ2

gm2 vπ2

vπ2
−

E2

RC

5.3 (a) Looking into the base of Q1 we see an equivalent resistance of rπ1 , so we can draw the following
equivalent circuit for finding Rin :
R1

R2

rπ1

Rin

Rin = R1 + R2 k rπ1
(b) Looking into the emitter of Q1 we see an equivalent resistance of
following equivalent circuit for finding Rin :

1
gm1

R1

1
gm1

k rπ1 , so we can draw the

k rπ1

Rin

Rin = R1 k

1
k rπ1
gm1

(c) Looking down from the emitter of Q1 we see an equivalent resistance of
the following equivalent circuit for finding Rin :

1
gm2

k rπ2 , so we can draw

VCC
Q1

Rin

1
gm2

Rin = rπ1 + (1 + β1 )



k rπ2

1
gm2

k rπ2



(d) Looking into the base of Q2 we see an equivalent resistance of rπ2 , so we can draw the following
equivalent circuit for finding Rin :

VCC
Q1

Rin

rπ2

Rin = rπ1 + (1 + β1 )rπ2

5.4 (a) Looking into the collector of Q1 we see an equivalent resistance of ro1 , so we can draw the following
equivalent circuit for finding Rout :

ro1

R1
Rout

Rout = ro1 k R1
(b) Let’s draw the small-signal model and apply a test source at the output.
RB
+
rπ1

vπ1

+
gm1 vπ1

ro1

it

vt
−

−

it = gm1 vπ1 +

vt
ro1

vπ1 = 0
vt
it =
ro1
vt
Rout =
= ro1
it
(c) Looking down from the emitter of Q1 we see an equivalent resistance of
can draw the following equivalent circuit for finding Rout :
Rout

Q1

1
gm2

k rπ2 k ro2



1
k rπ2 k ro2
Rout = ro1 + (1 + gm1 ro1 ) rπ1 k
gm2

1
gm2

k rπ2 k ro2 , so we

(d) Looking into the base of Q2 we see an equivalent resistance of rπ2 , so we can draw the following
equivalent circuit for finding Rout :
Rout

Q1

rπ2

Rout = ro1 + (1 + gm1 ro1 ) (rπ1 k rπ2 )

5.5 (a) Looking into the base of Q1 we see an equivalent resistance of rπ1 , so we can draw the following
equivalent circuit for finding Rin :
R1

R2

rπ1

Rin

Rin = R1 + R2 k rπ1
(b) Let’s draw the small-signal model and apply a test source at the input.
+
rπ1

gm1 vπ1

vπ1

R1

−
+
it

vt
−

vπ1
− gm1 vπ1
rπ1
vπ1 = −vt
vt
+ gm1 vt
it =
rπ1


1
it = vt gm1 +
rπ1
it = −

Rin =

vt
1
=
k rπ1
it
gm1

(c) From our analysis in part (b), we know that looking into the emitter we see a resistance of
1
gm2 k rπ2 . Thus, we can draw the following equivalent circuit for finding Rin :

VCC
Q1

Rin

1
gm2

Rin = rπ1 + (1 + β1 )



k rπ2

1
gm2

k rπ2



(d) Looking up from the emitter of Q1 we see an equivalent resistance of
the following equivalent circuit for finding Rin :

1
gm2

k rπ2 , so we can draw

VCC
1
gm2

k rπ2

Q1

Rin

Rin = rπ1 + (1 + β1 )



1
gm2

k rπ2



(e) We know that looking into the base of Q2 we see Rin = rπ2 if the emitter is grounded. Thus,
transistor Q1 does not affect the input impedance of this circuit.

5.6 (a) Looking into the collector of Q1 we see an equivalent resistance of ro1 , so we can draw the following
equivalent circuit for finding Rout :

ro1

RC
Rout

Rout = RC k ro1
(b) Looking into the emitter of Q2 we see an equivalent resistance of
the following equivalent circuit for finding Rout :

1
gm2

k rπ2 k ro2 , so we can draw

Rout

Q1

RE =

1
gm2

k rπ2 k ro2



1
Rout = ro1 + (1 + gm1 ro1 ) rπ1 k
k rπ2 k ro2
gm2

5.7 (a)
VCC − IB (100 kΩ) = VBE = VT ln(IC /IS )
1
VCC − IC (100 kΩ) = VT ln(IC /IS )
β
IC = 1.754 mA
VBE = VT ln(IC /IS ) = 746 mV
VCE = VCC − IC (500 Ω) = 1.62 V
Q1 is operating in forward active.
(b)
IE1 = IE2 ⇒ VBE1 = VBE2
VCC − IB1 (100 kΩ) = 2VBE1
1
VCC − IC1 (100 kΩ) = 2VT ln(IC1 /IS )
β
IC1 = IC2 = 1.035 mA
VBE1 = VBE2 = 733 mV
VCE2 = VBE2 = 733 mV
VCE1 = VCC − IC (1 kΩ) − VCE2
= 733 mV
Both Q1 and Q2 are at the edge of saturation.
(c)
VCC − IB (100 kΩ) = VBE + 0.5 V
1
VCC − IC (100 kΩ) = VT ln(IC /IS ) + 0.5 V
β
IC = 1.262 mA
VBE = 738 mV
VCE = VCC − IC (1 kΩ) − 0.5 V
= 738 mV
Q1 is operating at the edge of saturation.

5.8 See Problem 7 for the derivation of IC for each part of this problem.
(a)
IC1 = 1.754 mA
gm1 = IC1 /VT = 67.5 mS
rπ1 = β/gm1 = 1.482 kΩ
100 kΩ
+
rπ1

vπ1

gm1 vπ1

500 Ω

−

(b)
IC1 = IC2 = 1.034 mA
gm1 = gm2 = IC1 /VT = 39.8 mS
rπ1 = rπ2 = β/gm1 = 2.515 kΩ
100 kΩ
+
rπ1

vπ1

gm1 vπ1

1 kΩ

−

+
rπ2

vπ2

gm2 vπ2

−

(c)
IC1 = 1.26 mA
gm1 = IC1 /VT = 48.5 mS
rπ1 = β/gm1 = 2.063 kΩ
100 kΩ
+
rπ1

vπ1
−

gm1 vπ1

1 kΩ

5.9 (a)
VCC − VBE
VBE
IC
−
= IB =
34 kΩ
16 kΩ
β
VT ln(IC /IS )
VCC − VT ln(IC /IS )
−β
IC = β
34 kΩ
16 kΩ
IC = 677 µA
VBE = 726 mV
VCE = VCC − IC (3 kΩ) = 468 mV
Q1 is in soft saturation.
(b)
IE1 = IE2

VCC − 2VBE
9 kΩ

⇒ IC1 = IC2
⇒ VBE1 = VBE2 = VBE
2VBE
IC1
−
= IB1 =
16 kΩ
β
2VT ln(IC1 /IS )
VCC − 2VT ln(IC1 /IS )
−β
IC1 = β
9 kΩ
16 kΩ
IC1 = IC2 = 1.72 mA
VBE1 = VBE2 = VCE2 = 751 mV
VCE1 = VCC − IC1 (500 Ω) − VCE2 = 890 mV

Q1 is in forward active and Q2 is on the edge of saturation.
(c)
IC
VCC − VBE − 0.5 V VBE + 0.5 V
−
= IB =
12 kΩ
13 kΩ
β
VT ln(IC /IS ) + 0.5 V
VCC − VT ln(IC /IS ) − 0.5 V
−β
IC = β
12 kΩ
13 kΩ
IC = 1.01 mA
VBE = 737 mV
VCE = VCC − IC (1 kΩ) − 0.5 V = 987 mV
Q1 is in forward active.

5.10 See Problem 9 for the derivation of IC for each part of this problem.
(a)
IC = 677 µA
gm = IC /VT = 26.0 mS
rπ = β/gm = 3.84 kΩ

+
(34 kΩ) k (16 kΩ)

rπ

vπ

gm vπ

3 kΩ

−

(b)
IC1 = IC2 = 1.72 mA
gm1 = gm2 = IC1 /VT = 66.2 mS
rπ1 = rπ2 = β/gm1 = 1.51 kΩ

+
(9 kΩ) k (16 kΩ)

rπ1

vπ1
−

gm1 vπ1

500 Ω

+
rπ2

vπ2
−

gm2 vπ2

(c)
IC = 1.01 mA
gm = IC /VT = 38.8 mS
rπ = β/gm = 2.57 kΩ

(12 kΩ) k (13 kΩ)

rπ

+
vπ
−

gm vπ

1 kΩ

5.11 (a)

VCC

VCC

VCE ≥ VBE (in order to guarantee operation in the active mode)
− IC (2 kΩ) ≥ VBE

VCC − IC (2 kΩ) ≥ VT ln(IC /IS )
IC ≤ 886 µA
VCC − VBE
VBE
IC
−
= IB =
RB
3 kΩ
β
− VT ln(IC /IS ) VT ln(IC /IS )
IC
−
=
RB
3 kΩ
β


VT ln(IC /IS )
IC
= VCC − VT ln(IC /IS )
+
RB
β
3 kΩ
VCC − VT ln(IC /IS )
RB = I
VT ln(IC /IS )
C
β +
3 kΩ
RB ≥ 7.04 kΩ

(b)
VCC − VBE
VBE
IC
−
= IB =
RB
3 kΩ
β
VT ln(IC /IS )
VCC − VT ln(IC /IS )
−β
IC = β
RB
3 kΩ
IC = 1.14 mA
VBE = 735 mV
VCE = VCC − IC (2 kΩ) = 215 mV
VBC = VBE − VCE = 520 mV

5.13 We know the input resistance is Rin = R1 k R2 k rπ . Since we want the minimum values of R1 and
R2 such that Rin > 10 kΩ, we should pick the maximum value allowable for rπ , which means picking
the minimum value allowable for gm (since rπ ∝ 1/gm ), which is gm = 1/260 S.
1
S
260
IC = gm VT = 100 µA
VBE = VT ln(IC /IS ) = 760 mV
IC
IB =
= 1 µA
β
VBE
−
= IB
R2
VCC − VBE
R1 =
IB + VRBE
2
gm =

VCC − VBE
R1

β
= 26 kΩ
gm
= R1 k R2 k rπ

rπ =
Rin

=

VCC − VBE
IB + VRBE
2

> 10 kΩ
R2 > 23.57 kΩ
R1 > 52.32 kΩ

!

k R2 k rπ

5.14
IC
1
≥
S
VT
26
β
rπ =
= 2.6 kΩ
gm
Rin = R1 k R2 k rπ
≤ rπ
gm =

According to the above analysis, Rin cannot be greater than 2.6 kΩ. This means that the requirement
that Rin ≥ 10 kΩ cannot be met. Qualitatively, the requirement for gm to be large forces rπ to be
small, and since Rin is bounded by rπ , it puts an upper bound on Rin that, in this case, is below the
required 10 kΩ.

5.15
Rout = RC = R0
Av = −gm RC = −gm R0 = −

IC
R0 = −A0
VT

A0
VT
R0
VT
R0
rπ = β
=β
IC
A0

IC =

VBE = VT ln(IC /IS ) = VT ln
VBE
IC
VCC − VBE
−
= IB =
R1
R2
β
VCC − VBE
R1 = IC
VBE
β + R2



A0 VT
R0 IS



Rin = R1 k R2 k rπ



0 VT
VCC − VT ln A
R0 IS
R

  k R2 k β 0
=
IC
A0 VT
VT
A
0
β + R2 ln R0 IS

In order to maximize Rin , we can let R2 → ∞. This gives us


Rin,max = β

VCC − VT ln
IC



A0 VT
R0 IS



 k β R0
A0

5.16 (a)

VBE
VCC − VBE − IE RE
−
R1

IC = 0.25 mA
VBE = 696 mV
+ IE RE
IC
= IB =
R2
β
R1 =

VCC − VBE −
IC
β

+

1+β
β IC RE

VBE + 1+β
β IC RE
R2

= 22.74 kΩ
(b) First, consider a 5 % increase in RE .

VCC − VBE − IE RE
VBE
−
R1
VCC − VT ln(IC /IS ) −

1+β
β IC RE

R1

−

VT ln(IC /IS ) +
R2

RE = 210 Ω
+ IE RE
IC
= IB =
R2
β

1+β
β IC RE

IC
β
IC = 243 µA
= IB =

IC − IC,nom
× 100 = −2.6 %
IC,nom
Now, consider a 5 % decrease in RE .
RE = 190 Ω
IC = 257 µA
IC − IC,nom
× 100 = +2.8 %
IC,nom

5.17

VCC
VBE
VCC − VBE − IE RE
−
30 kΩ

VCE ≥ VBE (in order to guarantee operation in the active mode)
− IC RC ≥ VT ln(IC /IS )
IC ≤ 833 µA
IC
+ IE RE
= IB =
R2
β
VBE + IE RE
R2 = VCC −VBE −IE RE
−
30 kΩ
=

VT ln(IC /IS )

IC
β
+ 1+β
β IC RE

VCC −VT ln(IC /IS )− 1+β
β IC RE
30 kΩ

R2 ≤ 20.66 kΩ

−

IC
β

5.18 (a) First, note that VBE1 = VBE2 = VBE , but since IS1 = 2IS2 , IC1 = 2IC2 . Also note that
β1 = β2 = β = 100.
IB1 =

IC1
VBE + (IE1 + IE2 )RE
VCC − VBE − (IE1 + IE2 )RE
−
=
β
R1
R2

IC1 = β

VCC − VT ln(IC1 /IS1 ) −

3 1+β
2 β IC1 RE

R1

−

VT ln(IC1 /IS1 ) +

3 1+β
2 β IC1 RE

R2

IC1 = 707 µA
IC2 =

IC1
= 354 µA
2

(b) The small-signal model is shown below.

RC

+
R1

R2

rπ1

vπ1

+
rπ2

−

gm1 vπ1

vπ2
−

RE

We can simplify the small-signal model as follows:

RC

+
R1 k R2

rπ1 k rπ2

gm1 vπ

vπ
−

RE

gm2 vπ2

gm2 vπ2

gm1 = IC1 /VT = 27.2 mS
rπ1 = β1 /gm1 = 3.677 kΩ
gm2 = IC2 /VT = 13.6 mS
rπ2 = β2 /gm2 = 7.355 kΩ

5.19 (a)

VCC − 2VBE1
9 kΩ

IE1 = IE2 ⇒ VBE1 = VBE2
2VBE1
IC1
−
= IB1 =
16 kΩ
β1
2VT ln(IC1 /IS1 )
VCC − 2VT ln(IC1 /IS1 )
− β1
IC1 = β1
9 kΩ
16 kΩ
IC1 = IC2 = 1.588 mA
VBE1 = VBE2 = VT ln(IC1 /IS1 ) = 754 mV
VCE2 = VBE2 = 754 mV
VCE1 = VCC − IC1 (100 Ω) − VCE2 = 1.587 V

(b) The small-signal model is shown below.

(9 kΩ) k (16 kΩ)

rπ1

+
vπ1

gm1 vπ1

100 Ω

−

rπ2

+
vπ2

gm2 vπ2

−

IC1
= 61.1 mS
VT
β1
= 1.637 kΩ
=
gm1

gm1 = gm2 =
rπ1 = rπ2

5.22
VCC − IE (500 Ω) − IB (20 kΩ) − IE (400 Ω) = VBE
1+β
1
VCC −
IC (500 Ω + 400 Ω) − IC (20 kΩ) = VT ln(IC /IS )
β
β
IC = 1.584 mA
VBE = VT ln(IC /IS ) = 754 mV
VCE = VCC − IE (500 Ω) − IE (400 Ω)
1+β
IC (500 Ω + 400 Ω) = 1.060 V
= VCC −
β
Q1 is operating in forward active.

5.23

VCC − IE (1 kΩ) − IB RB − (VCC

VBC ≤ 200 mV
− IE (1 kΩ) − IC (500 Ω)) ≤ 200 mV
IC (500 Ω) − IB RB ≤ 200 mV
IB RB ≥ IC (500 Ω) − 200 mV
VCC − IE (1 kΩ) − IB RB = VBE = VT ln(IC /IS )

VCC −

1+β
IC (1 kΩ) − IC (500 Ω) + 200 mV ≤ VT ln(IC /IS )
β
IC ≥ 1.29 mA
RB ≥

IC (500 Ω) − 200 mV

≥ 34.46 kΩ

IC
β

5.25 (a)

VCC

IC1 = 1 mA
VCC − (IE1 + IE2 )(500 Ω) = VT ln(IC2 /IS2 )


1+β
1+β
−
IC1 +
IC2 (500 Ω) = VT ln(IC2 /IS2 )
β
β

IC2 = 2.42 mA
VB − (IE1 + IE2 )(500 Ω) = VT ln(IC1 /IS1 )


1+β
1+β
IC1 +
IC2 (500 Ω) = VT ln(IC1 /IS1 )
VB −
β
β
VB = 2.68 V
(b) The small-signal model is shown below.

200 Ω

rπ1

+
vπ1
−

rπ2

+
vπ2

gm1 vπ1

−

500 Ω

gm1 = IC1 /VT = 38.5 mS
rπ1 = β1 /gm1 = 2.6 kΩ
gm2 = IC2 /VT = 93.1 mS
rπ2 = β2 /gm2 = 1.074 kΩ

gm2 vπ2

5.26 (a)

VCC

VCC − IB (60 kΩ) = VEB
1
−
IC (60 kΩ) = VT ln(IC /IS )
βpnp
IC = 1.474 mA
VEB = VT ln(IC /IS ) = 731 mV
VEC = VCC − IC (200 Ω) = 2.205 V

Q1 is operating in forward active.
(b)

VCC − VT ln



βnpn
1 + βnpn

VCC − VBE1 − IB2 (80 kΩ) = VEB2
VCC − VT ln(IC1 /IS ) − IB2 (80 kΩ) = VT ln(IC2 /IS )
βnpn
IE1
IC1 =
1 + βnpn
βnpn
IE2
=
1 + βnpn
βnpn
1 + βpnp
=
·
IC2
1 + βnpn
βpnp

1
1 + βpnp IC2
−
·
·
IC2 (80 kΩ) = VT ln(IC2 /IS )
βpnp
IS
βpnp
IC2 = 674 µA
VBE2 = VT ln(IC2 /IS ) = 711 mV
IC1 = 680 µA
VBE1 = VT ln(IC1 /IS ) = 711 mV
VCE1 = VBE1 = 711 mV
VCE2 = VCC − VCE1 − IC2 (300 Ω)
= 1.585 V

Q1 is operating on the edge of saturation. Q2 is operating in forward active.

5.27 See Problem 26 for the derivation of IC for each part of this problem.
(a) The small-signal model is shown below.
60 kΩ
+
rπ

gm vπ

vπ
−

IC = 1.474 mA
IC
gm =
= 56.7 mS
VT
β
= 1.764 kΩ
rπ =
gm
(b) The small-signal model is shown below.
+
rπ1

vπ1

gm1 vπ1

−

−
rπ2

vπ2

gm2 vπ2

+

80 kΩ

300 Ω

IC1 = 680 µA
IC1
= 26.2 mS
gm1 =
VT
βnpn
rπ1 =
= 3.824 kΩ
gm1
IC2 = 674 µA
IC2
gm2 =
= 25.9 mS
VT
βpnp
rπ2 =
= 1.929 kΩ
gm2

200 Ω

5.30
VCC − IC (1 kΩ) = VEC = VEB (in order for Q1 to operate at the edge of saturation)
= VT ln(IC /IS )

VCC − VEB
RB

IC = 1.761 mA
VEB = 739 mV
VEB
IC
−
= IB =
5 kΩ
β
RB = 9.623 kΩ

First, let’s consider when RB is 5 % larger than its nominal value.
RB = 10.104 kΩ
IC
VCC − VT ln(IC /IS ) VT ln(IC /IS )
−
=
RB
5 kΩ
β
IC = 1.411 mA
VEB = 733 mV
VEC = VCC − IC (1 kΩ) = 1.089 V
VCB = −355 mV (the collector-base junction is reverse biased)
Now, let’s consider when RB is 5 % smaller than its nominal value.
RB = 9.142 kΩ
IC
VCC − VT ln(IC /IS ) VT ln(IC /IS )
−
=
RB
5 kΩ
β
IC = 2.160 mA
VEB = 744 mV
VEC = VCC − IC (1 kΩ) = 340 mV
VCB = 405 mV (the collector-base junction is forward biased)

5.31
VBC + IC (5 kΩ) VCC − VBC − IC (5 kΩ)
IC
−
= IB =
10 kΩ
10 kΩ
β
VBC = 300 mV
IC = 194 µA

VCC

VEB = VT ln(IC /IS ) = 682 mV
VCC − IE RE − IC (5 kΩ) = VEC = VEB + 300 mV
1+β
−
IC RE − IC (5 kΩ) = VEB + 300 mV
β
RE = 2.776 kΩ

Let’s look at what happens when RE is halved.
RE = 1.388 kΩ

β

VCC

VCC − IE RE − VEB
VCC − (VCC − IE RE − VEB )
IC
−
= IB =
10 kΩ
10 kΩ
β


1+β
VCC − VCC − β IC RE − VT ln(IC /IS )
− 1+β
β IC RE − VT ln(IC /IS )
−β
= IC
10 kΩ
10 kΩ
IC = 364 µA
VEB = 698 µV
VEC = 164 µV

Thus, when RE is halved, Q1 operates in deep saturation.

5.32

VCC

VCC − IB (20 kΩ) − IE (1.6 kΩ) = VBE = VT ln(IC /IS )
IC
1+β
−
(20 kΩ) −
IC (1.6 kΩ) = VBE = VT ln(IC /IS )
β
β
IC
i
IS = h
I
VCC − βC (20 kΩ)− 1+β
β IC (1.6 kΩ) /VT
e
IC = 1 mA
IS = 3 × 10−14 A

5.38 (a)
Av = −gm1



1
gm2

k rπ2



Rin = rπ1
Rout =

1
k rπ2
gm2

(b)


1
Av = −gm1 R1 +
k rπ2
gm2
Rin = rπ1
Rout = R1 +

1
k rπ2
gm2

(c)


1
k rπ2
Av = −gm1 RC +
gm2
Rin = rπ1
Rout = RC +

1
k rπ2
gm2

(d) Let’s determine the equivalent resistance seen looking up from the output by drawing a smallsignal model and applying a test source.
RC
+
vt

+
it

rπ2

gm2 vπ2

vπ2

−

−

vπ2
+ gm2 vπ2
rπ2
vπ2 = vt


1
it = vt
+ gm2
rπ2
1
vt
=
k rπ2
it
gm2
it =

Av = −gm1



1
gm2

Rin = rπ1
Rout =

1
k rπ2
gm2

k rπ2





1
k rπ2 . If we find
(e) From (d), we know the gain from the input to the collector of Q1 is −gm1 gm2
the gain from the collector of Q1 to vout , we can multiply these expressions to find the overall
gain. Let’s draw the small-signal model to find the gain from the collector of Q1 to vout . I’ll refer
to the collector of Q1 as node X in the following derivation.
RC
+
vX

rπ2
−

vout

+
vπ2

gm2 vπ2

−

vX − vout
= gm2 vπ2
RC
vπ2 = vX
vX − vout
= gm2 vX
RC


vout
1
− gm2 =
vX
RC
RC
vout
= 1 − gm2 RC
vX
Thus, we have
Av = −gm1



1
gm2


k rπ2 (1 − gm2 RC )

Rin = rπ1
To find the output resistance, let’s draw the small-signal model and apply a test source at the
output. Note that looking into the collector of Q1 we see infinite resistance, so we can exclude it
from the small-signal model.
RC
+
rπ2

vπ2
−

+
gm2 vπ2

it

vt
−

it = gm2 vπ2 +

vπ2
rπ2

rπ2
vt
rπ2 + RC


rπ2
1
vt
it = gm2 +
rπ2 rπ2 + RC
vt
Rout =
it


rπ2 + RC
1
=
k rπ2
gm2
rπ2
vπ2 =

5.39 (a)


1
Av = −gm1 ro1 k
k rπ2 k ro2
gm2
Rin = rπ1
Rout = ro1 k

1
k rπ2 k ro2
gm2

(b)



1
k rπ2 k ro2
Av = −gm1 ro1 k R1 +
gm2
Rin = rπ1
Rout = ro1 k



1
R1 +
k rπ2 k ro2
gm2

(c)


Av = −gm1 ro1



1
k RC +
k rπ2 k ro2
gm2

Rin = rπ1
Rout = ro1



1
k rπ2 k ro2
k RC +
gm2

(d) Let’s determine the equivalent resistance seen looking up from the output by drawing a smallsignal model and applying a test source.
RC
+
vt

+
it

−

X

rπ2

vπ2
−

gm2 vπ2

ro2

it =

vπ2
vt − vX
+
rπ2
RC

vX
vX − vt
+ gm2 vπ2 +
=0
RC
ro2
vπ2 = vt




1
1
1
vX
= vt
+
− gm2
RC
ro2
RC


1
vX = vt
− gm2 (ro2 k RC )
RC


1
vt
1
vt
+
−
vt
− gm2 (ro2 k RC )
it =
rπ2
RC
RC
RC




1
1
1
1
= vt
+
−
− gm2 (ro2 k RC )
rπ2
RC
RC RC




1
1
ro2
1
+
+ gm2 −
= vt
rπ2
RC
RC ro2 + RC
#
"
vt
1
ro2 + RC
= rπ2 k RC k
it
ro2
gm2 − R1C

Av = −gm1 ro1 k rπ2

"

1
ro2 + RC
k RC k
ro2
gm2 −

1
RC

#!

Rin = rπ1
Rout = ro1 k rπ2

"

ro2 + RC
1
k RC k
ro2
gm2 −

1
RC

#



C
(e) From (d), we know the gain from the input to the collector of Q1 is −gm1 ro1 k rπ2 k RC k ro2r+R
o2
If we find the gain from the collector of Q1 to vout , we can multiply these expressions to find the
overall gain. Let’s draw the small-signal model to find the gain from the collector of Q1 to vout .
I’ll refer to the collector of Q1 as node X in the following derivation.
RC
+
vX

rπ2
−

+
vπ2
−

vout
gm2 vπ2

ro2

1
gm2 − R1

C


.

vout − vX
vout
+ gm2 vπ2 +
=0
RC
ro2
vπ2 = vX
vout − vX
vout
+ gm2 vX +
=0
RC
ro2




1
1
1
= vX
+
− gm2
vout
RC
ro2
RC


vout
1
=
− gm2 (RC k ro2 )
vX
RC
Thus, we have
Av = −gm1 ro1 k rπ2

"

ro2 + RC
1
k RC k
ro2
gm2 −

1
RC

#! 


1
− gm2 (RC k ro2 )
RC

Rin = rπ1
To find the output resistance, let’s draw the small-signal model and apply a test source at the
output. Note that looking into the collector of Q1 we see ro1 , so we replace Q1 in the small-signal
model with this equivalent resistance. Also note that ro2 appears from the output to ground, so
we can remove it from this analysis and add it in parallel at the end to find Rout .
RC
+

+
ro1

rπ2

vπ2
−

it = gm2 vπ2 +

gm2 vπ2

vt

it
−

vπ2
rπ2 k ro1

rπ2 k ro1
vt
rπ2 k ro1 + RC


1
rπ2 k ro1
it = gm2 +
vt
rπ2 k ro1 rπ2 k ro1 + RC
vt
Rout = ro2 k
it



rπ2 k ro1 + RC
1
= ro2 k
k rπ2 k ro1
gm2
rπ2 k ro1
vπ2 =

5.43
Av = −

1
gm

= − VT
IC

RC
+ (200 Ω)
RC
+ (200 Ω)

= −100
VT
RC = 100
+ 100(200 Ω)
IC
IC RC − IE (200 Ω) = VCE = VBE = VT ln(IC /IS )


1+β
VT
+ 100(200 Ω) −
IC (200 Ω) = VT ln(IC /IS )
IC 100
IC
β
We can see that this equation has no solution. For example, if we let IC = 0, we see that according to
the left side, we should have VBE = 2.6 V, which is clearly an infeasible value. Qualitatively, we know
that in order to achieve a large gain, we need a large value for RC . However, increasing RC will result
in a smaller value of VCE , eventually driving the transistor into saturation. When Av = −100, there
is no value of RC that will provide such a large gain without driving the transistor into saturation.

5.46 (a)
Av = −

R1 +

1
gm2

1
gm1

k rπ2

+ RE

Rin = rπ1 + (1 + β1 )RE
Rout = R1 +

1
k rπ2
gm2

(b)
RC

Av = −

1
gm1

+

1
gm2

k rπ2

Rin = rπ1 + (1 + β1 )



1
gm2

k rπ2



k rπ2



Rout = RC
(c)
RC

Av = −

1
gm1

+

1
gm2

k rπ2

Rin = rπ1 + (1 + β1 )



1
gm2

Rout = RC
(d)
Av = −

1
gm1

+

RC
gm2 k rπ2 +
1

Rin = RB + rπ1 + (1 + β1 )

RB
1+β1



1
gm2

k rπ2



k rπ2



Rout = RC
(e)
Av = −

1
gm1

+

RC
gm2 k rπ2 +
1

Rin = RB + rπ1 + (1 + β1 )
Rout = RC

RB
1+β1



1
gm2

5.47 (a)
Av = −

RC +

1
gm2

1
gm1

k rπ2

+ RE

Rin = rπ1 + (1 + β1 ) RE
Rout = RC +

1
k rπ2
gm2

(b)
Av = −

1
gm2

RC +
1
gm1

= −

1
gm2
1
gm1

k rπ2

+ RE

·

1
gm2

RC +

k rπ2

1
gm2

k rπ2

k rπ2
+ RE

Rin = rπ1 + (1 + β1 ) RE
Rout =

1
k rπ2
gm2

(c)
Av = −

RC +
1

1
gm2

+

gm1

k rπ2

1
gm3 krπ3

Rin = rπ1 + (1 + β1 )
Rout = RC +

1
gm2



1
gm3 k rπ3

k rπ2

(d)
Av = −

RC k rπ2
1
gm1 + RE

Rin = rπ1 + (1 + β1 ) RE
Rout = RC k rπ2



5.49 (a) Looking into the emitter of Q2 we see an equivalent resistance of
the following equivalent circuit for finding Rout :

1
gm2

k rπ2 k ro2 , so we can draw

Rout

Q1

1
gm2

Rout

k rπ2 k ro2



1
= ro1 + (1 + gm1 ro1 ) rπ1 k
k rπ2 k ro2
gm2

+RB
(b) Looking into the emitter of Q2 we see an equivalent resistance of ro2 k rπ2
1+β2 (ro2 simply appears
in parallel with the resistance seen when VA = ∞), so we can draw the following equivalent circuit
for finding Rout :

Rout

Q1

ro2 k



rπ2 +RB
1+β2

Rout = ro1 + (1 + gm1 ro1 ) rπ1 k ro2

rπ2 + RB
k
1 + β2



(c) Looking down from the emitter of Q1 we see an equivalent resistance of R1 k rπ2 , so we can draw
the following equivalent circuit for finding Rout :

Rout

Q1

R1 k rπ2

Rout = ro1 + (1 + gm1 ro1 ) (rπ1 k R1 k rπ2 )

5.50 (a) Looking into the emitter of Q1 we see an equivalent resistance of
the following equivalent circuit for finding Rout :

1
gm1

k rπ1 k ro1 , so we can draw

VCC
1
gm1

k rπ1 k ro1

Q2

Rout



1
k rπ1 k ro1
Rout = ro2 + (1 + gm2 ro2 ) rπ2 k
gm1
(b) Looking into the emitter of Q1 we see an equivalent resistance of ro1 , so we can draw the following
equivalent circuit for finding Rout :
VCC
ro1

Q2

Rout

Rout = ro2 + (1 + gm2 ro2 ) (rπ2 k ro1 )
Comparing this to the solution to part (a), we can see that the output resistance is larger because
instead of a factor of 1/gm1 dominating the parallel resistors in the expression, rπ2 dominates
(assuming ro1 ≫ rπ2 ).

5.52 (a)

VCC

VCC − IB (100 kΩ) − IE (100 Ω) = VBE = VT ln(IC /IS )
1
1+β
− IC (100 kΩ) −
IC (100 Ω) = VT ln(IC /IS )
β
β
IC = 1.6 mA
1 kΩ
Av = − 1
gm + 100 Ω
gm = 61.6 mS
Av = −8.60

(b)
VCC − IB (50 kΩ) − IE (2 kΩ) = VT ln(IC /IS )
IC = 708 µA
Av = −

1 kΩ
1
gm

+

(1 kΩ)k(50 kΩ)
1+β

gm = 27.2 mS
Av = −21.54
(c)
IB =

IC
VCC − VBE − IE (2.5 kΩ) VBE + IE (2.5 kΩ)
=
−
β
14 kΩ
11 kΩ

IC = β

VCC − VT ln(IC /IS ) −

1+β
β IC (2.5

kΩ)

14 kΩ

IC = 163 µA
Av = −

10 kΩ
1
gm

+ 500 Ω +

gm = 6.29 mS
Av = −14.98

(1 kΩ)k(14 kΩ)k(11 kΩ)
1+β

−β

VT ln(IC /IS ) +

1+β
β IC (2.5

11 kΩ

kΩ)

5.53 (a)
VCC − 1.5 V
RC
= 4 mA

IC =

VBE = VT ln(IC /IS ) = 832 mV
VCC − VBE
= 66.7 µA
IB =
RB
IC
β=
= 60
IB
(b) Assuming the speaker has an impedance of 8 Ω, the gain of the amplifier is
Av = −gm (RC k 8 Ω)
IC
=−
(RC k 8 Ω)
VT
= −1.19
Thus, the circuit provides greater than unity gain.

5.54 (a)
Av = gm RC
IC
gm =
= 76.9 mS
VT
Av = 38.46
1
Rin =
k rπ
gm
β
= 1.3 kΩ
rπ =
gm
Rin = 12.87 Ω
Rout = RC = 500 Ω
(b) Since Av = gm RC and gm is fixed for a given value of IC , RC should be chosen as large as
possible to maximize the gain of the amplifier. Vb should be chosen as small as possible to
maximize the headroom of the amplifier (since in order for Q1 to remain in forward active, we
require Vb < VCC − IC RC ).

5.56 (a) Looking into the emitter of Q2 we see an equivalent resistance of
following equivalent circuit for finding Rin :

1
gm2

k rπ2 , so we can draw the

VCC
R1

Q1
1
gm2

k rπ2

Rin

Rin =

rπ1 +

1
gm2

k rπ2

1 + β1

(b) Looking right from the base of Q1 we see an equivalent resistance of R2 , so we can draw the
following equivalent circuit for finding Rin :
VCC
R1

Q1
R2

Rin

Rin =

rπ1 + R2
1 + β1

(c) Looking right from the base of Q1 we see an equivalent resistance of R2 k
draw the following equivalent circuit for finding Rin :

1
gm2

k rπ2 , so we can

VCC
R1

Q1
R2 k

1
gm2

k rπ2

Rin

Rin =

rπ1 + R2 k

1
gm2

k rπ2

1 + β1

(d) Looking right from the base of Q1 we see an equivalent resistance of R2 k rπ2 , so we can draw the
following equivalent circuit for finding Rin :
VCC
R1

Q1
R2 k rπ2

Rin

Rin =

rπ1 + R2 k rπ2
1 + β1

5.58 (a)
IB =

IC
VCC − VBE − IE (400 Ω) VBE + IE (400 Ω)
=
−
β
13 kΩ
12 kΩ

IC = β

VCC − VT ln(IC /IS ) −

1+β
β IC (400

Ω)

13 kΩ

−β

IC = 1.02 mA
VBE = VT ln(IC /IS ) = 725 mV
VCE = VCC − IC (1 kΩ) − IE (400 Ω) = 1.07 V
Q1 is operating in forward active.
(b)
Av = gm (1 kΩ)
gm = 39.2 mS
Av = 39.2

VT ln(IC /IS ) +

1+β
β IC (400

12 kΩ

Ω)

5.61 For small-signal analysis, we can draw the following equivalent circuit.
R1

vout
Q1
vin

Av = gm R1
Rin =

1
k rπ
gm

Rout = R1

5.61 For small-signal analysis, we can draw the following equivalent circuit.
R1

vout
Q1
vin

Av = gm R1
Rin =

1
k rπ
gm

Rout = R1

5.63 Since IS1 = 2IS2 and they’re biased identically, we know that IC1 = 2IC2 , which means gm1 = 2gm2 .
vout1
= gm1 RC = 2gm2 RC
vin
vout2
= gm2 RC
vin
vout1
vout2
⇒
=2
vin
vin

5.67
rπ + RS
1+β
βVT /IC + RS
=
1+β
≤5Ω
β
β
IC =
IE =
I1
1+β
1+β

Rout =

β(1+β)VT
βI1

+ RS

1+β

=

(1+β)VT
I1

+ RS

1+β

≤5Ω
I1 ≥ 8.61 mA

5.68 (a) Looking into the collector of Q2 we see an equivalent resistance of ro2 = ∞, so we can draw the
following equivalent circuit:
VCC
vin

Q1
vout
∞

Av = 1
Rin = ∞
Rout =

1
k rπ1
gm1

(b) Looking down from the emitter of Q1 we see an equivalent resistance of
the following equivalent circuit:

1
gm2

k rπ2 , so we can draw

VCC
vin

Q1
vout
1
gm2

Av =

1
gm2
1
gm1

+

k rπ2

k rπ2
1
gm2

k rπ2

Rin = rπ1 + (1 + β1 )
Rout =



1
gm2

k rπ2



1
1
k rπ1 k
k rπ2
gm1
gm2

(c) Looking into the emitter of Q2 we see an equivalent resistance of
following equivalent circuit:

rπ2 +RS
1+β2 ,

so we can draw the

VCC
vin

Q1
vout
rπ2 +RS
1+β2

Av =

Rin
Rout

rπ2 +RS
1+β2
rπ2 +RS
1
+
gm1
1+β2



rπ2 + RS
= rπ1 + (1 + β1 )
1 + β2


rπ2 + RS
1
=
k rπ1 k
gm1
1 + β2



(d) Looking down from the emitter of Q1 we see an equivalent resistance of RE +
can draw the following equivalent circuit:

1
gm2

k rπ2 , so we

VCC
vin

Q1
vout
RE +

Av =

RE +
1
gm1

Rin
Rout

1
gm2

+ RE +

1
gm2

k rπ2

k rπ2
1
gm2

k rπ2



1
= rπ1 + (1 + β1 ) RE +
gm2


1
1
=
k rπ1 k RE +
gm1
gm2

(e) Looking into the emitter of Q2 we see an equivalent resistance of
following equivalent circuit:

1
gm2

k rπ2 , so we can draw the

VCC
vin

Q1

RE
vout
1
gm2

Av =

RE +
1

+ RE +

gm1

1
gm2

=

1
gm1

Rin

1
gm2

k rπ2

k rπ2
1
gm2

·

k rπ2 RE +



+ RE +

1
gm1

k rπ2

1
gm2

k rπ2
1
gm2

k rπ2


= rπ1 + (1 + β1 ) RE +

Rout =

1
gm2

k rπ1 + RE



k

1
gm2 k rπ2
1
k rπ2
gm2



k rπ2

5.69 (a) Looking into the base of Q2 we see an equivalent resistance of rπ2 (assuming the emitter of Q2 is
grounded), so we can draw the following equivalent circuit for finding the impedance at the base
of Q1 :
VCC
Q1

Req

rπ2

Req = rπ1 + (1 + β1 )rπ2
1
k rπ1 (assuming the base of
(b) Looking into the emitter of Q1 we see an equivalent resistance of gm1
Q1 is grounded), so we can draw the following equivalent circuit for finding the impedance at the
emitter of Q2 :

VCC
1
gm1

k rπ1

Q2

Req

Req =

rπ2 +

1
gm1

k rπ1

1 + β2

(c)
β1 IB1 + β2 (1 + β1 )IB1
IC1 + IC2
=
IB1
IB1
= β1 + β2 (1 + β1 )
If we assume that β1 , β2 ≫ 1, then this simplifies to β1 β2 , meaning a Darlington pair has a current
gain approximately equal to the product of the current gains of the individual transistors.

5.70 (a)
RCS = ro2 + (1 + gm2 ro2 ) (rπ2 k RE )
(b)
Av =

ro2 + (1 + gm2 ro2 ) (rπ2 k RE )
gm1 + ro2 + (1 + gm2 ro2 ) (rπ2 k RE )
1

Rin = rπ1 + (1 + β1 ) [ro2 + (1 + gm2 ro2 ) (rπ2 k RE )]
Rout =

1
gm1

k rπ1 k [ro2 + (1 + gm2 ro2 ) (rπ2 k RE )]

5.72 (a) Looking into the base of Q2 we see an equivalent resistance of rπ2 , so we can draw the following
equivalent circuit for finding Rin :
VCC
Q1

Rin

RE k rπ2

Rin = rπ1 + (1 + β1 ) (RE k ro1 )
Looking into the collector of Q2 we see an equivalent resistance of ro2 . Thus,
Rout = RC k ro2
(b) Looking into the base of Q2 we see an equivalent resistance of rπ2 , so we can draw the following
equivalent circuit for finding vX /vin :
VCC
vin

Q1
vX
RE k rπ2

vX
=
vin

RE k rπ2 k ro1
gm1 + RE k rπ2 k ro1
1

We can find vout /vX by inspection.
vout
= −gm2 (RC k ro2 )
vX
vX vout
Av =
·
vin vX
= −gm2 (RC k ro2 )

RE k rπ2 k ro1
gm1 + RE k rπ2 k ro1
1

5.73 (a) Looking into the emitter of Q2 we see an equivalent resistance of
following equivalent circuit for finding Rin :

1
gm2

k rπ2 , so we can draw the

VCC
Q1

Rin

Rin

RE k

1
gm2

k rπ2



1
= rπ1 + (1 + β1 ) RE k
k rπ2
gm2

Looking into the collector of Q2 , we see an equivalent resistance of ∞ (because VA = ∞), so we
have
Rout = RC
(b) Looking into the emitter of Q2 we see an equivalent resistance of
following equivalent circuit for finding vX /vin :
VCC
vin

Q1
vX
RE k

vX
=
vin

1
gm2

1
gm2 k
1
1
gm1 +E k gm2

RE k

k rπ2

rπ2
k rπ2

We can find vout /vX by inspection.
vout
= gm2 RC
vX
vX vout
·
Av =
vin vX
= gm2 RC

1
gm2 k
1
1
gm1 +E k gm2

RE k

rπ2
k rπ2

1
gm2

k rπ2 , so we can draw the

5.74
Rout = RC = 1 kΩ
Av = −gm RC = −10
gm = 10 mS
IC = gm VT = 260 µA
VCC − VBE
IC
= IB =
RB
β
VCC − VT ln(IC /IS )
RB = β
IC
= 694 kΩ
Rin = RB k rπ = 9.86 kΩ > 5 kΩ
In sizing CB , we must consider the effect a finite impedance in series with the input will have on the
circuit parameters. Any series impedance will cause Rin to increase and will not impact Rout . However,
1
a series impedance can cause gain degradation. Thus, we must ensure that |ZB | = jωC
does not
B
degrade the gain significantly.
If we include |ZB | in the gain expression, we get:
Av = −
Thus, we want

1
1+β

|ZB | ≪

1
gm

RC
1
gm

+

(|ZB |)kRB
1+β

to ensure the gain is not significantly degraded.
1
1
1
≪
1 + β jωCB
gm
1
1
1 1
=
1 + β 2πf CB
10 gm
CB = 788 nF

5.75
Rout = RC ≤ 500 Ω
To maximize gain, we should maximize RC .
RC = 500 Ω
VCC − IC RC ≥ VBE − 400 mV = VT ln(IC /IS ) − 400 mV
IC ≤ 4.261 mA
To maximize gain, we should maximize IC .
IC = 4.261 mA
VCC − VBE
IC
=
IB =
β
RB
IC
VCC − VT ln(IC /IS )
=
=
β
RB
RB = 40.613 kΩ

5.76
Rout = RC = 1 kΩ
|Av | = gm RC
IC RC
=
VT
≥ 20
IC ≥ 520 µA
In order to maximize Rin = RB k rπ , we need to maximize rπ , meaning we should minimize IC (since
T
rπ = βV
IC ).
IC = 520 µA
IC
VCC − VBE
IB =
=
β
RB
VCC − VT ln(IC /IS )
=
RB
RB = 343 kΩ

5.77
Rout = RC = 2 kΩ
Av = −gm RC
IC RC
=−
VT
= −15
IC = 195 µA
VBE = VT ln(IC /IS ) = 689.2 mV
VCE ≥ VBE − 400 mV = 289.2 mV
To minimize the supply voltage, we should minimize VCE .
VCE = 289.2 mV
VCC − VCE
= IC
RC
VCC = 679.2 mV
Note that this value of VCC is less than the required VBE . This means that the value of VCC is
constrained by VBE , not VCE . In theory, we could pick VCC = VBE , but in this case, we’d have
to set RB = 0 Ω, which would short the input to VCC . Thus, let’s pick a reasonable value for RB ,
RB = 100 Ω .
IC
VCC − VBE
= IB =
RB
β
VCC = 689.4 mV

5.78
|Av | = gm RC
IC RC
=
VT
= A0
Rout = RC
IC Rout
A0 =
VT
A0 VT
IC =
Rout
P = IC VCC
=

A0 VT
VCC
Rout

Thus, we must trade off a small output resistance with low power consumption (i.e., as we decrease
Rout , power consumption increases and vice-versa).

5.79
P = (IB + IC )VCC
1+β
=
IC VCC
β
= 1 mW
IC = 396 µA
VCC − VBE
IC
= IB =
RB
β
VCC − VT ln(IC /IS )
RB = β
IC
= 453 kΩ
Av = −gm RC
IC RC
=−
VT
= −20
RC = 1.31 kΩ

5.81
Rout = RC ≥ 1 kΩ
To maximize gain, we should maximize Rout .
RC = 1 kΩ
VCC − IC RC − IE RE = VCE ≥ VBE − 400 mV
VCC − IC RC − 200 mV ≥ VT ln(IC /IS ) − 400 mV
IC ≤ 1.95 mA
To maximize gain, we should maximize IC .
IC = 1.95 mA
1+β
IC RE = 200 mV
IE RE =
β
VCC

RE = 101.5 Ω
− 10IB R1 − IE RE = VBE = VT ln(IC /IS )
R1 = 7.950 kΩ
9IB R2 − IE RE = VBE = VT ln(IC /IS )
R2 = 5.405 kΩ

5.82
P = (10IB + IC ) VCC


IC
= 10
+ IC VCC
β
= 5 mW

IC = 1.82 mA
1+β
IC RE = 200 mV
IE RE =
β
RE = 109 Ω
RC
Av = − 1
gm + RE
RC
IC + RE

= − VT
= −5

RC = 616 Ω
VCC − 10IB R1 − 200 mV = VBE = VT ln(IC /IS )
R1 = 8.54 kΩ
9IB R2 − 200 mV = VBE = VT ln(IC /IS )
R2 = 5.79 kΩ

5.83
1
= 50 Ω (since RE doesn’t affect Rin )
gm
= 20 mS

Rin =
gm

IC = gm VT = 520 µA
1+β
IC RE = 260 mV
IE RE =
β
RE = 495 Ω
Av = gm RC = 20
RC = 1 kΩ
VCC − 10IB R1 − IE RE = VBE = VT ln(IC /IS )
R1 = 29.33 kΩ
9IB R2 − IE RE = VBE = VT ln(IC /IS )
R2 = 20.83 kΩ
To pick CB , we must consider its effect on Av . If we assume the capacitor has an impedance ZB and
|ZB | ≪ R1 , R2 , then we have:
Av =
Thus, we should choose

1
1+β

|ZB | ≪

RC
1
gm

+

|ZB |
1+β

1
gm .

1
1 1
1
1
=
|ZB | =
1+β
1 + β 2πf CB
10 gm
CB = 1.58 µF

5.84
Rout = RC = 500 Ω
Av = gm RC = 8
gm = 16 mS
IC = gm VT = 416 µA
1+β
IC RE = 260 mV
IE RE =
β
RE = 619 Ω
VCC − 10IB R1 − IE RE = VBE = VT ln(IC /IS )
R1 = 36.806 kΩ
9IB R2 − IE RE = VBE = VT ln(IC /IS )
R2 = 25.878 kΩ
To pick CB , we must consider its effect on Av . If we assume the capacitor has an impedance ZB and
|ZB | ≪ R1 , R2 , then we have:
Av =
Thus, we should choose

1
1+β

|ZB | ≪

RC
1
gm

+

|ZB |
1+β

1
gm .

1
1 1
1
1
=
|ZB | =
1+β
1 + β 2πf CB
10 gm
CB = 1.26 µF

5.85
Rout = RC = 200 Ω
IC RC
= 20
Av = gm RC =
VT
IC = 2.6 mA
P = VCC (10IB + IC )


IC
= VCC 10
+ IC
β
= 7.15 mW

5.86
P = (IC + 10IB ) VCC


IC
= IC + 10
VCC
β
= 5 mW

IC = 1.82 mA
Av = gm RC
IC RC
=
VT
= 10
RC = 143 Ω
1+β
IE RE =
IC RE = 260 mV
β
VCC

RE = 141.6 Ω
− 10IB R1 − IE RE = VBE = VT ln(IC /IS )
R1 = 8.210 kΩ
9IB R2 − IE RE = VBE = VT ln(IC /IS )
R2 = 6.155 kΩ

To pick CB , we must consider its effect on Av . If we assume the capacitor has an impedance ZB and
|ZB | ≪ R1 , R2 , then we have:
Av =
Thus, we should choose

1
1+β

|ZB | ≪

RC
1
gm

+

|ZB |
1+β

1
gm .

1
1 1
1
1
=
|ZB | =
1+β
1 + β 2πf CB
10 gm
CB = 5.52 µF

5.87
1
= 50 Ω (since RE doesn’t affect Rin )
gm
= 20 mS

Rin =
gm

IC = gm VT = 520 µA
Av = gm RC = 20
RC = 1 kΩ
1+β
IE RE =
IC RE = 260 mV
β
RE = 495 Ω
To minimize the supply voltage, we should allow Q1 to operate in soft saturation, i.e., VBC = 400 mV.
VBE = VT ln(IC /IS ) = 715 mV
VCE = VBE − 400 mV = 315 mV
VCC − IC RC − IE RE = VCE
VCC

VCC = 1.095 V
− 10IB R1 − IE RE = VBE
R1 = 2.308 kΩ
9IB R2 − IE RE = VBE
R2 = 20.827 kΩ

To pick CB , we must consider its effect on Av . If we assume the capacitor has an impedance ZB and
|ZB | ≪ R1 , R2 , then we have:
Av =
Thus, we should choose

1
1+β

|ZB | ≪

RC
1
gm

+

|ZB |
1+β

1
gm .

1
1
1 1
1
=
|ZB | =
1+β
1 + β 2πf CB
10 gm
CB = 1.58 µF

5.90 As stated in the hint, let’s assume that IE RE ≫ VT . Given this assumption, we can assume that RE
does not affect the gain.
IE RE = 10VT = 260 mV
RL
Av = 1
= 0.8
gm + RL
gm = 80 mS
IC = gm VT = 2.08 mA
1+β
IC RE = 260 mV
β
RE = 124 Ω
VCC − IB R1 − IE RE = VBE = VT ln(IC /IS )
R1 = 71.6 kΩ
To pick C1 , we must consider its effect on Av . If we assume the capacitor has an impedance Z1 and
|Z1 | ≪ R1 , then we have:
Av =
Thus, we should choose

1
1+β

|Z1 | ≪

RE
1
gm

+ RE +

|Z1 |
1+β

1
gm .

1
1 1
1
1
=
|Z1 | =
1+β
1 + β 2πf C1
10 gm
C1 = 12.6 pF
To pick C2 , we must also consider its effect on Av . Since the capacitor appears in series with RL , we
need to ensure that |Z2 | ≪ RL , assuming the capacitor has impedance Z2 .
|Z2 | =

1
1
=
RL
2πf C2
10

C2 = 318 pF

6.4 (a)

Q(x)

Q(x) = W Cox (VGS − V (x) − VT H )
= W Cox (VGS − VT H ) − W Cox V (x)

W Cox (VGS − VT H )

Increasing VDS

L

x

The curve that intersects the axis at x = L (i.e., the curve for which the channel begins to pinch
off) corresponds to VDS = VGS − VT H .
(b)
1
µQ(x)

RLocal (x)

RLocal (x) ∝

Increasing VDS

L

x

Note that RLocal diverges at x = L when VDS = VGS − VT H .

6.15

ID

Increasing VDS

VT H

VGS

Initially, when VGS is small, the transistor is in cutoff and no current flows. Once VGS increases
beyond VT H , the curves start following the square-law characteristic as the transistor enters saturation.
However, once VGS increases past VDS + VT H (i.e., when VDS < VGS − VT H ), the transistor goes into
triode and the curves become linear. As we increase VDS , the transistor stays in saturation up to larger
values of VGS , as expected.

6.17
1
W
µn Cox
(VGS − VT H )α , α < 2
2
L
∂ID
,
∂VGS
α
W
α−1
= µn Cox
(VGS − VT H )
2
L
αID
=
VGS − VT H

ID =
gm

6.21 Since they’re being used as current sources, assume M1 and M2 are in saturation for this problem.
To find the maximum allowable value of λ, we should evaluate λ when 0.99ID2 = ID1 and 1.01ID2 =
ID1 , i.e., at the limits of the allowable values for the currents. However, note that for any valid λ
(remember, λ should be non-negative), we know that ID2 > ID1 (since VDS2 > VDS1 ), so the case
where 1.01ID2 = ID1 (which implies ID2 < ID1 ) will produce an invalid value for λ (you can check this
yourself). Thus, we need only consider the case when 0.99ID2 = ID1 .
W
1
2
(VB − VT H ) (1 + λVDS2 )
0.99ID2 = 0.99 µn Cox
2
L
= ID1
1
W
2
= µn Cox
(VB − VT H ) (1 + λVDS1 )
2
L
0.99 (1 + λVDS2 ) = 1 + λVDS1
λ = 0.02 V−1

5.27
VDD − ID RD = VGS = VT H +

s

2ID
µn Cox W
L

2ID
2
= (VDD − VT H − ID RD )
W
µn Cox L
i
W h
1
2
2 2
(VDD − VT H ) − 2ID RD (VDD − VT H ) + ID
RD
ID = µn Cox
2
L

We can rearrange this to the standard quadratic form as follows:




1
W 2
W
W
1
2
2
µn Cox RD ID − µn Cox RD (VDD − VT H ) + 1 ID + µn Cox
(VDD − VT H ) = 0
2
L
L
2
L
Applying the quadratic formula, we have:

 q

2

2
R
(V
−
V
)
+
1
±
µn Cox W
− 4 21 µn Cox W
µn Cox W
DD
TH
L D
L RD (VDD − VT H ) + 1
L RD (VDD − VT H )


ID =
2
2 21 µn Cox W
L RD
q

2

2
W
µn Cox W
− µn Cox W
µn Cox L RD (VDD − VT H ) + 1 ±
L RD (VDD − VT H ) + 1
L RD (VDD − VT H )
=
2
µn Cox W
L RD
q
µn Cox W
1 + 2µn Cox W
L RD (VDD − VT H ) + 1 ±
L RD (VDD − VT H )
=
2
µn Cox W
L RD
Note that mathematically, there are two possible solutions for ID . However, since M1 is diodeconnected, we know it will either be in saturation or cutoff. Thus, we must reject the value of ID
that does not match these conditions (for example, a negative value of ID would not match cutoff or
saturation, so it would be rejected in favor of a positive value).

6.33 (a) Assume M1 is operating in saturation.
VGS = 1 V
1
W
2
VDS = VDD − ID RD = VDD − µn Cox
(VGS − VT H ) (1 + λVDS ) RD
2
L
VDS = 1.35 V > VGS − VT H , which verifies our assumption
ID = 4.54 mA
W
(VGS − VT H ) = 13.333 mS
gm = µn Cox
L
1
= 2.203 kΩ
ro =
λID

+
vgs

gm vgs

ro

RD

−

(b) Since M1 is diode-connected, we know it is operating in saturation.
W
1
2
(VGS − VT H ) (1 + λVGS ) RD
VGS = VDS = VDD − ID RD = VDD − µn Cox
2
L
VGS = VDS = 0.546 V
ID = 251 µA
W
(VGS − VT H ) = 3.251 mS
gm = µn Cox
L
1
= 39.881 kΩ
ro =
λID

+
vgs

gm vgs

ro

RD

−

(c) Since M1 is diode-connected, we know it is operating in saturation.
ID = 1 mA
r

gm =
ro =

2µn Cox

W
ID = 6.667 mS
L

1
= 10 kΩ
λID

+
vgs

gm vgs

ro

−

(d) Since M1 is diode-connected, we know it is operating in saturation.
VGS = VDS
1
W
µn Cox
(VGS − VT H )2 (1 + λVGS ) (2 kΩ)
2
L
= 0.623 V

VDD − VGS = ID (2 kΩ) =
VGS = VDS

ID = 588 µA
W
gm = µn Cox
(VGS − VT H ) = 4.961 mS
L
1
ro =
= 16.996 kΩ
λID

+
gm vgs

vgs

ro

2 kΩ

−

(e) Since M1 is diode-connected, we know it is operating in saturation.
ID = 0.5 mA
r

gm =
ro =

2µn Cox

W
ID = 4.714 mS
L

1
= 20 kΩ
λID

+
vgs
−

gm vgs

ro

6.38 (a)
vout

+
vgs2

gm2 vgs2

ro2

gm1 vgs1

ro1

RD

−

vin

+
vgs1
−

(b)
vin

vout

+
vgs1

gm1 vgs1

ro1

RD

−

+
gm2 vgs2

ro2

vgs2
−

(c)
vin

vout

+
vgs1

gm1 vgs1

ro1

gm2 vgs2

ro2

−

+
vgs2
−

(d)

RD

vin

+
vgs1

gm1 vgs1

ro1

−
vout
+
vgs2

gm2 vgs2

ro2

−

(e)
vout

+
vgs1

gm1 vgs1

ro1

RD

−
vin
+
gm2 vgs2

ro2

vgs2
−

6.43 (a) Assume M1 is operating in triode (since |VGS | = 1.8 V is large).
|VGS | = 1.8 V
i
W h
1
2
2 (|VGS | − |VT H |) |VDS | − |VDS | (500 Ω)
µp Cox
2
L
|VDS | = 0.418 V < |VGS | − |VT H | , which verifies our assumption

VDD − |VDS | = |ID | (500 Ω) =

|ID | = 2.764 mA
(b) Since M1 is diode-connected, we know it is operating in saturation.
|VGS | = |VDS |
1
W
2
µp Cox
(|VGS | − |VT H |) (1 kΩ)
2
L
|VGS | = |VDS | = 0.952 V

VDD − |VGS | = |ID |(1 kΩ) =

|ID | = 848 µA
(c) Since M1 is diode-connected, we know it is operating in saturation.
|VGS | = |VDS |
|VGS | = VDD − |ID |(1 kΩ) = VDD − |ID |(1 kΩ) =
|VGS | = |VGS | = 0.952 V
|ID | = 848 µA

1
W
2
µp Cox
(|VGS | − |VT H |) (1 kΩ)
2
L

6.44 (a)

IX

Saturation

Cutoff

VDD − VT H

VDD

VX

VDD

VX

M1 goes from saturation to cutoff when VX = VDD − VT H = 1.4 V.
(b)

IX
1 + VT H

Saturation
M1 goes from saturation to triode when VX = 1 + VT H = 1.4 V.
(c)

Triode

IX
VDD − VT H

Saturation

VDD

VX

VDD

VX

Cutoff

M1 goes from saturation to cutoff when VX = VDD − VT H = 1.4 V.
(d)

IX

Saturation

Cutoff

VT H
M1 goes from cutoff to saturation when VX = VT H = 0.4 V.

7.1
VGS = VDD = 1.8 V
VDS > VGS − VT H (in order for M1 to operate in saturation)
VDS = VDD − ID (1 kΩ)
W
1
2
(VGS − VT H ) (1 kΩ)
= VDD − µn Cox
2
L
> VGS − VT H
W
< 2.04
L

7.3
VGS = VDD − ID (100 Ω)
VDS = VDD − ID (1 kΩ + 100 Ω)
> VGS − VT H (in order for M1 to operate in saturation)
VDD − ID (1 kΩ + 100 Ω) > VDD − ID (100 Ω) − VT H
ID (1 kΩ + 100 Ω) < ID (100 Ω) + VT H
ID (1 kΩ) < VT H
ID < 400 µA
Since gm increases with ID , we should pick the maximum ID to determine the maximum transconductance that M1 can provide.
ID,max = 400 µA
2ID,max
gm,max =
VGS − VT H
2ID,max
=
VDD − ID,max (100 Ω) − VT H
= 0.588 mS

7.5
ID1 = 0.5 mA
VGS = VT H +

s

2ID1
µn Cox W
L

= 0.612 V
1
ID1 R2
VGS =
10
R2 = 12.243 kΩ
11
1
VGS = VDD − ID1 R1 − ID1 RS
10
10
R1 = 21.557 kΩ

7.6
ID = 1 mA
2ID
1
gm =
=
VGS − VT H
100
VGS = 0.6 V
VGS = VDD − ID RD
RD = 1.2 kΩ

7.8 First, let’s analyze the circuit excluding RP .
20 kΩ
VDD = 1.2 V
10 kΩ + 20 kΩ
= VG − ID RS = VDS = VDD − ID (1 kΩ + 200 Ω)

VG =
VGS

ID = 600 µA
VGS = 1.08 V
2ID
W
=
= 12.9758 ≈ 13
L
µn Cox (VGS − VT H )2
Now, let’s analyze the circuit with RP .
VDD
10 kΩ

ID + IRP

1 kΩ

M1

20 kΩ

RS

IRP

RP

200 Ω

VG = 1.2 V
VDD − VDS
ID + IRP =
1 kΩ + 200 Ω
VGS = VG − (ID + IRP ) RS = VDS + VT H
VDD − VDS
VG −
RS = VDS + VT H
1 kΩ + 200 Ω
VDS = 0.6 V
VGS = 1 V
W
1
2
(VGS − VT H )
ID = µn Cox
2
L
= 467 µA
VDS
VDD − VDS
ID + IRP = ID +
=
RP
1 kΩ + 200 Ω
RP = 1.126 kΩ

7.9 First, let’s analyze the circuit excluding RP .
VGS = VDD = 1.8 V
VDS = VDD − ID (2 kΩ) = VGS − 100 mV
W
1
2
(VGS − VT H ) (2 kΩ) = VGS − 100 mV
VDD − µn Cox
2
L
W
= 0.255
L
Now, let’s analyze the circuit with RP .
VDD
30 kΩ

2 kΩ
RP
IRP
M1

VGS = VDD − IRP (30 kΩ)
VGS − VDS
50 mV
IRP =
=
RP
RP
VGS = VDD − (ID − IRP ) (2 kΩ) + 50 mV


W
1
2
µn Cox
(VGS − VT H ) − IRP (2 kΩ) + 50 mV
VDD − IRP (30 kΩ) = VDD −
2
L


1
W
2
VDD − IRP (30 kΩ) = VDD −
µn Cox
(VDD − IRP (30 kΩ) − VT H ) − IRP (2 kΩ) + 50 mV
2
L
IRP = 1.380 µA
50 mV
= 36.222 kΩ
RP =
IRP

7.12 Since we’re not given VDS for the transistors, let’s assume λ = 0 for large-signal calculations. Let’s
also assume the transistors operate in saturation, since they’re being used as current sources.
IX =

1
W1
2
(VB1 − VT H ) = 0.5 mA
µn Cox
2
L1

W1 = 3.47 µm
IY =

W2
1
(VB2 − VT H )2 = 0.5 mA
µn Cox
2
L2

W2 = 1.95 µm
1
= 20 kΩ
λIX
1
= 20 kΩ
=
λIY

Rout1 = ro1 =
Rout2 = ro2

Since IX = IY and λ is the same for each current source, the output resistances of the current sources
are the same.

7.13 Looking into the source of M1 we see a resistance of

1
gm .

Including λ in our analysis, we have

1
1
=
gm
(V
−
V
− |VT H |) (1 + λVX )
µp Cox W
X
B1
L
= 372 Ω

7.17 (a) Assume M1 is operating in saturation.
ID = 0.5 mA
VGS = VT H +

s

2ID
µn Cox W
L

= 0.573 V
VDS = VDD − ID RD = 0.8 volt > VGS − VT H , verifying that M1 is in saturation
(b)
Av = −gm RD
2ID
RD
=−
VGS − VT H
= −11.55

7.18 (a) Assume M1 is operating in saturation.
ID = 0.25 mA
s

VGS = VT H +

2ID
µn Cox W
L

= 0.55 V
VDS = VDD − ID RD = 1.3 V > VGS − VT H , verifying that M1 is in saturation
(b)
VGS = 0.55 V
VDS > VGS − VT H (to ensure M1 remains in saturation)
VDD − ID RD > VGS − VT H
W
1
2
(VGS − VT H ) RD > VGS − VT H
VDD − µn Cox
2
L
W
2 (VDD − VGS + VT H )
<
L
µn Cox (VGS − VT H )2 RD
= 366.67
20
= 3.3
0.18
Thus, W/L can increase by a factor of 3.3 while M1 remains in saturation.
Av = −gm RD
W
(VGS − VT H ) RD
L 

W
(VGS − VT H ) RD
= −µn Cox
L max

= −µn Cox
Av,max

= −22

7.19
P = VDD ID < 1 mW
ID < 556 µA
Av = −gm RD
r

= − 2µn Cox

= −5
W
20
<
L
0.18
RD > 1.006 kΩ

W
ID RD
L

7.20 (a)
ID1 = ID2 = 0.5 mA
Av = −gm1 (ro1 k ro2 )
s
 


W
1
1
= − 2µn Cox
ID1
k
L 1
λ1 ID1 λ2 ID2
= −10


W
L



= 7.8125

1

(b)
VDD − VB = VT H +
VB = 1.1 V

s

2 |ID2 |


µp Cox W
L 2

p
7.22 (a) If ID1 and ID2 remain constant while W and L double, then gm1 ∝ (W/L)1 ID1 will not change
1
1
(since it depends only on the ratio W/L), ro1 ∝ ID1
will not change, and ro2 ∝ ID2
will not
change. Thus, Av = −gm1 (ro1 k ro2 ) will not change .
p
√
1
(b) If ID1 , ID2 , W , and L double, then gm1 ∝ (W/L)1 ID1 will increase by a factor of 2, ro1 ∝ ID1
1
will halve, and ro2 ∝ ID2 will halve. This means that ro1 k ro2 will halve as well, meaning
√
Av = −gm1 (ro1 k ro2 ) will decrease by a factor of 2 .

7.26 (a)
ID1 = ID2 = 0.5 mA
s
2ID1


VGS1 = VT H +
µn Cox W
L 1
= 0.7 V

VDS1 = VGS1 − VT H (in order of M1 to operate at the edge of saturation)
= VDD − VGS2
s
2ID2


VGS2 = VDD − VGS1 + VT H = VT H +
µn Cox W
L 2
 
W
= 4.13
L 2
(b)
gm1
gm2
q
2µn Cox
= −q
2µn Cox
v
u W

u L

1
= −t W

Av = −

L

W
L 1 ID1




W
L 2 ID2




2

= −3.667

(c) Since (W/L)1 is fixed, we must minimize (W/L)2 in order to maximize the magnitude of the gain
(based on the expression derived in part (b)). If we pick the size of M2 so that M1 operates at the
edge of saturation, then if M2 were to be any smaller, VGS2 would have to be larger (given the
same ID2 ), driving M1 into triode. Thus, (W/L)2 is its smallest possible value (without driving
M1 into saturation) when M1 is at the edge of saturation, meaning the gain is largest in magnitude
with this choice of (W/L)2 .

7.27 (a)
gm1
gm2
q
2µn Cox
= −q
2µn Cox
v
u W

u
L 1


= −t W

Av = −

L

W
L 1 ID1




W
L 2 ID2




2

= −5


W
L



= 277.78

1

(b)

VDD − VT H

VDS1 > VGS1 − VT H (to ensure M1 is in saturation)
VDD − VGS2 > VGS1 − VT H
s
s
2ID2
2ID1

 >


−
W
µn Cox L 2
µn Cox W
L 1
ID1 = ID2 < 1.512 mA

7.28 For this problem, recall that looking into the drain of a transistor with a grounded gate and source
we see a resistance of ro , and looking into either terminal of a diode-connected transistor we see a
resistance of g1m k ro .
(a)


1
k ro2
Av = −gm1 ro1 k
gm2
(b)


1
k ro3
Av = −gm1 ro1 k ro2 k
gm3
(c)


1
Av = −gm1 ro1 k ro2 k
k ro3
gm3
(d)


1
k ro3
Av = −gm2 ro2 k ro1 k
gm3
(e)


1
k ro3
Av = −gm2 ro2 k ro1 k
gm3
(f) Let’s draw a small-signal model to find the equivalent resistance seen looking up from the output.
RD
+

+
vt

it
−

gm2 vgs2

vgs2

ro2

−

it = gm2 vgs2 +

vt − it RD
ro2

vgs2 = vt
vt − it RD
it = gm2 vt +
ro2




1
RD
= vt gm2 +
it 1 +
ro2
ro2
R

D
1 + ro2
vt
ro2 + RD
=
=
it
1 + gm2 ro2
gm2 + r1o2


ro2 + RD
Av = −gm1 ro1 k
1 + gm2 ro2

7.30 (a) Assume M1 is operating in saturation.
ID = 1 mA
ID RS = 200 mV
RS = 200 Ω
RD
Av = − 1
gm + RS
=−

RD
1
2µn Cox W
L ID

√

= −4
W
= 1000
L
s

VGS = VT H +

VDS

+ RS

2ID
µn Cox W
L

= 0.5 V
= VDD − ID RD − ID RS
= 0.6 V > VGS − VT H , verifying that M1 is in saturation

Yes , the transistor operates in saturation.
(b) Assume M1 is operating in saturation.
50
W
=
L
0.18
RS = 200 Ω
Av = −

√

RD
1
2µn Cox W
L ID

+ RS

= −4
RD = 1.179 kΩ
s

VGS = VT H +

VDS

2ID
µn Cox W
L

= 0.590 V
= VDD − ID RD − ID RS
= 0.421 V > VGS − VT H , verifying that M1 is in saturation

Yes , the transistor operates in saturation.

7.42 (a)
Rout = RD = 500 Ω
VG = VDD
VD > VG − VT H (in order for M1 to operate in saturation)
VDD − ID RD > VDD − VT H
ID < 0.8 mA
(b)
ID = 0.8 mA
1
Rin =
gm
1
=q
2µn Cox W
L ID
= 50 Ω
W
= 1250
L

(c)
Av = gm RD
1
gm =
S
50
RD = 500 Ω
Av = 10

7.43 (a)
ID = I1 = 1 mA
VG = VDD
VD = VG − VT H + 100 mV
VDD − ID RD = VG − VT H + 100 mV
RD = 300 Ω
(b)
RD = 300 Ω
Av = gm RD
r
=

2µn Cox

=5

W
= 694.4
L

W
ID RD
L

7.44 For this problem, recall that looking into the drain of a transistor with a grounded gate and source
we see a resistance of ro , and looking into either terminal of a diode-connected transistor we see a
resistance of g1m k ro .
(a) Referring to Eq. (7.109) with RD =

1
gm2

and gm = gm1 , we have

Av =

1
gm2
1
gm1

+ RS

(b) Let’s draw a small-signal model to find the equivalent resistance seen looking up from the output.
RD
+

+
vt

it

gm2 vgs2

vgs2

−

−

it = gm2 vgs2
vgs2 = vt
it = gm2 vt
1
vt
=
it
gm2
gm1
Av =
gm2
(c) Referring to Eq. (7.119) with RD =

1
gm2 ,

Av =

R3 = R1 , and gm = gm1 , we have
R1 k

1
gm1

RS + R1 k

1
gm1

gm1
gm2

(d)


1
k ro3
Av = gm1 RD +
gm2
(e)


1
Av = gm1 RD +
gm2

7.45 (a)


vX
1
= −gm1 RD1 k
vin
gm2
vout
= gm2 RD2
vX
vout
vX vout
=
vin
vin vX


1
= −gm1 gm2 RD2 RD1 k
gm2
(b)
lim

RD1 →∞



−gm1 gm2 RD2 RD1 k

1
gm2



= −gm1 RD2

This makes sense because the common-source stage acts as a transconductance amplifier with
a transconductance of gm1 . The common-gate stage acts as a current buffer with a current
gain of 1. Thus, the current gm1 vin flows through RD2 , meaning vout = −gm1 vin RD2 , so that
vout
vin = −gm1 RD2 .
This type of amplifier (with RD1 = ∞) is known as a cascode and will be studied in detail in
Chapter 9.

7.40
ID = 0.5 mA
1
Rin =
gm
1
=q
2µn Cox W
L ID

VDD

= 50 Ω
W
= 2000
L
VD > VG − VT H (in order for M1 to operate in saturation)
− ID RD > Vb − VT H
RD < 2.4 kΩ

Since |Av | ∝ RD , we need to maximize RD in order to maximize the gain. Thus, we should pick
RD = 2.4 kΩ . This corresponds to a voltage gain of Av = −gm RD = −48.

7.42 (a)
Rout = RD = 500 Ω
VG = VDD
VD > VG − VT H (in order for M1 to operate in saturation)
VDD − ID RD > VDD − VT H
ID < 0.8 mA
(b)
ID = 0.8 mA
1
Rin =
gm
1
=q
2µn Cox W
L ID
= 50 Ω
W
= 1250
L

(c)
Av = gm RD
1
gm =
S
50
RD = 500 Ω
Av = 10

7.43 (a)
ID = I1 = 1 mA
VG = VDD
VD = VG − VT H + 100 mV
VDD − ID RD = VG − VT H + 100 mV
RD = 300 Ω
(b)
RD = 300 Ω
Av = gm RD
r
=

2µn Cox

=5

W
= 694.4
L

W
ID RD
L

7.44 For this problem, recall that looking into the drain of a transistor with a grounded gate and source
we see a resistance of ro , and looking into either terminal of a diode-connected transistor we see a
resistance of g1m k ro .
(a) Referring to Eq. (7.109) with RD =

1
gm2

and gm = gm1 , we have

Av =

1
gm2
1
gm1

+ RS

(b) Let’s draw a small-signal model to find the equivalent resistance seen looking up from the output.
RD
+

+
vt

it

gm2 vgs2

vgs2

−

−

it = gm2 vgs2
vgs2 = vt
it = gm2 vt
1
vt
=
it
gm2
gm1
Av =
gm2
(c) Referring to Eq. (7.119) with RD =

1
gm2 ,

Av =

R3 = R1 , and gm = gm1 , we have
R1 k

1
gm1

RS + R1 k

1
gm1

gm1
gm2

(d)


1
k ro3
Av = gm1 RD +
gm2
(e)


1
Av = gm1 RD +
gm2

7.45 (a)


vX
1
= −gm1 RD1 k
vin
gm2
vout
= gm2 RD2
vX
vout
vX vout
=
vin
vin vX


1
= −gm1 gm2 RD2 RD1 k
gm2
(b)
lim

RD1 →∞



−gm1 gm2 RD2 RD1 k

1
gm2



= −gm1 RD2

This makes sense because the common-source stage acts as a transconductance amplifier with
a transconductance of gm1 . The common-gate stage acts as a current buffer with a current
gain of 1. Thus, the current gm1 vin flows through RD2 , meaning vout = −gm1 vin RD2 , so that
vout
vin = −gm1 RD2 .
This type of amplifier (with RD1 = ∞) is known as a cascode and will be studied in detail in
Chapter 9.

7.48 For small-signal analysis, we can short the capacitors, producing the following equivalent circuit.
R2 k R3 k RD

vout

M1
vin
R4

Av = gm (R2 k R3 k RD )

7.49
VGS = VDS
1
W
2
VGS = VDD − ID RS = VDD − µn Cox
(VGS − VT H ) (1 + λVGS ) RS
2
L
VGS = VDS = 0.7036 V
ID = 1.096 mA
Av =
gm =
ro =

1
gm

ro k RS
+ ro k RS

r

2µn Cox

W
ID = 6.981 mS
L

1
= 9.121 kΩ
λID

Av = 0.8628

7.50
Av =

1
gm

RS
+ RS

=
µn Cox W
L

RS
1
(VGS −VT H )

+ RS

= 0.8
VGS = 0.64 V
1
W
2
ID = µn Cox
(VGS − VT H )
2
L
= 960 µA
VG = VGS + VS = VGS + ID RS
= 1.12 V

7.55 For this problem, recall that looking into the drain of a transistor with a grounded gate and source
we see a resistance of ro , and looking into either terminal of a diode-connected transistor we see a
resistance of g1m k ro .
(a)
Av =

ro1 k (RS + ro2 )
gm1 + ro1 k (RS + ro2 )
1

(b) Looking down from the output we see an equivalent resistance of ro2 + (1 + gm2 ro2 ) RS by Eq.
(7.110).
ro1 k [ro2 + (1 + gm2 ro2 ) RS ]
Av = 1
gm1 + ro1 k [ro2 + (1 + gm2 ro2 ) RS ]
(c)
Av =

ro1 k
1
gm1

1
gm2

+ ro1 k

1
gm2

(d) Let’s draw a small-signal model to find the equivalent resistance seen looking down from the
output.
R1
+
R2

+
gm2 vgs2

vgs2

ro2

vt
vt
+ gm2 vgs2 +
R1 + R2
ro2
R2
=
vt
R1 + R2
R2
vt
vt
+ gm2
vt +
=
R1 + R2
R1 + R2
ro2


1
gm2 R2
1
= vt
+
+
R1 + R2
R1 + R2
ro2


R1 + R2
k ro2
= (R1 + R2 ) k
gm2 R2


1 +R2
k ro2
ro1 k (R1 + R2 ) k Rgm2
R2


=
R1 +R2
1
k ro2
+
r
k
(R
+
R
)
k
o1
1
2
gm1
gm2 R2

it =

it
it
vt
it
Av

vt
−

−

vgs2

it

(e)
Av =

ro2 k ro3 k
1
gm2

1
gm1

+ ro2 k ro3 k

1
gm1

(f) Looking up from the output we see an equivalent resistance of ro2 + (1 + gm2 ro2 ) ro3 by Eq.
(7.110).
ro1 k [ro2 + (1 + gm2 ro2 ) ro3 ]
Av = 1
gm1 + ro1 k [ro2 + (1 + gm2 ro2 ) ro3 ]

7.58
P = VDD ID = 2 mW
ID = 1.11 mA
RD ID = 1 V
RD = 900 Ω
Av = −gm RD
r
=−

2µn Cox

= −5

W
= 69.44
L

W
ID RD
L

7.60 Let’s let R1 and R2 consume exactly 5 % of the power budget (which means the branch containing RD ,
M1 , and RS will consume 95 % of the power budget). Let’s also assume Vov = VGS − VT H = 300 mV
exactly.
ID VDD = 0.95(2 mW)
ID = 1.056 mA
ID RS = 200 mV
RS = 189.5 Ω
Vov = VGS − VT H = 300 mV
W 2
1
ID = µn Cox Vov
2
L
W
= 117.3
L
RD
Av = − 1
gm + RS
=−

RD
1
√
2µn Cox W
L ID

+ RS

= −4
RD = 1.326 kΩ
2
VDD
= 0.05(2 mW)
R1 + R2
2
VDD
R1 + R2 =
0.1 mW
VG = VGS + ID RS = Vov + VT H + ID RS = 0.9 V
R2
VG =
VDD
R1 + R2
R2
= V2
= 0.9 V
DD

0.1 mW

R2 = 29.16 kΩ
R1 = 3.24 kΩ

7.61 Let’s let R1 and R2 consume exactly 5 % of the power budget (which means the branch containing
RD , M1 , and RS will consume 95 % of the power budget).
RD = 200 Ω
ID VDD = 0.95(6 mW)
ID = 3.167 mA
ID RS = Vov = VGS − VT H
Vov
RS =
ID
2ID
gm =
Vov
RD
Av = − 1
gm + RS
RD
Vov
2ID + ID

= − Vov
= −5

Vov = 84.44 mV
RS = 26.67 Ω
2ID
W
= 4441
=
2
L
µn Cox Vov
2
VDD
= 0.05(6 mW)
R1 + R2
2
VDD
R1 + R2 =
0.3 mW
VG = VGS + ID RS = Vov + VT H + ID RS = 0.5689 V
R2
VG =
VDD
R1 + R2
R2
= V2
= 0.5689 V
DD

0.3 mW

R2 = 6.144 kΩ
R1 = 4.656 kΩ

7.62
Rin = R1 = 20 kΩ
P = VDD ID = 2 mW
ID = 1.11 mA
VDS = VGS − VT H + 200 mV
VDD − ID RD = VDD − VT H + 200 mV
RD = 180 Ω
Av = −gm RD
r

2µn Cox

=−

W
ID RD
L

= −6

W
= 2500
L
s

VGS = VT H +

2ID
µn Cox W
L

= 0.467 V
VGS = VDD − ID RS
RS
1
2πf C1
1
2πf C1
f

= 1.2 kΩ
≪ R1
1
R1
10
= 1 MHz
=

C1 = 79.6 pF
1
1
k RS ≪
2πf CS
gm
1 1
1
=
2πf CS
10 gm
r
gm =

2µn Cox

CS = 52.9 nF

W
ID = 33.33 mS
L

7.64 (a)
Av = −gm1 (ro1 k RG k ro2 )

(b)
P = VDD ID1 = 3 mW
ID1 = |ID2 | = 1.67 mA
VDD
|VGS2 | = |VDS2 | = VDS =
2
 
W
1
(|VGS2 | − |VT H |)2 (1 + λp |VDS2 |)
|ID2 | = µp Cox
2
L 2
 
W
= 113
L 2
Av = −gm1 (ro1 k RG k ro2 )

RG = 10 (ro1 k ro2 )
1
ro1 =
= 6 kΩ
λn ID1
1
= 3 kΩ
ro2 =
λp |ID2 |
RG = 10 (ro1 k ro2 ) = 20 kΩ
s
 
W
Av = − 2µn Cox
ID1 (ro1 k RG k ro2 )
L 1
= −15


W
L



= 102.1
1

VIN = VGS1 = VT H +
= 0.787 V

s

µn Cox

2I

 D
(1 + λn VDS1 )

W
L 1

7.66
P = VDD ID1 = 1 mW
ID1 = |ID2 | = 556 µA
 
p
W
Vov1 = VGS1 − VT H = 2ID µn Cox
= 200 mV
L 1
 
W
= 138.9
L 1
gm1
Av = −
gm2
q


2µn Cox W
L 1 ID1
= −q


2µn Cox W
L 2 |ID2 |
v
u W

u L

1
= −t W
L

2

= −4



W
L



= 8.68

2

VIN = VGS1 = Vov1 + VT H = 0.6 V

7.67
P = VDD ID = 3 mW
ID = I1 = 1.67 mA
1
1
=q
Rin =
gm
2µ C
n

W
ox L ID

W
= 600
L

Av = gm RD =
RD = 250 Ω

1
RD = 5
50 Ω

= 50 Ω

7.68
P = VDD ID = 2 mW
ID = 1.11 mA
VD = VG − VT H + 100 mV
VDD − ID RD = VG − VT H + 100 mV
VG = VDD
2ID
RD = 4
VGS − VT H
VGS − VT H
RD = Av
2ID
VGS − VT H
= VDD − VT H + 100 mV
− ID Av
2ID
VGS = 0.55 V
Av = gm RD =

VDD

RD = 270 Ω
VS = VDD − VGS = ID RS
RS = 1.125 kΩ
2ID
W
=
2 = 493.8
L
µn Cox (VGS − VT H )

7.73
P = VDD ID1 = 3 mW
ID1 = ID2 = 1.67 mA
Av =
=

ro1 k ro2
gm1 + ro1 k ro2
1

ro1 k ro2
1
q
+ ro1
2µn Cox ( W
L ) ID1

k ro2

1

= 0.9
1
= 6 kΩ
ro1 = ro2 =
λID1
 
W
= 13.5
L 1
Let Vov2 = VGS2 − VT H = 0.3 V. Let’s assume that VOUT = VDS2 = Vov2 .



VGS2 = Vb = Vov2 + VT H = 0.7 V

2ID2
W
=
2
L 2
µn Cox (VGS2 − VT H ) (1 + λVDS2 )
= 161
VGS1 = VT H +

s

µn Cox

2ID1


W
L 1 (1

VDS1 = VDD − VDS2 = 1.5 V
VGS1 = 1.44 V

VIN = VGS1 + VDS2 = 1.74 V

+ λVDS1 )

Vout (V)

8.1

2

1

−5

−4

−3

−2

−1

0

0

1

2

3

4
Vin

−1

−2

5
(mV)

8.11
V− = V+ = Vin
R2
R4 k (R2 + R3 )
Vout = Vin
R1 + R4 k (R2 + R3 ) R2 + R3
−1

R2
R4 k (R2 + R3 )
=
R1 + R4 k (R2 + R3 ) R2 + R3

V− =
Vout
Vin

=

(R2 + R3 ) [R1 + R4 k (R2 + R3 )]
R2 [R4 k (R2 + R3 )]

If R1 → 0, we expect the result to be:
R2
Vout
R2 + R3
R2 + R3
R3
=
=1+
R2
R2

Vin =
Vout
Vin

R1 =0

Taking limit of the original expression as R1 → 0, we have:
(R2 + R3 ) [R1 + R4 k (R2 + R3 )]
(R2 + R3 ) [R4 k (R2 + R3 )]
=
R1 →0
R2 [R4 k (R2 + R3 )]
R2 [R4 k (R2 + R3 )]
R3
=1+
R2
lim

This agrees with the expected result. Likewise, if R3 → 0, we expect the result to be:
R2 k R4
Vout
R1 + R2 k R4
R1 + R2 k R4
=
R2 k R4
R1
=1+
R2 k R4

Vin =
Vout
Vin

R3 =0

Taking the limit of the original expression as R3 → 0, we have:
lim

R3 →0

(R2 + R3 ) [R1 + R4 k (R2 + R3 )]
R2 (R1 + R2 k R4 )
=
R2 [R4 k (R2 + R3 )]
R2 (R2 k R4 )
R1 + R2 k R4
=
R2 k R4
R1
=1+
R2 k R4

This agrees with the expected result.

8.14 We need to derive the closed-loop gain of the following circuit:
R1
R2

Rout
+

+
vin

−A0 vX

vX
−

vout

+
−

−

R2
+ vin
R1 + R2
R1 + R2
+ vin
= (−A0 vX − vin )
Rout + R1 + R2




R1 + R2
R2
+ vin − vin
+ vin
= −A0 (vout − vin )
R1 + R2
Rout + R1 + R2

vX = (vout − vin )
vout

Grouping terms, we have:





R2
R1 + R2
R2
R1 + R2
Rout + R1 + R2
vout 1 + A0
= vin
A0
− A0 − 1 +
R1 + R2 Rout + R1 + R2
Rout + R1 + R2
R1 + R2
R1 + R2



R1 + R2
R1
Rout + R1 + R2
= vin
− A0
−1
Rout + R1 + R2
R1 + R2
R1 + R2
1
[Rout + R1 + R2 − A0 R1 − R1 − R2 ]
= vin
Rout + R1 + R2


A0 R1 + R1 + R2
= vin 1 −
Rout + R1 + R2
R1 +R1 +R2
1 − AR0out
vout
+R1 +R2
=
0 R2
vin
1 + RoutA+R
1 +R2

=
=

Rout + R1 + R2 − A0 R1 − R1 − R2
Rout + R1 + R2 + A0 R2
Rout − A0 R1
Rout + R1 + (1 + A0 ) R2

To find the output impedance, we must find Zout =

vt
it

for the following circuit:

R1
R2

Rout
+

+

−

+
−A0 vX

vX
−

it

vt
−

vt
vt + A0 vX
+
Rout
R1 + R2
R2
vX =
vt
R1 + R2
2
vt
vt + A0 R1R+R
vt
2
+
it =
Rout
R1 + R2


A0 R2
1
1
+
+
= vt
Rout
Rout (R1 + R2 ) R1 + R2
R1 + (1 + A0 ) R2 + Rout
= vt
Rout (R1 + R2 )
it =

Zout =

Rout (R1 + R2 )
vt
=
it
R1 + (1 + A0 ) R2 + Rout

8.15 Refer to the analysis for Fig. 8.42.
Vout
R1
=
=4
Vin
R2
Rin ≈ R2 = 10 kΩ
R1 = 4R2 = 40 kΩ

From Eq. (8.99), we have
E =1−

A0 −
1+

A0 = 1000
Rout = 1 kΩ
E = 0.51 %

Rout
R2

Rout
R1

+ A0 +

R1
R2

8.17
V+ = V− (since A0 = ∞)
Vin
Vout
R3 k R4
=−
R2
R3 R1 + R3 k R4
Vout
R3 R1 + R3 k R4
= −
Vin
R2
R3 k R4
If R1 → 0 or R3 → 0, we expect the amplifier to reduce to the standard inverting amplifier.
Vout
Vin
Vout
Vin
The gain reduces to the expected expressions.

=−

R3
R2

=−

R1
R2

R1 →0

R3 →0

8.18
V+ = V− (since A0 = ∞)
R3
R2
VX =
Vout =
(Vout − Vin ) + Vin
R3 + R4
R1 + R2




R3
R2
R2
Vout
= Vin 1 −
−
R3 + R4
R1 + R2
R1 + R2




R1
R3 (R1 + R2 ) − R2 (R3 + R4 )
= Vin
Vout
(R1 + R2 ) (R3 + R4 )
R1 + R2
Vout
R1 (R3 + R4 )
=
Vin
R3 (R1 + R2 ) − R2 (R3 + R4 )

8.22 We must find the transfer function of the following circuit:
C1
R1

vout

+
vin

Rin
−

+
vX

+

−

−

−A0 vX

vout = −A0 vX


1
vX − vin
vX
vX = vout −
+
sC1 Rin
R1


vin
1
1
= vout +
+
vX 1 +
sRin C1
sR1 C1
sR1 C1
sR1 Rin C1 vout + Rin vin
vX =
sR1 Rin C1 + R1 + Rin
sR1 Rin C1 vout + Rin vin
vout = −A0
sR1 Rin C1 + R1 + Rin


Rin
sR1 Rin C1
= −A0 vin
vout 1 + A0
sR1 Rin C1 + R1 + Rin
sR1 Rin C1 + R1 + Rin
vout
−A0 Rin
sR1 Rin C1 + R1 + Rin
=
·
vin
sR1 Rin C1 + R1 + Rin sR1 Rin C1 + R1 + Rin + sR1 Rin C1 A0
−A0 Rin
=
sR1 Rin C1 + R1 + Rin + sR1 Rin C1 A0
−A0 Rin
=
sR1 Rin C1 (1 + A0 ) + R1 + Rin
−A0 Rin
=
1 (1+A0 )
1 + s R1 RRin1C+R
in
=

−A0 Rin / (R1 + Rin )
1 + s (R1 k Rin ) C1 (1 + A0 )

sp = −

1
(R1 k Rin ) C1 (1 + A0 )

Comparing this to the result in Eq. (8.37), we can see that we can simply replace R1 with R1 k Rin ,
effectively increasing the pole frequency (since R1 k Rin < R1 for finite Rin ).
We can also write the result as
sp = −

1
R1 C1 (1 + A0 )



1+

R1
Rin



In this form, it’s clear that the pole frequency increases by 1 + R1 /Rin .

8.23 We must find the transfer function of the following circuit:
C1
R1

Rout
+

+
vin

−A0 vX

vX
−

vout

+
−

−

vout = −A0 vX +

vin − vout
Rout
R1 + sC1 1

R1
(vout − vin )
R1 + sC1 1
"
#
R1
vin − vout
vout = −A0 vin +
(vout − vin ) +
Rout
R1 + sC1 1
R1 + sC1 1
"
"
#
#
A0 R1 + Rout
A0 R1 + Rout
vout 1 +
= vin −A0 +
R1 + sC1 1
R1 + sC1 1
vX = vin +

vout

R1 +

1
sC1

+ A0 R1 + Rout

R1 +

1
sC1

= vin

−A0 R1 − A0 sC1 1 + A0 R1 + Rout
R1 +

1
sC1

vout {1 + sC1 [(1 + A0 ) R1 + Rout ]} = −vin {A0 − sC1 Rout }
A0 − sC1 Rout
vout
= −
vin
1 + sC1 [(1 + A0 ) R1 + Rout ]
sp = −

1
C1 [(1 + A0 ) R1 + Rout ]

Comparing this to the result in Eq. (8.37), we can see that the pole gets reduced in magnitude due to
Rout .

8.26 We must find the transfer function of the following circuit:
R1
C1
vout
+
vin

+
Rin

+
−A0 vX

vX

−

−

−

vout = −A0 vX


vX − vout
vX = (vin − vX ) sC1 −
Rin
R1


Rin
Rin
= vin sRin C1 + vout
vX 1 + sRin C1 +
R1
R1
vX =

vin sRin C1 + vout RRin
1
1 + sRin C1 +

vout = −A0
"

vout 1 +

A0 RRin
1
1 + sRin C1 +

Rin
R1
Rin
R1

vin sRin C1 + vout RRin
1
1 + sRin C1 +

= −vin

sRin C1 A0
1 + sRin C1 + RRin
1

#

= −vin

sRin C1 A0
1 + sRin C1 + RRin
1

vout

1 + sRin C1 + (1 + A0 )

vout




Rin
1 + sRin C1 + (1 + A0 )
= −vin sRin C1 A0
R1

1 + sRin C1 +

Rin
R1

#

"

Rin
R1

Rin
R1

vout
sR1 Rin C1 A0
= −
vin
R1 + sR1 Rin C1 + (1 + A0 ) Rin
vout
lim
= −sR1 C1
A0 →∞ vin
Comparing this to Eq. (8.42), we can see that if we let A0 → ∞, the result actually reduces to Eq.
(8.42).

8.27 We must find the transfer function of the following circuit:
R1
C1

Rout
+

+
vin

−A0 vX

vX
−

−

−

vout = −A0 vX +
vX = vin +

vin − vout
Rout
R1 + sC1 1

1
sC1

R1 +
"

vout = −A0 vin +
"

vout 1 +
vout

R1 +

1
sC1

A0 sC1 1 + Rout
R1 +

1
sC1

#

+ A0 sC1 1 + Rout

R1 +

1
sC1

"

= vin −A0 +
= vin

vout

+

1
sC1

(vout − vin )
1
sC1

R1 +

1
sC1

(vout − vin ) +

A0 sC1 1 + Rout
R1 +

#

1
sC1

vin − vout
Rout
R1 + sC1 1

#

−A0 R1 − A0 sC1 1 + A0 sC1 1 + Rout
R1 +

1
sC1

vout {1 + A0 + sC1 (R1 + Rout )} = −vin {sC1 (A0 R1 − Rout )}
vout
sC1 (A0 R1 − Rout )
= −
vin
1 + A0 + sC1 (R1 + Rout )
vout
= −sR1 C1
lim
A0 →∞ vin
Comparing this to Eq. (8.42), we can see that if we let A0 → ∞, the result actually reduces to Eq.
(8.42).

8.28
vout = −A0 v−
v− = vin + (vout − vin ) 


1
sC1
1
sC1

k R1

vout = −A0 vin + (vout − vin ) 



1
sC1

k R1


+ sC1 2 k R2


k R1
 

k R1 + sC1 2 k R2

1
sC1



1
k R1
sC1 k R1
 
  = −vin A0 1 − 
 

vout 1 + A0 
1
1
1
1
+
+
k
R
k
R
k
R
k
R
1
2
1
2
sC1
sC2
sC1
sC2
 


 
 




1
1
1
1
1
1
k
R
k
R
k
R
k
R
k
R
k
R
+
+
A
+
−
1
2
1
1
2
1
0 sC1
sC1
sC2
sC1
sC2
sC1


 

 

= −vin A0
vout
1
1
1
1
sC1 k R1 + sC2 k R2
sC1 k R1 + sC2 k R2



 


1
1
1
vout (1 + A0 )
= −vin A0
k R1 +
k R2
k R2
sC1
sC2
sC2


1
sC1

1
k R2
vout

 sC2  
= −A0
vin
(1 + A0 ) sC1 1 k R1 + sC1 2 k R2

Unity gain occurs when the numerator and denominator are the same (note that we can drop the
negative sign since we only care about the magnitude of the gain):


 


1
1
1
k R2 = (1 + A0 )
k R1 +
k R2
A0
sC2
sC1
sC2




1
1
k R2 = (1 + A0 )
k R1
(A0 − 1)
sC2
sC1


1
sC2 k R2
A +1

 = 0
1
A
0−1
kR
sC1

1

It is possible to obtain unity gain by choosing the resistors and capacitors according to the above
formula.

8.31
vout = −A0 vX
v1 − vX
v2 − vX
vX − vout
+
=
R2
R1
RF
vout
v1
v2
vX
+
+
=
RF
R2
R1
R1 k R2 k RF


vout
v1
v2
vout = −A0 (R1 k R2 k RF )
+
+
RF
R2
R1




v2
v1
(R1 k R2 k RF )
= −A0 (R1 k R2 k RF )
+
vout 1 + A0
RF
R2
R1
v1
v2
R2 + R1
vout = −A0 (R1 k R2 k RF )
2 kRF )
1 + A0 (R1 kR
RF
= −A0 RF (R1 k R2 k RF )
= −



v1
v2
+
R2
R1



v1
R2

+

v2
R1

RF + A0 (R1 k R2 k RF )

[RF k A0 (R1 k R2 k RF )]

8.32 For A0 = ∞, we know that v+ = v− , meaning that no current flows through RP . Thus, RP will have
no effect on vout .
vout = −RF



v1
v2
+
R2
R1



, A0 = ∞

For A0 < ∞, we have to include the effects of RP .
vout = −A0 vX


v2 − vX
vout − vX
v1 − vX
RP
+
+
vX =
R2
R1
RF


1
v1
1
1
1
v2
vout
vX
=
+
+
+
+
+
RP
R1
R2
RF
R2
R1
RF


v2
vout
v1
(R1 k R2 k RF k RP )
+
+
vX =
R2
R1
RF


v1
v2
vout
vout = −A0
(R1 k R2 k RF k RP )
+
+
R2
R1
RF




v1
v2
A0
(R1 k R2 k RF k RP )
(R1 k R2 k RF k RP ) = −A0
+
vout 1 +
RF
R2
R1


v1
v2
(R1 k R2 k RF k RP )
vout = −A0
+
A0
R2
R1 1 + R
(R1 k R2 k RF k RP )
F


v1
v2
RF A0 (R1 k R2 k RF k RP )
=−
+
R2
R1 RF + A0 (R1 k R2 k RF k RP )


v1
v2
[RF k A0 (R1 k R2 k RF k RP )] , A0 < ∞
+
= −
R2
R1

8.33 We must find vout for the following circuit:
v1

R2

RF
Rout

v2

+

R1

vout

+
−A0 vX

vX
−

−




v1 − vX
v2 − vX
vout = −A0 vX +
Rout
+
R2
R1




v1
Rout
v2
Rout
+ Rout
+
+
= −vX A0 +
R1
R2
R2
R1


v1 − vX
v2 − vX
vX = vout +
RF
+
R2
R1


vout
1
1
v1
v2
1
=
+
+
+
+
vX
RF
R1
R2
RF
R2
R1


vout
v1
v2
vX =
(R1 k R2 k RF )
+
+
RF
R2
R1






v1
Rout
v1
v2
Rout
v2
vout
(R1 k R2 k RF ) A0 +
+ Rout
+
+
+
+
vout = −
RF
R2
R1
R1
R2
R2
R1
Grouping terms, we have:


(R1 k R2 k RF ) A0 +
vout 1 +
RF

Rout
R1 kR2






v1
Rout
v2
(R1 k R2 k RF ) A0 +
+ Rout
+
R1 k R2
R2
R1





Rout
v2
v1
(R1 k R2 k RF ) A0 +
+ Rout
+
=−
R2
R1
R1 k R2


Rout


v1
v2 Rout + (R1 k R2 k RF ) A0 + R1 kR2


= −RF
+
R2
R1 R + (R k R k R ) A + Rout
F
1
2
F
0
R1 kR2

=−

vout



v2
v1
+
R2
R1





8.34 We must find vout for the following circuit:
v1

R2

RF

vout
v2

+

R1

Rin

+
−A0 vX

vX
−

−

RP

vout = −A0 vX


v1 − vX 1 +
vX = 
R1

RP
Rin



+


v2 − vX 1 +
R2

Grouping terms, we have:




1
RP
1
vX
=
+ 1+
Rin
Rin R1 k R2 k RF


(R1 k R2 k RF ) + RP + Rin
vX
=
Rin (R1 k R2 k RF )
vX
vout

RP
Rin



+


vout − vX 1 +
RF

RP
Rin



 Rin

v1
v2
vout
+
+
R2
R1
RF
v1
v2
vout
+
+
R2
R1
RF


v2
vout
Rin (R1 k R2 k RF )
v1
+
+
=
R2
R1
RF (R1 k R2 k RF ) + RP + Rin


v1
v2
vout
Rin (R1 k R2 k RF )
= −A0
+
+
R2
R1
RF (R1 k R2 k RF ) + RP + Rin

Grouping terms, we have:




A0
Rin (R1 k R2 k RF )
v2
v1
A0 Rin (R1 k R2 k RF )
=−
vout 1 +
+
RF (R1 k R2 k RF ) + RP + Rin
R2
R1 (R1 k R2 k RF ) + RP + Rin




v1
v2
A0 Rin (R1 k R2 k RF )
RF [(R1 k R2 k RF ) + RP + Rin ] + A0 Rin (R1 k R2 k RF )
=−
+
vout
RF [(R1 k R2 k RF ) + RP + Rin ]
R2
R1 (R1 k R2 k RF ) + RP + Rin
Simplifying, we have:
vout = −



v1
v2
+
R2
R1



A0 RF Rin (R1 k R2 k RF )
RF [(R1 k R2 k RF ) + RP + Rin ] + A0 Rin (R1 k R2 k RF )

8.35
ID1 =

(

Vin
R1

0

Vin > 0
Vin < 0

Plotting ID1 (t), we have

ID1 (t)

V0 /R1

0

0

−π/ω

0
t

π/ω

−V0

Vin (t) = V0 cos(ωt) (Dotted)

V0

8.36
ID1 =

(

Vin
R1

0

Vin > 0
Vin < 0

Plotting ID1 (t), we have

ID1 (t)

V0 /R1

0

0

−π/ω

0
t

π/ω

−V0

Vin (t) = V0 cos(ωt) (Dotted)

V0

8.37
VY =

(

Vin − VD,on
VDD

Vin < 0
Vout
Vin > 0

=

(

Vin
0

Vin < 0
ID1 =
Vin > 0

(

Vin
R1

0

Vin < 0
Vin > 0

Plotting VY (t) and Vout (t), we have

Vin (t) = V0 cos(ωt)
VY (t)
Vout (t)
VDD

V0

−π/ω

0

0

π/ω
t

−V0

Plotting ID1 (t), we have:

ID1 (t)
V0 /R1

0

−π/ω
0
t
π/ω
V0

0

−V0

Vin (t) = V0 cos(ωt) (Dotted)

8.38 Since the negative feedback loop is never broken (even when the diode is off, RP provides negative
feedback), V+ = V− will always hold, meaning VX = Vin .
We must determine when D1 turns on/off to determine VY . We know that for Vin < 0, the diode will
be off, and VX will follow Vin . As Vin begins to go positive, the diode will remain off until
Vin

RP
> VD,on
R1

Once the diode turns on, VY will be fixed at Vin + VD,on . Thus, we can write:
VX = Vin


(
R1
P
Vin < VD,on R
Vin 1 + R
R1
P
VY =
R1
Vin + VD,on
Vin > VD,on R
P
Plotting VY (t) and Vout (t), we have

Vin (t) = V0 cos(ωt)
VX (t)
VY (t)
V0 + VD,on
V0

−π/ω

0

0

π/ω
t

−V0

−V0 (1 + RP /R1 )

8.40 Note that although in theory the output is unbounded (i.e., by Eq. (8.66), we can take the logarithm
of an arbitrarily small positive number), in reality the output will be limited by the positive supply
rail, as shown in the following plot.

Vout
VX

VDD

−1

0
−1

0

R1 IS

1
Vin (V)

8.42 When Vin > 0, the feedback loop will be broken, and the output will go to the positive rail.
When Vin < 0, we have:
Vin
= IS eVBE /VT = IS e−Vout /VT
R1


Vin
= −VT ln −
R1 IS

IC = −
Vout

Vout (V)

This gives us the following plot of Vout vs. Vin :

−1

−R1 IS

VDD

0

0

Note that this circuit fails to behave as a non-inverting logarithmic amplifier.

1
Vin (V)

8.44 (a)



Vin
Vout = −VT ln
R1 IS


1V
−0.2 V = −VT ln
R1 IS
R1 IS = 456 µV
(b)
dVout
dVin

Vin =1 V

VT
=−
Vin

Vin =1 V

Av =

= −0.026

8.45 When Vin < VT H , the output goes to the positive rail. When Vin > VT H , we have:
ID =

Vin − VT H
R1

VGS = −Vout = VT H +

Vout = −VT H −

s

s

W
L

2ID
µn Cox

2 (Vin − VT H )
R1 W
L µn Cox

s
R1 W
2
1
dVout
L µn Cox
=−
dVin
2 2 (Vin − VT H ) R1 W
L µn Cox
s
1
= −
, Vin > VT H
2R1 W
µ
C
L n ox (Vin − VT H )

8.46 When Vin > 0, the output goes to the negative rail. When Vin < 0, we have:
ID = −

Vin
R1

VSG = Vout = |VT H | +

Vout = VT H +

s

−

s

2 |ID |
µp Cox

W
L

2Vin
, Vin < 0
R1 W
L µp Cox

8.49 We model an input offset with a series voltage source at one of the inputs.
R1
R2
−

Vout
Vin

+
−

+

Vos +
−

Vin − Vos
Vout = Vin −
(R1 + R2 )
R2


R1 + R2
R1 + R2
+ Vos
= Vin 1 −
R2
R2


R1
R1
= − Vin + 1 +
Vos
R2
R2
Note that even when Vin = 0, Vout = (1 + R1 /R2 ) Vos .

8.54 Let Vin = 0.
V+ = −IB1 (R1 k R2 ) = − (IB2 + ∆I) (R1 k R2 ) = V−


V−
Vout = V− + IB2 +
R1
R2


(IB2 + ∆I) (R1 k R2 )
R1
= − (IB2 + ∆I) (R1 k R2 ) + IB2 −
R2


R1
= − (IB2 + ∆I) (R1 k R2 ) 1 +
+ IB2 R1
R2
= −∆IR1

If the magnitude of the error must be less than ∆V , we have:
∆IR1 < ∆V
R1 <
Note that this does not depend on R2 .

∆V
∆I

8.57
Vout = −

A0
V−
1 + ωs0

V− = Vin +

Vout − Vin
R1
R1 + sC1 1

Vout − Vin
Vin +
R1
Vout
R1 + sC1 1
"
#
#
"
A0
A0
R1
R1
Vout 1 +
=
−1
Vin
1 + ωs0 R1 + sC1 1
1 + ωs0
R1 + sC1 1



1 + ωs0
R1 + sC1 1 + A0 R1
A0 sC1 1





= −Vin 
Vout
1 + ωs0
1 + ωs0
R1 + sC1 1
R1 + sC1 1
A0
=−
1 + ωs0

A0 sC1 1


s
R1 + sC1 1 + A0 R1
ω0

Vout
= −
Vin
1+
= −

A0

s
ω0

1+

=−



(1 + sR1 C1 ) + sA0 R1 C1
A0



1 + s R1 C1 +
= −

If ω0 ≫

1
R1 C1 ,

1
ω0


1
+ A0 R1 C1 + s2 Rω1 C
0

A0
h
1 + s (1 + A0 ) R1 C1 +

we have:
Vout
=−
Vin
=−
≈−

1
h

1
A0

+s

1
A0


+s 1+

1+

1
A0

1
A0



1


R1 C1 +

1
ω0

i

1 C1
+ s2 R
A0 ω 0

1 C1
R1 C1 + s2 R
A0 ω 0

1
(assuming A0 ≫ 1)
1 C1
sR1 C1 + s2 R
A0 ω 0

= −

1


sR1 C1 1 +

s
A0 ω 0



!

1
ω0

i

1
+ s2 Rω1 C
0

8.61 Let E refer to the gain error.
R1
=8
R2
R1 = 8 kΩ
R2 = 1 kΩ
A0 − RRout
R1
vout
1
=−
vin
R2 1 + RRout + A0 +
2

R1
(1 − E)
R2
A0 − RRout
1
E =1−
+
A
1 + RRout
0+
2

R1
R2

(Eq. 8.99)

=−

R1
R2

= 0.1 %
A0 = 9103
Note that we can pick any R1 , R2 such that their ratio is 8 (i.e., this solution is not unique). However,
A0 will change depending on the values chosen.

8.66

dVout
dVin




Vin
R1 IS
R1 IS 1
= −VT
Vin R1 IS
VT
=−
Vin

Vout = −VT ln

out
T
No, it is not possible to satisfy both requirements. As shown above, dV
, meaning for a
= VVin
dVin
specified temperature and input, the gain is fixed. Assuming we could fix the temperature as part of
the design, we could still only meet one of the two constraints, since the temperatures at which the
constraints are met are not equal.

9.7 Let R2 be the resistance seen looking into the collector of Q2 .
Rout = ro1 + (1 + gm1 ro1 ) (rπ1 k R2 )
Note that this expressoin is maximized as R2 → ∞. This gives us
Rout,max = ro1 + (1 + gm1 ro1 ) rπ1

9.9
1 VA βVA VT
(Eq. 9.9)
IC1 VT VA + βVT
1 VA
βVT
=
IC1 VT
βVA
=
IC1

Rout ≈

= βro
This resembles Eq. (9.12) because the assumption that
VA ≫ βVT
can be equivalently expressed as
VT
VA
≫β
IC
IC
ro ≫ rπ
This is the same assumption used in arriving at Eq. (9.12).

9.12
ID = 0.5 mA
Rout = ro1 + (1 + gm1 ro1 ) ro2
!
r
1
1
1
W
=
+ 1 + 2 µn Cox ID
λID
L
λID λID
≥ 50 kΩ
λ ≤ 0.558 V−1

9.15 (a)
VD1 = VDD − ID RD = 1.3 V > VG1 − VT H = Vb1 − VT H
Vb1 < 1.7 V
(b)
Vb1 = 1.7 V
VGS1 = Vb1 − VX
s
= VT H +

= 0.824 V
VX = 0.876 V

2I

 D
µn Cox

W
L 1

9.16 (a) Looking down from the source of M1 , we see an equivalent resistance of
Rout = gm1 ro1



1
gm2

k ro2

1
gm2

k ro2 . Thus, we have



(b)
Rout = gm1 ro1 ro2
(c) Putting two transistors in parallel, their transconductances will add and their output resistances
will be in parallel (i.e., we can treat M1 and M3 as a single transistor with gm = gm1 + gm3 and
ro = ro1 k ro3 ). This can be seen from the small-signal model.
Rout = (gm1 + gm3 ) (ro1 k ro3 ) ro2
(d) Let’s draw the small-signal model and apply a test source to find Rout .
+

+

vgs1

gm1 vgs1

ro1

−

vgs2
−

gm2 vgs2

+
it

vt
−

ro2

vgs1
vgs2 + vgs1
= gm1 vgs1 +
ro2
ro1
vgs1 = gm2 ro2 vt − it ro2
vt + gm2 ro2 vt − it ro2
it = gm1 (gm2 ro2 vt − it ro2 ) +
ro1




ro2
1 + gm2 ro2
it 1 + gm1 ro2 +
= vt gm1 gm2 ro2 +
ro1
ro1
it = gm2 vgs2 −

it (gm1 ro1 ro2 ) = vt (gm1 gm2 ro1 ro2 )
Rout =

1
vt
=
it
gm2

9.17
ID = 0.5 mA
Rout = ro1 + (1 + gm1 ro1 ) ro2
s  
!
W
1
1
1
=
µp Cox ID
+ 1+ 2
λID
L 1
λID λID


W
L



1

= 40 kΩ
 
W
=
= 8
L 2

9.20 (a)
Gm = gm1
Rout =

1

k ro1

gm2

Av = −gm1



1
gm2

k ro1



(b)
Gm = −gm2
Rout =

1

k ro2 k ro1

gm2

Av = gm2



1
gm2

k ro2 k ro1



(c) Let’s draw the small-signal model to find Gm .
iout
vin

rπ1

+
vπ1

gm1 vπ1

−

RE

ro1

vin − vπ1
vπ1
+
rπ1
RE
vπ1 = vin + (iout − gm1 vπ1 ) ro1
vπ1 (1 + gm1 ro1 ) = vin + iout ro1
vin + iout ro1
vπ1 =
1 + gm1 ro1
vin + iout ro1
vin
vin + iout ro1
−
+
iout = −
rπ1 (1 + gm1 ro1 ) RE
RE (1 + gm1 ro1 )




ro1
1
1
ro1
1
iout 1 +
= vin
−
+
−
rπ1 (1 + gm1 ro1 ) RE (1 + gm1 ro1 )
RE
rπ1 (1 + gm1 ro1 ) RE (1 + gm1 ro1 )
rπ1 RE (1 + gm1 ro1 ) + ro1 RE + ro1 rπ1
rπ1 (1 + gm1 ro1 ) − RE − rπ1
iout
= vin
rπ1 RE (1 + gm1 ro1 )
rπ1 RE (1 + gm1 ro1 )
iout [rπ1 RE (1 + gm1 ro1 ) + ro1 RE + ro1 rπ1 ] = vin [rπ1 (1 + gm1 ro1 ) − RE − rπ1 ]
iout
Gm =
vin
iout = −

rπ1 (1 + gm1 ro1 ) − RE − rπ1
rπ1 RE (1 + gm1 ro1 ) + ro1 RE + ro1 rπ1
gm1
(if rπ1 , ro1 are large)
≈
1 + gm1 RE
= ro2 k [ro1 + (1 + gm1 ro1 ) (rπ1 k RE )]
=

Rout

Av = −

rπ1 RE (1 + gm1 ro1 ) − RE − rπ1
{ro2 k [ro1 + (1 + gm1 ro1 ) (rπ1 k RE )]}
rπ1 RE (1 + gm1 ro1 ) + ro1 RE + ro1 rπ1

(d)
Gm = gm2
Rout = ro2 k [ro1 + (1 + gm1 ro1 ) (rπ1 k RE )]
Av = −gm2 {ro2 k [ro1 + (1 + gm1 ro1 ) (rπ1 k RE )]}
(e) Let’s draw the small-signal model to find Gm .
iout
+
vgs1

gm1 vgs1

−

RS

vin

ro1

Since the gate and drain are both at AC ground, the dependent current source looks like a resistor
with value 1/gm1 . Thus, we have:
Gm =

1
iout
=−
1
vin
k ro1
RS + gm1

=−

1
RS +

ro1
1+gm1 ro1

1 + gm1 ro1
ro1 + RS + gm1 ro1 RS
gm1
≈−
(if ro1 is large)
1 + gm1 RS
= [ro2 + (1 + gm2 ro2 ) RE ] k [ro1 + (1 + gm1 ro1 ) RS ]
= −

Rout

Av =

1 + gm1 ro1
{[ro2 + (1 + gm2 ro2 ) RE ] k [ro1 + (1 + gm1 ro1 ) RS ]}
ro1 + RS + gm1 ro1 RS

(f) We can use the result from part (c) to find Gm here. If we simply let rπ → ∞ (and obviously we
replace the subscripts as appropriate) in the expression for Gm from part (c), we’ll get the result
we need here.
rπ2 RE (2 + gm2 ro2 ) − RE − rπ2
rπ2 RE (2 + gm2 ro2 ) + ro2 RE + ro2 rπ2
gm2 ro2
=
ro2 + RE + gm2 ro2 RE
gm2
≈
(if ro2 is large)
1 + gm2 RE
= [ro2 + (1 + gm2 ro2 ) RE ] k [ro1 + (1 + gm1 ro1 ) RS ]

Gm = lim

rπ2 →∞

Rout

Av = −

gm2 ro2
{[ro2 + (1 + gm2 ro2 ) RE ] k [ro1 + (1 + gm1 ro1 ) RS ]}
ro2 + RE + gm2 ro2 RE

(g) Once again, we can use the result from part (c) to find Gm here (replacing subscripts as appropriate).
rπ2 RE (1 + gm2 ro2 ) − RE − rπ2
rπ2 RE (1 + gm2 ro2 ) + ro2 RE + ro2 rπ2
gm2
≈
(if rπ2 , ro2 are large)
1 + gm2 RE
= RC k [ro2 + (1 + gm2 ro2 ) (rπ2 k RE )]

Gm =

Rout

Av = −

rπ2 RE (1 + gm2 ro2 ) − RE − rπ2
{RC k [ro2 + (1 + gm2 ro2 ) (rπ2 k RE )]}
rπ2 RE (1 + gm2 ro2 ) + ro2 RE + ro2 rπ2

9.22
Av = −gm1 [ro2 + (1 + gm2 ro2 ) (rπ2 k ro1 )]
IC1 ≈ IC2 = I1
VA1 = VA2 = VA




I1 VA
VA
βVT VA
Av ≈ −
+ 1+
k
VT I1
VT
I1
I1
= −500

VA1 = VA2 = 0.618 V−1

9.23 (a) Although the output resistance of this stage is the same as that of a cascode, the transconductance
of this stage is lower than that of a cascode stage. A cascode has Gm = gm , where as this stage
m2
has Gm = 1+ggm2
ro1 .
(b)
Gm =

gm2
1 + gm2 ro1

Rout = ro2 + (1 + gm2 ro2 ) (rπ2 k ro1 )
Av = −Gm Rout
= −

gm2
[ro2 + (1 + gm2 ro2 ) (rπ2 k ro1 )]
1 + gm2 ro1

9.24
Gm = −gm1
Rout = ro2 + (1 + gm2 ro2 ) (rπ2 k ro1 )
Av = gm1 [ro2 + (1 + gm2 ro2 ) (rπ2 k ro1 )]

9.25 (a)
Gm = gm2

RP k rπ1
gm1 + RP k rπ1
1

Rout = ro1 + (1 + gm1 ro1 ) (rπ1 k ro2 k RP )
Av = −gm2

RP k rπ1
[ro1 + (1 + gm1 ro1 ) (rπ1 k ro2 k RP )]
gm1 + RP k rπ1
1

(b)
Gm = gm2
Rout = ro1 k RP + [1 + gm1 (ro1 k RP )] (rπ1 k ro2 )
Av = −gm2 {ro1 k RP + [1 + gm1 (ro1 k RP )] (rπ1 k ro2 )}
(c)
gm2
1 + gm2 RE
= ro1 + (1 + gm1 ro1 ) [rπ1 k (ro2 + (1 + gm2 ro2 ) (rπ2 k RE ))]

Gm =
Rout

Av = −

gm2
{ro1 + (1 + gm1 ro1 ) [rπ1 k (ro2 + (1 + gm2 ro2 ) (rπ2 k RE ))]}
1 + gm2 RE

(d)
Gm = gm2
Rout = ro1 + (1 + gm1 ro1 ) (rπ1 k ro2 k ro3 )
Av = −gm2 [ro1 + (1 + gm1 ro1 ) (rπ1 k ro2 k ro3 )]

9.26
Av = −gm1 {[ro2 + (1 + gm2 ro2 ) (rπ2 k ro1 )] k [ro3 + (1 + gm3 ro3 ) (rπ3 k ro4 )]}


 




IC
VA,N
VA,P
VA,N
βN VT VA,N
VA,P
βP VT VA,P
=−
k
+ 1+
k
+ 1+
k
VT
IC
VT
IC
IC
IC
VT
IC
IC

i h

i
h


VA,P
VA,N
VA,N
VA,P
VA,P
VA,N
βN VT
βP VT
k IC
k IC
IC + 1 + VT
IC
IC + 1 + VT
IC
IC

i h

i


h
=−
V
V
V
VA,P
βN VT
βP VT
VT VA,N + 1 + VA,N
+ IA,P
k A,N
+ 1 + VA,P
IC
VT
IC
IC
IC k IC
C
T
#"
#
"




βN VT VA,N
βP VT VA,P
VA,P
VA,N
VA,P
VA,N
”
”
“
“
V
VA,P
βN VT
βP VT
IC + 1 + VT
IC + 1 + VT
2
2
IC
IC
+ A,N
IC
IC
IC + IC
IC
# "
#
"
=−




VT V
β
V
V
β
V
V
V
V
V
N
T
A,N
P
T
A,P
A,P
A,N
A,P
A,N
” +
”
“
“
VA,N
VA,P
βN VT
βP VT
IC + 1 + VT
IC + 1 + VT
2
2
IC

h



IC

+

IC



IC

IC

+

IC

ih


i
V
βN VT VA,N
βP VT VA,P
VA,P
1
VA,N + 1 + VA,N
V
+
1
+
2
A,P
β
V
+V
V
β
V
+V
IC
IC
T
N T
A,N
T
P T
A,P
h


i
h


i
=−
VT 1 VA,N + 1 + VA,N βN VT VA,N + 1 VA,P + 1 + VA,P βP VT VA,P
IC
VT
βN VT +VA,N
IC
VT
βP VT +VA,P
h


ih


i
VA,N
VA,P
βN VT VA,N
βP VT VA,P
V
+
1
+
V
+
1
+
A,N
A,P
VT
βN VT +VA,N
VT
βP VT +VA,P
1
h


i h


i
= −
VA,N
VA,P
βN VT VA,N
βP VT VA,P
VT V
+
1
+
+
V
+
1
+
A,N
A,P
VT
βN VT +VA,N
VT
βP VT +VA,P

The result does not depend on the bias current.

9.28

|Av |

Av ≈ −gm1 gm2 ro1 ro2 (Eq. 9.69)
s  
s  

2
W
W
1
=− 2
µn Cox ID 2
µn Cox ID
L 1
L 2
λID
s    
2
W
1
W
= −2µn Cox ID
L 1 L 2 λID
s   
1 1
W
W
= −2µn Cox
2
ID λ
L 1 L 2

ID

9.30 From Problem 28, we have
1 1
Av = −2µn Cox
ID λ2

s

W
L

 
1

W
L



2

If we increase the transistor widths by a factor of N , we will get a new voltage gain A′v :
s
   
W
W
1 1
′
2
N
Av = −2µn Cox
ID λ2
L 1 L 2
s   
1 1
W
W
= −2N µn Cox
2
ID λ
L 1 L 2
= N Av
Thus, the gain increases by a factor of N .

9.31 From Problem 28, we have
1 1
Av = −2µn Cox
ID λ2

s

W
L

 
1

W
L



2

If we decrease the transistor widths by a factor of N , we will get a new voltage gain A′v :
s
   
W
W
1 1
1
′
Av = −2µn Cox
ID λ2 N 2 L 1 L 2
s   
1
W
1 1
W
= −2 µn Cox
2
N
ID λ
L 1 L 2
=

1
Av
N

Thus, the gain decreases by a factor of N .

9.32
Gm = −gm2
Rout = ro2 k [ro3 + (1 + gm3 ro3 ) ro4 ]
Av = gm2 {ro2 k [ro3 + (1 + gm3 ro3 ) ro4 ]}

9.33
Av = −gm1 {[ro2 + (1 + gm2 ro3 ) ro1 ] k [ro3 + (1 + gm3 ro3 ) ro4 ]}
= −500
s  
W
gm1 = gm2 = 2
µn Cox ID
L
s  
W
µp Cox ID
gm3 = gm4 = 2
L
1
λn ID
1
=
λp ID

ro1 = ro1 =
ro3 = ro4

ID = 1.15 mA

9.34 (a)
Gm = gm1
Rout = [(ro2 k RP ) + (1 + gm2 (ro2 k RP )) ro1 ] k [ro3 + (1 + gm3 ro3 ) ro4 ]
Av = −gm1 {[(ro2 k RP ) + (1 + gm2 (ro2 k RP )) ro1 ] k [ro3 + (1 + gm3 ro3 ) ro4 ]}
(b)
Gm = gm1

ro1 k RP
gm2 + ro1 k RP
1

Rout = [ro2 + (1 + gm2 ro2 ) (ro1 k RP )] k [ro3 + (1 + gm3 ro3 ) ro4 ]
Av = −gm1

ro1 k RP
{[ro2 + (1 + gm2 ro2 ) (ro1 k RP )] k [ro3 + (1 + gm3 ro3 ) ro4 ]}
gm2 + ro1 k RP
1

(c)
Gm = gm5
Rout = [ro2 + (1 + gm2 ro2 ) (ro1 k ro5 )] k [ro3 + (1 + gm3 ro3 ) ro4 ]
Av = −gm5 {[ro2 + (1 + gm2 ro2 ) (ro1 k ro5 )] k [ro3 + (1 + gm3 ro3 ) ro4 ]}
(d)
Gm = gm5
Rout = [ro2 + (1 + gm2 ro2 ) ro1 ] k [ro3 + (1 + gm3 ro3 ) (ro4 k ro5 )]
Av = −gm5 {[ro2 + (1 + gm2 ro2 ) ro1 ] k [ro3 + (1 + gm3 ro3 ) (ro4 k ro5 )]}

9.36
2

1
R2
W
I1 = µn Cox
(Eq. 9.85)
VDD − VT H
2
L R1 + R2


∂I1
R2
R2
W
=
VDD − VT H
µn Cox
∂VDD
L
R1 + R2
R1 + R2
=

R2
gm
R1 + R2

∂I1
. Since VGS is
Intuitively, we know that gm is the derivative of I1 with respect to VGS , or gm = ∂V
GS
∂VGS
is a
linearly dependent on VDD by the relationship established by the voltage divider (meaning ∂V
DD
∂I1
∂I1
∂VGS
∂I1
∂VGS
constant), we’d expect ∂VDD to also be proportional to gm , since ∂VDD = ∂VDD · ∂VGS = ∂VDD gm .

9.37
2
R2
(Eq. 9.85)
VDD − VT H
R1 + R2


R2
W
= −µn Cox
VDD − VT H
L R1 + R2

1
W
I1 = µn Cox
2
L
∂I1
∂VT H



The sensitivity of I1 to VT H becomes a more serious issue at low supply voltages because as VDD
becomes smaller with respect to VT H , VT H has more control over the sensitivity. When VDD is large
enough, it dominates the last term of the expression, reducing the control of VT H over the sensitivity.

9.38 As long as VREF > 0, the circuit operates in negative feedback, so that V+ = V− = 0 V.
VREF
IC1 = IS1 e−V1 /VT =
R

1
VREF
V1 = −VT ln
= VBE2
R1 IS1
If VREF > R1 IS1 , then we have VBE2 < 0, and IX = 0. If VREF < R1 IS1 , then we have:
IX = IS2 e

−VT ln

“

VREF
R1 IS1

”
/VT

“
”
V
− ln REF

R1 IS1
= IS2 e
R1 IS1
= IS2
VREF

Thus, if VREF > R1 IS1 (which will typically be true, since IS1 is typically very small), then we get no
output, i.e., IX = 0. When VREF < R1 IS1 , we get an inverse relationship between IX and VREF .

9.39 As long as VREF > 0, the circuit operates in negative feedback, so that V+ = V− = 0 V.
VREF
IC1 = IS1 e−V1 /VT =
R

1
VREF
V1 = −VT ln
= −VBE2
R1 IS1
If VREF < R1 IS1 , then we have VBE2 < 0, and IX = 0. If VREF > R1 IS1 , then we have:
V ln

“

VREF

R1 IS1
IX = IS2 e T
VREF
= IS2
R1 IS1
IS2 VREF
=
IS1 R1
IS2
=
IC1
IS1

”
/VT

Thus, if VREF < R1 IS1 , then we get no output, i.e., IX = 0. When VREF > R1 IS1 (which will typically
be true, since IS1 is typically very small), we get a current mirror relationship between Q1 and Q2
(ensured by the op-amp).
(with IX copying IC1 ), where the reference current for Q1 is VREF
R1

9.46 (a)
Icopy = 5IC,REF
IREF = IC,REF + IB,REF + IB1
IC,REF
Icopy
= IC,REF +
+
β
β
IC,REF
5IC,REF
= IC,REF +
+
β
β


5
1
= IC,REF 1 + +
β
β


Icopy 6 + β
=
5
β


β
Icopy =
5IREF
6+β

(b)
IC,REF
5
= IC,REF + IB,REF + IB1
IC,REF
Icopy
= IC,REF +
+
β
β
IC,REF
IC,REF
+
= IC,REF +
β
5β


1
1
= IC,REF 1 + +
β
5β


6 + 5β
= 5Icopy
5β


IREF
5β
=
6 + 5β
5

Icopy =
IREF

Icopy

(c)
3
IC,REF
2
5
I2 = IC,REF
2
IREF = IC,REF + IB,REF + IB1 + IB2
IC,REF
Icopy
I2
= IC,REF +
+
+
β
β
β
3IC,REF
5IC,REF
IC,REF
+
+
= IC,REF +
β
2β
2β


3
5
1
+
= IC,REF 1 + +
β
2β
2β


2
10 + 2β
= Icopy
3
2β


3
2β
IREF
Icopy =
10 + 2β 2
Icopy =

9.49
VGS,REF = VT H +

s

2IREF
µn Cox W
L

VGS1 = VGS,REF − I1 RP
s
2IREF
− I1 RP
= VT H +
µn Cox W
L
s
2IREF
IREF
RP
−
= VT H +
2
µn Cox W
L
!2
s
W
IREF
1
2IREF
RP
−
I1 = µn Cox
2
L
2
µn Cox W
L
IREF
s2
IREF
IREF
RP =
−
2
µn Cox W
L
s
s
IREF
2IREF
IREF
RP =
−
W
2
µn Cox W
µ
n Cox L
L
s

√
IREF
=
2−1
µn Cox W
L
√


2 2−1
RP = q
IREF µn Cox W
L
=

s

2IREF
µn Cox W
L

Given this choice of RP , I1 does not change if the threshold voltages of the transistors change by the
same amount ∆V . Looking at the expression for I1 in the derivation above, we can see that it has no
dependence on VT H (note that RP does not depend on VT H either).

9.54
IC1 = 1 mA
1 + βn
IE1 RE =
IC1 RE = 0.5 V
βn
RE == 0.5 V
RE = 495.05 Ω
Rout,a = ro1 + (1 + gm1 ro1 ) (rπ1 k RE )
Rout,b

= 85.49 kΩ
= ro1 + (1 + gm1 ro1 ) (rπ1 k ro2 )
= 334.53 kΩ

The output impedance of the circuit in Fig. 9.72(b) is significantly larger than the output impedance
of the circuit in Fig. 9.72(a) (by a factor of about 4).

9.56 (a)
Rout = ro1 + (1 + gm1 ro1 ) ro2 = 200 kΩ
1
ro1 = ro2 =
λID
r
W
gm1 = gm2 = 2 µn Cox ID
L
 
 
W
W
=
= 1.6
L 1
L 2
(b)
Vb2 = VGS2 = VT H +
= 2.9 V

s

W
L

2ID
µn Cox

9.57 (a) Assume IC1 ≈ IC2 , since β ≫ 1.
Av = −gm1 [ro2 + (1 + gm2 ro2 ) (rπ2 k ro1 )]
I1
gm1 = gm2 =
VT
VA
ro1 = ro2 =
I1
VT
rπ1 = rπ2 = β
I
" 1 
#

β VIT1 VIA1
VA
I1 VA
+ 1+
Av = −
VT I1
VT β VIT + VIA
1
1




1
VA
βVT VA
=−
VA + 1 +
VT
VT βVT + VA
= −500

VA = 0.618 V
(b)
Vin = VBE1 = VT ln



I1
IS1



= 714 mV
(c)
Vb1 = VBE2 + VCE1
= VBE2 + 500 mV


I1
+ 500 mV
= VT ln
IS2
= 1.214 V

9.58 Assume all of the collector currents are the same, since β ≫ 1.
P = IC VCC = 2 mW
IC = 0.8 mA
 
IC
Vin = VT ln
= 726 mV
IS

Vb1 = VBE2 + VCE1
 
IC
= VT ln
+ VBE1 − VBC1
IS
= 1.252 V
Vb3 = VCC − VT ln



IC
IS



= 1.774 V

Vb2 = VCC − VEC4 − VEB3
= VCC − (VEB4 − VCB4 ) − VT ln



IC
IS



= 1.248 V
Av = −gm1 {[ro2 + (1 + gm2 ro2 ) (rπ2 k ro1 )] k [ro3 + (1 + gm3 ro3 ) (rπ3 k ro4 )]}
= 4887

9.62
Rout = RC = 500 Ω
IC RC
= 20
Av = gm2 RC =
VT
IC = 1.04 mA
P = (IC + IREF ) VCC = 3 mW
IREF = 0.16 mA
AE1
IC =
IREF
AE,REF
AE1
= 6.5

AE,REF

AE,REF = AE
AE1 = 6.5AE

9.63
Icopy = nIC,REF
IREF = IC,REF + IB,REF + IB1
IC,REF
Icopy
= IC,REF +
+
β
β
IC,REF
nIC,REF
= IC,REF +
+
β
β


n
1
= IC,REF 1 + +
β
β


Icopy n + 1 + β
=
n
β


β
nIREF
Icopy =
n+1+β
β
Since nIREF is the nominal value of Icopy , the error term, n+1+β
, must be between 0.99 and 1.01 so
that the actual value of Icopy is within 1 % of the nominal value. Since the upper constraint (that the
error term must be less than 1.01) results in a negative value of n (meaning that we can only get less
than the nominal current if we include the error term), we only care about the lower error bound.

β
≥ 0.99
n+1+β
n ≤ 0.0101
IREF ≥ 50 mA
We can see that in order to decrease the error term, we must use a smaller value for n (in the ideal
β
). However, the smaller value of
case, we have n approaching zero and the error term approaching 1+β
n we use, the larger value we must use for IREF , meaning the more power we must consume. Thus,
we have a direct trade-off between accuracy and power consumption.

9.64
IC,M =

AE,M
IC,REF 1
AE,REF 1

IREF 1 = IC,REF 1 + IB,REF 1 + IB,M
IC,REF 1
IC,M
= IC,REF 1 +
+
βn
βn
AE,M IC,REF 1
IC,REF 1
+
= IC,REF 1 +
βn
AE,REF 1 βn


1
AE,M
= IC,REF 1 1 +
+
βn
AE,REF 1 βn


AE,REF 1
AE,REF 1 βn + AE,REF 1 + AE,M
=
IC,M
AE,M
AE,REF 1 βn


AE,M
AE,REF 1 βn
IREF
IC,M =
AE,REF 1 βn + AE,REF 1 + AE,M AE,REF 1
Using a similar derivation to find IC2 , we have:


AE2
AE,REF 2 βp
IC,M
IC1 = IC2 =
AE,REF 2 βp + AE,REF 2 + AE2 AE,REF 2



AE2
AE,REF 2 βp
AE,M
AE,REF 1 βp
·
IREF
=
AE,REF 1 βp + AE,REF 1 + AE,M
AE,REF 2 βp + AE,REF 2 + AE2 AE,REF 1 AE,REF 2
We want the error term to be between 0.90 and 1.10 so that IC2 is within 10 % of its nominal value.
Since the error term cannot exceed 1 (since we only lose current through the base), we only have to
worry about the lower bound.



AE,REF 1 βn
AE,REF 2 βp
≥ 0.90
AE,REF 1 βn + AE,REF 1 + AE,M
AE,REF 2 βp + AE,REF 2 + AE2
Let’s let the reference transistors QREF 1 and QREF 2 have unit size AE . Then we have:
!
!
βn
βp
> 0.90
A
βp + 1 + AAE2
βn + 1 + AE,M
E
E

We can pick any AE,M and AE2 such that this constraint is satisfied. One valid solution is AE,M = AE ,
AE2 = 3.466AE , and IREF = 0.2885 mA. This gives a nominal value for IC2 of 1 mA with an error
of 10 %. This solution is not unique (for example, another solution would be AE,M = AE2 = AE and
IREF = 1 mA, which gives a nominal current of 1 mA and an error of 5.73 %).

9.68
Av = gm1 ro3 = gm1

1
= 20
λp ID1

1
k ro2
gm1
ro2
=
1 + gm1 ro2

Rin =

=

gm1

1
λn ID1

1 + gm1 λn 1ID1

= 50 Ω
= 19.5 mS

ID1 = 4.88 mA
s
 
W
ID1
gm1 = 2µn Cox
L 1
 
W
= 390
L 1
We need to size the rest of the transistors to ensure they provide the correct bias current to the amplifier
and to ensure they are all in saturation. VG3 will be important in determining how we should bias
VG5 , since in order for M5 to be in saturation, we require VG3 > VG5 − VT Hn , and VG3 is fixed by the
previously calculated value of ID1 .
!
s
2ID1


VG3 = VDD − VSG3 = VDD − |VT Hp | +
µp Cox W
L 3
= 0.363 V

Let’s let IREF = ID5 = 1 mA (which ensures we meet our power constraint, since P = (IREF + ID5 + ID1 ) VDD =
12.4 mW) and VGS,REF = VGS5 = 0.5 V (which ensures M5 operates in saturation). Then we have
 
W
1
2
(VGS,REF − VT H )
IREF = µn Cox
2
L REF
 
 
W
W
360
=
=
L REF
L 5
0.18
(W/L)3
ID3
=
(W/L)4
ID4
 
8.2
W
=
L 4
0.18
(W/L)2
ID2
=
(W/L)REF
IREF
 
W
1756
=
L 2
0.18

10.3 (a) Looking into the collector of Q1 , we see an infinite impedance (assuming IEE is an ideal source).
Thus, the gain from VCC to Vout is 1 .
(b) Looking into the drain of M1 , we see an impedance of ro1 + (1 + gm1 ro1 ) RS . Thus, the gain from
VCC to Vout is
ro1 + (1 + gm1 ro1 ) RS
RD + ro1 + (1 + gm1 ro1 ) RS
(c) Let’s draw the small-signal model.
+
rπ1

gm1 vπ1

vπ1

ro1

vcc

−

vout

vout = −vπ1


vcc − vout
rπ1
vout = gm1 vπ1 +
ro1


vcc − vout
= −gm1 vout +
rπ1
ro1


rπ1
rπ1
= vcc
vout 1 + gm1 rπ1 +
ro1
ro1
vout
rπ1


=
vcc
1 + β + rπ1
r
o1

=

ro1

rπ1
ro1 (1 + β) + rπ1

(d) Let’s draw the small-signal model.
+
vgs1

gm1 vgs1

−

vout
RS

ro1

vcc

vout = −vgs1


vcc − vout
RS
vout = gm1 vgs1 +
ro1


vcc − vout
RS
= −gm1 vout +
ro1


RS
RS
vout 1 + gm1 RS +
= vcc
ro1
ro1
vout
RS


=
vcc
ro1 1 + gm1 RS + RS
ro1

=

RS
ro1 (1 + gm1 RS ) + RS

10.8

VX (t)
VY (t)
2I0 RC
1.8I0 RC

I0 RC
0.8I0 RC

π/ω

−π/ω
−0.2I0 RC
X and Y are not true differential signals, since their common-mode values differ.

t

10.9 (a)
VX = VCC − I1 RC
VY = VCC − (I2 + IT ) RC

VX (t)
VY (t)
VCC

VCC − IT RC
VCC − I0 RC

VCC − (I0 + IT ) RC
VCC − 2I0 RC

VCC − (2I0 + IT ) RC
π/ω

−π/ω

t
(b)
VX = VCC − (I1 − IT ) RC
VY = VCC − (I2 + IT ) RC

VX (t)
VY (t)
VCC + IT RC

VCC − (I0 − IT ) RC
VCC − IT RC
VCC − (2I0 − IT ) RC
VCC − (I0 + IT ) RC

VCC − (2I0 + IT ) RC
−π/ω

π/ω
t

(c)


VX − VY
VX = VCC − I1 +
RC
RP




VY
RC
= VCC − I1 −
RC
VX 1 +
RP
RP


VY
RC
VCC − I1 − R
P
VX =
RC
1+ R
P
VCC RP − (I1 RP − VY ) RC
R + RC

 P
VY − VX
RC
VY = VCC − I2 +
RP




VX
RC
= VCC − I2 −
RC
VY 1 +
RP
RP


VX
VCC − I2 − R
RC
P
VY =
RC
1+ R
P
=

=
VX =
=
VX

1−

2
RC

(RP + RC )2
2

VX

2
(RP + RC ) − RC
RP + RC

VCC RP − (I2 RP − VX ) RC
RP + RC


2 RP −VX )RC
RC
VCC RP − I1 RP − VCC RP −(I
RP +RC
RP + RC

VCC RP − I1 RP RC +

VCC RP RC −I2 RP R2C +VX R2C
RP +RC

RP + RC

!

=

!

= VCC RP − I1 RP RC +

VCC RP − I1 RP RC +

VCC RP RC −I2 RP R2C
RP +RC

RP + RC
2
VCC RP RC − I2 RP RC
RP + RC



2
VX RP2 + 2RP RC = VCC RP (RP + RC ) − I1 RP RC (RP + RC ) + VCC RP RC − I2 RP RC

2
VCC RP (RP + RC ) − I1 RP RC (RP + RC ) + VCC RP RC − I2 RP RC
2
RP + 2RP RC
VCC RP (2RC + RP ) − RP RC [I1 (RP + RC ) + I2 RC ]
=
RP (2RC + RP )

VX =

Substituting I1 and I2 , we have:
VCC RP (2RC + RP ) − RP RC [(I0 + I0 cos (ωt)) (RP + RC ) + (I0 − I0 cos (ωt)) RC ]
RP (2RC + RP )
VCC RP (2RC + RP ) − RP RC [I0 (2RC + RP ) + I0 cos (ωt) RP ]
=
RP (2RC + RP )
RC RP
= VCC − I0 RC + I0 cos (ωt)
2RC + RP

VX =

By symmetry, we can write:
VY = VCC − I0 RC − I0 cos (ωt)

RC RP
2RC + RP

VX (t)
VY (t)
RP
VCC − I0 RC + I0 2RRCC+R
P

VCC − I0 RC

RP
VCC − I0 RC − I0 2RRCC+R
P

π/ω

−π/ω

t
(d)
VX = VCC − I1 RC


VY
RC
VY = VCC − I2 +
RP


RC
VY 1 +
= VCC − I2 RC
RP
VCC − I2 RC
VY =
C
1+ R
RP
=

VCC RP − I2 RC RP
RP + RC

VX (t)
VY (t)
VCC

VCC − I0 RC
P
VCC RPR+R
C

VCC RP −I0 RC RP
RP +RC

VCC − 2I0 RC
VCC RP −2I0 RC RP
RP +RC

−π/ω

π/ω
t

10.11 Note that since the circuit is symmetric and IEE is an ideal source, no matter what value of VCC we
have, the current through Q1 and Q2 must be IEE /2. That means if the supply voltage increases by
some amount ∆V , VX and VY must also increase by the same amount to ensure the current remains
the same.
∆VX = ∆V
∆VY = ∆V
∆ (VX − VY ) = 0
We can say that this circuit rejects supply noise because changes in the supply voltage (i.e., supply
noise) do not show up as changes in the differential output voltage VX − VY .

10.23 If the temperature increases from 27 ◦ C to 100 ◦ C, then VT will increase from 25.87 mV to 32.16 mV.
Will will cause the curves to stretch horizontally, since the differential input will have to be larger in
magnitude in order to drive the current to one side of the differential pair. This stretching is shown in
the following plots.

IC1 , T
IC1 , T
IC2 , T
IC2 , T

= 27 ◦ C
= 100 ◦ C
= 27 ◦ C
= 100 ◦ C

IEE

IEE
2

Vin1 − Vin2

Vout1 , T
Vout1 , T
Vout2 , T
Vout2 , T

= 27 ◦ C
= 100 ◦ C
= 27 ◦ C
= 100 ◦ C

VCC

VCC − IEE RC /2

VCC − IEE RC

Vin1 − Vin2

Vout1 − Vout2 , T = 27 ◦ C
Vout1 − Vout2 , T = 100 ◦ C
IEE RC

Vin1 − Vin2

−IEE RC

10.33 (a) Treating node P as a virtual ground, we can draw the small-signal model to find Gm .
iout
+
vin

rπ

vπ

gm vπ

ro

−

RE

vin − vπ
vπ
+
rπ
RE
vπ = vin − (−iout + gm vπ ) ro

iout = −

vπ (1 + gm ro ) = vin + iout ro
vin + iout ro
vπ =
1 + gm ro
vin + iout ro
vin + iout ro
vin
iout = −
−
+
rπ (1 + gm ro ) RE
RE (1 + gm ro )




ro
1
1
ro
1
= vin
iout 1 +
−
+
−
rπ (1 + gm ro ) RE (1 + gm ro )
RE
rπ (1 + gm ro ) RE (1 + gm ro )




rπ (1 + gm ro ) − RE − rπ
rπ RE (1 + gm ro ) + ro (rπ + RE )
= vin
iout
rπ RE (1 + gm ro )
rπ RE (1 + gm ro )
iout
rπ (1 + gm ro ) − RE − rπ
Gm =
=
vin
rπ RE (1 + gm ro ) + ro (rπ + RE )
Rout = RC k [ro + (1 + gm ro ) (rπ k RE )]

Av = −

rπ (1 + gm ro ) − RE − rπ
{RC k [ro + (1 + gm ro ) (rπ k RE )]}
rπ RE (1 + gm ro ) + ro (rπ + RE )

(b) The result is identical to the result from part (a), except R1 appears in parallel with ro .
Av = −

rπ (1 + gm (ro k R1 )) − RE − rπ
{RC k [(ro k R1 ) + (1 + gm (ro k R1 )) (rπ k RE )]}
rπ RE (1 + gm (ro k R1 )) + (ro k R1 ) (rπ + RE )

10.36
VDD −

ISS RD
> VCM − VT H,n
2
VDD > VCM − VT H,n +
VDD > 1 V

ISS RD
2

10.38 Let JD be the current density of a MOSFET, as defined in the problem statement.
ID
11
=
µn Cox (VGS − VT H )2
W
2
L
s
2ID
=
W
L µn Cox
s
2JD
=
1
µ
L n Cox

JD =
(VGS − VT H )equil

The equilibrium overdrive voltage increases as the square root of the current density.

10.39 Let id1 , id2 , and vP denote the changes in their respective values given a small differential input of vin
(+vin to Vin1 and −vin to Vin2 ).
id1 = gm (vin − vP )
id2 = gm (−vin − vP )
vP = (id1 + id2 ) RSS
= −2gm vP RSS
⇒ vP = 0
Note that we can justify the last step by noting that if vP 6= 0, then we’d have 2gm RSS = −1, which
makes no sense, since all the values on the left side must be positive. Thus, since the voltage at P does
not change with a small differential input, node P acts as a virtual ground.

10.41
P = ISS VDD = 2 mW
ISS = 1 mA
VCM,out = VDD −

ISS RD
= 1.6 V
2

RD = 800 Ω
|Av | = gm RD
s  
W
= 2
µn Cox ID RD
L 1


W
L



1

=5
 
W
=
= 390.625
L 2

Let’s formulate the trade-off between VDD and W/L, let’s assume we’re trying to meet an output
common-mode level of VCM,out . Then we have:
ISS =

P
VDD

ISS RD
2
P RD
= VDD −
2V

 DD
VDD − VCM,out
RD = 2VDD
P

VCM,out = VDD −

|Av | = gm RD
r
W
µn Cox ISS RD
=
L
r



W
VDD − VCM,out
P
=
2VDD
µn Cox
L
VDD
P
To meet a certain gain, W/L and VDD must be adjusted according to the above equation. We can see
that if we decrease VDD , we’d have to increase W/L in order to meet the same gain.

10.55 Let’s draw the half circuit.
vout
Q3

vin

Gm = gm1 RP
2

= gm1

RP
2

Q1

k ro1 k rπ3

k ro1 k rπ3 +
RP
2 k
gm3 R2P

gm3

RP /2

1
gm3

ro1 k rπ3






k ro1 k rπ3


RP
Rout = ro3 + (1 + gm3 ro3 ) rπ3 k
k ro1
2

 


RP
gm3 2 k ro1 k rπ3
RP


Av = −gm1
k
r
r
+
(1
+
g
r
)
r
k
o1
o3
m3 o3
π3
2
1 + gm3 R2P k ro1 k rπ3
1+

10.60 Assume IC =

IEE
2

for all of the transistors (since β ≫ 1).

Av = −gm1 {[ro3 + (1 + gm3 ro3 ) (rπ3 k ro1 )] k [ro5 + (1 + gm5 ro5 ) (rπ5 k ro7 )]}
h


ih


i
VA,n
VA,p
βn VT VA,n
βp VT VA,p
V
+
1
+
V
+
1
+
A,n
A,p
VT
βn VT +VA,n
VT
βp VT +VA,p
1
h


i h


i
=−
V
V
β
V
V
βp VT VA,p
A,n
A,p
n
T
A,n
VT VA,n + 1 +
+ VA,p + 1 +
VT

= −800

VA,n = 2.16 V
VA,p = 1.08 V

βn VT +VA,n

VT

βp VT +VA,p

10.61




1
Av = −gm1 [ro3 + (1 + gm3 ro3 ) (rπ3 k ro1 )] k ro5 + (1 + gm5 ro5 ) rπ5 k
k rπ7 k ro7
gm7
This topology is not a telescopic cascode. The use of NPN transistors for Q7 and Q8 drops the output
resistance of the structure from that of the typical telescopic cascode.

10.73 (a)
VN = VDD − VSG3
s
= VDD −

I

 SS
W
L 3 µp Cox

− |VT Hp |

(b) By symmetry, we know that ID for M3 and M4 is the same, and we also know that their VSG
values are the same. Thus, their VSD values must also be equal, meaning VY = VN .
(c) If VDD changes by ∆V , then both VY and VN will change by ∆V .

10.83
P = VCC IEE = 1 mW
IEE = 0.4 mA
Av = −gm1 (ro1 k ro3 k R1 )
= −100
R1 = R2 = 59.1 kΩ

10.92
P = VCC IEE = 3 mW
IEE = 1.2 mA
Av = gm,n (ro,n k ro,p )
= 200
VA,n = 15.6 V
VA,p = 7.8 V

11.1


Vout
1
(jω) = −gm RD k
Vin
jωCL
gm RD
=−
1 + jωCL RD
gm RD
Vout
(jω) = q
Vin
1 + (ωCL RD )2

gm RD
q
= 0.9gmRD
2
1 + (ω−1 dB CL RD )

ω−1 dB = 4.84 × 108 rad/s
ω−1 dB
f−1 dB =
= 77.1 MHz
2π

11.3 (a)
ω−3 dB = 

1
1
gm2


k rπ2 CL

(b)
ω−3 dB = 

1
rπ2 +RB
1+β



CL

≈

1
1
gm2

+

RB
1+β

(c)
ω−3 dB =

1
(ro1 k ro2 ) CL

(d)
ω−3 dB = 
ro1 k

1
1
gm2


k ro2 CL



CL

Vout
(jω)
Vin

11.4 Since all of these circuits are have one pole, all of the Bode plots will look qualitatively identical, with
some DC gain at low frequencies that rolls off at 20 dB/dec after hitting the pole at ω−3 dB . This is
shown in the following plot:

20 log |Av |
Slope = −20 dB/dec

ω−3 dB

ω

For each circuit, we’ll derive |Av | and ω−3 dB , from which the Bode plot can be constructed as in the
figure.
(a)


|Av | = gm1
ω−3 dB = 

1
gm2

k rπ2



1
1
gm2


k rπ2 CL

(b)
|Av | = gm1
ω−3 dB = 



rπ2 + RB
1+β
1

rπ2 +RB
1+β



CL



≈ gm1

≈



1
gm2

RB
1+β

1
1
gm2

+

RB
1+β

(c)
|Av | = gm1 (ro1 k ro2 )
ω−3 dB =

+

1
(ro1 k ro2 ) CL



CL



(d)


1
|Av | = gm1 ro1 k
k ro2
gm2
ω−3 dB = 

1
ro1 k

1
gm2


k ro2 CL

Assuming the transfer function is of the form

we get the following Bode plot:

Vout
(jω) = 
Vin

Av
1 + j ωωp1

2

Vout
(jω)
Vin

11.5

20 log |Av |

ωp1

ω
Slope = −40 dB/dec

Vout
(jω)
Vin

11.6

20 log |Av |
Slope = −20 dB/dec

Slope = −20 dB/dec

2π × 108

2π × 109 2π × 1010

ω

Vout
(jω)
Vin

11.7 The gain at arbitrarily low frequencies approaches infinity.

Slope = −20 dB/dec

ω

Vout
(jω)
Vin

11.8 The gain at arbitrarily high frequencies approaches infinity.

ω

Slope = +20 dB/dec

11.16 Using Miller’s theorem, we can split the resistor RF as follows:
VCC
RC
Vout
Vin

RB

RF
1+ gm1R

Q1

C

RF
1+gm RC

Av = −gm

RF
1+gm RC
rπ k 1+gRmFRC

rπ k
RB +

!

RF
RC k
1 + gm1RC

!

11.17 Using Miller’s theorem, we can split the resistor RF as follows:
VDD
Vin

RS

M1

RF
gm RL
1− 1+g
mR

Vout

L

RL



Av = 

RF
g

R

m L
1− 1+g
mR

L

RS +

RF
gm RL
1− 1+g
m RL

1−

RF
1+gm RL
gm RL


 
RF
m RL

 gm RL k 1− 1+g
gm RL





RF
1 + gm RL k
1+gm RL




1−

gm RL

11.18 Using Miller’s theorem, we can split the resistor ro as follows:
VCC
RC
ro
1− gm1R

Vout
C

Q1

Vb

RB

Vin

ro
1−gm RC

Av = gm

1
gm

RB +

ro
1−gm RC
rπ k 1−grmo RC

k rπ k
1
gm

k

!

RC k

ro
1−

1
gm RC

!

11.20 Using Miller’s theorem, we can split the capacitor CF as follows (note that the DC gain is Av =

gm ro
1+gm ro ):

VDD
M1

CF 1 −

gm ro
1+gm ro




CF 1 −

1+gm ro
gm ro



Thus, we have
Cin = CF


1−

gm ro
1 + gm ro



As λ → 0, ro → ∞, meaning the gain approaches 1. When this happens, the input capacitance goes
to zero.

11.26 At high frequencies (such as fT ), we can neglect the effects of rπ and ro , since the low impedances of
the capacitors will dominate at high frequencies. Thus, we can draw the following small-signal model
to find fT (for BJTs):
Cµ

Iin

+
vπ
−

Cπ

Iout
gm vπ

Iin = jωvπ (Cπ + Cµ )
Iin
Iπ =
jω (Cπ + Cµ )
Iout = gm vπ − jωCµ vπ

Iout
Iin
Iout
Iin
q
2 + (ω C )2
gm
T µ
ωT (Cπ + Cµ )

= vπ (gm − jωCµ )
Iin
=
(gm − jωCµ )
jω (Cπ + Cµ )
gm − jωCµ
=
jω (Cπ + Cµ )
q
2 + (ωC )2
gm
µ
=
ω (Cπ + Cµ )
=1

2
gm
+ ωT2 Cµ2 = ωT2 Cπ2 + 2Cπ Cµ + Cµ2


2
gm
= ωT2 Cπ2 + 2Cπ Cµ
gm
ωT = p
2
Cπ + 2Cπ Cµ

fT =

gm
p
2
2π Cπ + 2Cπ Cµ




The derivation of fT for a MOSFET is identical to the derivation of fT for a BJT, except we have CGS
instead of Cπ and CGD instead of Cµ . Thus, we have:
fT =

2π

gm
p
2 + 2C
CGS
GS CGD

11.37 Using Miller’s theorem to split Cµ1 , we have:
ωp,in =
ωp,out =

1
(RS k rπ1 ) {Cπ1 + Cµ1 [1 + gm1 (ro1 k ro2 )]}

(ro1

1
h
n
k ro2 ) Cµ2 + CCS1 + CCS2 + Cµ1 1 +



rπ1
g
m1
rπ1 +RS (ro1 k ro2 )
Vout


(s) = − 
s
s
Vin
1 + ωp,out
1 + ωp,in

1
gm1 (ro1 kro2 )

io

11.39 (a)
1
= 3.125 × 1010 rad/s
RS [CGS + CGD (1 + gm RD )]
1
i = 3.846 × 1010 rad/s

h
=
1
RD CDB + CGD 1 + gm RD

ωp,in =
ωp,out

(b)
(CGD s − gm ) RD
Vout
(s) =
VT hev
as2 + bs + 1
a = RS RD (CGS CGD + CDB CGD + CGS CDB ) = 2.8 × 10−22
b = (1 + gm RD ) CGD RS + RS CGS + RD (CGD + CDB ) = 5.7 × 10−11
Setting the denominator equal to zero and solving for s, we have:
√
−b ± b2 − 4a
s=
2a
|ωp1 | = 1.939 × 1010 rad/s
|ωp2 | = 1.842 × 1011 rad/s
We can see substantial differences between the poles calculated with Miller’s approximation and
the poles calculated from the transfer function directly. We can see that Miller’s approximation
does a reasonably good job of approximating the input pole (which corresponds to |ωp1 |). However,
the output pole calculated with Miller’s approximation is off by nearly an order of magnitude when
compared to ωp2 .

11.40 (a) Note that the DC gain is Av = −∞ if we assume VA = ∞.
ωp,in =

1
= 0
(RS k rπ ) [Cπ + Cµ (1 − Av )]

ωp,out = 0
(b)
Vout
(Cµ s − gm ) RL
(s) = lim
RL →∞ as2 + bs + 1
VT hev
a = (RS k rπ ) RL (Cπ Cµ + CCS Cµ + Cπ CCS )
b = (1 + gm RL ) Cµ (RS k rπ ) + (RS k rπ ) Cπ + RL (Cµ + CCS )
lim

RL →∞

Cµ s − gm
(Cµ s − gm ) RL
=
as2 + bs + 1
[(RS k rπ ) (Cπ Cµ + CCS Cµ + Cπ CCS )] s2 + [gm Cµ (RS k rπ ) + Cµ + CCS ] s
Cµ s − gm
=
s {(RS k rπ ) (Cπ Cµ + CCS Cµ + Cπ CCS ) s + [gm Cµ (RS k rπ ) + Cµ + CCS ]}
|ωp1 | = 0
|ωp2 | =

gm Cµ (RS k rπ ) + Cµ + CCS
(RS k rπ ) (Cπ Cµ + CCS Cµ + Cπ CCS )

We can see that the Miller approximation correctly predicts the input pole to be at DC. However,
it incorrectly estimates the output pole to be at DC as well, when in fact it is not, as we can see
from the direct analysis.

11.41
1
= 0
RL →∞ (1 + gm RL ) Cµ (RS k rπ ) + (RS k rπ ) Cπ + RL (Cµ + CCS )
(RS k rπ ) RL (Cπ Cµ + CCS Cµ + Cπ CCS )
|ωp2 | = lim
RL →∞ (1 + gm RL ) Cµ (RS k rπ ) + (RS k rπ ) Cπ + RL (Cµ + CCS )
|ωp1 | = lim

=

(RS k rπ ) (Cπ Cµ + CCS Cµ + Cπ CCS )
gm Cµ (RS k rπ ) + Cµ + CCS

The dominant-pole approximation gives the same results as analyzing the transfer function directly, as
in Problem 40(b).

11.49
ωp1 =

≈

1
io

h
n
1
k rπ2
(RB k rπ1 ) Cπ1 + Cµ1 1 + gm1 gm2
1
n
h
(RB k rπ1 ) Cπ1 + Cµ1 1 +

IC1 = 4IC2 ⇒ gm1 = 4gm2
ωp1 =
ωp2 ≈

ωp3 =

io

1
(RB k rπ1 ) (Cπ1 + 5Cµ1 )
1
gm2

=

gm1
gm2

1
h

CCS1 + CCS3 + Cµ3 + Cπ2 + Cµ1 1 +

gm2
CCS1 + CCS3 + Cµ3 + Cπ2 + 45 Cµ1

gm2
gm1

i

1
RC (CCS2 + Cµ2 )

Miller’s effect is more significant here than in a standard cascode. This is because the gain in the
common-emitter stage is increased to four in this topology, where it is about one in a standard cascode.
This means that the capacitor Cµ1 will be multiplied by a larger factor when using Miller’s theorem.

11.58
ID =

1
2



W
L



2
µn Cox Vov
= 0.5 mA

1

(W/L)1 = (W/L)2 = 250
W1 = W2 = 45 µm
W
µn Cox Vov = 5 mS
gm1 = gm2 =
L
CGD1 = CGD2 = C0 W = 9 fF
2
CGS1 = CGS2 = W LCox = 64.8 fF
3
1
o = 2π × 5 GHz

n
ωp,in =
m1
RG CGS1 + CGD1 1 + ggm2
RG = 384 Ω
1
= 2π × 10 GHz
ωp,out =
RD CGD2
RD = 1.768 kΩ
Av = −gm1 RD = −8.84

12.1 (a)
Y = A1 (X − KA2 Y )
Y (1 + KA1 A2 ) = A1 X
A1
Y
=
X
1 + KA1 A2
(b)
Y = X − KY − A1 (X − KY )
Y (1 + K − A1 K) = X (1 − A1 )
1 − A1
Y
=
X
1 + K (1 − A1 )
(c)
Y = A2 X − A1 (X − KY )
Y (1 − A1 K) = X (A2 − A1 )
Y
A2 − A1
=
X
1 − A1 K
(d)
Y = X − (KY − Y ) − A1 [X − (KY − Y )]
Y = X − KY + Y − A1 X + KA1 Y − A1 Y
Y [A1 (1 − K) + K] = X (1 − A1 )
1 − A1
Y
=
X
A1 (1 − K) + K

12.5 The loop gains calculated in Problem 4 are used.
(a)
AOL = A1
Aloop = KA1



R2
R1 + R2



A1
Y


=
X
2
1 + KA1 R1R+R
2
(b)
AOL = −A1
Aloop = gm3 RD A1



R2
R1 + R2



Y
A1


= −
X
2
1 + gm3 RD A1 R1R+R
2
(c)
AOL = −A1
Aloop = gm3 RD A1
Y
A1
= −
X
1 + gm3 RD A1
(d)



gm1 R2
AOL = A1
1 + gm1 R2


gm1 R2
Aloop = A1
1 + gm1 R2


gm1 R2
A
1
1+gm1 R2
Y


=
gm1 R2
X
1 + A1 1+g
m1 R2

12.8
AOL = −gm ro
r
1
W
= − 2 µn Cox ID
L
λID
r
1
W
2 µn Cox
=− √
L
λ ID
AOL
Vout
=
Vin
1 + KAOL
We want to look at the maximum and minimum deviations that VVout
will have from the base value
in
given the variations in λ and µn Cox . First, let’s consider what happens when λ decreases
by 20 % and
√
1.1
µn Cox increases by 10 %. This causes AOL to increase in magnitude by a factor of 0.8 = 1.311. We
want VVout
to change by less than 5 % given this deviation in AOL .
in
AOL
1.311AOL
< 1.05
1 + 1.311KAOL
1 + KAOL
KAOL > 3.982
Next, let’s consider what happens when λ increases
by 20 % and µn Cox decreases by 10 %. This causes
√
0.9
AOL to decrease in magnitude by a factor of 1.2
= 0.7906. We want VVout
to change by less than 5 %
in
given this deviation in AOL .
AOL
0.7906AOL
< 0.95
1 + 0.7906KAOL
1 + KAOL
KAOL > 4.033
Thus, to satisfy the constraints on both the maximum and minimum deviations, we require KAOL >
4.033 .

12.10

AOL = −gm ro k

1
sCL



gm ro
1 + sro CL
AOL
=
1 + KAOL
gm ro
− 1+sr
o CL
=
gm ro
1 − K 1+sr
o CL
gm ro
=−
1 + sro CL − Kgm ro
=−

Vout
Vin

Setting the denominator equal to zero and solving for s gives us the bandwidth B.
B=

Kgm ro − 1
ro CL

K=

1 + Bro CL
gm ro

12.11 (a)

•
•
•
•

Feedforward system: M1 and RD (which act as a common-gate amplifier)
Sense mechanism: C1 and C2 (which act as a capacitive divider)
Feedback network: C1 and C2
Comparison mechanism: M1 (which amplifies the difference between the fed back signal and
the input)

(b)
AOL = gm RD

Aloop = gm RD

C1
C1 + C2



gm RD
vout


=
vin
1
1 + gm RD C1C+C
2
(c)
Rin,open =

Rin,closed =

1
gm
1 + gm RD



gm

C1
C1 +C2



Rout,open = RD
Rout,closed =

RD


1
1 + gm RD C1C+C
2

12.15 (a)
1/gm2

vin

vout

+
−gm1 RD vin
−

(b)
1/gm2

iin
1
gm1

vout

+
−RD iin
−

(c)
vin
gm vin

(d)
iin
1
gm1

−iin

ro

12.18 (a)

• Sense mechanism: Voltage at the source of M3
• Return mechanism: Voltage at the gate of M2

(b)

• Sense mechanism: Voltage at the source of M3
• Return mechanism: Voltage at the gate of M2

(c)

• Sense mechanism: Current Ωowing through R1
• Return mechanism: Voltage at the gate of M2

(d)

• Sense mechanism: Current Ωowing through R1
• Return mechanism: Voltage at the gate of M2

(e)

• Sense mechanism: Voltage divider formed by R1 and R2
• Return mechanism: Voltage at the gate of M2

(f)

• Sense mechanism: Voltage at the source of M3
• Return mechanism: Voltage at the gate of M2

12.20 (a)

• Sense mechanism: Voltage at the gate of M2
• Return mechanism: Current through M2

(b)

• Sense mechanism: Voltage at the gate of M2
• Return mechanism: Current through M2

(c)

• Sense mechanism: Voltage at the source of M2
• Return mechanism: Current through M2

(d)

• Sense mechanism: Voltage at the gate of M2
• Return mechanism: Current through M2

21.

22.

12.23 If Iin increases, then the voltage at the gate of M1 will increase, meaning ID1 will increase. This will
cause the drain voltage of M1 to decrease, meaning ID2 will decrease and Vout will increase. This will
cause the voltage at the gate of M1 to decrease, which counters the original increase, meaning there is
negative feedback .

12.24
Fig. 12.83 (a) Vin ↑, VS1 ↑, VG3 ↑, Vout ↓, VG3 ↓⇒ negative feedback .
(b) Vin ↑, VS1 ↑, VG3 ↑, Vout ↓, VG3 ↑⇒ positive feedback .
(c) Same as (b), positive feedback .
(d) Same as (a), negative feedback .
(e) Vin ↑, VS1 ↑, Vout ↑, VG2 ↑, Vout ↓⇒ negative feedback .
(f) Vin ↑, VS1 ↑, VG3 ↑, Vout ↓, VG3 ↑⇒ positive feedback .
Fig. 12.84 (a) Vin ↑, VG2 ↑, Vout ↑, VG2 ↓⇒ negative feedback .
(b) Vin ↑, VG2 ↑, Vout ↓, VG2 ↑⇒ positive feedback .
(c) Same as (b), positive feedback .
(d) Vin ↑, VG2 ↑, Vout ↓, VG2 ↑⇒ positive feedback .
Fig. 12.85 (a) Iin ↑, VG1 ↑ (consider Iin Ωows through an equivalent small-signal resistance of 1/gm2 at the
gate of M1 ), Vout ↓, VG1 ↓⇒ negative feedback .
(b) Iin ↑, VG1 ↑, Vout ↓, VG1 ↑⇒ positive feedback .
(c) Iin ↑, VG1 ↑, Vout ↓, VG1 ↓⇒ negative feedback .
(d) Iin ↑, VS1 ↑, Vout ↑, VS1 ↓⇒ negative feedback .
Fig. 12.86 (a) Iin ↑, VBE1 ↑, Vout ↓, VBE1 ↓⇒ negative feedback .
(b) Iin ↑, VG3 ↑, Vout ↑, VG3 ↑⇒ positive feedback .

25.

26.

12.27
AOL = gm1 (ro2 k ro4 )



gm5 ro5
1 + gm5 ro5



K = 1 (since the output is fed back directly to the inverting input)


gm5 ro5
gm1 (ro2 k ro4 ) 1+g
r
vout
m5 o5


=
gm5 ro5
vin
1 + gm1 (ro2 k ro4 ) 1+gm5 ro5
Rout,open =
Rout,closed =

1
k ro5
gm5

1
gm5

1 + gm1 (ro2

k ro5


gm5 ro5
k ro4 ) 1+g
m5 ro5

Let’s recall the gain and output impedance of a simple source follower, as shown in the following
diagram.
VCC
vin

M1
vout
Ibias

gm1 ro1
1 + gm1 ro1
1
=
k ro1
gm1

Av =
Rout

We can see that the gain of the circuit in Fig. 12.90 is the gain of a simple source follower multiplied
by a factor of
gm1 (ro2 k ro4 )


gm5 ro5
1 + gm1 (ro2 k ro4 ) 1+g
m5 ro5

This factor is less than 1, which means that the gain is reduced. However, we do get an improvement
in output resistance, which is reduced by a factor of


gm5 ro5
1 + gm1 (ro2 k ro4 )
1 + gm5 ro5

12.28 (a) Vin ↑, VG5 ↑, Vout ↓, VG5 ↑⇒ positive feedback.
(b)
Aloop = −gm1 gm5 (ro2 k ro4 ) ro5
Since the loop gain is negative, the feedback is positive.

12.29
AOL = gm1 gm5 (ro1 k ro3 ) ro5
K=1
vout
gm1 gm5 (ro1 k ro3 ) ro5
=
vin
1 + gm1 gm5 (ro1 k ro3 ) ro5
Rin,open = Rin,closed = ∞
Rout,open = ro5
Rout,closed =

ro5
1 + gm1 gm5 (ro1 k ro3 ) ro5

Like the circuit in Problem 12.27, the closed loop gain is approximately (but slightly less than) 1.
Looking at the equations, the closed loop gain of this circuit will typically be larger than the closed
loop gain of the circuit in Problem 12.27.
The output impedance of this circuit is not quite as small as the output impedance of the circuit in
Problem 12.27. Despite the loop gain being larger, the open loop output impedance is significantly
higher than that of Problem 12.27, so that overall, the output impedance is slightly higher in this
circuit.

12.30 (a) Iin ↑, VG2 ↑, Vout ↑, VS1 ↑, VG2 ↑⇒ positive feedback.
(b)
Aloop



1
gm1 gm2 RD RF + gm1

i
= −h
1
1 + gm2 RF + gm1
(1 + gm1 RF )

Since the loop gain is negative, the feedback is positive.

31.

32.

33.

34.

35.

36.

37.

38.

39.

12.40 (a)
AOL = gm2 (RC k rπ2 )
gm1 gm2 (RF k RM ) (RC k rπ2 )
1 + gm1 RF
gm2 (RF k RM ) (RC k rπ2 )
=
1
gm1 + RF

Aloop =

RF k RM
(since RF is very large)
RF

≈ gm2 (RC k rπ2 )

gm2 (RC k rπ2 )
iout
=
M
iin
1 + gm2 (RC k rπ2 ) RFRkR
F
Rin,open =
Rin,closed =

1
k rπ1
gm1
1
gm2

k rπ1

M
1 + gm2 (RC k rπ2 ) RFRkR
F

Rout,open = Rout,closed = ∞ (since VA = ∞)
(b)
AOL = −gm2 RM (RC k rπ2 )
Aloop ≈ gm2 (RC k rπ2 )

RF k RM
(same as (a))
RF

−gm2 RM (RC k rπ2 )
vout
= −
M
iin
1 + gm2 (RC k rπ2 ) RFRkR
F
Rin,open =
Rin,closed =

1
k rπ1
gm1
1
gm1

k rπ1

M
1 + gm2 (RC k rπ2 ) RFRkR
F

Rout,open = RM k RF
Rout,closed =

RM k RF
M
1 + gm2 (RC k rπ2 ) RFRkR
F

41.

12.42 We can break the feedback network as shown here:
VCC
RC
vout
R1

Q1
vin

R1

R2
R2

AOL =
K=

RC k (R1 + R2 )
1
gm1

+

R1 kR2
1+β

R2
R1 + R2

vout
=
vin
1+

RC k(R1 +R2 )
R1 kR2
1
+ 1+β
g
m1

RC k(R1 +R2 )
R2
R1 +R2 1 + R1 kR2
gm1

1+β

rπ1 + R1 k R2
1+β
!


R2
RC k (R1 + R2 )
rπ1 + R1 k R2
1+
=
1
1+β
R1 + R2
+ R1 kR2

Rin,open =
Rin,closed

gm1

Rout,open = RC k (R1 + R2 )
Rout,closed =

RC k (R1 + R2 )
1+

RC k(R1 +R2 )
R2
R1 +R2 1 + R1 kR2
gm1

1+β

1+β

12.43 We can break the feedback network as shown here:
VCC

vout
R1

Q1
vin

R1

R2
R2

AOL =

R1 + R2
1
gm1

K=

+

R1 kR2
1+β

R2
R1 + R2
R1 +R2

R1 kR2
1
vout
gm1 + 1+β
=
vin
1 + 1 RR2 1 kR2
gm1

+

1+β

rπ1 + R1 k R2
1+β


rπ1 + R1 k R2
1+
=
1+β

Rin,open =
Rin,closed

Rout,open = R1 + R2
Rout,closed =

R1 + R2
1 + 1 RR2 1 kR2
gm1

+

1+β

R2
1
gm1

+

R1 kR2
1+β

!

12.44 We can break the feedback network as shown here:
VDD

RD1
vout
M2
vin

R1

M1
R2

R1

R2

gm1 gm2 RD1 (R1 + R2 )
1 + gm1 (R1 k R2 )
R2
K=
R1 + R2

AOL =

gm1 gm2 RD1 (R1 +R2 )

vout
1+gm1 (R1 kR2 )
=
gm1 gm2 RD1 R2
vin
1 + 1+g
m1 (R1 kR2 )
Rin,open = Rin,closed = ∞
Rout,open = R1 + R2
Rout,closed =

1+

R1 + R2
gm1 gm2 RD1 R2
1+gm1 (R1 kR2 )

12.45 We can break the feedback network as shown here:
VDD

RD1
vout
Q2
vin

R1

Q1
RP

R1

R2

R2

gm1 gm2 [RD1 k (rπ2 + (1 + β) RP )] (R1 + R2 )
[1 + gm1 (R1 k R2 )] (1 + gm2 RP )
R2
K=
R1 + R2

AOL =

gm1 gm2 [RD1 k(rπ2 +(1+β)RP )](R1 +R2 )
[1+gm1 (R1 kR2 )](1+gm2 RP )
gm1 gm2 [RD1 k(rπ2 +(1+β)RP )]R2
[1+gm1 (R1 kR2 )](1+gm2 RP )

vout
=
vin
1+

Rin,open = rπ1 + (1 + β) (R1 k R2 )
Rin,closed



gm1 gm2 [RD1 k (rπ2 + (1 + β) RP )] R2
= {rπ1 + (1 + β) (R1 k R2 )} 1 +
[1 + gm1 (R1 k R2 )] (1 + gm2 RP )

Rout,open = R1 + R2
Rout,closed =

1+

R1 + R2
gm1 gm2 [RD1 k(rπ2 +(1+β)RP )]R2
[1+gm1 (R1 kR2 )](1+gm2 RP )

12.46 We can break the feedback network as shown here:
VCC

vout
vin

Q1

R1

Q2

R1

R2
R2

AOL =
K=

vout
=
vin

Rin,open

Rin,closed

rπ2 +R1 kR2
1+β2
rπ2 +R1 kR2
1
+
gm1
1+β2

R2
R1 + R2
 rπ2 +R1 kR2
1+

!

1
gm2



1+β2
rπ2 +R1 kR2
1
gm1 +
1+β2
rπ2 +R1 kR2
1+β2
rπ2 +R1 kR2
1
gm1 +
1+β2



R1 + R2

1

gm2


+

rπ2 +R1 kR2
1+β2

R1 +R2
r
+R kR
+ π21+β1 2

1
gm2

2



R2
+

rπ2 +R1 kR2
1+β2

!




rπ2 + R1 k R2
= rπ1 + (1 + β1 )
1 + β2
 "


rπ2 + R1 k R2
= rπ1 + (1 + β1 )
1+
1 + β2


Rout,open = R1 + R2
Rout,closed =
1+



R1 + R2


rπ2 +R1 kR2
1+β2
rπ2 +R1 kR2
1
+
gm1
1+β2

R2
1
gm2

+

rπ2 +R1 kR2
1+β2



rπ2 +R1 kR2
1+β2
rπ2 +R1 kR2
1
gm1 +
1+β2

!

R2
1
gm2

+

rπ2 +R1 kR2
1+β2

!#

12.47 We can break the feedback network as shown here:
VDD

RD1
vout
M2
M1

R1

Vb
RF

iin

R2
R2

RF

R1

AOL = −gm2 [ro2 k (R1 + R2 k RF )] RD1

RF + R1 k R2
gm1 + RF + R1 k R2
1

To find the feedback factor K, we can use the following diagram:
R1
vx +
−

R2

ix

RF

K=

ix
R2
R2 k RF
=−
=−
vx
(R1 + R2 k RF ) (R2 + RF )
RF (R1 + R2 k RF )

vout
= −
iin

Rin,open =
Rin,closed =

gm2 [ro2 k (R1 + R2 k RF )] RD1

1 + gm2 [ro2 k (R1 + R2 k RF )] RD1

RF +R1 kR2
+RF +R1 kR2

1
gm1

RF +R1 kR2
1
+RF +R1 kR2
g
m1

1
k (RF + R1 k R2 )
gm1

n

R2 kRF
RF (R1 +R2 kRF )

1
gm1 k (RF + R1 k R2 )

n
o
R2 kRF
F +R1 kR2
1 + gm2 [ro2 k (R1 + R2 k RF )] RD1 1 R+R
RF (R1 +R2 kRF )
+R kR
gm1

F

1

2

Rout,open = ro2 k (R1 + R2 k RF )
Rout,closed =

ro2 k (R1 + R2 k RF )
n

o
R2 kRF
F +R1 kR2
1 + gm2 [ro2 k (R1 + R2 k RF )] RD1 1 R+R
RF (R1 +R2 kRF )
+R kR
gm1

F

1

2

o

48.

49.

50.

51.

52.

53.

12.54 We can break the feedback network as shown here:
VCC
Q2

iout
Q1

Vb

iin
RF

AOL = −β2
K = −1 (by inspection)
β2
iout
= −
iin
1 + β2
Rin,open =
Rin,closed =

1
gm1
1
gm1

k rπ1
k rπ1

1 + β2

Rout,open = Rout,closed = ∞ (since VA = ∞)

12.55 We can break the feedback network as shown here:
VCC
RC
Q2
Q1
iout
iin

RF

We can find AOL = iiout
by using current dividers to determine how much of iin goes to iout . Let’s
in
assume the device has some small-signal resistance RL .
RC
RC + rπ2 + (1 + β2 ) (RL + RF )
K = −1 (by inspection)

AOL = −β1 β2

RC
β1 β2 RC +rπ2 +(1+β
iout
2 )(RL +RF )
= −
RC
iin
1 + β1 β2 RC +rπ2 +(1+β
2 )(RL +RF )

Rin,open = rπ1
Rin,closed =

RC
1 + β1 β2 RC +rπ2 +(1+β
2 )(RL +RF )

rπ2 + RC
+ RF
1 + β2
RC
1
+
+ RF
≈
gm2
1 + β2



RC
rπ2 + RC
1 + β1 β2
+ RF
=
1 + β2
RC + rπ2 + (1 + β2 ) (RL + RF )

Rout,open =

Rout,closed

rπ1

12.56 (a) We can break the feedback network as shown here:
VDD
I1
M2

M2

vout

M1
iin

AOL = −gm1 ro1



1
gm2

k ro2



To find the feedback factor K, we can use the following diagram:
VDD
M2

vx

ix

K=

vx
= −gm2
ix



1
k ro2
gm1 ro1 gm2
vout


= −
1
iin
1 + gm1 gm2 ro1 gm2
k ro2
Rin,open =
Rin,closed =

1
k ro2
gm2

1
gm2

1 + gm1 gm2 ro1

Rout,open = ro1
Rout,closed =

k ro2


1
gm2

k ro2



1

k ro2



ro1
1 + gm1 gm2 ro1



gm2

(b) We can break the feedback network as shown here:
VDD
I1
vout
M2

M2

M1
iin

AOL



1
= −gm1 ro2 ro1 k
k ro2
gm2

To find the feedback factor K, we can use the following diagram:
vx
M2

ix

K=
vout
iin

vx
= −gm2
ix



1
gm1 ro2 ro1 k gm2
k ro2


= −
1
k ro2
1 + gm1 gm2 ro2 ro1 k gm2

Rin,open = ro2
Rin,closed =

ro2

1 + gm1 gm2 ro2 ro1 k

Rout,open = ro1 k
Rout,closed =

1

gm2

1
gm2

k ro2



k ro2



k ro2
ro1 k

1
gm2

k ro2



1 + gm1 gm2 ro2 ro1 k

1
gm2

(c) We can break the feedback network as shown here:
VDD
I1
M2

M2

vout

M1

Vb

iin

AOL = gm1 ro1



1
gm1

k ro2



To find the feedback factor K, we can use the following diagram:

VDD
M2

vx

ix

K=

ix
= gm2
vx
gm1 ro1



1

k ro2



gm1
vout


=
1
iin
k ro2
1 + gm1 gm2 ro1 gm1

Rin,open =
Rin,closed =

1

gm1

k ro2
1
gm1

k ro2


1 + gm1 gm2 ro1

1

gm1

Rout,open = ro1 + (1 + gm1 ro1 ) ro2
Rout,closed =

k ro2



ro1 + (1 + gm1 ro1 ) ro2


1
k ro2
1 + gm1 gm2 ro1 gm1

57.

58.

12.59 Let’s draw the small-signal model and find

vout
vin (s).

CF

RG

vout

+
vin

vgs

gm vgs

RD

−

vin − vgs
= (vgs − vout ) sCF
RG
vout
(vgs − vout ) sCF = gm vgs +
ro1


1
+ sCF
vgs (sCF − gm ) = vout
ro1
1 + sCF ro1
vgs =
ro1 (sCF − gm )


vin
1
= vgs
+ sCF − vout sCF
RG
RG




1 + sCF ro1
1
vin
= vout
+ sCF − sCF
RG
ro1 (sCF − gm )
RG



1 + sCF ro1
vin = vout
(1 + sCF RG ) − sCF RG
ro1 (sCF − gm )


(1 + sCF ro1 ) (1 + sCF RG ) − sCF RG ro1 (sCF − gm )
vin = vout
ro1 (sCF − gm )
ro1 (sCF − gm )
vout
(s) =
vin
(1 + sCF ro1 ) (1 + sCF RG ) − sCF RG ro1 (sCF − gm )
From the transfer function, we can see that we’ll have one zero and two poles (since the numerator is
of degree 1 and the denominator is of degree 2).

0◦
−45◦
−90◦

Slope = −20 dB/dec

−135◦
−180◦
−225◦
−270◦

ωp1

ωp2

ω

ωz

H(jω) (Dotted)

Slope = −40 dB/dec

6

|H(jω)| (Solid)

Slope = −20 dB/dec

60.

61.

62.

12.63 We’ll drop the negative sign in H(s) as done in Example 12.38.
(gm RD )3
H(s) = 
3
1 + ωsp
 
ω
−1
∠H(jω) = −3 tan
ωp


ωP X
−1
= −180
−3 tan
ωp
√
ωP X = 3ωp
3

(gm RD )
<1
 3

ωP X
1 + ωp

|KH(jωP X )| = 0.1 r
gm RD <

√
3
80 = 4.31

64.

12.65
H(s) =

A0
1 + ωs0

|KH (ωGX )| = r

A0


ωGX
ω0

2 = 1

1+
q
ωGX = ω0 A20 − 1


ωGX
∠H(jωGX ) = − tan−1
ω0
!
p
ω0 A20 − 1
−1
= − tan
ω0
q

−1
2
A0 − 1
= − tan

Phase Margin = ∠H(jωGX ) + 180◦
q

◦
−1
2
A0 − 1
= 180 − tan
The phase margin can be anything from 90◦ to 180◦ , depending on the value of A0 (smaller A0 means
larger phase margin).

12.66
H(s) =

A0
1 + ωs0

|KH (ωGX )| = 0.5 r

A0


ωGX
ω0

2 = 1

1+
s 
2
A0
ωGX = ω0
−1
2


ωGX
∠H(jωGX ) = − tan−1
ω0

 q


A0 2
−
1
ω
0
2

= − tan−1 
ω0
s

 2
A
0
= − tan−1 
− 1
2

Phase Margin = ∠H(jωGX ) + 180◦
s

= 180◦ − tan−1 

A0
2

2



− 1

The phase margin can be anything from 90◦ to 180◦ , depending on the value of A0 (smaller A0 means
larger phase margin).

67.

12.68 With a factor of K = 0.5, the magnitude Bode plot of KH will simply be the magnitude plot of H
shifted down by 6 dB (since 20 log 0.5 = −6 dB). Since the slope of the magnitude plot between ωp1
6
and ωp2 is −20 dB/dec, this means that ωGX will be shifted left by 20
= 0.3 decades, or a factor of
0.3
10 = 2.
′
′
Thus, the new value of ωGX , which we’ll call ωGX
, is ωGX
=

ωGX
2

=

ωp2
2 .

Now, we need to find ∠H(jωGX ).





ω
ω
−1
∠H(jω) = − tan
− tan
ωp1
ωp2
 ω 
p2
∠H(jωGX ) = ∠H j
2



ωp2
ωGX
−1
−1
= − tan
− tan
2ωp1
2ωp2
−1

= −90◦ − tan−1 (0.5)
= −116◦

Phase Margin = 180◦ + ∠H(jωGX )
= 180◦ − 116◦
= 63◦

69.

0◦

0 dB

−45◦
−90◦
−135◦
Slope = −40 dB/dec
−180◦
−225◦
−270◦
1
RD (C1 +CC )

ω

1
RD C1

H(jω) (Dotted)

Slope = −20 dB/dec

6

|H(jω)| (Solid)

12.70 The compensation capacitor allowsus to push the pole associated with that node to a lower frequency
(while the other poles do not change). This will cause the gain to start dropping sooner, so that ωGX
decreases. By adjusting CC properly, we can reduce ωGX enough so that the phase is at −135◦ at
ωGX . This results in the following Bode plots:

12.71
AOL ≈ gm1 (ro2 k ro4 )


2
2
= gm1
k
λn ISS λp ISS

= 50
gm1 = 3.75 mS
R2
R2
=
K=
R1 + R2
10 (ro2 k ro4 )
vout
gm1 (ro2 k ro4 )
=
R2
vin
1 + gm1 10(ro2
kro4 ) (ro2 k ro4 )
=4
R2 = 30.667 kΩ
R1 = 102.667 kΩ

72.

12.73
AOL = −gm2 RD1 RD2 = −10 kΩ
1
K=−
RF
vout
gm2 RD1 RD2
=−
RD2
iin
1 + gm2 RRD1
F
=−

10 kΩ
1 + 10RkΩ
F

= −1 kΩ
RF = 1.111 kΩ
1
Rin,open =
gm1

−1
1
gm2 RD1 RD2
Rin,closed =
1+
gm1
RF
−1

10 kΩ
1
1+
=
gm1
1.111 kΩ
= 50 Ω

gm1 = 2 mS
Rout,open = RD2
Rout,closed =
=

1+

RD2
gm2 RD1 RD2
RF
RD2

1+

10 kΩ
1.111 kΩ

= 200 Ω
RD2 = 2 kΩ
AOL
gm2 =
RD1 RD2
= 5 mS

74.

75.

12.76 See Problem 44 for derivations of the following expressions.
AOL =

gm1 gm2 RD1 (R1 + R2 )
= 20
1 + gm1 (R1 k R2 )
gm1 gm2 RD1 (R1 +R2 )

vout
1+gm1 (R1 kR2 )
=
gm1 gm2 RD1 R2
vin
1 + 1+g
m1 (R1 kR2 )
=
1 + 20

20


R2
R1 +R2



=4
R2
= 0.2
R1 + R2
Rout,open = R1 + R2 = 2 kΩ
R2 = 400 Ω
R1 = 1.6 kΩ

Lacking any additional constraints, we can pick any gm1 , gm2 , and RD1 so that AOL = 20. Let’s pick
gm1 = gm2 = 2 mS . This gives us RD1 = 4.1 kΩ .
If we are also required to minimize the power consumption of the amplifier, we need to minimize the
current consumption of each stage. This requires minimizing gm1 and gm2 and maximizing RD1 while
keeping all transistors in saturation.

12.77 See Problem 46 for derivations of the following expressions.


1
2
gm1 gm2 gm2
+ Rβ12kR
+1

 (R1 + R2 ) = 2
AOL =
R1 kR2
1
1 + gm1 gm2 + β2 +1
gm1 = gm2 =

1
ISS
=
S
2VT
52

R2
R1 + R2
R1 + R2
=
1 + KAOL
R1 + R2
=
2
1 + R12R
+R2

K=
Rout,closed

2

=

(R1 + R2 )
1 + 3R2

Looking at this expression for Rout,closed , we can see that it will be minimized for very small values of
R1 . This will force R2 to be larger in order to meet the required AOL , but since Rout depends more
strongly on R1 than R2 , we should focus on minimizing R1 .
In fact, we can actually set R1 = 0 . We can then solve the AOL equation to find R2 = 208 Ω , which
means Rout = 69.33 Ω.

12.78 See Problem 50 for derivations of the following expressions. Assume β = 100.
gm1 gm2 (RF k rπ1 ) RF {RC k [rπ2 + (1 + β) RF ]}
1 + gm2 RF
AOL
=
= −1 kΩ
1 − AROL
F

AOL = −
vout
iin

RF k rπ1
= 50 Ω
1 − AROL
F
1
Ω
= gm2 =
26
β
= rπ2 =
= 2.6 kΩ
gm

Rin =
gm1
rπ1

= −1 kΩ and Rin = 50 Ω) and two unknowns (RF and AOL ). Solving,
We have two equations ( viout
in
we get:
RF = 1.071 kΩ
AOL = 15167
RC = 535.2 Ω

Razavi 1e – Fundamentals of Microelectronics
CHAPTER 16 SOLUTIONS MANUAL

***For Chapter 16 solutions, please refer to
Chapter 7 as the questions are identical in each
chapter.



---

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