BoylestadnashelIM Solution Manual Electronic Devices And Circuit Theory 10th Edition

Instructor’s Resource Manual

to accompany

Electronic Devices and

Circuit Theory

Tenth Edition

Robert L. Boylestad

Louis Nashelsky

Upper Saddle River, New Jersey

Columbus, Ohio

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Instructors of classes using Boylestad/Nashelsky, Electronic Devices and Circuit Theory, 10th edition, may

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10 9 8 7 6 5 4 3 2 1

ISBN-13: 978-0-13-503865-9

ISBN-10: 0-13-503865-0

iii

Contents

Solutions to Problems in Text 1

Solutions for Laboratory Manual 185

1

Chapter 1

1. Copper has 20 orbiting electrons with only one electron in the outermost shell. The fact that

the outermost shell with its 29th electron is incomplete (subshell can contain 2 electrons) and

distant from the nucleus reveals that this electron is loosely bound to its parent atom. The

application of an external electric field of the correct polarity can easily draw this loosely

bound electron from its atomic structure for conduction.

Both intrinsic silicon and germanium have complete outer shells due to the sharing (covalent

bonding) of electrons between atoms. Electrons that are part of a complete shell structure

require increased levels of applied attractive forces to be removed from their parent atom.

2. Intrinsic material: an intrinsic semiconductor is one that has been refined to be as pure as

physically possible. That is, one with the fewest possible number of impurities.

Negative temperature coefficient: materials with negative temperature coefficients have

decreasing resistance levels as the temperature increases.

Covalent bonding: covalent bonding is the sharing of electrons between neighboring atoms to

form complete outermost shells and a more stable lattice structure.

3. −

4. W = QV = (6 C)(3 V) = 18 J

5. 48 eV = 48(1.6 × 10−19 J) = 76.8 × 10−19 J

Q = W

V =

19

76.8 10 J

12 V

−

× = 6.40 × 10−19 C

6.4 × 10−19 C is the charge associated with 4 electrons.

6. GaP Gallium Phosphide Eg = 2.24 eV

ZnS Zinc Sulfide Eg = 3.67 eV

7. An

n-type semiconductor material has an excess of electrons for conduction established by

doping an intrinsic material with donor atoms having more valence electrons than needed to

establish the covalent bonding. The majority carrier is the electron while the minority carrier

is the hole.

A

p-type semiconductor material is formed by doping an intrinsic material with acceptor

atoms having an insufficient number of electrons in the valence shell to complete the covalent

bonding thereby creating a hole in the covalent structure. The majority carrier is the hole

while the minority carrier is the electron.

8. A donor atom has five electrons in its outermost valence shell while an acceptor atom has

only 3 electrons in the valence shell.

9. Majority carriers are those carriers of a material that far exceed the number of any other

carriers in the material.

Minority carriers are those carriers of a material that are less in number than any other carrier

of the material.

2

10. Same basic appearance as Fig. 1.7 since arsenic also has 5 valence electrons (pentavalent).

11. Same basic appearance as Fig. 1.9 since boron also has 3 valence electrons (trivalent).

12. −

13. −

14. For forward bias, the positive potential is applied to the p-type material and the negative

potential to the n-type material.

15. TK = 20 + 273 = 293

k = 11,600/n = 11,600/2 (low value of VD) = 5800

I

D = Is1

D

K

kV

T

e

⎛⎞

−

⎜⎟

⎝⎠

= 50 × 10−9

(5800)(0.6)

293 1e

⎛⎞

−

⎜⎟

⎝⎠

= 50 × 10−9 (e11.877 − 1) = 7.197 mA

16. k = 11,600/n = 11,600/2 = 5800 (n = 2 for VD = 0.6 V)

T

K = TC + 273 = 100 + 273 = 373

(5800)(0.6 V)

/9.33

373

K

kV T

ee e== = 11.27 × 103

I = /

(1)

K

kV T

s

Ie − = 5

μ

A(11.27 × 103 − 1) = 56.35 mA

17. (a)

TK = 20 + 273 = 293

k = 11,600/n = 11,600/2 = 5800

I

D = Is1

D

K

kV

T

e

⎛⎞

−

⎜⎟

⎝⎠

= 0.1

μ

A

(5800)( 10 V)

293 1e

−

⎛⎞

−

⎜⎟

⎝⎠

= 0.1 × 10−6(e−197.95 − 1) = 0.1 × 10−6(1.07 × 10−86 − 1)

≅ 0.1 × 10−6 0.1

μ

A

ID = Is = 0.1

μ

A

(b) The result is expected since the diode current under reverse-bias conditions should equal

the saturation value.

18. (a)

x y = ex

0 1

1 2.7182

2 7.389

3 20.086

4 54.6

5 148.4

(b)

y = e0 = 1

(c) For V = 0 V, e0 = 1 and I = Is(1 − 1) = 0 mA

3

19. T = 20°C: Is = 0.1

μ

A

T = 30°C: Is = 2(0.1

μ

A) = 0.2

μ

A (Doubles every 10°C rise in temperature)

T = 40°C: Is = 2(0.2

μ

A) = 0.4

μ

A

T = 50°C: Is = 2(0.4

μ

A) = 0.8

μ

A

T = 60°C: Is = 2(0.8

μ

A) = 1.6

μ

A

1.6

μ

A: 0.1

μ

A ⇒ 16:1 increase due to rise in temperature of 40°C.

20. For most applications the silicon diode is the device of choice due to its higher temperature

capability. Ge typically has a working limit of about 85 degrees centigrade while Si can be

used at temperatures approaching 200 degrees centigrade. Silicon diodes also have a higher

current handling capability. Germanium diodes are the better device for some RF small signal

applications, where the smaller threshold voltage may prove advantageous.

21. From 1.19:

−75°C 25°C 125°C

VF

@ 10 mA

Is

1.1 V

0.01 pA

0.85 V

1 pA

0.6 V

1.05

μ

A

VF decreased with increase in temperature

1.1 V: 0.6 V ≅ 1.83:1

Is increased with increase in temperature

1.05

μ

A: 0.01 pA = 105 × 103:1

22. An “ideal” device or system is one that has the characteristics we would prefer to have when

using a device or system in a practical application. Usually, however, technology only

permits a close replica of the desired characteristics. The “ideal” characteristics provide an

excellent basis for comparison with the actual device characteristics permitting an estimate of

how well the device or system will perform. On occasion, the “ideal” device or system can be

assumed to obtain a good estimate of the overall response of the design. When assuming an

“ideal” device or system there is no regard for component or manufacturing tolerances or any

variation from device to device of a particular lot.

23. In the forward-bias region the 0 V drop across the diode at any level of current results in a

resistance level of zero ohms – the “on” state – conduction is established. In the reverse-bias

region the zero current level at any reverse-bias voltage assures a very high resistance level −

the open circuit or “off” state − conduction is interrupted.

24. The most important difference between the characteristics of a diode and a simple switch is

that the switch, being mechanical, is capable of conducting current in either direction while

the diode only allows charge to flow through the element in one direction (specifically the

direction defined by the arrow of the symbol using conventional current flow).

25. VD ≅ 0.66 V, ID = 2 mA

R

DC = 0.65 V

2 mA

D

V

I= = 325 Ω

4

26. At ID = 15 mA, VD = 0.82 V

RDC = 0.82 V

15 mA

D

V

I= = 54.67 Ω

As the forward diode current increases, the static resistance decreases.

27. VD = −10 V, ID = Is = −0.1

μ

A

RDC = 10 V

0.1 A

D

V

I

μ

= = 100 MΩ

V

D = −30 V, ID = Is= −0.1

μ

A

R

DC = 30 V

0.1 A

D

V

I

μ

= = 300 MΩ

As the reverse voltage increases, the reverse resistance increases directly (since the diode

leakage current remains constant).

28. (a) rd = 0.79 V 0.76 V 0.03 V

15 mA 5 mA 10 mA

d

V

I

Δ−

==

Δ− = 3 Ω

(b) rd = 26 mV 26 mV

10 mA

D

I= = 2.6 Ω

(c) quite close

29. ID = 10 mA, VD = 0.76 V

RDC = 0.76 V

10 mA

D

V

I= = 76 Ω

rd = 0.79 V 0.76 V 0.03 V

15 mA 5 mA 10 mA

d

V

I

Δ−

≅=

Δ− = 3 Ω

R

DC >> rd

30. ID = 1 mA, rd = 0.72 V 0.61 V

2 mA 0 mA

d

V

I

Δ−

=

Δ−

= 55 Ω

I

D = 15 mA, rd = 0.8 V 0.78 V

20 mA 10 mA

d

V

I

Δ−

=

Δ−

= 2 Ω

31. ID = 1 mA, rd = 26 mV

2

D

I

⎛⎞

⎜⎟

⎝⎠

= 2(26 Ω) = 52 Ω vs 55 Ω (#30)

I

D = 15 mA, rd = 26 mV 26 mV

15 mA

D

I= = 1.73 Ω vs 2 Ω (#30)

32. rav = 0.9 V 0.6 V

13.5 mA 1.2 mA

d

V

I

Δ−

=

Δ−

= 24.4 Ω

5

33. rd = 0.8 V 0.7 V 0.09 V

7 mA 3 mA 4 mA

d

V

I

Δ−

≅=

Δ− = 22.5 Ω

(relatively close to average value of 24.4 Ω (#32))

34. rav = 0.9 V 0.7 V 0.2 V

14 mA 0 mA 14 mA

d

V

I

Δ−

==

Δ− = 14.29 Ω

35. Using the best approximation to the curve beyond VD = 0.7 V:

rav = 0.8 V 0.7 V 0.1 V

25 mA 0 mA 25 mA

d

V

I

Δ−

≅=

Δ− = 4 Ω

36. (a) VR = −25 V: CT ≅ 0.75 pF

VR = −10 V: CT ≅ 1.25 pF

1.25 pF 0.75 pF 0.5 pF

10 V 25 V 15 V

T

R

C

V

Δ−

==

Δ− = 0.033 pF/V

(b) VR = −10 V: CT ≅ 1.25 pF

VR = −1 V: CT ≅ 3 pF

1.25 pF 3 pF 1.75 pF

10 V 1 V 9 V

T

R

C

V

Δ−

==

Δ− = 0.194 pF/V

(c) 0.194 pF/V: 0.033 pF/V = 5.88:1 ≅ 6:1

Increased sensitivity near VD = 0 V

37. From Fig. 1.33

VD = 0 V, CD = 3.3 pF

V

D = 0.25 V, CD = 9 pF

38. The transition capacitance is due to the depletion region acting like a dielectric in the reverse-

bias region, while the diffusion capacitance is determined by the rate of charge injection into

the region just outside the depletion boundaries of a forward-biased device. Both

capacitances are present in both the reverse- and forward-bias directions, but the transition

capacitance is the dominant effect for reverse-biased diodes and the diffusion capacitance is

the dominant effect for forward-biased conditions.

6

39. VD = 0.2 V, CD = 7.3 pF

X

C = 11

2 2 (6 MHz)(7.3 pF)fC

ππ

= = 3.64 kΩ

V

D = −20 V, CT = 0.9 pF

X

C = 11

2 2 (6 MHz)(0.9 pF)fC

ππ

= = 29.47 kΩ

40. If = 10 V

10 kΩ = 1 mA

ts + tt = trr = 9 ns

t

s + 2ts = 9 ns

ts = 3 ns

t

t = 2ts = 6 ns

41.

42. As the magnitude of the reverse-bias potential increases, the capacitance drops rapidly from a

level of about 5 pF with no bias. For reverse-bias potentials in excess of 10 V the capacitance

levels off at about 1.5 pF.

43. At

VD = −25 V, ID = −0.2 nA and at VD = −100 V, ID ≅ −0.45 nA. Although the change in IR is

more than 100%, the level of IR and the resulting change is relatively small for most

applications.

44. Log scale: TA = 25°C, IR = 0.5 nA

TA = 100°C, IR = 60 nA

The change is significant.

60 nA: 0.5 nA = 120:1

Yes, at 95°C IR would increase to 64 nA starting with 0.5 nA (at 25°C)

(and double the level every 10°C).

7

45. IF = 0.1 mA: rd ≅ 700 Ω

I

F = 1.5 mA: rd ≅ 70 Ω

I

F = 20 mA: rd ≅ 6 Ω

The results support the fact that the dynamic or ac resistance decreases rapidly with

increasing current levels.

46. T = 25°C: Pmax = 500 mW

T = 100°C: Pmax = 260 mW

P

max = VFIF

IF = max 500 mW

0.7 V

F

P

V= = 714.29 mA

I

F = max 260 mW

0.7 V

F

P

V= = 371.43 mA

714.29 mA: 371.43 mA = 1.92:1 ≅ 2:1

47. Using the bottom right graph of Fig. 1.37:

IF = 500 mA @ T = 25°C

At

IF = 250 mA, T ≅ 104°C

48.

49. TC = +0.072% =

10

100%

()

Z

V

VT T

Δ×

−

0.072 =

1

0.75 V 100

10 V( 25)T×

−

0.072 =

1

7.5

25

T−

T1 − 25° = 7.5

0.072 = 104.17°

T

1 = 104.17° + 25° = 129.17°

50. TC =

10

()

Z

V

VT T

Δ

− × 100%

= (5 V 4.8 V)

5 V(100 25 )

−

°− ° × 100% = 0.053%/°C

8

51. (20 V 6.8 V)

(24 V 6.8 V)

−

− × 100% = 77%

The 20 V Zener is therefore ≅ 77% of the distance between 6.8 V and 24 V measured from

the 6.8 V characteristic.

At

IZ = 0.1 mA, TC ≅ 0.06%/°C

(5 V 3.6 V)

(6.8 V 3.6 V)

−

− × 100% = 44%

The 5 V Zener is therefore ≅ 44% of the distance between 3.6 V and 6.8 V measured from the

3.6 V characteristic.

At

IZ = 0.1 mA, TC ≅ −0.025%/°C

52.

53. 24 V Zener:

0.2 mA: ≅ 400 Ω

1 mA: ≅ 95 Ω

10 mA: ≅ 13 Ω

The steeper the curve (higher dI/dV) the less the dynamic resistance.

54. VT ≅ 2.0 V, which is considerably higher than germanium (≅ 0.3 V) or silicon (≅ 0.7 V). For

germanium it is a 6.7:1 ratio, and for silicon a 2.86:1 ratio.

55. Fig. 1.53 (f) IF ≅ 13 mA

Fig. 1.53 (e) VF ≅ 2.3 V

56. (a) Relative efficiency @ 5 mA ≅ 0.82

@ 10 mA ≅ 1.02

1.02 0.82

0.82

− × 100% = 24.4% increase

ratio:

1.02

0.82 = 1.24

(b) Relative efficiency @ 30 mA ≅ 1.38

@ 35 mA ≅ 1.42

1.42 1.38

1.38

− × 100% = 2.9% increase

ratio:

1.42

1.38 = 1.03

(c) For currents greater than about 30 mA the percent increase is significantly less than for

increasing currents of lesser magnitude.

9

57. (a)

0.75

3.0 = 0.25

From Fig. 1.53 (i) ( ≅ 75°

(b) 0.5 ⇒ ( = 40 °

58. For the high-efficiency red unit of Fig. 1.53:

0.2 mA 20 mA

C

x

=

°

x = 20 mA

0.2 mA/ C° = 100°C

10

Chapter 2

1. The load line will intersect at ID = 8 V

330

E

R

=

Ω

= 24.24 mA and VD = 8 V.

(a)

Q

D

V ≅ 0.92 V

Q

D

I ≅ 21.5 mA

VR = E − Q

D

V = 8 V − 0.92 V = 7.08 V

(b)

Q

D

V ≅ 0.7 V

Q

D

I ≅ 22.2 mA

VR = E − Q

D

V = 8 V − 0.7 V = 7.3 V

(c)

Q

D

V ≅ 0 V

Q

D

I ≅ 24.24 mA

VR = E − Q

D

V = 8 V − 0 V = 8 V

For (a) and (b), levels of Q

D

V and Q

D

I are quite close. Levels of part (c) are reasonably close

but as expected due to level of applied voltage E.

2. (a)

ID = 5 V

2.2 k

E

R=Ω = 2.27 mA

The load line extends from ID = 2.27 mA to VD = 5 V.

Q

D

V ≅ 0.7 V, Q

D

I ≅ 2 mA

(b) ID = 5 V

0.47 k

E

R

=Ω = 10.64 mA

The load line extends from ID = 10.64 mA to VD = 5 V.

Q

D

V ≅ 0.8 V, Q

D

I ≅ 9 mA

(c) ID = 5 V

0.18 k

E

R=Ω = 27.78 mA

The load line extends from ID = 27.78 mA to VD = 5 V.

Q

D

V ≅ 0.93 V, Q

D

I ≅ 22.5 mA

The resulting values of Q

D

V are quite close, while Q

D

I extends from 2 mA to 22.5 mA.

3. Load line through Q

D

I = 10 mA of characteristics and VD = 7 V will intersect ID axis as

11.25 mA.

ID = 11.25 mA = 7 VE

R

=

with R = 7 V

11.25 mA = 0.62 kΩ

11

4. (a) ID = IR = 30 V 0.7 V

2.2 k

D

EV

R

−−

=Ω = 13.32 mA

VD = 0.7 V, VR = E − VD = 30 V − 0.7 V = 29.3 V

(b) ID = 30 V 0 V

2.2 k

D

EV

R

−−

=Ω = 13.64 mA

VD = 0 V, VR = 30 V

Yes, since E  VT the levels of ID and VR are quite close.

5. (a) I = 0 mA; diode reverse-biased.

(b) V20Ω = 20 V − 0.7 V = 19.3 V (Kirchhoff’s voltage law)

I = 19.3 V

20

Ω

= 0.965 A

(c) I = 10 V

10 Ω = 1 A; center branch open

6. (a) Diode forward-biased,

Kirchhoff’s voltage law (CW): −5 V + 0.7 V − Vo = 0

Vo = −4.3 V

IR = ID = 4.3 V

2.2 k

o

V

R=Ω = 1.955 mA

(b) Diode forward-biased,

ID = 8 V 0.7 V

1.2 k 4.7 k

−

Ω+ Ω = 1.24 mA

Vo = V4.7 kΩ + VD = (1.24 mA)(4.7 kΩ) + 0.7 V

= 6.53 V

7. (a) Vo = 2 k (20 V 0.7 V 0.3V)

2 k 2 k

Ω−−

Ω+ Ω

= 1

2(20 V – 1 V) = 1

2(19 V) = 9.5 V

(b) I = 10 V 2V 0.7 V) 11.3 V

1.2 k 4.7 k 5.9 k

+− =

Ω+ Ω Ω = 1.915 mA

V ′ = IR = (1.915 mA)(4.7 kΩ) = 9 V

V

o = V ′ − 2 V = 9 V − 2 V = 7 V

12

8. (a) Determine the Thevenin equivalent circuit for the 10 mA source and 2.2 kΩ resistor.

ETh = IR = (10 mA)(2.2 kΩ) = 22 V

RTh = 2. 2kΩ

Diode forward-biased

ID = 22 V 0.7 V

2.2 k 1.2 k

−

Ω

+Ω

= 6.26 mA

Vo = ID(1.2 kΩ)

= (6.26 mA)(1.2 kΩ)

=

7.51 V

(b) Diode forward-biased

ID = 20 V + 5 V 0.7 V

6.8 k

−

Ω = 2.65 mA

Kirchhoff’s voltage law (CW):

+Vo − 0.7 V + 5 V = 0

Vo = −4.3 V

9. (a)

1

o

V = 12 V – 0.7 V = 11.3 V

2

o

V = 0.3 V

(b)

1

o

V = −10 V + 0.3 V + 0.7 V = −9 V

I = 10 V 0.7 V 0.3 V 9 V

1.2 k 3.3 k 4.5 k

−−=

Ω+ Ω Ω = 2 mA, 2

o

V = −(2 mA)(3.3 kΩ) = −6.6 V

10. (a) Both diodes forward-biased

IR = 20 V 0.7 V

4.7 k

−

Ω = 4.106 mA

Assuming identical diodes:

ID = 4.106 mA

22

R

I= = 2.05 mA

Vo = 20 V − 0.7 V = 19.3 V

(b) Right diode forward-biased:

ID = 15 V + 5 V 0.7 V

2.2 k

−

Ω = 8.77 mA

Vo = 15 V − 0.7 V = 14.3 V

11. (a) Ge diode “on” preventing Si diode from turning “on”:

I = 10 V 0.3 V 9.7 V

1 k 1 k

−=

ΩΩ

= 9.7 mA

(b) I = 16 V 0.7 V 0.7 V 12 V 2.6 V

4.7 k 4.7 k

−−−

=

ΩΩ

= 0.553 mA

V

o = 12 V + (0.553 mA)(4.7 kΩ) = 14.6 V

13

12. Both diodes forward-biased:

1

o

V = 0.7 V, 2

o

V = 0.3 V

I1 kΩ = 20 V 0.7 V

1 k

−

Ω = 19.3 V

1 k

Ω

= 19.3 mA

I0.47 kΩ = 0.7 V 0.3 V

0.47 k

−

Ω = 0.851 mA

I(Si diode) = I1 kΩ − I0.47 kΩ

= 19.3 mA − 0.851 mA

=

18.45 mA

13. For the parallel Si − 2 kΩ branches a Thevenin equivalent will result (for “on” diodes) in a

single series branch of 0.7 V and 1 kΩ resistor as shown below:

I2 kΩ = 6.2 V

2 kΩ = 3.1 mA

I

D = 2 k 3.1 mA

22

IΩ= = 1.55 mA

14. Both diodes “off”. The threshold voltage of 0.7 V is unavailable for either diode.

Vo = 0 V

15. Both diodes “on”, Vo = 10 V − 0.7 V = 9.3 V

16. Both diodes “on”.

Vo = 0.7 V

17. Both diodes “off”, Vo = 10 V

18. The Si diode with −5 V at the cathode is “on” while the other is “off”. The result is

Vo = −5 V + 0.7 V = −4.3 V

19. 0 V at one terminal is “more positive” than −5 V at the other input terminal. Therefore

assume lower diode “on” and upper diode “off”.

The result:

Vo = 0 V − 0.7 V = −0.7 V

The result supports the above assumptions.

20. Since all the system terminals are at 10 V the required difference of 0.7 V across either diode

cannot be established. Therefore, both diodes are “off” and

Vo = +10 V

as established by 10 V supply connected to 1 kΩ resistor.

14

21. The Si diode requires more terminal voltage than the Ge diode to turn “on”. Therefore, with

5 V at both input terminals, assume Si diode “off” and Ge diode “on”.

The result: Vo = 5 V − 0.3 V = 4.7 V

The result supports the above assumptions.

22. Vdc = 0.318 Vm ⇒Vm = dc 2 V

0.318 0.318

V= = 6.28 V

Im = 6.28 V

2.2 k

m

V

R=Ω = 2.85 mA

23. Using Vdc ≅ 0.318(Vm − VT)

2 V = 0.318(Vm − 0.7 V)

Solving: Vm = 6.98 V ≅ 10:1 for Vm:VT

24. Vm = dc 2 V

0.318 0.318

V= = 6.28 V

max

L

I = 6.28 V

6.8 kΩ = 0.924 mA

15

Imax(2.2 kΩ) = 6.28 V

2.2 kΩ = 2.855 mA

max max

DL

II= + Imax(2.2 kΩ) = 0.924 mA + 2.855 mA = 3.78 mA

25. Vm = 2(110 V) = 155.56 V

V

dc = 0.318Vm = 0.318(155.56 V) = 49.47 V

26. Diode will conduct when vo = 0.7 V; that is,

vo = 0.7 V = 10 k ( )

10 k 1 k

i

vΩ

Ω+ Ω

Solving:

vi = 0.77 V

For

vi ≥ 0.77 V Si diode is “on” and vo = 0.7 V.

For

vi < 0.77 V Si diode is open and level of vo is determined

by voltage divider rule:

vo = 10 k ( )

10 k 1 k

i

vΩ

Ω+ Ω = 0.909 vi

For

vi = −10 V:

vo = 0.909(−10 V)

= −9.09 V

When

vo = 0.7 V, max max

R

i

vv= − 0.7 V

= 10 V − 0.7 V = 9.3 V

max

9.3 V

1 k

R

I=

Ω

= 9.3 mA

Imax(reverse) = 10 V

1 k 10 kΩ+ Ω = 0.909 mA

16

27. (a) Pmax = 14 mW = (0.7 V)ID

ID = 14 mW

0.7 V = 20 mA

(b) 4.7 kΩ || 56 kΩ = 4.34 kΩ

VR = 160 V − 0.7 V = 159.3 V

I

max = 159.3 V

4.34 kΩ = 36.71 mA

(c)

Idiode = max 36.71 mA

22

I= = 18.36 mA

(d) Yes, ID = 20 mA > 18.36 mA

(e)

Idiode = 36.71 mA  Imax = 20 mA

28. (a)

Vm = 2 (120 V) = 169.7 V

m

L

V = m

i

V − 2VD

= 169.7 V − 2(0.7 V) = 169.7 V − 1.4 V

= 168.3 V

Vdc = 0.636(168.3 V) = 107.04 V

(b) PIV =

Vm(load) + VD = 168.3 V + 0.7 V = 169 V

(c)

ID(max) = 168.3 V

1 k

m

L

V

R=Ω = 168.3 mA

(d) Pmax = VDID = (0.7 V)Imax

= (0.7 V)(168.3 mA)

=

117.81 mW

29.

17

30. Positive half-cycle of vi:

Voltage-divider rule:

max

o

V = max

2.2 k ( )

2.2 k 2.2 k

i

VΩ

Ω+ Ω

= max

1()

2i

V

= 1(100 V)

2

= 50 V

Polarity of vo across the 2.2 kΩ

resistor acting as a load is the same.

Voltage-divider rule:

max

o

V = max

2.2 k ( )

2.2 k 2.2 k

i

VΩ

Ω+ Ω

= max

1()

2i

V

= 1(100 V)

2

= 50 V

Vdc = 0.636Vm = 0.636 (50 V)

= 31.8 V

31. Positive pulse of vi:

Top left diode “off”, bottom left diode “on”

2.2 kΩ || 2.2 kΩ = 1.1 kΩ

p

eak

o

V = 1.1 k (170 V)

1.1 k 2.2 k

Ω

+Ω

= 56.67 V

Negative pulse of vi:

Top left diode “on”, bottom left diode “off”

p

eak

o

V = 1.1 k (170 V)

1.1 k 2.2 k

Ω

+Ω

= 56.67 V

Vdc = 0.636(56.67 V) = 36.04 V

32. (a) Si diode open for positive pulse of vi and vo = 0 V

For −20 V < vi ≤ −0.7 V diode “on” and vo = vi + 0.7 V.

For vi = −20 V, vo = −20 V + 0.7 V = −19.3 V

For vi = −0.7 V, vo = −0.7 V + 0.7 V = 0 V

18

(b) For vi ≤ 5 V the 5 V battery will ensure the diode is forward-biased and vo = vi − 5 V.

At vi = 5 V

vo = 5 V − 5 V = 0 V

At vi = −20 V

vo = −20 V − 5 V = −25 V

For vi > 5 V the diode is reverse-biased and vo = 0 V.

33. (a) Positive pulse of vi:

Vo = 1.2 k (10 V 0.7 V)

1.2 k 2.2 k

Ω−

Ω+ Ω = 3.28 V

Negative pulse of vi:

diode “open”, vo = 0 V

(b) Positive pulse of vi:

Vo = 10 V − 0.7 V + 5 V = 14.3 V

Negative pulse of vi:

diode “open”, vo = 0 V

34. (a) For vi = 20 V the diode is reverse-biased and vo = 0 V.

For vi = −5 V, vi overpowers the 2 V battery and the diode is “on”.

Applying Kirchhoff’s voltage law in the clockwise direction:

−5 V + 2 V − vo = 0

vo = −3 V

(b) For vi = 20 V the 20 V level overpowers the 5 V supply and the diode is “on”. Using the

short-circuit equivalent for the diode we find vo = vi = 20 V.

For vi = −5 V, both vi and the 5 V supply reverse-bias the diode and separate vi from vo.

However, vo is connected directly through the 2.2 kΩ resistor to the 5 V supply and

vo = 5 V.

19

35. (a) Diode “on” for vi ≥ 4.7 V

For vi > 4.7 V, Vo = 4 V + 0.7 V = 4.7 V

For vi < 4.7 V, diode “off” and vo = vi

(b) Again, diode “on” for vi ≥ 4.7 V but vo

now defined as the voltage across the diode

For vi ≥ 4.7 V, vo = 0.7 V

For vi < 4.7 V, diode “off”, ID = IR = 0 mA and V2.2 kΩ = IR = (0 mA)R = 0 V

Therefore, vo = vi − 4 V

At vi = 0 V, vo = −4 V

vi = −8 V, vo = −8 V − 4 V = −12 V

36. For the positive region of vi:

The right Si diode is reverse-biased.

The left Si diode is “on” for levels of vi greater than

5.3 V + 0.7 V = 6 V. In fact, vo = 6 V for vi ≥ 6 V.

For vi < 6 V both diodes are reverse-biased and vo = vi.

For the negative region of vi:

The left Si diode is reverse-biased.

The right Si diode is “on” for levels of vi more negative than 7.3 V + 0.7 V = 8 V. In

fact, vo = −8 V for vi ≤ −8 V.

For vi > −8 V both diodes are reverse-biased and vo = vi.

iR: For −8 V < vi < 6 V there is no conduction through the 10 kΩ resistor due to the lack of a

complete circuit. Therefore, iR = 0 mA.

For vi ≥ 6 V

vR = vi − vo = vi − 6 V

For vi = 10 V, vR = 10 V − 6 V = 4 V

and iR = 4 V

10 kΩ = 0.4 mA

For vi ≤ −8 V

vR = vi − vo = vi + 8 V

20

For vi = −10 V

vR = −10 V + 8 V = −2 V

and iR = 2V

10 k

−

Ω = −0.2 mA

37. (a) Starting with vi = −20 V, the diode is in the “on” state and the capacitor quickly charges

to −20 V+. During this interval of time vo is across the “on” diode (short-current

equivalent) and vo = 0 V.

When vi switches to the +20 V level the diode enters the “off” state (open-circuit

equivalent) and vo = vi + vC = 20 V + 20 V = +40 V

(b) Starting with vi = −20 V, the diode is in the “on” state and the capacitor quickly charges

up to −15 V+. Note that vi = +20 V and the 5 V supply are additive across the capacitor.

During this time interval vo is across “on” diode and 5 V supply and vo = −5 V.

When vi switches to the +20 V level the diode enters the “off” state and vo = vi + vC =

20 V + 15 V = 35 V.

21

38. (a) For negative half cycle capacitor charges to peak value of 120 V − 0.7 V = 119.3 V with

polarity . The output vo is directly across the “on” diode resulting in

vo = −0.7 V as a negative peak value.

For next positive half cycle vo = vi + 119.3 V with peak value of

vo = 120 V + 119.3 V = 239.3 V.

(b) For positive half cycle capacitor charges to peak value of 120 V − 20 V − 0.7 V = 99.3 V

with polarity . The output vo = 20 V + 0.7 V = 20.7 V

For next negative half cycle vo = vi − 99.3 V with negative peak value of

vo = −120 V − 99.3 V = −219.3 V.

Using the ideal diode approximation the vertical shift of part (a) would be 120 V rather

than 119.3 V and −100 V rather than −99.3 V for part (b). Using the ideal diode

approximation would certainly be appropriate in this case.

39. (a)

τ

= RC = (56 kΩ)(0.1

μ

F) = 5.6 ms

5

τ

= 28 ms

(b) 5

τ

= 28 ms  2

T = 1 ms

2 = 0.5 ms, 56:1

(c) Positive pulse of vi:

Diode “on” and vo = −2 V + 0.7 V = −1.3 V

Capacitor charges to 10 V + 2 V − 0.7 V = 11.3 V

Negative pulse of vi:

Diode “off” and vo = −10 V − 11.3 V = −21.3 V

22

40. Solution is network of Fig. 2.176(b) using a 10 V supply in place of the 5 V source.

41. Network of Fig. 2.178 with 2 V battery reversed.

42. (a) In the absence of the Zener diode

VL = 180 (20 V)

180 220

Ω

Ω+ Ω = 9 V

V

L = 9 V < VZ = 10 V and diode non-conducting

Therefore, IL = IR = 20 V

220 180Ω+ Ω = 50 mA

with IZ = 0 mA

and VL = 9 V

(b) In the absence of the Zener diode

VL = 470 (20 V)

470 220

Ω

Ω+ Ω = 13.62 V

V

L = 13.62 V > VZ = 10 V and Zener diode “on”

Therefore, VL = 10 V and

s

R

V = 10 V

/ 10 V/220

ss

RRs

IVR== Ω = 45.45 mA

IL = VL/RL = 10 V/470 Ω = 21.28 mA

and IZ =

s

R

I − IL = 45.45 mA − 21.28 mA = 24.17 mA

(c)

max

Z

P = 400 mW = VZIZ = (10 V)(IZ)

IZ = 400 mW

10 V = 40 mA

min

L

I = maxs

RZ

II− = 45.45 mA − 40 mA = 5.45 mA

RL =

min

10 V

5.45 mA

L

V

I= = 1,834.86 Ω

Large RL reduces IL and forces more of

s

R

I to pass through Zener diode.

(d) In the absence of the Zener diode

VL = 10 V = (20 V)

220

L

R

R+Ω

10RL + 2200 = 20RL

10RL = 2200

RL = 220 Ω

23

43. (a) VZ = 12 V, RL = 12 V

200 mA

L

V

I= = 60 Ω

VL = VZ = 12 V = 60 (16 V)

60

Li

Ls s

RV

R

RR

Ω

=

+Ω+

720 + 12Rs = 960

12Rs = 240

Rs = 20 Ω

(b)

max

Z

P = VZmax

Z

I

= (12 V)(200 mA)

= 2.4 W

44. Since IL = LZ

LL

VV

R

= is fixed in magnitude the maximum value of

s

R

I will occur when IZ is a

maximum. The maximum level of

s

R

I will in turn determine the maximum permissible level

of Vi.

max

400 mW

8 V

Z

P

IV

== = 50 mA

IL = 8 V

220

LZ

LL

VV

RR

== Ω = 36.36 mA

s

R

I = IZ + IL = 50 mA + 36.36 mA = 86.36 mA

s

iZ

R

s

VV

IR

−

=

or Vi = s

R

s

IR + VZ

= (86.36 mA)(91 Ω) + 8 V = 7.86 V + 8 V = 15.86 V

Any value of vi that exceeds 15.86 V will result in a current IZ that will exceed the maximum

value.

45. At 30 V we have to be sure Zener diode is “on”.

∴ VL = 20 V = 1 k (30 V)

1 k

Li

Ls s

RV

R

RR

Ω

=

+Ω+

Solving, Rs = 0.5 kΩ

At 50 V, 50 V 20 V

0.5 k

S

R

I−

=Ω = 60 mA, IL = 20 V

1 k

Ω

= 20 mA

IZM = S

R

I − IL = 60 mA − 20 mA = 40 mA

46. For vi = +50 V:

Z1 forward-biased at 0.7 V

Z2 reverse-biased at the Zener potential and 2

Z

V = 10 V.

Therefore, Vo = 12

Z

VV+ = 0.7 V + 10 V = 10.7 V

24

For vi = −50 V:

Z1 reverse-biased at the Zener potential and 1

Z

V = −10 V.

Z2 forward-biased at −0.7 V.

Therefore, Vo = 12

Z

VV+ = −10.7 V

For a 5 V square wave neither Zener diode will reach its Zener potential. In fact, for either

polarity of vi one Zener diode will be in an open-circuit state resulting in vo = vi.

47. Vm = 1.414(120 V) = 169.68 V

2Vm = 2(169.68 V) = 339.36 V

48. The PIV for each diode is 2Vm

∴PIV = 2(1.414)(Vrms)

25

Chapter 3

1. −

2. A bipolar transistor utilizes holes and electrons in the injection or charge flow process, while

unipolar devices utilize either electrons or holes, but not both, in the charge flow process.

3. Forward- and reverse-biased.

4. The leakage current ICO is the minority carrier current in the collector.

5. −

6. −

7. −

8. IE the largest

I

B the smallest

IC ≅ IE

9. IB = 1

100 C

I ⇒ IC = 100IB

IE = IC + IB = 100IB + IB = 101IB

IB = 8 mA

101 101

E

I= = 79.21

μ

A

IC = 100IB = 100(79.21

μ

A) = 7.921 mA

10. −

11. IE = 5 mA, VCB = 1 V: VBE = 800 mV

V

CB = 10 V: VBE = 770 mV

VCB = 20 V: VBE = 750 mV

The change in VCB is 20 V:1 V = 20:1

The resulting change in VBE is 800 mV:750 mV = 1.07:1 (very slight)

12. (a) rav = 0.9 V 0.7 V

8 mA 0

V

I

Δ−

=

Δ−

= 25 Ω

(b) Yes, since 25 Ω is often negligible compared to the other resistance levels of the

network.

13. (a) IC ≅ IE = 4.5 mA

(b) IC ≅ IE = 4.5 mA

(c) negligible: change cannot be detected on this set of characteristics.

(d) IC ≅ IE

26

14. (a) Using Fig. 3.7 first, IE ≅ 7 mA

Then Fig. 3.8 results in IC ≅ 7 mA

(b) Using Fig. 3.8 first, IE ≅ 5 mA

Then Fig. 3.7 results in VBE ≅ 0.78 V

(c) Using Fig. 3.10(b) IE = 5 mA results in VBE ≅ 0.81 V

(d) Using Fig. 3.10(c) IE = 5 mA results in VBE = 0.7 V

(e) Yes, the difference in levels of VBE can be ignored for most applications if voltages of

several volts are present in the network.

15. (a) IC =

α

IE = (0.998)(4 mA) = 3.992 mA

(b) IE = IC + IB ⇒ IC = IE − IB = 2.8 mA − 0.02 mA = 2.78 mA

α

dc = 2.78 mA

2.8 mA

C

E

I

I= = 0.993

(c) IC =

β

IB = 0.98

110.98

B

I

α

⎛⎞⎛ ⎞

=

⎜⎟⎜ ⎟

−−

⎝⎠⎝ ⎠

(40

μ

A) = 1.96 mA

I

E = 1.96 mA

0.993

C

I

α

= = 2 mA

16. −

17. Ii = Vi/Ri = 500 mV/20 Ω = 25 mA

I

L ≅ Ii = 25 mA

VL = ILRL = (25 mA)(1 kΩ) = 25 V

Av = 25 V

0.5 V

o

i

V

V= = 50

18. Ii = 200 mV 200 mV

20 100 120

i

is

V

RR

==

+Ω+ΩΩ

= 1.67 mA

I

L = Ii = 1.67 mA

V

L = ILR = (1.67 mA)(5 kΩ) = 8.35 V

A

v = 8.35 V

0.2 V

o

i

V

V= = 41.75

19. −

20. (a) Fig. 3.14(b): IB ≅ 35

μ

A

Fig. 3.14(a): IC ≅ 3.6 mA

(b) Fig. 3.14(a): VCE ≅ 2.5 V

Fig. 3.14(b): VBE ≅ 0.72 V

27

21. (a)

β

= 2 mA

17 A

C

B

I

μ

= = 117.65

(b)

α

= 117.65

1117.651

β

=

++

= 0.992

(c) ICEO = 0.3 mA

(d) ICBO = (1 −

α

)ICEO

= (1 − 0.992)(0.3 mA) = 2.4

μ

A

22. (a) Fig. 3.14(a): ICEO ≅ 0.3 mA

(b) Fig. 3.14(a): IC ≅ 1.35 mA

β

dc = 1.35 mA

10 A

C

B

I

μ

= = 135

(c)

α

= 135

1 136

β

=

+ = 0.9926

ICBO ≅ (1 −

α

)ICEO

= (1 − 0.9926)(0.3 mA)

=

2.2

μ

A

23. (a)

β

dc = 6.7 mA

80 A

C

B

I

μ

= = 83.75

(b)

β

dc = 0.85 mA

5 A

C

B

I

μ

= = 170

(c)

β

dc = 3.4 mA

30 A

C

B

I

μ

= = 113.33

(d)

β

dc does change from pt. to pt. on the characteristics.

Low IB, high VCE → higher betas

High IB, low VCE → lower betas

24. (a)

β

ac = 5 V

C

CE

B

I

V

I

Δ

=

Δ = 7.3 mA 6 mA 1.3 mA

90 A 70 A 20 A

μ

μμ

−=

− = 65

(b)

β

ac = 15 V

C

CE

B

I

V

I

Δ

=

Δ = 1.4 mA 0.3 mA 1.1 mA

10 A 0 A 10 A

μ

μμ

−

=

− = 110

(c)

β

ac = 10 V

C

CE

B

I

V

I

Δ

=

Δ = 4.25 mA 2.35 mA 1.9 mA

40 A 20 A 20 A

μ

μμ

−=

− = 95

(d)

β

ac does change from point to point on the characteristics. The highest value was

obtained at a higher level of VCE and lower level of IC. The separation between IB curves

is the greatest in this region.

28

(e) VCE IB

β

dc

β

ac IC

β

dc/

β

ac

5 V

80

μ

A 83.75 65 6.7 mA 1.29

10 V

30

μ

A 113.33 95 3.4 mA 1.19

15 V

5

μ

A 170 110 0.85 mA 1.55

As IC decreased, the level of

β

dc and

β

ac increased. Note that the level of

β

dc and

β

ac in

the center of the active region is close to the average value of the levels obtained. In

each case

β

dc is larger than

β

ac, with the least difference occurring in the center of the

active region.

25.

β

dc = 2.9 mA

25 A

C

B

I

μ

= = 116

α

= 116

11161

β

=

++

= 0.991

IE = IC/

α

= 2.9 mA/0.991 = 2.93 mA

26. (a)

β

= 0.987 0.987

1 1 0.987 0.013

α

==

−− = 75.92

(b)

α

= 120 120

1 120 1 121

β

==

++ = 0.992

(c) IB = 2 mA

180

C

I

β

= = 11.11

μ

A

IE = IC + IB = 2 mA + 11.11

μ

A

= 2.011 mA

27. −

28. Ve = Vi − Vbe = 2 V − 0.1 V = 1.9 V

A

v = 1.9 V

2 V

o

i

V

V= = 0.95 ≅ 1

Ie = 1.9 V

1 k

E

V

R=Ω = 1.9 mA (rms)

29. Output characteristics:

Curves are essentially the

same with new scales as shown.

Input characteristics:

Common-emitter input characteristics may be used directly for common-collector

calculations.

29

30. max

C

P = 30 mW = VCEIC

IC = max

C

I, VCE = max

max

C

P

I = 30 mW

7 mA = 4.29 V

V

CE = max

CE

V, IC = max

max

C

CE

P

V = 30 mW

20 V = 1.5 mA

V

CE = 10 V, IC = max 30 mW

10 V

C

CE

P

V= = 3 mA

I

C = 4 mA, VCE = max

C

P

I=30 mW

4 mA = 7.5 V

V

CE = 15 V, IC = max 30 mW

15 V

C

CE

P

V= = 2 mA

31. IC = max

C

I, VCE = max

max

30 mW

6 mA

C

P

I= = 5 V

V

CB = max

CB

V, IC = max

max

30 mW

15 V

C

CB

P

V= = 2 mA

IC = 4 mA, VCB = max 30 mW

4 mA

C

P

I= = 7.5 V

VCB = 10 V, IC = max 30 mW

10 V

C

CB

P

V= = 3 mA

30

32. The operating temperature range is −55°C ≤ TJ ≤ 150°C

°F = 9C

5° + 32°

= 9(55C)

5−° + 32° = −67°F

°F = 9(150 C)

5° + 32° = 302°F

∴ −67°F ≤ TJ ≤ 302°F

33. max

C

I = 200 mA, max

CE

V = 30 V, max

D

P = 625 mW

IC = max

C

I, CE

V = max

max

D

C

P

I = 625 mW

200 mA = 3.125 V

VCE = max

CE

V, IC = max

max

625 mW

30 V

D

CE

P

V= = 20.83 mA

IC = 100 mA, VCE = max 625 mW

100 mA

D

C

P

I= = 6.25 V

VCE = 20 V, IC = max 625 mW

20 V

D

CE

P

V= = 31.25 mA

34. From Fig. 3.23 (a) ICBO = 50 nA max

β

avg = min max

2

β

+

= 50 150

2

+ = 200

2

= 100

∴ICEO ≅

β

ICBO = (100)(50 nA)

= 5

μ

A

31

35. hFE (

β

dc) with VCE = 1 V, T = 25°C

IC = 0.1 mA, hFE ≅ 0.43(100) = 43

↓

IC = 10 mA, hFE ≅ 0.98(100) = 98

hfe(

β

ac) with VCE = 10 V, T = 25°C

IC = 0.1 mA, hfe ≅ 72

↓

IC = 10 mA, hfe ≅ 160

For both

hFE and hfe the same increase in collector current resulted in a similar increase

(relatively speaking) in the gain parameter. The levels are higher for hfe but note that VCE is

higher also.

36. As the reverse-bias potential increases in magnitude the input capacitance Cibo decreases (Fig.

3.23(b)). Increasing reverse-bias potentials causes the width of the depletion region to

increase, thereby reducing the capacitance A

Cd

⎛⎞

=∈

⎜⎟

⎝⎠

.

37. (a) At

IC = 1 mA, hfe ≅ 120

At

IC = 10 mA, hfe ≅ 160

(b) The results confirm the conclusions of problems 23 and 24 that beta tends to increase

with increasing collector current.

39. (a)

β

ac = 3 V

C

CE

B

I

V

I

Δ

=

Δ = 16 mA 12.2 mA 3.8 mA

80 A 60 A 20 A

μ

μμ

−

=

− = 190

(b)

β

dc = 12 mA

59.5 A

C

B

I

μ

= = 201.7

(c)

β

ac = 4 mA 2 mA 2 mA

18 A 8 A 10 A

μ

μμ

−=

− = 200

(d)

β

dc = 3 mA

13 A

C

B

I

μ

= = 230.77

(e) In both cases

β

dc is slightly higher than

β

ac (≅ 10%)

(f)(g)

In general

β

dc and

β

ac increase with increasing IC for fixed VCE and both decrease for

decreasing levels of VCE for a fixed IE. However, if IC increases while VCE decreases

when moving between two points on the characteristics, chances are the level of

β

dc or

β

ac may not change significantly. In other words, the expected increase due to an

increase in collector current may be offset by a decrease in VCE. The above data reveals

that this is a strong possibility since the levels of

β

are relatively close.

32

Chapter 4

1. (a) 16 V 0.7 V 15.3 V

470 k 470 k

Q

CC BE

B

VV

IR

−−

== =

Ω

Ω = 32.55

μ

A

(b)

QQ

CB

II

β

= = (90)(32.55

μ

A) = 2.93 mA

(c)

QQ

CE CC C C

VVIR=− = 16 V − (2.93 mA)(2.7 kΩ) = 8.09 V

(d)

VC = Q

CE

V = 8.09 V

(e)

VB = VBE = 0.7 V

(f)

VE = 0 V

2. (a)

IC =

β

IB = 80(40

μ

A) = 3.2 mA

(b)

RC = 12 V 6 V 6 V

3.2 mA 3.2 mA

C

RCC C

CC

VVV

II

−−

== =

= 1.875 kΩ

(c)

RB = 12 V 0.7 V 11.3 V

40 A 40 A

B

R

B

V

I

μ

−

==

= 282.5 kΩ

(d)

VCE = VC = 6 V

3. (a)

IC = IE − IB = 4 mA − 20

μ

A = 3.98 mA ≅ 4 mA

(b)

VCC = VCE + ICRC = 7.2 V + (3.98 mA)(2.2 kΩ)

= 15.96 V ≅ 16 V

(c)

β

= 3.98 mA

20 A

C

B

I

μ

= = 199 ≅ 200

(d) RB = 15.96 V 0.7 V

20 A

B

RCC BE

BB

VVV

II

μ

−−

== = 763 kΩ

4. sat

C

I = 16 V

2.7 k

CC

C

V

R=Ω = 5.93 mA

5. (a) Load line intersects vertical axis at IC = 21 V

3 k

Ω

= 7 mA

and horizontal axis at VCE = 21 V.

(b)

IB = 25

μ

A: RB = 21 V 0.7 V

25 A

CC BE

B

VV

I

μ

−

= = 812 kΩ

(c) Q

C

I ≅ 3.4 mA, Q

CE

V ≅ 10.75 V

33

(d)

β

= 3.4 mA

25 A

C

B

I

μ

= = 136

(e)

α

= 136 136

11361137

β

==

++ = 0.992

(f)

sat

C

I = 21 V

3 k

CC

C

V

R=Ω = 7 mA

(g)

−

(h)

PD = QQ

CE C

VI = (10.75 V)(3.4 mA) = 36.55 mW

(i)

Ps = VCC(IC + IB) = 21 V(3.4 mA + 25

μ

A) = 71.92 mW

(j)

PR = Ps − PD = 71.92 mW − 36.55 mW = 35.37 mW

6. (a) 20 V 0.7 V

( 1) 510 k (101)1.5 k

Q

CC BE

B

BE

VV

IRR

β

−

==

++ Ω+ Ω

= 19.3 V

661.5 k

Ω

= 29.18

μ

A

(b)

QQ

CB

II

β

= = (100)(29.18

μ

A) = 2.92 mA

(c)

Q

CE

V = VCC − IC(RC + RE) = 20 V − (2.92 mA)(2.4 kΩ + 1.5 kΩ)

= 20 V − 11.388 V

= 8.61 V

(d) VC = VCC − ICRC = 20 V − (2.92 mA)(2.4 kΩ) = 20 V − 7.008 V

= 13 V

(e) VB = VCC − IBRB = 20 V − (29.18

μ

A)(510 kΩ)

= 20 V − 14.882 V = 5.12 V

(f) VE = VC − VCE = 13 V − 8.61 V = 4.39 V

7. (a)

RC = 12 V 7.6 V 4.4 V

2 mA 2 mA

CC C

C

VV

I

−

== = 2.2 kΩ

(b)

IE ≅ IC: RE = 2.4 V

2 mA

E

V

I= = 1.2 kΩ

(c)

RB = 12 V 0.7 V 2.4 V 8.9 V

2 mA/80 25 A

B

RCC BE E

BB

VVVV

II

μ

−− −−

== =

= 356 kΩ

(d) VCE = VC − VE = 7.6 V − 2.4 V = 5.2 V

(e)

VB = VBE + VE = 0.7 V + 2.4 V = 3.1 V

34

8. (a)

IC ≅ IE = 2.1 V

0.68 k

E

V

R=Ω = 3.09 mA

β

= 3.09 mA

20 A

C

B

I

μ

= = 154.5

(b) VCC = C

R

V + VCE + VE

= (3.09 mA)(2.7 kΩ) + 7.3 V + 2.1 V = 8.34 V + 7.3 V + 2.1 V

= 17.74 V

(c)

RB = 17.74 V 0.7 V 2.1 V

20 A

B

RCC BE E

BB

VVVV

II

μ

−− −−

==

= 14.94 V

20 A

μ

= 747 kΩ

9. sot

20 V 20 V

2.4 k 1.5 k 3.9 k

CC

C

CE

V

IRR

== =

+Ω+ΩΩ

= 5.13 mA

10. (a)

sat

C

I = 6.8 mA = 24 V

1.2 k

CC

CE C

V

RR R

=

++Ω

RC + 1.2 kΩ = 24 V

6.8 mA = 3.529 kΩ

RC = 2.33 kΩ

(b)

β

= 4 mA

30 A

C

B

I

μ

= = 133.33

(c) RB = 24 V 0.7 V (4 mA)(1.2 k )

30 A

B

RCC BE E

BB

VVVV

II

μ

−−

−

−Ω

==

= 18.5 V

30 A

μ

= 616.67 kΩ

(d)

PD = QQ

CE C

VI

= (10 V)(4 mA) = 40 mW

(e)

P = 2

CC

IR = (4 mA)2(2.33 kΩ)

= 37.28 mW

11. (a) Problem 1: Q

C

I = 2.93 mA, Q

CE

V = 8.09 V

(b)

Q

B

I = 32.55

μ

A (the same)

QQ

CB

II

β

= = (135)(32.55

μ

A) = 4.39 mA

QQ

CE CC C C

VVIR=− = 16 V − (4.39 mA)(2.7 kΩ) = 4.15 V

35

(c) %ΔIC = 4.39 mA 2.93 mA

2.93 mA

− × 100% = 49.83%

%

ΔVCE = 4.15 V 8.09 V

8.09 V

− × 100% = 48.70%

Less than 50% due to level of accuracy carried through calculations.

(d) Problem 6: Q

C

I = 2.92 mA, Q

CE

V = 8.61 V ( Q

B

I = 29.18

μ

A)

(e) 20 V 0.7 V

( 1) 510 k (150 1)(1.5 k )

Q

CC BE

B

BE

VV

IRR

β

−

==

++ Ω+ + Ω

= 26.21

μ

A

QQ

CB

II

β

= = (150)(26.21

μ

A) = 3.93 mA

Q

CE

V = VCC − IC(RC + RE)

= 20 V − (3.93 mA)(2.4 kΩ + 1.5 kΩ) = 4.67 V

(f) %ΔIC = 3.93 mA 2.92 mA

2.92 mA

− × 100% = 34.59%

%

ΔVCE = 4.67 V 8.61 V

8.61 V

− × 100% = 46.76%

(g) For both

IC and VCE the % change is less for the emitter-stabilized.

12.

β

RE

?

≥ 10R2

(80)(0.68 kΩ) ≥ 10(9.1 kΩ)

54.4 kΩ ≥ 91 kΩ (No!)

(a) Use exact approach:

RTh = R1 || R2 = 62 kΩ || 9.1 kΩ = 7.94 kΩ

E

Th = 2

21

(9.1 k )(16 V)

9.1 k 62 k

CC

RV

RR

Ω

=

+Ω+Ω

= 2.05 V

2.05 V 0.7 V

( 1) 7.94 k (81)(0.68 k )

Q

Th BE

B

Th E

EV

IRR

β

−−

==

++ Ω+ Ω

= 21.42

μ

A

(b)

QQ

CB

II

β

= = (80)(21.42

μ

A) = 1.71 mA

(c)

Q

CE

V = VCC − Q

C

I(RC + RE)

= 16 V − (1.71 mA)(3.9 kΩ + 0.68 kΩ)

= 8.17 V

(d) VC = VCC − ICRC

= 16 V − (1.71 mA)(3.9 kΩ)

= 9.33 V

(e) VE = IERE ≅ ICRE = (1.71 mA)(0.68 kΩ)

= 1.16 V

(f) VB = VE + VBE = 1.16 V + 0.7 V

= 1.86 V

36

13. (a)

IC = 18 V 12 V

4.7 k

CC C

C

VV

R

−−

=Ω = 1.28 mA

(b)

VE = IERE ≅ ICRE = (1.28 mA)(1.2 kΩ) = 1.54 V

(c) VB = VBE + VE = 0.7 V + 1.54 V = 2.24 V

(d)

R1 = 1

1

:

R

V

I 1

R

V = VCC − VB = 18 V − 2.24 V = 15.76 V

1

R

I ≅ 2

2

2.24 V

5.6 k

B

R

V

IR

==

Ω

= 0.4 mA

R1 = 1

1

R

V

I = 15.76 V

0.4 mA = 39.4 kΩ

14. (a)

IC =

β

IB = (100)(20

μ

A) = 2 mA

(b)

IE = IC + IB = 2 mA + 20

μ

A

= 2.02 mA

VE = IERE = (2.02 mA)(1.2 kΩ)

= 2.42 V

(c)

VCC = VC + ICRC = 10.6 V + (2 mA)(2.7 kΩ)

= 10.6 V + 5.4 V

= 16 V

(d)

VCE = VC − VE = 10.6 V − 2.42 V

= 8.18 V

(e)

VB = VE + VBE = 2.42 V + 0.7 V = 3.12 V

(f)

12

R

RB

III=+

= 3.12 V

8.2 kΩ + 20

μ

A = 380.5

μ

A + 20

μ

A = 400.5

μ

A

R1 =

1

16 V 3.12 V

400.5 A

CC B

R

VV

I

μ

−−

= = 32.16 kΩ

15. sat

C

I = 16 V

3.9 k 0.68 k

CC

CE

V

RR

=

+Ω+Ω

= 16 V

4.58 k

Ω

= 3.49 mA

37

16. (a)

β

RE ≥ 10R2

(120)(1 kΩ) ≥ 10(8.2 kΩ)

120 kΩ ≥ 82 kΩ (checks)

∴VB = 2

12

(8.2 k )(18 V)

39 k 8.2 k

CC

RV

RR

Ω

=

+Ω+Ω

= 3.13 V

VE = VB − VBE = 3.13 V − 0.7 V = 2.43 V

I

C ≅ IE = 2.43 V

1 k

E

V

R=Ω = 2.43 mA

(b)

VCE = VCC − IC(RC + RE)

= 18 V − (2.43 mA)(3.3 kΩ + 1 kΩ)

= 7.55 V

(c)

IB = 2.43 mA

120

C

I

β

= = 20.25

μ

A

(d)

VE = IERE ≅ ICRE = (2.43 mA)(1 kΩ) = 2.43 V

(e)

VB = 3.13 V

17. (a)

RTh = R1 || R2 = 39 kΩ || 8.2 kΩ = 6.78 kΩ

ETh =

12

8.2 k (18 V)

39 k 8.2 k

CCC

RV

RR

Ω

=

+Ω+Ω

= 3.13 V

I

B = 3.13 V 0.7 V

( 1) 6.78 k (121)(1 k )

Th BE

Th E

EV

RR

β

−−

=

++ Ω+ Ω

= 2.43 V

127.78 kΩ = 19.02

μ

A

IC =

β

IB = (120)(19.02

μ

A) = 2.28 mA (vs. 2.43 mA #16)

(b) VCE = VCC − IC(RC + RE) = 18 V − (2.28 mA)(3.3 kΩ + 1 kΩ)

= 18 V − 9.8 V = 8.2 V (vs. 7.55 V #16)

(c)

19.02

μ

A (vs. 20.25

μ

A #16)

(d)

VE = IERE ≅ ICRE = (2.28 mA)(1 kΩ) = 2.28 V (vs. 2.43 V #16)

(e)

VB = VBE + VE = 0.7 V + 2.28 V = 2.98 V (vs. 3.13 V #16)

The results suggest that the approximate approach is valid if Eq. 4.33 is satisfied.

18. (a)

VB = 2

12

9.1 k (16 V)

62 k 9.1 k

CC

RV

RR

Ω

=

+Ω+Ω

= 2.05 V

VE = VB − VBE = 2.05 V − 0.7 V = 1.35 V

I

E = 1.35 V

0.68 k

E

V

R=Ω = 1.99 mA

Q

C

I ≅ IE = 1.99 mA

38

Q

CE

V = VCC − IC (RC + RE)

= 16 V − (1.99 mA)(3.9 kΩ + 0.68 kΩ)

= 16 V − 9.11 V

= 6.89 V

Q

B

I = 1.99 mA

80

Q

C

I

β

= = 24.88

μ

A

(b) From Problem 12:

Q

C

I = 1.71 mA, Q

CE

V = 8.17 V, Q

B

I = 21.42

μ

A

(c) The differences of about 14% suggest that the exact approach should be employed when

appropriate.

19. (a)

sat

24 V 24 V

7.5 mA 34

CC

C

CE EE E

V

I

R

RRR R

== = =

++

RE = 24 V 24 V

4(7.5 mA) 30 mA

= = 0.8 kΩ

R

C = 3RE = 3(0.8 kΩ) = 2.4 kΩ

(b) V

E = IERE ≅ ICRE = (5 mA)(0.8 kΩ) = 4 V

(c)

VB = VE + VBE = 4 V + 0.7 V = 4.7 V

(d)

VB = 2

21

CC

RV

R

R+, 4.7 V = 2

2

(24 V)

24 k

R

+

Ω

R2 = 5.84 kΩ

(e)

β

dc = 5 mA

38.5 A

C

B

I

μ

= = 129.8

(f)

β

RE ≥ 10R2

(129.8)(0.8 kΩ) ≥ 10(5.84 kΩ)

103.84 kΩ ≥ 58.4 kΩ (checks)

20. (a) From problem 12b, IC = 1.71 mA

From problem 12c, VCE = 8.17 V

(b)

β

changed to 120:

From problem 12a, ETh = 2.05 V, RTh = 7.94 kΩ

IB = 2.05 V 0.7 V

( 1) 7.94 k + (121)(0.68 k )

Th BE

Th E

EV

RR

β

−−

=

++ Ω Ω

= 14.96

μ

A

IC =

β

IB = (120)(14.96

μ

A) = 1.8 mA

V

CE = VCC − IC(RC + RE)

= 16 V − (1.8 mA)(3.9 kΩ + 0.68 kΩ)

= 7.76 V

39

(c) 1.8 mA 1.71 mA

%1.71 mA

C

I−

Δ= × 100% = 5.26%

7.76 V 8.17 V

%8.17 V

CE

V−

Δ= × 100% = 5.02%

(d) 11c 11f 20c

%ΔIC 49.83% 34.59% 5.26%

%ΔVCE 48.70% 46.76% 5.02%

Fixed-bias Emitter

feedback

Voltage-

divider

(e) Quite obviously, the voltage-divider configuration is the least sensitive to changes in

β

.

21. I.(a) Problem 16: Approximation approach: Q

C

I = 2.43 mA, Q

CE

V = 7.55 V

Problem 17: Exact analysis: Q

C

I = 2.28 mA, Q

CE

V = 8.2 V

The exact solution will be employed to demonstrate the effect of the change of

β

. Using

the approximate approach would result in %ΔIC = 0% and %ΔVCE = 0%.

(b) Problem 17: ETh = 3.13 V, RTh = 6.78 kΩ

IB = 3.13 V 0.7 V 2.43 V

( 1) 6.78 k (180 1)1 k 187.78 k

TH BE

Th E

EV

RR

β

−

==

++ Ω+ + Ω Ω

= 12.94

μ

A

IC =

β

IB = (180)(12.94

μ

A) = 2.33 mA

V

CE = VCC − IC(RC + RE) = 18 V − (2.33 mA)(3.3 kΩ + 1 kΩ)

= 7.98 V

(c) %ΔIC = 2.33 mA 2.28 mA

2.28 mA

− × 100% = 2.19%

%

ΔVCE = 7.98 V 8.2 V

8.2 V

− × 100% = 2.68%

For situations where

β

RE > 10R2 the change in IC and/or VCE due to significant change in

β

will be relatively small.

(d) %ΔIC = 2.19% vs. 49.83% for problem 11.

%

ΔVCE = 2.68% vs. 48.70% for problem 11.

(e) Voltage-divider configuration considerably less sensitive.

II. The resulting %ΔIC and %ΔVCE will be quite small.

40

22. (a)

IB = 16 V 0.7 V

( ) 470 k + (120)(3.6 k 0.51 k )

CC BE

BCE

VV

RRR

β

−

=

++ Ω Ω+Ω

= 15.88

μ

A

(b)

IC =

β

IB = (120)(15.88

μ

A)

= 1.91 mA

(c)

VC = VCC − ICRC

= 16 V − (1.91 mA)(3.6 kΩ)

= 9.12 V

23. (a)

IB = 30 V 0.7 V

( ) 6.90 k 100(6.2 k 1.5 k )

CC BE

BCE

VV

RRR

β

−

=

+ + Ω+ Ω+ Ω = 20.07

μ

A

IC =

β

IB = (100)(20.07

μ

A) = 2.01 mA

(b) VC = VCC − ICRC

= 30 V − (2.01 mA)(6.2 kΩ) = 30 V − 12.462 V = 17.54 V

(c)

VE = IERE ≅ ICRE = (2.01 mA)(1.5 kΩ) = 3.02 V

(d) VCE = VCC − IC(RC + RE) = 30 V − (2.01 mA)(6.2 kΩ + 1.5 kΩ)

= 14.52 V

24. (a)

IB = 22 V 0.7 V

( ) 470 k (90)(9.1 k 9.1 k )

CC BE

BCE

VV

RRR

β

−−

=

+ + Ω+ Ω+ Ω

= 10.09

μ

A

IC =

β

IB = (90)(10.09

μ

A) = 0.91 mA

V

CE = VCC − IC(RC + RE) = 22 V − (0.91 mA)(9.1 kΩ + 9.1 kΩ)

= 5.44 V

(b)

β

= 135, IB = 22 V 0.7 V

( ) 470 k (135)(9.1 k 9.1 k )

CC BE

BCE

VV

RRR

β

−−

=

++ Ω+ Ω+Ω

= 7.28

μ

A

IC =

β

IB = (135)(7.28

μ

A) = 0.983 mA

V

CE = VCC − IC(RC + RE) = 22 V − (0.983 mA)(9.1 kΩ + 9.1 kΩ)

= 4.11 V

(c) 0.983 mA 0.91 mA

%0.91 mA

C

I−

Δ= × 100% = 8.02%

4.11 V 5.44 V

%5.44 V

CE

V−

Δ= × 100% = 24.45%

(d) The results for the collector feedback configuration are closer to the voltage-divider

configuration than to the other two. However, the voltage-divider configuration

continues to have the least sensitivities to change in

β

.

41

25. 1 MΩ = 0 Ω, RB = 150 kΩ

IB = 12 V 0.7 V

( ) 150 k (180)(4.7 k 3.3 k )

CC BE

BCE

VV

RRR

β

−

=

+ + Ω+ Ω+ Ω

= 7.11

μ

A

IC =

β

IB = (180)(7.11

μ

A) = 1.28 mA

V

C = VCC −ICRC = 12 V − (1.28 mA)(4.7 kΩ)

= 5.98 V

Full 1 MΩ: RB = 1,000 kΩ + 150 kΩ = 1,150 kΩ = 1.15 MΩ

IB = 12 V 0.7 V

( ) 1.15 M (180)(4.7 k 3.3 k )

CC BE

BCE

VV

RRR

β

−−

=

+ + Ω+ Ω+ Ω

= 4.36

μ

A

IC =

β

IB = (180)(4.36

μ

A) = 0.785 mA

V

C = VCC − ICRC = 12 V − (0.785 mA)(4.7 kΩ)

= 8.31 V

VC ranges from 5.98 V to 8.31 V

26. (a)

VE = VB − VBE = 4 V − 0.7 V = 3.3 V

(b) IC ≅ IE = 3.3 V

1.2 k

E

V

R=Ω = 2.75 mA

(c)

VC = VCC − ICRC = 18 V − (2.75 mA)(2.2 kΩ)

= 11.95 V

(d) VCE = VC − VE = 11.95 V − 3.3 V = 8.65 V

(e) IB = 11.95 V 4 V

330 k

B

RCB

BB

VVV

RR

−−

== Ω = 24.09

μ

A

(f)

β

= 2.75 mA

24.09 A

C

B

I

μ

= = 114.16

27. (a)

IB = 6 V + 6 V 0.7 V

( 1) 330 k (121)(1.2 k )

CC EE BE

BE

VVV

RR

β

+− −

=

++ Ω+ Ω

= 23.78

μ

A

IE = (

β

+ 1)IB = (121)(23.78

μ

A)

= 2.88 mA

−VEE + IERE − VE = 0

VE = −VEE + IERE = −6 V + (2.88 mA)(1.2 kΩ)

= −2.54 V

28. (a)

IB = 12 V 0.7 V

( 1) 9.1 k (120 1)15 k

EE BE

BE

VV

RR

β

−

=

++ Ω+ + Ω

= 6.2

μ

A

(b)

IC =

β

IB = (120)(6.2

μ

A) = 0.744 mA

(c) VCE = VCC + VEE − IC(RC + RE)

= 16 V + 12 V − (0.744 mA)(27 kΩ)

= 7.91 V

(d)

VC = VCC − ICRC = 16 V − (0.744 mA)(12 kΩ) = 7.07 V

42

29. (a) IE = 8 V 0.7 V 7.3 V

2.2 k 2.2 k

−=

ΩΩ

= 3.32 mA

(b) VC = 10 V − (3.32 mA)(1.8 kΩ) = 10 V − 5.976

= 4.02 V

(c) VCE = 10 V + 8 V − (3.32 mA)(2.2 kΩ + 1.8 kΩ)

= 18 V − 13.28 V

= 4.72 V

30. (a)

β

RE > 10R2 not satisfied ∴Use exact approach:

Network redrawn to determine the Thevenin equivalent:

RTh = 510 k

2

Ω

= 255 kΩ

I = 18 V + 18 V

510 k 510 k

Ω

+Ω

= 35.29

μ

A

E

Th = −18 V + (35.29

μ

A)(510 kΩ)

= 0 V

IB = 18 V 0.7 V

255 k (130 1)(7.5 k )

−

Ω

++ Ω

=

13.95

μ

A

(b)

IC =

β

IB = (130)(13.95

μ

A) = 1.81 mA

(c) VE = −18 V + (1.81 mA)(7.5 kΩ)

= −18 V + 13.58 V

= −4.42 V

(d)

VCE = 18 V + 18 V − (1.81 mA)(9.1 kΩ + 7.5 kΩ)

= 36 V − 30.05 V = 5.95 V

31. (a) IB = 8 V 0.7 V

560 k

B

RCBE

BB

VVV

RR

−−

==

Ω

= 13.04

μ

A

(b) IC = CC C

C

VV

R

− = 18 V 8 V 10 V

3.9 k 3.9 k

−=

ΩΩ

= 2.56 mA

(c)

β

= 2.56 mA

13.04 A

C

B

I

μ

= = 196.32

(d) VCE = VC = 8 V

43

32. IB = 2.5 mA

80

C

I

β

= = 31.25

μ

A

R

B = 12 V 0.7 V

31.25 A

B

RCC BE

BB

VVV

II

μ

−−

== = 361.6 kΩ

R

C = 12 V 6 V 6 V

2.5 mA 2.5 mA

Q

C

Q

CC CE

RCC C

CC C

VV

VVV

II I

−

−−

== = =

= 2.4 kΩ

Standard values:

RB = 360 kΩ

R

C = 2.4 kΩ

33. sat

C

I = CC

CE

V

R

R+ = 10 mA

20 V

4

E

R

R+ = 10 mA ⇒ 20 V

5

E

R

= 10 mA ⇒ 5RE = 20 V

10 mA = 2 kΩ

RE = 2 k

5

Ω = 400 Ω

R

C = 4RE = 1.6 kΩ

I

B = 5 mA

120

C

I

β

= = 41.67

μ

A

R

B = VRB/IB = 20 V 0.7 V 5 mA(0.4 k ) 19.3 2 V

41.67 A 41.67 A

μ

−− Ω −

=

= 415.17 kΩ

Standard values: RE = 390 Ω, RC = 1.6 kΩ, RB = 430 kΩ

34. RE = 3 V

4 mA

EE

EC

VV

II

≅= = 0.75 kΩ

R

C = ()

Q

CCC CE E

RCC C

CC C

VVV

II I

−+

−

==

= 24 V (8 V + 3 V) 24 V 11 V 13 V

4 mA 4 mA 4 mA

−−

==

= 3.25 kΩ

VB = VE + VBE = 3 V + 0.7 V = 3.7 V

V

B = 22

21 21

(24 V)

3.7 V =

CC

RV R

RR RR

⎫

⇒⎬

++

⎭ 2 unknowns!

∴ use

β

RE ≥ 10R2 for increased stability

(110)(0.75 kΩ) = 10R2

R2 = 8.25 kΩ

Choose R2 = 7.5 kΩ

44

Substituting in the above equation:

3.7 V =

1

7.5 k (24 V)

7.5 k

R

Ω

Ω+

R1 = 41.15 kΩ

Standard values:

RE = 0.75 kΩ, RC = 3.3 kΩ, R2 = 7.5 kΩ, R1 = 43 kΩ

35. VE = 1

5CC

V = 1(28 V)

5 = 5.6 V

R

E = 5.6 V

5 mA

E

V

I= = 1.12 kΩ (use 1.1 kΩ)

V

C = 28 V

22

CC

E

VV

+= + 5.6 V = 14 V + 5.6 V = 19.6 V

C

R

V = VCC − VC = 28 V − 19.6 V = 8.4 V

RC = 8.4 V

5 mA

C

R

C

V

I= = 1.68 kΩ (use 1.6 kΩ)

V

B = VBE + VE = 0.7 V + 5.6 V = 6.3 V

V

B = 2

21

CC

RV

R

R+ ⇒ 6.3 V = 2

21

(28 V)R

R

R+ (2 unknowns)

β

= 5 mA

37 A

C

B

I

μ

= = 135.14

β

RE = 10R2

(135.14)(1.12 kΩ) = 10(R2)

R2 = 15.14 kΩ (use 15 kΩ)

Substituting: 6.3 V =

1

(15.14 k )(28 V)

15.14 k

R

Ω

Ω+

Solving,

R1 = 52.15 kΩ (use 51 kΩ)

Standard values:

RE = 1.1 kΩ

R

C = 1.6 kΩ

R

1 = 51 kΩ

R

2 = 15 kΩ

36. I2 kΩ = 18 V 0.7 V

2 k

−

Ω = 8.65 mA ≅ I

37. For current mirror:

I(3 kΩ) = I(2.4 kΩ) = I = 2 mA

38. Q

DDSS

II= = 6 mA

45

39. VB ≅ 4.3 k (18 V)

4.3 k 4.3 k

Ω−

Ω+ Ω = −9 V

V

E = −9 V − 0.7 V = −9.7 V

I

E = 18 V ( 9.7 V)

1.8 k

−−−

Ω = 4.6 mA = I

40. IE = 5.1 V 0.7 V

1.2 k

ZBE

E

VV

R

−−

=Ω = 3.67 mA

41. sat

10 V

2.4 k

CC

C

V

IR

== Ω = 4.167 mA

From characteristics

max

B

I ≅ 31

μ

A

IB = 10 V 0.7 V

180 k

iBE

B

VV

R

−−

=Ω = 51.67

μ

A

51.67

μ

A  31

μ

A, well saturated

Vo = 10 V − (0.1 mA)(2.4 kΩ)

= 10 V − 0.24 V

= 9.76 V

42. sat

C

I = 8 mA = 5 V

C

R

RC = 5 V

8 mA = 0.625 kΩ

max

B

I = sat 8 mA

100

C

I

β

= = 80

μ

A

Use 1.2 (80

μ

A) = 96

μ

A

RB = 5 V 0.7 V

96 A

μ

− = 44.79 kΩ

Standard values:

RB = 43 kΩ

R

C = 0.62 kΩ

46

43. (a) From Fig. 3.23c:

IC = 2 mA: tf = 38 ns, tr = 48 ns, td = 120 ns, ts = 110 ns

ton = tr + td = 48 ns + 120 ns = 168 ns

toff = ts + tf = 110 ns + 38 ns = 148 ns

(b)

IC = 10 mA: tf = 12 ns, tr = 15 ns, td = 22 ns, ts = 120 ns

t

on = tr + td = 15 ns + 22 ns = 37 ns

toff = ts + tf = 120 ns + 12 ns = 132 ns

The turn-on time has dropped dramatically

168 ns:37 ns =

4.54:1

while the turn-off time is only slightly smaller

148 ns:132 ns =

1.12:1

44. (a) Open-circuit in the base circuit

Bad connection of emitter terminal

Damaged transistor

(b) Shorted base-emitter junction

Open at collector terminal

(c) Open-circuit in base circuit

Open transistor

45. (a) The base voltage of 9.4 V reveals that the 18 kΩ resistor is not making contact with the

base terminal of the transistor.

If operating properly:

VB ≅ 18 k (16 V)

18 k 91 k

Ω

Ω+ Ω = 2.64 V vs. 9.4 V

As an emitter feedback bias circuit:

IB =

1

16 V 0.7 V

( 1) 91 k (100 1)1.2 k

CC BE

E

VV

RR

β

−

=

++ Ω+ + Ω

= 72.1

μ

A

VB = VCC − IB(R1) = 16 V − (72.1

μ

A)(91 kΩ)

= 9.4 V

47

(b) Since

VE > VB the transistor should be “off”

With

IB = 0

μ

A, VB = 18 k (16 V)

18 k 91 k

Ω

+Ω

= 2.64 V

∴ Assume base circuit “open”

The 4 V at the emitter is the voltage that would exist if the transistor were shorted

collector to emitter.

VE = 1.2 k (16 V)

1.2 k 3.6 k

Ω

Ω+ Ω = 4 V

46. (a)

RB↑, IB↓, IC↓, VC↑

(b)

β

↓, IC↓

(c) Unchanged,

sat

C

I not a function of

β

(d)

VCC↓, IB↓, IC↓

(e)

β

↓, IC↓, C

R

V↓, E

R

V↓, VCE↑

47. (a)

IB = (1)

Th BE Th BE

Th E Th E

E

VEV

R

RR R

ββ

−−

≅

++ +

IC =

β

IB = Th BE Th BE

Th

Th E

E

VEV

R

RR

R

ββ

β

⎡⎤

−−

=

⎢⎥

+

⎣⎦

+

As

β

↑, Th

R

β

↓, IC↑, C

R

V↑

VC = VCC − C

R

V

and

VC↓

(b)

R2 = open, IB↑, IC↑

V

CE = VCC − IC(RC + RE)

and

VCE↓

(c) VCC↓, VB↓, VE↓, IE↓, IC↓

(d) IB = 0

μ

A, IC = ICEO and IC(RC + RE) negligible

with VCE ≅ VCC = 20 V

(e) Base-emitter junction = short IB↑ but transistor action lost and IC = 0 mA with

VCE = VCC = 20 V

48. (a)

RB open, IB = 0

μ

A, IC = ICEO ≅ 0 mA

and

VC ≅ VCC = 18 V

(b)

β

↑, IC↑, C

R

V ,

E

R

V , VCE↓

(c) RC↓, IB↑, IC↑, VE↑

(d) Drop to a relatively low voltage ≅ 0.06 V

(e) Open in the base circuit

↑↑

48

49. IB = 12 V 0.7 V 11.3 V

510 k 510 k

CC BE

B

VV

R

−−

==

ΩΩ

= 22.16

μ

A

IC =

β

IB = (100)(22.16

μ

A) = 2.216 mA

VC = −VCC + ICRC = −12 V + (2.216 mA)(3.3 kΩ)

= −4.69 V

VCE = VC = −4.69 V

50.

β

RE ≥ 10R2

(220)(0.75 kΩ) ≥ 10(16 kΩ)

165 kΩ ≥ 160 kΩ (checks)

Use approximate approach:

VB ≅ 16 k ( 22 V)

16 k + 82 k

Ω−

ΩΩ

= −3.59 V

V

E = VB + 0.7 V = −3.59 V + 0.7 V = −2.89 V

IC ≅ IE = VE/RE = 2.89/0.75 kΩ = 3.85 mA

I

B = 3.85 mA

220

C

I

β

= = 17.5

μ

A

VC = −VCC + ICRC

= −22 V + (3.85 mA)(2.2 kΩ)

= −13.53 V

51. IE = 8 V 0.7 V 7.3 V

3.3 k 3.3 k

BE

E

VV

R

−−

==

ΩΩ

= 2.212 mA

V

C = −VCC + ICRC = −12 V + (2.212 mA)(3.9 kΩ)

= −3.37 V

52. (a)

S(ICO) =

β

+ 1 = 91

(b)

S(VBE) = 90

470 k

B

R

β

−−

=Ω = −1.92 ×10−4 S

(c)

S(

β

) = 1

1

2.93 mA

90

C

I

β

= = 32.56 × 10−6 A

(d)

ΔIC = S(ICO)ΔICO + S(VBE)ΔVBE + S(

β

)Δ

β

= (91)(10

μ

A − 0.2

μ

A) + (−1.92 × 10−4S)(0.5 V − 0.7 V) + (32.56 × 10−6A)(112.5 − 90)

= (91)(9.8

μ

A) + (1.92 × 10−4S)(0.2 V) + (32.56 × 10−6A)(22.5)

= 8.92 × 10−4A + 0.384 × 10−4A + 7.326 × 10−4A

= 16.63 × 10−4A

≅ 1.66 mA

49

53. For the emitter-bias:

(a)

S(ICO) = (

β

+ 1) (1 / ) (1 510 k / 1.5 k )

(100 1)

( 1) / (100 1) 510 k /1.5 k

BE

RR

β

++ΩΩ

=+

++ ++ Ω Ω

= 78.1

(b)

S(VBE) = 100

( 1) 510 k (100 1)1.5 k

BE

RR

β

−−

=

++ Ω+ + Ω

= −1.512 × 10−4S

(c)

S(

β

) = 1

12

(1 / ) 2.92 mA(1 + 340)

(1 / ) 100(1 125 340)

CBE

BE

IRR

RR

ββ

+=

++ + +

= 21.37 × 10−6A

(d)

ΔIC = S(ICO)ΔICO + S(VBE)ΔVBE + S(

β

)Δ

β

= (78.1)(9.8

μ

A) + (−1.512 × 10−14S)(−0.2 V) + (21.37 × 10−6 A)(25)

= 0.7654 mA + 0.0302 mA + 0.5343 mA

= 1.33 mA

54. (a)

RTh = 62 kΩ || 9.1 kΩ = 7.94 kΩ

S(ICO) = (

β

+ 1) 1/ (1 7.94 k / 0.68 k )

(80 1)

(1) / (801)7.94 k/0.68 k

Th E

RR

β

++Ω Ω

=+

++ ++ Ω Ω

= (81)(1 11.68)

81 11.68

+

+ = 11.08

(b)

S(VBE) = 80

( 1) 7.94 k (81)(0.68 k )

Th E

RR

β

−−

=

++ Ω+ Ω

= 80

7.94 k 55.08 k

−

Ω+ Ω = −1.27 × 10−3S

(c)

S(

β

) = 1

12

(1 / ) 1.71 mA(1 + 7.94 k / 0.68 k )

(1 / ) 80(1 100 7.94 k / 0.68 k )

CThE

Th E

IRR

RR

ββ

+

Ω

=

++ + + Ω Ω

= 1.71 mA(12.68)

80(112.68) = 2.41 × 10−6 A

(d)

ΔIC = S(ICO)ΔICO + S(VBE) ΔVBE + S(

β

)Δ

β

= (11.08)(10

μ

A − 0.2

μ

A) + (−1.27 × 10−3S)(0.5 V − 0.7 V) + (2.41 × 10−6A)(100 − 80)

= (11.08)(9.8

μ

A) + (−1.27 × 10−3S)(−0.2 V) + (2.41 × 10−6A)(20)

= 1.09 × 10−4A + 2.54 × 10−4A + 0.482 × 10−4A

= 4.11 × 10−4A = 0.411 mA

50

55. For collector-feedback bias:

(a)

S(ICO) = (

β

+ 1) (1 / ) (1 560 k / 3.9 k )

(196.32 1)

( 1) / (196.32 1) 560 k / 3.9 k

BC

RR

β

++ΩΩ

=+

++ ++ Ω Ω

= (197.32) 1 143.59

(197.32 143.59)

+

= 83.69

(b) S(VBE) = 196.32

( 1) 560 k (196.32 1)3.9 k

BC

RR

β

−−

=

++ Ω+ + Ω

= −1.477 × 10−4S

(c) S(

β

) = 1

12

() 2.56 mA(560 k 3.9 k )

( ( 1)) 196.32(560 k 3.9 k (245.4 1))

CB C

BC

IR R

RR

ββ

+Ω+ Ω

=

++ Ω+Ω+

= 4.83 × 10−6A

(d) ΔIC = S(ICO)ΔICO + S(VBE) ΔVBE + S(

β

)Δ

β

= (83.69)(9.8

μ

A) + (−1.477 × 10−4S)(−0.2 V) + (4.83 × 10−6A)(49.1)

= 8.20 × 10−4A + 0.295 × 10−4A + 2.372 × 10−4A

= 10.867 × 10−4A = 1.087 mA

56. Type S(ICO) S(VBE) S(

β

)

Collector feedback 83.69

−1.477 × 10−4S 4.83 × 10−6 A

Emitter-bias 78.1

−1.512 × 10−4S 21.37 × 10−6 A

Voltage-divider 11.08

−12.7 × 10−4S 2.41 × 10−6A

Fixed-bias 91

−1.92 × 10−4S 32.56 × 10−6 A

S(ICO): Considerably less for the voltage-divider configuration compared to the other three.

S(VBE): The voltage-divider configuration is more sensitive than the other three (which have

similar levels of sensitivity).

S(

β

): The voltage-divider configuration is the least sensitive with the fixed-bias

configuration very sensitive.

In general, the voltage-divider configuration is the least sensitive with the fixed-bias the most

sensitive.

57. (a) Fixed-bias:

S(ICO) = 91, ΔIC = 0.892 mA

S(VBE) = −1.92 × 10−4S, ΔIC = 0.0384 mA

S(

β

) = 32.56 × 10−6A, ΔIC = 0.7326 mA

(b) Voltage-divider bias:

S(ICO) = 11.08, ΔIC = 0.1090 mA

S(VBE) = −1.27 × 10−3S, ΔIC = 0.2540 mA

S(

β

) = 2.41 × 10−6A, ΔIC = 0.0482 mA

51

(c) For the fixed-bias configuration there is a strong sensitivity to changes in ICO and

β

and

less to changes in VBE.

For the voltage-divider configuration the opposite occurs with a high sensitivity to

changes in VBE and less to changes in ICO and

β

.

In total the voltage-divider configuration is considerably more stable than the fixed-bias

configuration.

52

Chapter 5

1. (a) If the dc power supply is set to zero volts, the amplification will be zero.

(b) Too low a dc level will result in a clipped output waveform.

(c) Po = I2R = (5 mA)22.2 kΩ = 55 mW

Pi = VCCI = (18 V)(3.8 mA) = 68.4 mW

(ac) 55 mW

(dc) 68.4 mW

o

i

P

η

== = 0.804 ⇒ 80.4%

2. −

3. xC = 11

22(1 kHz)(10F)fC

π

πμ

= = 15.92 Ω

f = 100 kHz: xC = 0.159 Ω

Yes, better at 100 kHz

4. −

5. (a) Zi = 10 mV

0.5 mA

i

V

I=

= 20 Ω (=re)

(b) Vo = IcRL

=

α

IcRL

= (0.98)(0.5 mA)(1.2 kΩ)

= 0.588 V

(c) Av = 0.588 V

10 mV

o

i

V

V=

= 58.8

(d) Zo = ∞ Ω

(e) Ai = o

i

I

I = e

e

I

α

=

α

= 0.98

(f) Ib = Ie − Ic

= 0.5 mA − 0.49 mA

= 10

μ

A

53

6. (a) re = 48 mV

3.2 mA

i

V

I= = 15 Ω

(b) Zi = re = 15 Ω

(c) IC =

α

Ie = (0.99)(3.2 mA) = 3.168 mA

(d) Vo = ICRL = (3.168 mA)(2.2 kΩ) = 6.97 V

(e) Av = 6.97 V

48 mV

o

i

V

V= = 145.21

(f) Ib = (1 −

α

)Ie = (1 − 0.99)Ie = (0.01)(3.2 mA)

= 32

μ

A

7. (a) re = 26 mV 26 mV

(dc) 2 mA

E

I= = 13 Ω

Zi =

β

re = (80)(13 Ω)

= 1.04 kΩ

(b) Ib = 11

Ce ee

II II

β

α

βββ ββ

== ⋅=

++

= 2 mA

81 = 24.69

μ

A

(c)

Ai = oL

ib

II

=

I

L = ()

ob

oL

rI

rR

β

+

A

i =

o

b

oL o

boL

rI

rR r

IrR

β

⋅

+=⋅

+

= 40 k (80)

40 k 1.2 k

Ω

Ω+ Ω

= 77.67

(d)

Av = 1.2 k 40 k

13

Lo

e

Rr

r

ΩΩ

−=− Ω

= 1.165 k

13

Ω

−Ω

= −89.6

54

8. (a)

Zi =

β

re = (140)re = 1200

re = 1200

140 = 8.571 Ω

(b) Ib = 30 mV

1.2 k

i

V

Z=Ω = 25

μ

A

(c) Ic =

β

Ib = (140)(25

μ

A) = 3.5 mA

(d)

IL = (50 k )(3.5 mA)

50 k 2.7 k

oc

oL

rI

rR

Ω

=

+Ω+Ω

= 3.321 mA

A

i = 3.321 mA

25 A

L

i

I

μ

= = 132.84

(e)

Av = oiL

ii

VAR

VZ

−

= = (2.7 k )

(132.84) 1.2 k

Ω

−

Ω

= −298.89

9. (a)

re: IB = 12 V 0.7 V

220 k

CC BE

B

VV

R

−−

=Ω = 51.36

μ

A

IE = (

β

+ 1)IB = (60 + 1)(51.36

μ

A)

= 3.13 mA

re = 26 mV 26 mV

3.13 mA

E

I= = 8.31 Ω

Z

i = RB ||

β

re = 220 kΩ || (60)(8.31 Ω) = 220 kΩ || 498.6 Ω

= 497.47 Ω

ro ≥ 10RC ∴ Zo = RC = 2.2 kΩ

(b)

Av = 2.2 k

8.31

C

e

R

r

−Ω

−= Ω = −264.74

(c) Zi = 497.47 Ω (the same)

Z

o = ro || RC = 20 kΩ || 2.2 kΩ

= 1.98 kΩ

(d)

Av = 1.98 k

8.31

Co

e

Rr

r

−−Ω

=Ω = −238.27

A

i = −AvZi/RC

= −(−238.27)(497.47 Ω)/2.2 kΩ

= 53.88

55

10. Av = C

e

R

r

− ⇒ re = 4.7 k

(200)

C

v

R

A

Ω

−=−

− = 23.5 Ω

r

e = 26 mV

E

I ⇒ IE = 26 mV 26 mV

23.5

e

r=

Ω

= 1.106 mA

I

B = 1.106 mA

191

E

I

β

=

+ = 12.15

μ

A

I

B = CC BE

B

VV

R

− ⇒ VCC = IBRB + VBE

= (12.15

μ

A)(1 MΩ) + 0.7 V

= 12.15 V + 0.7 V

= 12.85 V

11. (a) IB = 10 V 0.7 V

390 k

CC BE

B

VV

R

−−

=Ω = 23.85

μ

A

IE = (

β

+ 1)IB = (101)(23.85

μ

A) = 2.41 mA

r

e = 26 mV 26 mV

2.41 mA

E

I= = 10.79 Ω

I

C =

β

IB = (100)(23.85

μ

A) = 2.38 mA

(b) Zi = RB ||

β

re = 390 kΩ || (100)(10.79 Ω) = 390 kΩ || 1.08 kΩ

= 1.08 kΩ

ro ≥ 10RC ∴Zo = RC = 4.3 kΩ

(c) Av = 4.3 k

10.79

C

e

R

r

−Ω

−= Ω = −398.52

(d) Av = (4.3 k)(30 k) 3.76 k

10.79 10.79

Co

e

Rr

r

ΩΩ

Ω

−=− =−

Ω

Ω = −348.47

12. (a) Test

β

RE ≥ 10R2

(100)(1.2 kΩ)

?

≥ 10(4.7 kΩ)

120 kΩ > 47 kΩ (satisfied)

Use approximate approach:

VB = 2

12

4.7 k (16 V)

39 k 4.7 k

CC

RV

RR

Ω

=

+Ω+Ω

= 1.721 V

V

E = VB − VBE = 1.721 V − 0.7 V = 1.021 V

I

E = 1.021 V

1.2 k

E

V

R=Ω = 0.8507 mA

r

e = 26 mV 26 mV

0.8507 mA

E

I= = 30.56 Ω

56

(b)

Zi = R1 || R2 ||

β

re

= 4.7 kΩ || 39 kΩ || (100)(30.56 Ω)

= 1.768 kΩ

ro ≥ 10RC ∴ Zo ≅ RC = 3.9 kΩ

(c) Av = 3.9 k

30.56

C

e

R

r

Ω

−=− Ω = −127.6

(d)

ro = 25 kΩ

(b) Zi(unchanged) = 1.768 kΩ

Z

o = RC || ro = 3.9 kΩ || 25 kΩ = 3.37 kΩ

(c) Av = ( ) (3.9 k)(25 k) 3.37 k

30.56 30.56

Co

e

Rr

r

ΩΩ

Ω

−=− =−

Ω

= −110.28 (vs. −127.6)

13.

β

RE

?

≥ 10R2

(100)(1 kΩ) ≥ 10(5.6 kΩ)

100 kΩ > 56 kΩ (checks!) & ro ≥ 10RC

Use approximate approach:

Av = 3.3 k

160

CC

e

ev

RR

r

rA

Ω

− ⇒ =− =− − = 20.625 Ω

re = 26 mV 26 mV 26 mV

20.625

E

Ee

I

Ir

⇒= =

Ω

= 1.261 mA

I

E = E

E

V

R⇒VE = IERE = (1.261 mA)(1 kΩ) = 1.261 V

V

B = VBE + VE = 0.7 V + 1.261 V = 1.961 V

V

B = 5.6 k

5.6 k 82 k

CC

VΩ

Ω+ Ω = 1.961 V

5.6 kΩ VCC = (1.961 V)(87.6 kΩ)

VCC = 30.68 V

14. Test

β

RE ≥ 10R2

(180)(2.2 kΩ)

?

≥ 10(56 kΩ)

396 kΩ < 560 kΩ (not satisfied)

Use exact analysis:

(a)

RTh = 56 kΩ || 220 kΩ = 44.64 kΩ

E

Th = 56 k (20 V)

220 k 56 k

Ω

+Ω

= 4.058 V

I

B = 4.058 V 0.7 V

( 1) 44.64 k (181)(2.2 k )

Th BE

Th E

EV

RR

β

−−

=

++ Ω+ Ω

57

= 7.58

μ

A

IE = (

β

+ 1)IB = (181)(7.58

μ

A)

= 1.372 mA

re = 26 mV 26 mV

1.372 mA

E

I= = 18.95 Ω

(b) VE = IERE = (1.372 mA)(2.2 kΩ) = 3.02 V

V

B = VE + VBE = 3.02 V + 0.7 V

= 3.72 V

VC = VCC − ICRC

= 20 V −

β

IBRC = 20 V − (180)(7.58

μ

A)(6.8 kΩ)

= 10.72 V

(c)

Zi = R1 || R2 ||

β

re

= 56 kΩ || 220 kΩ || (180)(18.95 kΩ)

= 44.64 kΩ || 3.41 kΩ

= 3.17 kΩ

ro < 10RC ∴ Av = Co

e

R

r

−

= (6.8 k)(50 k)

18.95

Ω

−Ω

= −315.88

15. (a)

IB = 20 V 0.7 V

( 1) 390 k (141)(1.2 k )

CC BE

BE

VV

RR

β

−

=

++ Ω+ Ω

= 19.3 V

559.2 kΩ = 34.51

μ

A

IE = (

β

+ 1)IB = (140 + 1)(34.51

μ

A) = 4.866 mA

r

e = 26 mV 26 mV

4.866 mA

E

I= = 5.34 Ω

(b)

Zb =

β

re + (

β

+ 1)RE

= (140)(5.34 kΩ) + (140 + 1)(1.2 kΩ) = 747.6 Ω + 169.9 kΩ

= 169.95 kΩ

Zi = RB || Zb = 390 kΩ || 169.95 kΩ = 118.37 kΩ

Z

o = RC = 2.2 kΩ

(c)

Av = C

b

R

Z

β

− = (140)(2.2 k )

169.95 k

Ω

−

Ω

= −1.81

(d)

Zb =

β

re + (1) /

1( )/

Co

E

CEo

Rr

R

RRr

β

⎡⎤

++

⎢⎥

++

⎣⎦

= 747.6 Ω (141) 2.2 k / 20 k

1 (3.4 k ) / 20 k

⎡⎤

+ΩΩ

⎢⎥

+ΩΩ

⎣⎦

1.2 kΩ

58

= 747.6 Ω + 144.72 kΩ

= 145.47 kΩ

Zi = RB || Zb = 390 kΩ || 145.47 kΩ = 105.95 kΩ

Z

o = RC = 2.2 kΩ (any level of ro)

A

v =

1

CeC

boo

o

C

i

o

R

rR

Z

rr

V

R

V

r

β

⎡⎤

−++

⎢⎥

⎣⎦

=

+

=

(140)(2.2 k ) 5.34 2.2 k

1

145.47 k 20 k 20 k

2.2 k

120 k

−ΩΩΩ

⎡⎤

++

⎢⎥

ΩΩΩ

⎣⎦

Ω

+Ω

= 2.117 0.11

1.11

−+

= −1.81

16. Even though the condition ro ≥ 10RC is not met it is sufficiently close to permit the use of the

approximate approach.

Av = CCC

bEE

R

RR

Z

RR

β

−=−=−

= −10

∴ RE = 8.2 k

10 10

C

RΩ

= = 0.82 kΩ

IE = 26 mV 26 mV

3.8

e

r=Ω = 6.842 mA

V

E = IERE = (6.842 mA)(0.82 kΩ) = 5.61 V

V

B = VE + VBE = 5.61 V + 0.7 V = 6.31 V

I

B = 6.842 mA

( 1) 121

E

I

β

=

+ = 56.55

μ

A

and RB = 20 V 6.31 V

56.55 A

B

RCC B

BB

VVV

II

μ

−−

== = 242.09 kΩ

17. (a) dc analysis the same

∴ re = 5.34 Ω (as in #15)

(b) Zi = RB || Zb = RB ||

β

re = 390 kΩ || (140)(5.34 Ω) = 746.17 Ω vs. 118.37 kΩ in #15

Zo = RC = 2.2 kΩ (as in #15)

(c) Av = 2.2 k

5.34

C

e

R

r

−−Ω

=Ω = −411.99 vs −1.81 in #15

(d) Zi = 746.17 Ω vs. 105.95 kΩ for #15

Z

o = RC || ro = 2.2 kΩ || 20 kΩ = 1.98 kΩ vs. 2.2 kΩ in #15

59

A

v = 1.98 k

5.34

Co

e

Rr

r

Ω

−=−

Ω

= −370.79 vs. −1.81 in #15

Significant difference in the results for Av.

18. (a) IB = (1)

CC BE

B

E

VV

R

β

−

++

= 22 V 0.7 V 21.3 V

330 k (81)(1.2 k 0.47 k ) 465.27 k

−=

Ω+ Ω+ Ω Ω

= 45.78

μ

A

IE = (

β

+ 1)IB = (81)(45.78

μ

A) = 3.71 mA

r

e = 26 mV 26 mV

3.71 mA

E

I= = 7 Ω

(b) ro < 10(RC + RE)

∴Zb =

β

re + (1) /

1( )/

Co

E

CEo

Rr

R

RRr

β

⎡⎤

++

⎢⎥

++

⎣⎦

= (80)(7 Ω) + (81) 5.6 k / 40 k

1 6.8 k / 40 k

+

ΩΩ

⎡⎤

⎢⎥

+Ω Ω

⎣⎦

1.2 kΩ

= 560 Ω + 81 0.14

10.17

+

⎡⎤

⎢⎥

+

⎣⎦

1.2 kΩ

(note that (

β

+ 1) = 81  RC/ro = 0.14)

= 560 Ω + [81.14 /1.17]1.2 kΩ = 560 Ω + 83.22 kΩ

= 83.78 kΩ

Zi = RB || Zb = 330 kΩ || 83.78 kΩ = 66.82 kΩ

A

v =

1

CeC

boo

C

o

R

rR

Z

rr

R

r

β

⎛⎞

−++

⎜⎟

⎝⎠

+

=

(80)(5.6 k ) 7 5.6 k

1

83.78 k 40 k 40 k

15.6 k/40 k

−ΩΩΩ

⎛⎞

++

⎜⎟

ΩΩΩ

⎝⎠

+ΩΩ

= (5.35) 0.14

10.14

−+

+

= −4.57

19. (a) IB = 16 V 0.7 V 15.3 V

( 1) 270 k (111)(2.7 k ) 569.7 k

CC BE

BE

VV

RR

β

−−

==

++ Ω+ Ω Ω

60

= 26.86

μ

A

IE = (

β

+ 1)IB = (110 + 1)(26.86

μ

A)

= 2.98 mA

r

e = 26 mV 26 mV

2.98 mA

E

I= = 8.72 Ω

β

re = (110)(8.72 Ω) = 959.2 Ω

(b) Zb =

β

re + (

β

+ 1)RE

= 959.2 Ω + (111)(2.7 kΩ)

= 300.66 kΩ

Zi = RB || Zb = 270 kΩ || 300.66 kΩ

= 142.25 kΩ

Zo = RE || re = 2.7 kΩ || 8.72 Ω = 8.69 Ω

(c) Av = 2.7 k

2.7 k 8.69

E

Ee

R

Rr

Ω

=

+Ω+Ω

≅ 0.997

20. (a) IB = 8 V 0.7 V

( 1) 390 k (121)5.6 k

CE BE

BE

VV

RR

β

−

=

++ Ω+ Ω

= 6.84

μ

A

IE = (

β

+ 1)IB = (121)(6.84

μ

A) = 0.828 mA

r

e = 26 mV 26 mV

0.828 mA

E

I= = 31.4 Ω

r

o < 10RE:

Z

b =

β

re + (1)

1/

E

o

R

r

β

+

= (120)(31.4 Ω) + (121)(5.6 k )

1 5.6 k /40 k

Ω

+ΩΩ

= 3.77 kΩ + 594.39 kΩ

= 598.16 kΩ

Zi = RB || Zb = 390 kΩ || 598.16 kΩ

= 236.1 kΩ

Zo ≅ RE || re

= 5.6 kΩ || 31.4 Ω

= 31.2 Ω

(b) Av = (1)/

1/

E

b

Eo

R

Z

R

r

β

+

= (121)(5.6 k ) / 598.16 k

1 5.6 k / 40 k

ΩΩ

+ΩΩ

= 0.994

(c) Av = 0

i

V

V = 0.994

V

o = AvVi = (0.994)(1 mV) = 0.994 mV

61

21. (a) Test

β

RE

?

≥ 10R2

(200)(2 kΩ) ≥ 10(8.2 kΩ)

400 kΩ ≥ 82 kΩ (checks)!

Use approximate approach:

VB = 8.2 k (20 V)

8.2 k 56 k

Ω

Ω+ Ω = 2.5545 V

VE = VB − VBE = 2.5545 V − 0.7 V ≅ 1.855 V

I

E = 1.855 V

2 k

E

V

R=Ω = 0.927 mA

I

B = 0.927 mA

( 1) (200 + 1)

E

I

β

=

+ = 4.61

μ

A

I

C =

β

IB = (200)(4.61

μ

A) = 0.922 mA

(b) re = 26 mV 26 mV

0.927 mA

E

I= = 28.05 Ω

(c) Zb =

β

re + (

β

+ 1)RE

= (200)(28.05 Ω) + (200 + 1)2 kΩ

= 5.61 kΩ + 402 kΩ = 407.61 kΩ

Zi = 56 kΩ || 8.2 kΩ || 407.61 kΩ

= 7.15 kΩ || 407.61 kΩ

= 7.03 kΩ

Zo = RE || re = 2 kΩ || 28.05 Ω = 27.66 Ω

(d) Av = 2 k

2 k 28.05

E

Ee

R

Rr

Ω

=

+Ω+Ω

= 0.986

22. (a) IE = 6 V 0.7 V

6.8 k

EE BE

E

VV

R

−−

=Ω = 0.779 mA

re = 26 mV 26 mV

0.779 mA

E

I= = 33.38 Ω

(b) Zi = RE || re = 6.8 kΩ || 33.38 Ω

= 33.22 Ω

Zo = RC = 4.7 kΩ

(c) Av = (0.998)(4.7 k )

33.38

C

e

R

r

α

Ω

=Ω

= 140.52

23.

α

= 75

176

β

=

+ = 0.9868

62

IE = 5 V 0.7 V 4.3 V

3.9 k 3.9 k

EE BE

E

VV

R

−−

==

ΩΩ

= 1.1 mA

r

e = 26 mV 26 mV

1.1 mA

E

I= = 23.58 Ω

A

v =

α

(0.9868)(3.9 k )

23.58

C

e

R

r

Ω

=Ω = 163.2

24. (a) IB = 12 V 0.7 V

220 k 120(3.9 k )

CC BE

FC

VV

RR

β

−

=

+Ω+Ω

= 16.42

μ

A

IE = (

β

+ 1)IB = (120 + 1)(16.42

μ

A)

= 1.987 mA

re = 26 mV 26 mV

1.987 mA

E

I= = 13.08 Ω

(b) Zi =

β

re ||

F

v

R

A

Need Av!

A

v = 3.9 k

13.08

C

e

R

r

−−Ω

=Ω = −298

Zi = (120)(13.08 Ω) || 220 k

298

Ω

= 1.5696 kΩ || 738 Ω

= 501.98 Ω

Zo = RC || RF = 3.9 kΩ || 220 kΩ

= 3.83 kΩ

(c) From above,

Av = −298

25. Av = C

e

R

r

− = −160

R

C = 160(re) = 160(10 Ω) = 1.6 kΩ

Ai = F

F

C

R

β

+ = 19 ⇒ 19 = 200

200(1.6 k )

F

R

+

Ω

19RF + 3800RC = 200RF

RF = 3800

181

C

R

= 3800(1.6 k )

181

Ω

= 33.59 kΩ

IB = CC BE

F

C

VV

R

β

−

+

I

B(RF +

β

RC) = VCC − VBE

63

and VCC = VBE + IB(RF +

β

RC)

with

IE = 26 mV 26 mV

10

e

r=Ω = 2.6 mA

IB = 2.6 mA

1 200 1

E

I

β

=

++

= 12.94

μ

A

∴VCC = VBE + IB(RF +

β

RC)

= 0.7 V + (12.94

μ

A)(33.59 kΩ + (200)(1.6 kΩ))

= 5.28 V

26.

(a)

Av: Vi = Ib

β

re + (

β

+ 1)IbRE

Io + I′ = IC =

β

Ib

but

Ii = I′ + Ib

and

I′ = Ii − Ib

Substituting,

Io + (Ii − Ib) =

β

Ib

and

Io = (

β

+ 1)Ib − Ii

Assuming (

β

+ 1)Ib  Ii

Io ≅ (

β

+ 1)Ib

and

Vo = −IoRC = −(

β

+ 1)IbRC

Therefore, (1)

(1)

obC

ibe bE

VIR

VIr IR

β

ββ

−+

=++

bC

be b E

IR

Ir IR

β

ββ

≅+

and

Av = oCC

ieE E

VR R

VrR R

≅− ≅−

+

(b)

Vi ≅

β

Ib(re + RE)

For re  RE

Vi ≅

β

IbRE

Now

Ii = I′ + Ib

= io

b

F

VV I

R

−+

64

Since

Vo  Vi

Ii = o

b

F

VI

R

−+

or

Ib = Ii + o

F

V

R

and

Vi =

β

IbRE

Vi =

β

REIi + o

E

F

V

R

β

but Vo = AvVi

and

Vi =

β

REIi + vi E

F

AVR

R

β

or

Vi − vEi

F

ARV

R

β

=

β

REIi

Vi

[]

1vE

E

i

F

AR

R

I

R

ββ

⎡⎤

−=

⎢⎥

⎣⎦

so Zi = ()

1

iEEF

vE

iFvE

F

VRRR

AR

IRAR

R

β

ββ

==

+−

−

Zi = i

i

V

I = x || y where x =

β

RE and y = RF/|Av|

with Zi = ()(/)

/

E

Fv

E

Fv

R

RA

xy

x

yRRA

β

⋅=

++

Zi ≅

β

EF

E

vF

RR

RA R

+

Zo: Set Vi = 0

Vi = Ib

β

re + (

β

+ 1)IbRE

Vi ≅

β

Ib(re + RE) = 0

since

β

, re + RE ≠ 0 Ib = 0 and

β

Ib = 0

65

∴ Io = 11

oo

o

CF CF

VV

V

R

RRR

⎡⎤

+= +

⎢⎥

⎣⎦

and Zo = 1

11

oCF

CF

VRR

IRR

RR

==

+

= RC || RF

(c) Av ≅ 2.2 k

1.2 k

C

E

R

Ω

−=− Ω = −1.83

Z

i ≅ (90)(1.2 k )(120 k )

(90)(1.2 k )(1.83) 120 k

EF

Ev F

RR

RA R

β

ΩΩ

=

+Ω+Ω

= 40.8 kΩ

Zo ≅ RC || RF

= 2.2 kΩ || 120 kΩ

= 2.16 kΩ

27. (a) IB = 9 V 0.7 V

(39 k 22 k ) (80)(1.8 k )

CC BE

FC

VV

RR

β

−−

=

+Ω+Ω+Ω

= 8.3 V 8.3 V

61 k 144 k 205 k

=

Ω+ Ω Ω = 40.49

μ

A

IE = (

β

+ 1)IB = (80 + 1)(40.49

μ

A) = 3.28 mA

r

e = 26 mV 26 mV

3.28 mA

E

I= = 7.93 Ω

Z

i = 1

F

R

|| e

r

β

= 39 kΩ || (80)(7.93 Ω) = 39 kΩ || 634.4 Ω = 0.62 kΩ

Zo = RC || 2

F

R

= 1.8 kΩ || 22 kΩ = 1.66 kΩ

(b) Av = 21.8 k 22 k

7.93

CF

ee

RR

R

rr

−

Ω

′

−==−

Ω

= 1.664 k

7.93

−Ω

Ω = −209.82

28. Ai ≅

β

= 60

29. Ai ≅

β

= 100

30. Ai = −AvZi/RC = −(−127.6)(1.768 kΩ)/3.9 kΩ = 57.85

31. (c)

Ai = (140)(390 k )

390 k 0.746 k

B

Bb

R

RZ

β

Ω

=

+Ω+Ω

= 139.73

(d) Ai = i

v

C

Z

A

R

− = −(−370.79)(746.17 Ω)/2.2 kΩ

= 125.76

66

32. Ai = −AvZi/RE = −(0.986)(7.03 kΩ)/2 kΩ = −3.47

33. Ai = oe

ie

II

α

= =

α

= 0.9868 ≅ 1

34. Ai = −AvZi/RC = −(−298)(501.98 Ω)/3.9 kΩ = 38.37

35. Ai = ( 209.82)(0.62 k )

1.8 k

i

v

C

Z

AR

−− Ω

−= Ω = 72.27

36. (a)

IB = 18 V 0.7 V

680 k

CC BE

B

VV

R

−−

=Ω = 25.44

μ

A

IE = (

β

+ 1)IB = (100 + 1)(25.44

μ

A)

= 2.57 mA

re = 26 mV

2.57 mA = 10.116 Ω

3.3 k

10.116

NL

C

v

e

R

Ar

Ω

=− =− Ω = −326.22

Zi = RB ||

β

re = 680 kΩ || (100)(10.116 Ω)

= 680 kΩ || 1,011.6 Ω

= 1.01 kΩ

Zo = RC = 3.3 kΩ

(b) −

(c)

L

v

A = 4.7 k

4.7 k 3.3 k

NL

L

v

Lo

RA

RR

Ω

=

+Ω+Ω

(−326.22)

= −191.65

(d)

L

i

A = (1.01 k )

( 191.65) 4.7 k

L

i

v

L

Z

AR

Ω

−=−−

Ω

= 41.18

(e)

L

v

A = ()

100(1.939 k )

( ) 100(10.116 )

bC L

o

ibe

IRR

V

VIr

β

−

Ω

==

Ω

= −191.98

Zi = RB ||

β

re = 1.01 kΩ

I

L = ()

Cb

CL

R

I

R

β

+ = 41.25Ib

Ib = Bi

B

e

R

I

R

r

β

+ = 0.9985Ii

L

i

A = ob

LL

iibi

II

IIII

==⋅

= (41.25)(0.9985)

= 41.19

Zo = RC = 3.3 kΩ

67

37. (a)

N

L

v

A = −326.22

L

v

A =

N

L

v

Lo

RA

RR+

R

L = 4.7 kΩ:

L

v

A = 4.7 k ( 326.22)

4.7 k 3.3 k

Ω

−

Ω

+Ω = −191.65

RL = 2.2 kΩ:

L

v

A = 2.2 k ( 326.22)

2.2 k 3.3 k

Ω−

Ω+ Ω = −130.49

R

L = 0.5 kΩ:

L

v

A = 0.5 k ( 326.22)

0.5 k 2.3 k

Ω−

Ω+ Ω = −42.92

As RL↓,

L

v

A↓

(b) No change for Zi, Zo, and

N

L

v

A!

38. (a)

IB = 12 V 0.7 V

1 M

CC BE

B

VV

R

−−

=Ω = 11.3

μ

A

IE = (

β

+ 1)IB = (181)(11.3

μ

A) = 2.045 mA

r

e = 26 mV 26 mV

2.045 mA

E

I= = 12.71 Ω

N

L

V

A = 3 k

12.71

C

e

R

r

Ω

−=−

Ω

= −236

Zi = RB ||

β

re = 1 MΩ || (180)(12.71 Ω) = 1 MΩ || 2.288 kΩ

= 2.283 kΩ

Zo = RC = 3 kΩ

(b) −

(c) No-load:

Av =

N

L

v

A = −236

(d) 2.283 k ( 236)

2.283 k 0.6 k

sNL

i

vv

is

Z

AA

ZR

Ω−

==

+Ω+Ω

= −186.9

(e)

Vo = −IoRC = −

β

IbRC

Vi = Ib

β

re

Av = 3 k

12.71

obCC

ibee

VIRR

VIrr

β

Ω

=− =− =−

Ω

= −236

s

ooi

v

s

is

VVV

AVVV

==⋅

Vi = (1 M ) 2.288 k ( )

(1 M ) 2.288 k 0.6 k

es s

es

rV V

rR

β

ΩΩ

=

Ω+ Ω+Ω

= 0.792 Vs

s

v

A = (−236)(0.792)

= −186.9 (same results)

68

(f) No change!

(g) 2.283 k ( 236)

()

2.283 k 1 k

sNL

i

vv

is

Z

AA

ZR

Ω−

==

+Ω+Ω

= −164.1

Rs↑,

s

v

A↓

(h) No change!

39. (a)

IB = 24 V 0.7 V

500 k

CC BE

B

VV

R

−−

=Ω = 41.61

μ

A

IE = (

β

+ 1)IB = (80 + 1)(41.61

μ

A) = 3.37 mA

r

e = 26 mV 26 mV

3.37 mA

E

I= = 7.715 Ω

N

L

v

A = 4.3 k

7.715

L

e

R

r

Ω

−=− Ω = −557.36

Zi = RB ||

β

re = 560 kΩ || (80)(7.715 Ω)

= 560 kΩ || 617.2 Ω

= 616.52 Ω

Zo = RC = 4.3 kΩ

(b) −

(c)

L

v

A = 2.7 k ( 557.36)

2.7 k 4.3 k

NL

oL

v

iLo

VRA

VRR

Ω−

==

+Ω+Ω

= −214.98

s

ooi

v

s

is

VVV

AVVV

==⋅

Vi = 616.52

616.52 1 k

is s

is

ZV V

ZR

Ω

=

+Ω+Ω

= 0.381 Vs

( 214.98)(0.381)

s

v

A=−

= −81.91

(d) 1 k 616.52

(81.91) 2.7 k

ss

si

iv

L

RZ

AAR

⎛⎞

+

Ω

+Ω

⎛⎞

=− =− −

⎜⎟ ⎜⎟

Ω

⎝⎠

=

49.04

(e)

L

v

A = 5.6 k ( 557.36)

5.6 k 4.3 k

NL

oL

v

iLo

VRA

VRR

Ω−

==

+Ω+Ω

= −315.27

i

s

V

V the same = 0.381

s

oi

v

is

VV

AVV

=⋅ = (−315.27)(0.381) = −120.12

As

RL↑,

s

v

A↑

69

(f)

L

v

A the same = −214.98

616.52

616.52 0.5 k

ii

sis

VZ

VZR

Ω

==

+Ω+Ω

= 0.552

s

oi

v

is

VV

AVV

=⋅ = (−214.98)(0.552) = −118.67

As

Rs↓,

s

v

A↑

(g) No change!

40. (a) Exact analysis:

ETh = 2

12

16 k (16 V)

68 k 16 k

CC

RV

RR

Ω

=

+Ω+Ω

= 3.048 V

R

Th = R1 || R2 = 68 kΩ || 16 kΩ = 12.95 kΩ

IB = 3.048 V 0.7 V

( 1) 12.95 k (101)(0.75 k )

Th BE

Th E

EV

RR

β

−−

=

++ Ω+ Ω

= 26.47

μ

A

IE = (

β

+ 1)IB = (101)(26.47

μ

A)

= 2.673 mA

re = 26 mV 26 mV

2.673 mA

E

I= = 9.726 Ω

2.2 k

9.726

NL

C

v

e

R

Ar

−Ω

==− Ω = −226.2

Zi = 68 kΩ || 16 kΩ ||

β

re

= 12.95 kΩ || (100)(9.726 Ω)

= 12.95 kΩ || 972.6 Ω

= 904.66 Ω

Zo = RC = 2.2 kΩ

(b) −

(c)

L

v

A = 5.6 k ( 226.2)

()

5.6 k 2.2 k

NL

L

v

Lo

RA

RZ

Ω−

=

+Ω+Ω

= −162.4

(d)

L

i

A = L

i

v

L

Z

A

R

−

= −(−162.4) (904.66 )

5.6 k

Ω

= 26.24

70

(e)

L

v

A = 2.2 k 5.6 k

9.726

Ce

e

RR

r

−−ΩΩ

=Ω

= −162.4

Zi = 68 kΩ || 16 kΩ || 972.6 Ω

β

re

= 904.66 Ω

L

i

A = L

i

v

L

Z

A

R

−

= ( 162.4)(904.66 )

5.6 k

−Ω

Ω

= 26.24

Zo = RC = 2.2 kΩ

Same results!

41. (a)

L

v

A =

N

L

v

Lo

RA

RZ+

RL = 4.7 kΩ:

L

v

A = 4.7 k (226.4)

4.7 k 2.2 k

Ω−

Ω+ Ω = −154.2

R

L = 2.2 kΩ:

L

v

A = 2.2 k ( 226.4)

2.2 k 2.2 k

Ω−

Ω+ Ω = −113.2

R

L = 0.5 kΩ:

L

v

A = 0.5 k (226.4)

0.5 k 2.2 k

Ω

−

Ω+ Ω = −41.93

R

L↓,

L

v

A↓

(b) Unaffected!

42. (a)

IB = 18 V 0.7 V

( 1) 680 k (111)(0.82 k )

CC BE

BE

VV

RR

β

−−

=

++ Ω+ Ω

= 22.44

μ

A

IE = (

β

+ 1)IB = (110 + 1)(22.44

μ

A)

= 2.49 mA

re = 26 mV 26 mV

2.49 mA

E

I= = 10.44 Ω

3 k

10.44 0.82 k

NL

C

v

eE

R

ArR

Ω

=− =−

+Ω+Ω

= −3.61

Zi ≅ RB || Zb = 680 kΩ || (

β

re + (

β

+ 1)RE)

= 680 kΩ || (610)(10.44 Ω) + (110 + 1)(0.82 kΩ)

= 680 kΩ || 92.17 kΩ

= 81.17 kΩ

Zo ≅ RC = 3 kΩ

(b) −

71

(c)

L

v

A= 4.7 k ( 3.61)

4.7 k 3 k

NL

oL

v

iLo

VRA

VRR

Ω

−

==

+Ω+Ω

= −2.2

s

ooi

v

s

is

VVV

AVVV

==⋅

Vi = 81.17 k ( )

81.17 k 0.6 k

is s

is

ZV V

ZR

Ω

=

+

Ω+ Ω = 0.992 Vs

s

v

A = (−2.2)(0.992)

= −2.18

(d) None!

(e)

L

v

A – none!

81.17 k

81.17 k 1 k

ii

sis

VZ

VZR

Ω

==

+Ω+Ω

= 0.988

s

v

A = (−2.2)(0.988)

= −2.17

Rs↑,

s

v

A↓, (but only slightly for moderate changes in Rs since Zi is typically much larger

than Rs)

43. Using the exact approach:

IB = (1)

Th BE

Th E

EV

R

β

−

++ ETh = 2

12

CC

RV

RR+

= 2.33 V 0.7 V

10.6 k (121)(1.2 k )

−

Ω+ Ω = 12 k (20 V)

91 k 12 k

Ω

Ω+ Ω = 2.33 V

= 10.46

μ

A RTh = R1 || R2 = 91 kΩ || 12 kΩ = 10.6 kΩ

IE = (

β

+ 1)IB = (121)(10.46

μ

A)

= 1.266 mA

re = 26 mV 26 mV

1.266 mA

E

I= = 20.54 Ω

(a) 1.2 k

20.54 1.2 k

NL

E

v

eE

R

ArR

Ω

≅=

+Ω+Ω

= 0.983

Zi = R1 || R2 || (

β

re + (

β

+1)RE)

= 91 kΩ || 12 kΩ || ((120)(20.54 Ω) + (120 + 1)(1.2 kΩ))

= 10.6 kΩ || (2.46 kΩ + 145.2 kΩ)

= 10.6 kΩ || 147.66 kΩ

= 9.89 kΩ

Zo = RE || re = 1.2 kΩ || 20.54 Ω

= 20.19 Ω

72

(b) −

(c)

L

v

A = 2.7 k (0.983)

2.7 k 20.19

NL

L

v

Lo

RA

RZ

Ω

=

+Ω+Ω

= 0.976

9.89 k (0.976)

9.89 k 0.6 k

sL

i

vv

is

Z

AA

ZR

Ω

==

+Ω+Ω

= 0.92

(d)

L

v

A = 0.976 (unaffected by change in Rs)

9.89 k (0.976)

9.89 k 1 k

sL

i

vv

is

Z

AA

ZR

Ω

==

+Ω+Ω

= 0.886 (vs. 0.92 with Rs = 0.6 kΩ)

As

Rs↑,

s

v

A↓

(e) Changing Rs will have no effect on

N

L

v

A, Zi, or Zo.

(f)

L

v

A =

()

5.6 k (0.983)

5.6 k 20.19

NL

L

v

Lo

RA

RZ

Ω

=

+Ω+Ω

= 0.979 (vs. 0.976 with RL = 2.7 kΩ)

9.89 k (0.979)

()

9.89 k 0.6 k

sL

i

vv

is

Z

AA

ZR

Ω

==

+Ω+Ω

= 0.923 (vs. 0.92 with RL = 2.7 kΩ)

As

RL↑,

L

v

A↑,

s

v

A↑

44. (a)

IE = 6 V 0.7 V

2.2 k

EE BE

E

VV

R

−−

=Ω

= 2.41 mA

re = 26 mV 26 mV

2.41 mA

E

I= = 10.79 Ω

N

L

v

A = 4.7 k

10.79

C

e

R

r

Ω

=Ω = 435.59

Zi = RE || re = 2.2 kΩ || 10.79 Ω = 10.74 Ω

Zo = RC = 4.7 kΩ

(b) −

(c)

L

v

A = 5.6 k (435.59)

5.6 k 4.7 k

NL

L

v

Lo

RA

RR

Ω

=

+Ω+Ω

= 236.83

V

i = 10.74 ( )

10.74 100

is

s

is

ZV

V

ZR

Ω

=

+Ω+Ω

= 0.097 Vs

s

oi

v

is

VV

AVV

=⋅ = (236.83)(0.097)

= 22.97

73

(d)

Vi = Ie ⋅ re

Vo = −IoRL

Io = 4.7 k ( )

4.7 k 5.6 k

e

I−Ω

Ω+ Ω = −0.4563Ie

L

v

A = o

i

V

V = (0.4563 ) 0.4563(5.6 k )

10.79

eL

ee

IR

Ir

+

Ω

=

⋅Ω

= 236.82 (vs. 236.83 for part c)

:

s

v

A 2.2 kΩ || 10.79 Ω = 10.74 Ω

Vi = 10.74 ( )

10.74 100

is

s

is

ZV

V

ZR

Ω

⋅=

+Ω+Ω

= 0.097 Vs

s

oi

v

is

VV

AVV

=⋅ = (236.82)(0.097)

= 22.97 (same results)

(e)

L

v

A = 2.2 k (435.59)

2.2 k 4.7 k

NL

L

v

Lo

RA

RR

Ω

=

+Ω+Ω

= 138.88

010.74

,10.74 500

s

ii i

v

is s i s

VV V Z

AVV V Z R

Ω

=⋅ = =

+

Ω+ Ω = 0.021

s

v

A = (138.88)(0.021) = 2.92

s

v

A very sensitive to increase in Rs due to relatively small Zi; Rs↑,

s

v

A↓

L

v

A sensitive to RL; RL↓,

L

v

A↓

(f)

Zo = RC = 4.7 kΩ unaffected by value of Rs!

(g)

Zi = RE || re = 10.74 Ω unaffected by value of RL!

45. (a)

1

1 k ( 420)

1 k 3.3 k

NL

Lv

v

Lo

RA

ARR

Ω−

==

+Ω+Ω

= −97.67

2

2.7 k ( 420)

2.7 k 3.3 k

NL

Lv

v

Lo

RA

ARR

Ω−

==

+Ω+Ω

= −189

(b)

L

v

A= 12

vv

AA⋅ = (−97.67)(−189) = 18.46 × 103

11

21

s

oi

oo

v

s

iis

VV

AVV VV

==⋅ ⋅

= 21

i

vv

s

V

AA

V

⋅⋅

Vi = 1 k ( )

1 k 0.6 k

is s

is

ZV V

ZR

Ω

=

+Ω+Ω

= 0.625

s

v

A = (−189)(−97.67)(0.625)

= 11.54 × 103

74

(c)

1

(97.67)(1 k )

1 k

vi

i

L

AZ

AR

−− Ω

=− = Ω = 97.67

2

( 189)(1 k )

2.7 k

vi

i

L

AZ

AR

−−− Ω

== Ω = 70

(d) 12L

iii

AAA=⋅ = (97.67)(70) = 6.84 × 103

(e) No effect!

(f) No effect!

(g) In phase

46. (a) 2

11

21

1.2 k (1)

1.2 k 20

NL

i

vv

io

Z

AA

ZZ

Ω

==

+Ω+Ω

= 0.984

2

v

A = 2

2

2.2 k ( 640)

2.2 k 4.6 k

NL

L

v

Lo

RA

RZ

Ω

=−

+Ω+Ω

= −207.06

(b) 12L

vvv

AAA=⋅ = (0.984)(−207.06)

= −203.74

s

L

i

vv

is

Z

AA

ZR

=+

= 50 k ( 203.74)

50 k 1 k

Ω−

Ω+ Ω

= −199.75

(c) 1

11

2

i

iv

i

Z

AA

Z

=−

= (50 k )

(0.984) 1.2 k

Ω

−Ω

= −41

2

22

i

iv

L

Z

AA

R

=−

= (1.2 k )

( 207.06) 2.2 k

Ω

−− Ω

= 112.94

(d) 1

LL

i

iv

L

Z

AA

R

=−

= −(−203.74) (50 k )

2.2 k

Ω

= 4.63 × 103

75

(e) A load on an emitter-follower configuration will contribute to the emitter resistance (in

fact, lower the value) and therefore affect Zi (reduce its magnitude).

(f) The fact that the second stage is a CE amplifier will isolate Zo from the first stage and Rs.

(g) The emitter-follower has zero phase shift while the common-emitter amplifier has a

180° phase shift. The system, therefore, has a total phase shift of 180° as noted by the

negative sign in front of the gain for T

v

A in part b.

47. For each stage:

VB = 6.2 k

24 k 6.2 k

Ω

Ω+ Ω(15 V) = 3.08 V

V

E = VB − 0.7 V = 3.08 V − 0.7 V = 2.38 V

I

E ≅ IC = 2.38 V

1.5 k

E

V

R=Ω = 1.59 mA

V

C = VCC − ICRC = 15 V − (1.59 mA)(5.1 kΩ)

= 6.89 V

48. re = 26 mV 26 mV

1.59 mA

E

I= = 16.35 Ω

2

i

R

= R1 || R2 ||

β

re = 6.2 kΩ || 24 kΩ || (150)(16.35 Ω)

= 1.64 kΩ

2

1

5.1 k 1.64 k

16.35

Ci

v

e

RR

Ar

ΩΩ

=− = Ω = −75.8

2

5.1 k

16.35

C

v

e

R

Ar

−Ω

=− = Ω = −311.9

Av = 12

vv

AA = (−75.8)(−311.9) = 23,642

49. 1

3.9 k (20 V)

3.9 k 6.2 k 7.5 k

B

VΩ

=Ω+ Ω+ Ω = 4.4 V

2

6.2 k 3.9 k (20 V)

3.9 k 6.2 k 7.5 k

B

VΩ+ Ω

=Ω+ Ω+ Ω = 11.48 V

11

E

B

VV= − 0.7 V = 4.4 V − 0.7 V = 3.7 V

1

11

3.7 V

1 k

E

CE

E

V

IIR

≅= =Ω = 3.7 mA ≅ 22

E

C

II

≅

2

C

V = VCC − ICRC = 20 V − (3.7 mA)(1.5 kΩ)

= 14.45 V

76

50. re = 26 mV 26 mV

3.7 mA

E

I= = 7 Ω

1

e

v

e

r

Ar

=− = −1

2

1.5 k

7

E

v

e

R

Ar

Ω

==Ω ≅ 214

12T

vvv

AAA= = (−1)(214) = −214

Vo = T

vi

AV = (−214)(10 mV) = −2.14 V

51. Ro = RD = 1.5 kΩ (Vo (from problem 50) = −2.14 V)

Vo(load) = 10 k

() (2.14 V)

10 k 1.5 k

L

o

oL

RV

RR

Ω

=−

+Ω+Ω

= −1.86 V

52. IB = (16 V 1.6 V)

(6000)(510 ) 2.4 M

CC BE

DE B

VV

RR

β

−−

=

+Ω+Ω

= 14.4 V

5.46 MΩ = 2.64

μ

A

IC ≅ IE =

β

DIB = 6000(2.64

μ

A) = 15.8 mA

V

E = IERE = (15.8 mA)(510 Ω) = 8.06 V

53. From problem 69, IE = 15.8 mA

re = 26 26 V

15.8 mA

E

I= = 1.65 Ω

A

v = 510

1.65 510

E

eE

R

rR

Ω

=

+Ω+Ω

= 0.997 ≈ 1

54. dc: 16 V 1.6 V

2.4 M (6000)(510 )

CC BE

B

BDE

VV

IRR

β

−−

≅=

+Ω+Ω

= 2.64

μ

A

(6000)(2.64 A)

CDB

II

β

μ

== = 15.84 mA

2

26 mV 26 mV

15.84 mA

e

E

rI

== = 1.64 Ω

ac: 2(6000)(1.64 ) 9.84 k

iDe

Zr

β

≅= Ω= Ω

19.84 k

i

b

V

I=Ω

1

( )( ) (6000) (200 )

9.84 k

= 121.95

i

oDbC

V

VIR

Vi

β

⎛⎞

=− =− Ω

⎜⎟

Ω

⎝⎠

−

and

o

v

i

V

AV

=≅ −121.95

77

55. IB = 1

12

16 V 0.7 V

1.5 M (160)(200)(100 )

CC EB

BE

VV

RR

ββ

−−

=

+Ω+Ω

= 3.255

μ

A

IC ≅

β

1

β

2IB = (160)(200)(3.255

μ

A) ≅ 104.2 mA

2

C

V = VCC − ICRC = 16 V − (104.2 mA)(100 Ω) = 5.58 V

1

B

V = IBRB = (3.255

μ

A)(1.5 MΩ) = 4.48 V

56. From problem 55: 1

E

I = 0.521 mA

1

26 mV 26 mV

(mA) 0.521 mA

e

E

rI

== = 49.9 Ω

1i

ie

R

r

β

= = 160(49.9 Ω) = 7.98 kΩ

Av =

1

12

(160)(200)(100 )

(160)(200)(100 ) 7.98 k

C

Ci

R

RR

β

ββ

Ω

=

+Ω+Ω

= 0.9925

Vo = AvVi = 0.9975 (120 mV)

= 119.7 mV

57. re =

(dc)

26 mV 26 mV

1.2 mA

E

I= = 21.67 Ω

β

re = (120)(21.67 Ω) = 2.6 kΩ

58. −

59. −

60. −

61. −

62. (a)

Av = o

i

V

V = −160

Vo = −160 Vi

(b)

Ib = (1 )

ireo irevi i rev

ie ie ie

V hV V hAV V hA

hh h

−− −

==

=

(

)

4

1 (2 10 )(160)

1 k

i

V−

−×

Ω

Ib = 9.68 × 10−4Vi

(c)

Ib = 1 k

i

V

Ω = 1 × 10−3Vi

78

(d) % Difference =

34

3

110 9.6810

110

ii

i

VV

V

−−

−

×−×

× × 100%

= 3.2 %

(e) Valid first approximation

63. % difference in total load = 1/

LL oe

L

R

Rh

R

− × 100%

=

2.2 k (2.2 k 50 k )

2.2 k

Ω− Ω Ω

Ω × 100%

=

2.2 k 2.1073 k

2.2 k

Ω− Ω

Ω × 100%

=

4.2 %

In this case the effect of 1/hoe can be ignored.

64. (a)

Vo = −180Vi (hie = 4 kΩ, hre = 4.05 × 10−4)

(b) Ib =

4

(4.05 10 )(180 )

4 k

ii

VV

−

−×

Ω

= 2.32 × 10−4Vi

(c) Ib = 4 k

ii

ie

VV

h=Ω = 2.5 × 10−4Vi

(d) % Difference =

44

4

2.5 10 2.32 10

2.5 10

ii

i

VV

V

−−

−

×−×

× × 100% = 7.2%

(e) Yes, less than 10%

65. From Fig. 5.18

min max

hoe: 1

μ

S 30

μ

S Avg = (1 30) S

2

μ

+

= 15.5

μ

S

66. (a)

hfe =

β

= 120

hie ≅

β

re = (120)(4.5 Ω) = 540 Ω

h

oe = 11

40 k

o

r=Ω = 25

μ

S

(b) re ≅ 1 k

90

ie

h

β

Ω

= = 11.11 Ω

β

= hfe = 90

r

o = 11

20 S

oe

h

μ

=

= 50 kΩ

79

67. (a)

re = 8.31 Ω (from problem 9)

(b)

f

e

h =

β

= 60

ie

h =

β

re = (60)(8.31 Ω) = 498.6 Ω

(c) Zi = RB || hie = 220 kΩ || 498.6 Ω = 497.47 Ω

Z

o = RC = 2.2 kΩ

(d) Av = (60)(2.2 k )

498.6

fe C

ie

hR

h

−−Ω

=Ω = −264.74

A

i ≅ hfe = 60

(e) Zi = 497.47 Ω (the same)

Z

o = ro || RC, ro = 1

25 S

μ

= 40 kΩ

= 40 kΩ || 2.2 kΩ

= 2.09 kΩ

(f) Av = ()

(60)(2.085 k )

498.6

fe o C

ie

hrR

h

−−Ω

=Ω = −250.90

A

i = −AvZi/RC = −(−250.90)(497.47 Ω)/2.2 kΩ = 56.73

68. (a) 68 kΩ || 12 kΩ = 10.2 kΩ

Zi = 10.2 kΩ || hie = 10.2 kΩ || 2.75 kΩ

= 2.166 kΩ

Zo = RC || ro

= 2.2 kΩ || 40 kΩ

=

2.085 kΩ

(b)

Av =

f

eC

ie

hR

h

′

− C

R

′ = RC || ro = 2.085 kΩ

=

(180)(2.085 k )

2.75 k

−Ω

Ω = −136.5

Ai = ooi

iii

III

′

=

⋅

′

= 10.2 k

1 10.2 k 2.68 k

e

f

oe L

h

hR

⎛⎞ Ω

⎛⎞

⎜⎟

+Ω+Ω

⎝⎠

= 180 (0.792)

1(25 S)(2.2 k)

μ

⎛⎞

⎜⎟

+Ω

⎝⎠

= 135.13

80

69. (a)

Zi = RE || hib

= 1.2 kΩ || 9.45 Ω

= 9.38 Ω

Zo = RC || 1

ob

h = 2.7 kΩ ||

6

1

A

110 V

−

×

= 2.7 kΩ || 1 MΩ ≅ 2.7 kΩ

(b) Av = (1/) ( 0.992)( 2.7 k )

9.45

fb C ob

ib

hR h

h

−−− ≅ Ω

=Ω

= 284.43

Ai ≅ −1

(c)

α

= −hfb = −(−0.992) = 0.992

β

= 0.992

110.992

α

=

−− = 124

r

e = hib = 9.45 Ω

r

o = 11

1 A/V

ob

h

μ

= = 1 MΩ

70. (a)

4

(180)(2 10 )(2.2 k )

2.75 k

1(125S)(2.2 k)

fe re L

iie

oe L

hhR

Zh hR

μ

−

×

Ω

′=− = Ω−

++Ω

2.68 k=Ω

Zi = 10.2 kΩ || i

Z

′ = 2.12 kΩ

4

11

( / ) 25 S (180)(2 10 ) / 2.75 k

83.75 k

o

oe fe re ie

Zhhhh

μ

−

′==

−−×Ω

=Ω

Zo = 2.2 kΩ || 83.75 kΩ = 2.14 kΩ

(b)

()

4

(180)(2.2 k )

()

2.75 k (2.75 k )(25 S) (180)(2 10 ) 2.2 k

fe L

v

ie ie oe fe re L

hR

AhhhhhR

μ

−

−−Ω

==

+−

Ω

+Ω−× Ω

=140.3−

(c) (180) 170.62

11(25S)(2.2 k)

fe

i

oe L

h

AhR

μ

′== =

++ Ω

Ai = 10.2 k

(170.62) 10.2 k 2.68 k

ooi

ii

i

III

II

I

′Ω

⎛⎞

=⋅= ⎜⎟

Ω

+Ω

′⎝⎠

=135.13

81

71. (a) Zi = hie = 1

f

ere L

oe L

hhR

hR

−

+

= 0.86 kΩ −

4

(140)(1.5 10 )(2.2 k )

1 (25 S)(2.2 k )

μ

−

×

Ω

+Ω

= 0.86 kΩ − 43.79 Ω

= 816.21 Ω

i

Z

′ = RB || Zi

= 470 kΩ || 816.21 Ω

= 814.8 Ω

(b) Av = ()

fe L

ie ie oe fe re L

hR

hhhhhR

−

+−

= 4

(140)(2.2 k )

0.86 k ((0.86 k )(25 S) (140)(1.5 10 ))2.2 k

μ

−

Ω

Ω+ Ω − × Ω

= −357.68

(c) Ai = 140

1 1 (25 S)(2.2 k )

fe

o

ioeL

h

I

IhR

μ

==

++ Ω

= 132.70

ooi

i

ii

III

AI

II

⎛⎞

′⎜⎟

==

⎜⎟

′′

⎝⎠

Ii = 470 k

470 k 0.816 k

i

I′

Ω

+Ω

= (132.70)(0.998) i

i

I

′

= 0.998

= 132.43

(d)

Zo =

()

1

/( )

oe fe re ie s

hhhhR−+

=

()

4

1

25 S (140)(1.5 10 ) /(0.86 k 1 k )

μ

−

×Ω+Ω

=

1

13.71 S

μ

≅ 72.9 kΩ

72. (a)

4

( 0.997)(1 10 )(2.2 k )

9.45

1 1 (0.5 A/V)(2.2 k )

fb rb L

iib

ob L

hhR

Zh hR

μ

−

×Ω

′=− = Ω−

++Ω

9.67 =Ω

Zi = 1.2 kΩ || i

Z

′ = 1.2 kΩ || 9.67 Ω = 9.59 Ω

(b) 0.997 0.996

1 1 (0.5 A/V)(2.2 k )

fb

i

ob L

h

AhR

μ

−

′== =−

++ Ω

1.2 k

(0.996)1.2 k 9.67 k

oi

i

II

AI

I

′Ω

⎛⎞

=⋅=− ⎜⎟

Ω

+Ω

′⎝⎠

≅0.988−

82

(c)

()

4

()

( 0.997)(2.2 k )

9.45 (9.45 )(0.5 A/V) ( 0.997)(1 10 ) (2.2 k )

fb L

v

ib ib ob fb rb L

hR

AhhhhhR

μ

−

=+−

−− Ω

=Ω+ Ω − − × Ω

=226.61

(d)

4

1

/

1

0.5 A/V ( 0.997)(1 10 ) / 9.45

90.5 k

o

ob fb rb ib

Zhhhh

μ

−

′=⎡⎤

−⎣⎦

=⎡⎤

−− × Ω

⎣⎦

=Ω

Zo = 2.2 kΩ || o

Z

′ = 2.15 kΩ

73. −

74. (a) hfe (0.2 mA) ≅ 0.6 (normalized)

hfe (1 mA) = 1.0

% change = (0.2 mA) (1 mA)

(0.2 mA)

fe fe

fe

hh

h

− × 100%

= 0.6 1

0.6

− × 100%

= 66.7%

(b) hfe(1 mA) = 1.0

h

fe(5 mA) ≅ 1.5

% change = (1 mA) (5 mA)

(1 mA)

fe fe

fe

hh

h

− × 100%

= 11.5

1

− × 100%

= 50%

75. Log-log scale!

(a) Ic = 0.2 mA, hie = 4 (normalized)

I

c = 1 mA, hie = 1(normalized)

% change = 41

4

− × 100% = 75%

(b) Ie = 5 mA, hie = 0.3 (normalized)

% change = 10.3

1

− × 100% = 70%

83

76. (a) hoe = 20

μ

S @ 1 mA

Ic = 0.2 mA, hoe = 0.2(hoe @ 1 mA)

= 0.2(20

μ

S)

= 4

μ

S

(b) ro = 11

4 S

oe

h

μ

= = 250 kΩ  6.8 kΩ

Ignore 1/hoe

77. (a) Ic = 10 mA, hoe = 10(20

μ

S) = 200

μ

S

(b) ro = 11

200 S

oe

h

μ

= = 5 kΩ vs. 8.6 kΩ

Not a good approximation

78. (a) hre(0.1 mA) = 4(hre(1 mA))

= 4(2 × 10−4)

= 8 × 10−4

(b) hreVce = hreAv ⋅ Vi

= (8 × 10−4)(210)Vi

= 0.168 Vi

In this case hreVce is too large a factor to be ignored.

79. (a) hfe

(b) hoe

(c) hoe ≅ 30 (normalized) to

h

oe ≅ 0.1 (normalized) at low levels of Ic

(d) mid-region

80. (a) hie is the most temperature-sensitive parameter of Fig. 5.33.

(b) hoe exhibited the smallest change.

(c) Normalized: hfe(max) = 1.5, hfe(min) = 0.5

For hfe = 100 the range would extend from 50 to 150—certainly significant.

(d) On a normalized basis re increased from 0.3 at −65°C to 3 at 200°C—a significant

change.

(e) The parameters show the least change in the region 0° → 100°C.

84

81. (a) Test:

β

RE ≥ 10R2

70(1.5 kΩ) ≥ 10(39 kΩ)

105 kΩ

?

≥ 390 kΩ

No!

RTh = 39 kΩ || 150 kΩ = 30.95 kΩ

E

Th = 39 k (14 V)

39 k 150 k

Ω

Ω+ Ω = 2.89 V

I

B = 2.89 V 0.7 V

( 1) 30.95 k (71)(1.5 k )

Th BE

Th E

EV

RR

β

−−

=

++ Ω+ Ω

= 15.93

μ

A

VB = ETh − IBRTh

= 2.89 V − (15.93

μ

A)(30.95 kΩ)

= 2.397 V

VE = 2.397 V − 0.7 V = 1.697 V

and IE = 1.697 V

1.5 k

E

V

R=Ω = 1.13 mA

VCE = VCC − IC(RC + RE)

= 14 V − 1.13 mA(2.2 kΩ + 1.5 kΩ)

= 9.819 V

Biasing OK

(b) R2 not connected at base:

I

B = 014 V 0.7 V

( 1) 150 k + (71)(1.5 k )

CC

BE

V

RR

β

−

=

++ Ω Ω

= 51.85

μ

A

V

B = VCC − IBRB = 14 V − (51.85

μ

A)(150 kΩ)

= 6.22 V as noted in Fig. 5.187.

85

Eq. (6.1) is valid!

Chapter 6

1. −

2. From Fig. 6.11:

VGS = 0 V, ID = 8 mA

V

GS = −1 V, ID = 4.5 mA

VGS = −1.5 V, ID = 3.25 mA

VGS = −1.8 V, ID = 2.5 mA

VGS = −4 V, ID = 0 mA

V

GS = −6 V, ID = 0 mA

3. (a) VDS ≅ 1.4 V

(b)

rd = 1.4 V

6 mA

V

I= = 233.33 Ω

(c) VDS ≅ 1.6 V

(d) rd = 1.6 V

3 mA

V

I= = 533.33 Ω

(e) VDS ≅ 1.4 V

(f) rd = 1.4 V

1.5 mA

V

I= = 933.33 Ω

(g) ro = 233.33 Ω

r

d =

[][]

22

233.33 233.33

0.5625

1 1 ( 1 V) ( 4 V)

o

GS P

r

VV

Ω

==

−−−−

= 414.81 Ω

(h) rd =

[]

2

233.33 233.33

0.25

1 ( 2V) ( 4V)

ΩΩ

=

−− − = 933.2 Ω

(i) 533.33 Ω vs. 414.81 Ω

933.33 Ω vs 933.2 Ω

4. (a) VGS = 0 V, ID = 8 mA (for VDS > VP)

VGS = −1 V, ID = 4.5 mA

ΔID = 3.5 mA

(b) VGS = −1 V, ID = 4.5 mA

V

GS = −2 V, ID = 2 mA

ΔID = 2.5 mA

86

(c) VGS = −2 V, ID = 2 mA

V

GS = −3 V, ID = 0.5 mA

ΔID = 1.5 mA

(d) VGS = −3 V, ID = 0.5 mA

V

GS = −4 V, ID = 0 mA

ΔID = 0.5 mA

(e) As VGS becomes more negative, the change in ID gets progressively smaller for the same

change in VGS.

(f) Non-linear. Even though the change in VGS is fixed at 1 V, the change in ID drops from a

maximum of 3.5 mA to a minimum of 0.5 mA—a 7:1 change in ΔID.

5. The collector characteristics of a BJT transistor are a plot of output current versus the output

voltage for different levels of input current. The drain characteristics of a JFET transistor are

a plot of the output current versus input voltage. For the BJT transistor increasing levels of

input current result in increasing levels of output current. For JFETs, increasing magnitudes

of input voltage result in lower levels of output current. The spacing between curves for a

BJT are sufficiently similar to permit the use of a single beta (on an approximate basis) to

represent the device for the dc and ac analysis. For JFETs, however, the spacing between the

curves changes quite dramatically with increasing levels of input voltage requiring the use of

Shockley’s equation to define the relationship between ID and VGS. sat

C

Vand VP define the

region of nonlinearity for each device.

6. (a) The input current IG for a JFET is effectively zero since the JFET gate-source junction is

reverse-biased for linear operation, and a reverse-biased junction has a very high resistance.

(b) The input impedance of the JFET is high due to the reverse-biased junction between the

gate and source.

(c) The terminology is appropriate since it is the electric field established by the applied

gate to source voltage that controls the level of drain current. The term “field” is

appropriate due to the absence of a conductive path between gate and source (or drain).

7. VGS = 0 V, ID = IDSS = 12 mA

V

GS = VP = −6 V, ID = 0 mA

Shockley’s equation: VGS = −1 V, ID = 8.33 mA; VGS = −2 V, ID = 5.33 mA; VGS = −3 V,

ID = 3 mA; VGS = −4 V, ID = 1.33 mA; VGS = −5 V, ID = 0.333 mA.

8. For a

p-channel JFET, all the voltage polarities in the network are reversed as compared to an

n-channel device. In addition, the drain current has reversed direction.

87

9. (b)

IDSS = 10 mA, VP = −6 V

10. VGS = 0 V, ID = IDSS = 12 mA

V

GS = VP = −4 V, ID = 0 mA

V

GS = 2

P

V = −2 V, ID = 4

DSS

I = 3 mA

V

GS = 0.3VP = −1.2 V, ID = 6 mA

V

GS = −3 V, ID = 0.75 mA (Shockley’s equation)

11. (a) ID = IDSS = 9 mA

(b)

ID = IDSS(1 − VGS/VP)2

= 9 mA(1 − (−2 V)/(−3.5 V))2

=

1.653 mA

(c)

VGS = VP = −3.5 V, ID = 0 mA

(d) VGS < VP = −3.5 V, ID = 0 mA

12. VGS = 0 V, ID = 16 mA

V

GS = 0.3VP = 0.3(−5 V) = −1.5 V, ID = IDSS/2 = 8 mA

V

GS = 0.5VP = 0.5(−5 V) = −2.5 V, ID = IDSS/4 = 4 mA

V

GS = VP = −5 V, ID = 0 mA

13. VGS = 0 V, ID = IDSS = 7.5 mA

V

GS = 0.3VP = (0.3)(4 V) = 1.2 V, ID = IDSS/2 = 7.5 mA/2 = 3.75 mA

V

GS = 0.5VP = (0.5)(4 V) = 2 V, ID = IDSS/4 = 7.5 mA/4 = 1.875 mA

V

GS = VP = 4 V, ID = 0 mA

14. (a)

ID = IDSS(1 − VGS/VP)2 = 6 mA(1 − (−2 V)/(−4.5 V))2

= 1.852 mA

ID = IDSS(1 − VGS/VP)2 = 6 mA(1 − (−3.6 V)/(−4.5 V))2

= 0.24 mA

(b)

VGS = 3 mA

1(4.5 V)1

6 mA

D

P

DSS

I

VI

⎛⎞ ⎛⎞

−=− −

⎜⎟ ⎜⎟

⎜⎟

⎜⎟ ⎝⎠

⎝⎠

= −1.318 V

VGS = 5.5 mA

1(4.5 V)1

6 mA

D

P

DSS

I

VI

⎛⎞ ⎛⎞

−=− −

⎜⎟ ⎜⎟

⎜⎟

⎜⎟ ⎝⎠

⎝⎠

= −0.192 V

15. ID = IDSS(1 − VGS/VP)2

3 mA = IDSS(1 − (−3 V)/(−6 V))2

3 mA = IDSS(0.25)

IDSS = 12 mA

88

16. From Fig. 6.22:

−0.5 V < VP < −6 V

1 mA < IDSS < 5 mA

For IDSS = 5 mA and VP = −6 V:

VGS = 0 V, ID = 5 mA

V

GS = 0.3VP = −1.8 V, ID = 2.5 mA

V

GS = VP/2 = −3 V, ID = 1.25 mA

V

GS = VP = −6 V, ID = 0 mA

For IDSS = 1 mA and VP = −0.5 V:

VGS = 0 V, ID = 1 mA

V

GS = 0.3VP = −0.15 V, ID = 0.5 mA

V

GS = VP/2 = −0.25 V, ID = 0.25 mA

V

GS = VP = −0.5 V, ID = 0 mA

17. VDS = max

DS

V = 25 V, ID = max

max

120 mW

25 V

D

DS

P

V= = 4.8 mA

I

D = IDSS = 10 mA, VDS = max 120 mW

10 mA

D

DSS

P

I= = 12 V

I

D = 7 mA, VDS = max 120 mW

7 mA

D

P

I= = 17.14 V

89

2.5 mA

18. VGS = −0.5 V, ID = 6.5 mA

V

GS = −1 V, ID = 4 mA

Determine ΔID above 4 mA line:

2.5 mA

0.5 V 0.3 V

x

= ⇒ x = 1.5 mA

ID = 4 mA + 1.5 mA = 5.5 mA corresponding with values determined from a purely

graphical approach.

19. Yes, all knees of VGS curves at or below |VP| = 3 V.

20. From Fig 6.25, IDSS ≅ 9 mA

At VGS = −1 V, ID = 4 mA

ID = IDSS(1 − VGS/VP)2

D

DSS

I

I = 1 − VGS/VP

GS

P

V

V = 1 − D

DSS

I

VP = 1 V

4 mA

1

19 mA

GS

D

DSS

V

I

−

=

−

= −3 V (an exact match)

21. ID = IDSS(1 − VGS/VP)2

= 9 mA(1 − (−1 V)/(−3 V))2

= 4 mA, which compares very well with the level obtained using Fig. 6.25.

22. (a) VDS ≅ 0.7 V @ ID = 4 mA (for VGS = 0 V)

r = 0.7 V 0 V

4 mA 0 mA

DS

D

V

I

Δ−

=

Δ−

= 175 Ω

(b) For VGS = −0.5 V, @ ID = 3 mA, VDS = 0.7 V

r = 0.7 V

3 mA = 233 Ω

(c) rd = 22

175

(1 / ) (1 ( 0.5 V)/( 3 V)

o

GS P

r

VV

Ω

=

−−−−

= 252 Ω vs. 233 Ω from part (b)

23. −

24. The construction of a depletion-type MOSFET and an enchancement-type MOSFET are

identical except for the doping in the channel region. In the depletion MOSFET the channel is

established by the doping process and exists with no gate-to-source voltage applied. As the

gate-to-source voltage increases in magnitude the channel decreases in size until pinch-off

occurs. The enhancement MOSFET does not have a channel established by the doping

sequence but relies on the gate-to-source voltage to create a channel. The larger the

magnitude of the applied gate-to-source voltage, the larger the available channel.

90

D

I

Δ

=3.34 mA

D

I

Δ

=4.67 mA

D

I

Δ

=6 mA

25. −

26. At

VGS = 0 V, ID = 6 mA

At

VGS = −1 V, ID = 6 mA(1 − (−1 V)/(−3 V))2 = 2.66 mA

At

VGS = +1 V, ID = 6 mA(1 − (+1 V)/(−3 V))2 = 6 mA(1.333)2 = 10.667 mA

At

VGS = +2 V, ID = 6 mA(1 − (+2 V)/(−3 V))2 = 6 mA(1.667)2 = 16.67 mA

From −1 V to 0 V, ΔID = 3.34 mA

while from +1 V to +2 V, ΔID = 6 mA − almost a 2:1 margin.

In fact, as VGS becomes more and more positive, ID will increase at a faster and faster

rate due to the squared term in Shockley’s equation.

27. VGS = 0 V, ID = IDSS = 12 mA; VGS = −8 V, ID = 0 mA; VGS = 2

P

V = −4 V, ID = 3 mA;

V

GS = 0.3VP = −2.4 V, ID = 6 mA; VGS = −6 V, ID = 0.75 mA

28. From problem 20:

VP = 1 V 1 V 1 V

1 1.21395

14 mA 1 1.473

1

19.5 mA

GS

D

DSS

V

I

+++

===

−

= 1

0.21395− ≅ −4.67 V

29. ID = IDSS(1 − VGS/VP)2

I

DSS =

()

22

4 mA

(1 ( 2 V) ( 5 V))

1

D

GS P

I

VV

=−− −

− = 11.11 mA

30. From problem 14(b):

VGS = 20 mA

1(5 V)1

2.9 mA

D

P

DSS

I

VI

⎛⎞

−=−−

⎜⎟

⎝⎠

= (−5 V)(1 − 2.626) = (−5 V)(−1.626)

= 8.13 V

31. From Fig. 6.34, max

D

P = 200 mW, ID = 8 mA

P = VDSID

and VDS = max 200 mW

8 mA

D

P

I= = 25 V

VGS ID

−1 V 2.66 mA

0 6.0 mA

+1 V 10.67 mA

+2 V 16.67 mA

91

32. (a) In a depletion-type MOSFET the channel exists in the device and the applied voltage

VGS controls the size of the channel. In an enhancement-type MOSFET the channel is

not established by the construction pattern but induced by the applied control voltage

VGS.

(b) −

(c) Briefly, an applied gate-to-source voltage greater than VT will establish a channel

between drain and source for the flow of charge in the output circuit.

33. (a) ID = k(VGS − VT)2 = 0.4 × 10−3(VGS − 3.5)2

VGS ID

3.5 V 0

4 V 0.1 mA

5 V 0.9 mA

6 V 2.5 mA

7 V 4.9 mA

8 V 8.1 mA

(b) ID = 0.8 × 10−3(VGS − 3.5)2

VGS ID

3.5 V 0

4 V 0.2 mA

5 V 1.8 mA

6 V 5.0 mA

7 V 9.8 mA

8 V 16.2 mA

For same levels of VGS, ID attains

twice the current level as part (a).

Transfer curve has steeper slope.

For both curves, ID = 0 mA for

VGS < 3.5 V.

34. (a) k =

()

(on)

22

(on)

4mA

(6 V 4 V)

D

GS T

I

VV

=−

− = 1 mA/V2

ID = k(VGS − VT)2 = 1 × 10−3(VGS − 4 V)2

(b) VGS ID For VGS < VT = 4 V, ID = 0 mA

4 V 0 mA

5 V 1 mA

6 V 4 mA

7 V 9 mA

8 V 16 mA

(c) VGS ID (VGS < VT)

2 V 0 mA

5 V 1 mA

10 V 36 mA

92

35. From Fig. 6.58, VT = 2.0 V

At ID = 6.5 mA, VGS = 5.5 V: ID = k(VGS − VT)2

6.5 mA = k(5.5 V − 2 V)2

k = 5.31 × 10−4

I

D = 5.31 × 10−4(VGS − 2)2

36. ID =

(

)

2

(on)GS T

kV V−

and

()

2

(on)

D

GS T

I

VV

k

−=

(on)

D

GS T

I

VV

k

−=

VT = (on)GS

V − D

I

k

= 4 V − 3

3mA

0.4 10−

× = 4 V − 7.5 V

= 4 V − 2.739 V

= 1.261 V

37. ID = k(VGS − VT)2

D

I

k = (VGS − VT)2

D

I

k = VGS − VT

VGS = VT + D

I

k = 5 V + 3

30 mA

0.06 10

−

×

= 27.36 V

38. Enhancement-type MOSFET:

ID = 2

()

GS T

kV V−

2( ) ( )

D

GS T GS T

GS GS

dI d

kV V V V

dV dV

⎡⎤

=− −

⎢⎥

⎣⎦

D

GS

dI

dV = 2k(VGS − VT)

1

93

Depletion-type MOSFET:

ID = IDSS(1 − VGS/VP)2

D

GS

dI

dV =

2

1GS

DSS

GS P

V

d

IdV V

⎛⎞

−

⎜⎟

⎝⎠

21 0

GS GS

DSS

PGS P

Vd V

IVdV V

⎡⎤⎡ ⎤

=− −

⎢⎥⎢ ⎥

⎣⎦⎣ ⎦

1

P

V

−

= 1

21

GS

DSS

PP

V

IVV

⎛⎞⎛⎞

−−

⎜⎟⎜⎟

⎝⎠⎝⎠

= 21

DSS GS

PP

IV

VV

⎛⎞

−−

⎜⎟

⎝⎠

= 21

DSS GS

P

PP P

IV

V

VV V

⎛⎞⎛ ⎞

−−

⎜⎟⎜ ⎟

⎝⎠⎝ ⎠

D

GS

dI

dV = 2

2()

DSS

GS P

P

IVV

V−

For both devices D

GS

dI

dV = k1(VGS − K2)

revealing that the drain current of each will increase at about the same rate.

39. ID = k(VGS − VT)2 = 0.45 × 10−3(VGS − (−5 V))2

= 0.45 × 10−3(VGS + 5 V)2

VGS = −5 V, ID = 0 mA; VGS = −6 V, ID = 0.45 mA; VGS = −7 V, ID = 1.8 mA;

V

GS = −8 V, ID = 4.05 mA; VGS = −9 V, ID = 7.2 mA; VGS = −10 V, ID = 11.25 mA

41. −

42. (a) −

(b) For the “on” transistor: R = 0.1 V

4 mA

V

I= = 25 ohms

For the “off” transistor: R = 4.9 V

0.5 A

V

I

μ

= = 9.8 MΩ

Absolutely, the high resistance of the “off” resistance will ensure Vo is very close to 5 V.

43. −

94

Chapter 7

1. (a) VGS = 0 V, ID = IDSS = 12 mA

VGS = VP = −4 V, ID = 0 mA

V

GS = VP/2 = −2 V, ID = IDSS/4 = 3 mA

V

GS = 0.3VP = −1.2 V, ID = IDSS/2 = 6 mA

(b)

(c)

Q

D

I ≅ 4.7 mA

Q

DS

V = VDD − Q

DD

IR = 12 V − (4.7 mA)(1.2 kΩ)

= 6.36 V

(d)

Q

D

I = IDSS(1 − VGS/VP)2 = 12 mA(1 − (−1.5 V)/(−4 V))2

= 4.69 mA

Q

DS

V = VDD − Q

DD

IR = 12 V − (4.69 mA)(1.2 kΩ)

= 6.37 V

excellent comparison

2. (a)

Q

D

I = IDSS

2

(1 / )

GS P

VV−

= 10 mA

()

2

1 ( 3 V) ( 4.5 V)−− −

= 10 mA(0.333)2

Q

D

I = 1.11 mA

(b)

Q

GS

V = −3 V

(c) VDS = VDD − ID(RD + RS)

= 16 V − (1.11 mA)(2.2 kΩ)

= 16 V − 2.444 V

=

13.56 V

VD = VDS = 13.56 V

V

G = Q

GS

V = −3 V

V

S = 0 V

95

3. (a) 14 V 9 V

1.6 k

Q

DD D

D

VV

IR

−

==

Ω

= 3.125 mA

(b) VDS = VD − VS = 9 V − 0 V = 9 V

(c) ID = IDSS(1 − VGS/VP)2 ⇒ VGS = 1D

P

DSS

I

VI

⎛⎞

−

⎜⎟

⎝⎠

V

GS = (−4 V) 3.125 mA

18 mA

⎛⎞

−

⎜⎟

⎝⎠

= −1.5 V

∴VGG = 1.5 V

4. Q

GS

V = 0 V, ID = IDSS = 5 mA

VD = VDD − IDRD

= 20 V − (5 mA)(2.2 kΩ)

= 20 V − 11 V

= 9 V

5. VGS = VP = −4 V

∴Q

D

I = 0 mA

and

VD = VDD − Q

DD

IR = 18 V − (0)(2 kΩ)

= 18 V

6. (a)(b)

VGS = 0 V, ID = 10 mA

V

GS = VP = −4 V, ID = 0 mA

V

GS = 2

P

V = −2 V, ID = 2.5 mA

V

GS = 0.3VP = −1.2 V, ID = 5 mA

V

GS = −IDRS

ID = 5 mA:

V

GS = −(5 mA)(0.75 kΩ)

= −3.75 V

(c)

Q

D

I ≅ 2.7 mA

Q

GS

V ≅ −1.9 V

(d) VDS = VDD − ID(RD + RS)

= 18 V − (2.7 mA)(1.5 kΩ + 0.75 kΩ)

=

11.93 V

VD = VDD − IDRD

= 18 V − (2.7 mA)(1.5 kΩ)

=

13.95 V

VG = 0 V

V

S = ISRS = IDRS

= (2.7 mA)(0.75 kΩ)

= 2.03 V

96

7. ID = IDSS(1 − VGS/VP)2 =

22

2

1DS DS

DSS

PP

IR IR

IVV

⎛⎞

++

⎜⎟

⎝⎠

2

21

DSS S DSS S

DDDSS

PP

IR IR

III

VV

⎛⎞⎛ ⎞

+−+

⎜⎟⎜ ⎟

⎝⎠⎝ ⎠ = 0

Substituting: 351.56 2

D

I − 4.75ID + 10 mA = 0

ID =

24

2

bb ac

a

−± − = 10.91 mA, 2.60 mA

Q

D

I = 2.6 mA (exact match #6)

VGS = −IDRS = −(2.60 mA)(0.75 kΩ)

= −1.95 V vs. −2 V (#6)

8. VGS = 0 V, ID = IDSS = 6 mA

V

GS = VP = −6 V, ID = 0 mA

V

GS = 2

P

V = −3 V, ID = 1.5 mA

V

GS = 0.3VP = −1.8 V, ID = 3 mA

V

GS = −IDRS

ID = 2 mA:

V

GS = −(2 mA)(1.6 kΩ)

= −3.2 V

(a)

Q

D

I = 1.7 mA

Q

GS

V = −2.8 V

(b)

VDS = VDD − ID(RD + RS)

= 12 V − (1.7 mA)(2.2 kΩ + 1.6 kΩ)

=

5.54 V

VD = VDD − IDRD

= 12 V − (1.7 mA)(2.2 kΩ)

=

8.26 V

VG = 0 V

V

S = ISRS = IDRS

= (1.7 mA)(1.6 kΩ)

= 2.72 V (vs. 2.8 V from VS = ()

Q

GS

V)

9. (a)

Q

D

I = IS = 1.7 V

0.51 k

S

V

R=Ω = 3.33 mA

(b)

QQ

GS D S

VIR=− = −(3.33 mA)(0.51 kΩ)

≅ −1.7 V

97

(c) ID = IDSS(1 − VGS/VP)2

3.33 mA = IDSS(1 − (−1.7 V)/(−4 V))2

3.33 mA = IDSS(0.331)

IDSS = 10.06 mA

(d)

VD = VDD − Q

D

IR

= 18 V − (3.33 mA)(2 kΩ) = 18 V − 6.66 V

= 11.34 V

(e)

VDS = VD − VS = 11.34 V − 1.7 V

= 9.64 V

10. (a)

VGS = 0 V

∴ID = IDSS = 4.5 mA

(b)

VDS = VDD − ID(RD + RS)

= 20 V − (4.5 mA)(2.2 kΩ + 0.68 kΩ)

= 20 V − 12.96

=

7.04 V

(c)

VD = VDD − IDRD

= 20 V − (4.5 mA)(2.2 kΩ)

= 10.1 V

(d)

VS = ISRS = IDRS

= (4.5 mA)(0.68 kΩ)

= 3.06 V

11. Network redrawn:

VGS = 0 V, ID = IDSS = 6 mA

V

GS = VP = −6 V, ID = 0 mA

V

GS = 2

P

V = −3 V, ID = 1.5 mA

V

GS = 0.3VP = −1.8 V, ID = 3 mA

V

GS = −IDRS = −ID(0.39 kΩ)

For ID = 5 mA, VGS = −1.95 V

98

From graph Q

D

I ≅ 3.55 mA

Q

GS

V ≅ −1.4 V

VS =

()

Q

GS

V− = −(−1.4 V)

= +1.4 V

12. (a)

VG = 2

12

DD

RV

RR+ = 110 k (20 V)

910 k 110 k

Ω

+Ω

= 2.16 V

V

GS = 0 V, ID = IDSS = 10 mA

V

GS = VP = −3.5 V, ID = 0 mA

V

GS = 2

P

V = −1.75 V, ID = 2.5 mA

V

GS = 0.3VP = −1.05 V, ID = 5 mA

Q

GS

V = VG − IDRS

Q

GS

V = 2.16 − ID(1.1 kΩ)

ID = 0: Q

GS

V = VG = 2.16 V

Q

GS

V = 0 V, ID = 2.16 V

1.1 kΩ = 1.96 mA

(b)

Q

D

I ≅ 3.3 mA

Q

GS

V ≅ −1.5 V

(c) VD = VDD − Q

DD

IR

= 20 V − (3.3 mA)(2.2 kΩ)

=

12.74 V

VS = ISRS = IDRS

= (3.3 mA)(1.1 kΩ)

= 3.63 V

(d)

Q

D

S

V = VDD − ()

Q

D

DS

IR R+

= 20 V − (3.3 mA)(2.2 k Ω + 1.1 kΩ)

= 20 V − 10.89 V

= 9.11 V

99

110 k (20 V)

110 k 910 k

G

V

Ω

=

Ω

+Ω

=2.16 V

13. (a) ID = IDSS = 10 mA, VP = −3.5 V

V

GS = 0 V, ID = IDSS = 10 mA

V

GS = VP = −3.5 V, ID = 0 mA

V

GS = 3.5 V

22

P

V−

= = −1.75 V, ID = 2.5 mA

V

GS = 0.3VP = −1.05 V, ID = 5 mA

Q

D

I ≅ 5.8 mA vs. 3.3 mA (#12)

Q

GS

V ≅ −0.85 V vs. −1.5 V (#12)

(b) As

RS decreases, the intersection on the vertical axis increases. The maximum occurs at

ID = IDSS = 10 mA.

∴min

2.16 V

10 mA

G

S

DSS

V

RI

== = 216 Ω

14. (a)

ID = 18 V 9 V 9 V

2 k 2 k

D

RDD D

DD

VVV

RR

−−

== =

Ω

Ω = 4.5 mA

(b)

VS = ISRS = IDRS = (4.5 mA)(0.68 kΩ)

= 3.06 V

VDS = VDD − ID(RD + RS)

= 18 V − (4.5 mA)(2 kΩ + 0.68 kΩ)

= 18 V − 12.06 V

=

5.94 V

(c)

VG = 2

12

91 k (18 V)

750 k 91 k

DD

RV

RR

Ω

=

+Ω+Ω

= 1.95 V

V

GS = VG − VS = 1.95 V − 3.06 V = −1.11 V

(d)

VP = 1.11 V

4.5 mA

11

8 mA

GS

D

DSS

V

I

−

==

−−

4.44 V−

= −1.48 V

100

15. (a) VGS = 0 V, ID = IDSS = 6 mA

VGS = VP = −6 V, ID = 0 mA

V

GS = VP/2 = −3 V, ID = 1.5 mA

V

GS = 0.3VP = −1.8 V, ID = 3 mA

VGS = VSS − IDRS

VGS = 4 V − ID(2.2 kΩ)

V

GS = 0 V, ID = 4 V

2.2 kΩ = 1.818 mA

I

D = 0 mA, VGS = 4 V

Q

D

I ≅ 2.7 mA

Q

GS

V ≅ −2 V

(b)

VDS = VDD + VSS − ID(RD + RS)

= 16 V + 4 V − (2.7 mA)(4.4 kΩ)

=

8.12 V

VS = −VSS + IDRS = −4 V + (2.7 mA)(2.2 kΩ)

= 1.94 V

or VS =

()

Q

GS

V− = −(−2 V) = +2 V

16. (a)

ID = 12 V+ 3 V 4 V 11 V

3 k 2 k 5 k

DD SS DS

DS

VVV

V

RRR

+− −

== =

+Ω+ΩΩ

= 2.2 mA

(b)

VD = VDD − IDRD = 12 V − (2.2 mA)(3 kΩ)

= 5.4 V

VS = ISRS + VSS = IDRS + VSS

= (2.2 mA)(2 kΩ) + (−3 V)

= 4.4 V − 3 V

= 1.4 V

(c)

VGS = VG − VS

= 0 V − 1.4 V

=

−1.4 V

17. (a) Q

D

I = 4 mA

(b) 12 V 4 mA(1.8 k ) 12 V 7.2 V

Q

D

V=− Ω=− = 4.8 V

Q

D

S

V = 4.8 V

(c) Ps = (12 V)(4 mA) = 48 mW

Pd = (4.8 V)(4 mA) = 19.2 mW

101

18. VGS = 0 V, ID = IDSS = 6 mA

V

GS = VP = −4 V, ID = 0 mA

V

GS = VP/2 = −2 V, ID = IDSS/4 = 1.5 mA

V

GS = 0.3VP = −1.2 V, ID = IDSS/2 = 3 mA

VGS = −IDRS = −ID(0.43 kΩ)

ID = 4 mA, VGS = −1.72 V

(b) VDS = VDD − ID(RD + RS)

= 14 V − 2.9 mA(1.2 kΩ + 0.43 kΩ)

=

9.27 V

VD = VDD − IDRD

= 14 V − (2.9 mA)(1.2 kΩ)

= 10.52 V

19. (a) VGS = 0 V, ID = IDSS = 8 mA

V

GS = VP = −8 V, ID = 0 mA

V

GS = 2

P

V = −4 V, ID = 2 mA

V

GS = 0.3VP = −2.4 V, ID = 4 mA

V

GS = +1 V, ID = 10.125 mA

V

GS = +2 V, ID = 12.5 mA

V

GS = −VSS − IDRS

= −(−4 V) − ID(0.39 kΩ)

V

GS = 4 − ID(0.39 kΩ)

ID = 0: VGS = +4 V

V

GS = 0: ID = 4

0.39 kΩ = 10.26 mA

Q

D

I ≅ 9 mA

Q

GS

V ≅ +0.5 V

102

(b) VDS = VDD − ID(RD + RS) + VSS

= 18 V − 9 mA(1.2 kΩ + 0.39 kΩ) + 4 V

= 22 V − 14.31 V

=

7.69 V

V

S =

()

Q

GS

V− = −0.5 V

20. ID = k(VGS − VT)2

k =

()

(on)

2

(on)

D

GS Th

I

VV− = 22

5 mA 5 mA

(7 V 4 V) 9 V

=

−

K = 0.556 × 10−3 A/V2

and ID = 0.556 × 10−3(VGS − 4 V)2

VDS = VDD − ID(RD + RS)

VDS = 0 V; ID = DD

DS

V

R

R+

=

22 V

1.2 k 0.51 k

Ω

+Ω

= 12.87 mA

ID = 0 mA, VDS = VDD

= 22 V

(c) VD = VDD − IDRD

= 22 V − (8.25 mA)(1.2 kΩ)

=

12.1 V

VS = ISRS = IDRS

= (8.25 mA)(0.51 kΩ)

=

4.21 V

(d) VDS = VD − VS

= 12.1 V − 4.21 V

=

7.89 V

vs. 7.9 V obtained graphically

103

21. (a) VG = 2

12

6.8 M (24 V)

10 M 6.8 M

DD

RV

RR

Ω

=

+Ω+Ω

= 9.71 V

VGS = VG − IDRS

VGS = 9.71 − ID(0.75 kΩ)

At ID = 0 mA, VGS = 9.71 V

At VGS = 0 V, ID = 9.71 V

0.75 k

Ω

= 12.95 mA

k =

()

(on)

222

(on) ( )

5 mA 5 mA

(6 V 3 V) (3 V)

D

GS GS Th

I

VV

==

−

= 0.556 × 10−3A/V2

∴ID = 0.556 × 10−3(VGS − 3 V)2

VGS ID

3 V 0 mA

4 V 0.556 mA

5 V 2.22 mA

6 V 5 mA

7 V 8.9 mA

Q

D

I ≅ 5 mA

Q

GS

V ≅ 6 V

104

(b) VD = VDD − IDRD = 24 V − (5 mA)(2.2 kΩ)

= 13 V

VS = ISRS = IDRS

= (5 mA)(0.75 kΩ)

= 3.75 V

22. (a) VG = 2

12

18 k (20 V)

91 k 18 k

CC

RV

RR

Ω

=

+Ω+Ω

= 3.3 V

(b) VGS = 0 V, ID = IDSS = 6 mA

V

GS = VP = −6 V, ID = 0 mA

V

GS = 2

P

V = −3 V, ID = 1.5 mA

V

GS = VP = −1.8 V, ID = 3 mA

Q

D

I ≅ 3.75 mA

Q

GS

V ≅ −1.25 V

(c) IE = ID = 3.75 mA

(d) IB = 3.75 mA

160

C

I

β

= = 23.44

μ

A

(e) VD = VE = VB − VBE = VCC − IBRB − VBE = 20 V − (23.44

μ

A)(330 kΩ) − 0.7 V

= 11.56 V

(f) VC = VCC − ICRC = 20 V − (3.75 mA)(1.1 kΩ)

= 15.88 V

23. Testing:

β

RE ≥ 10R2

(100)(1.2 kΩ) ≥ 10(10 kΩ)

120 kΩ > 100 kΩ (satisfied)

(a) VB = VG = 2

12

10 k (16 V)

40 k 10 k

DD

RV

RR

Ω

=

+Ω+Ω

= 3.2 V

(b) VE = VB − VBE = 3.2 V − 0.7 V = 2.5 V

(c) IE = 2.5 V

1.2 k

E

V

R=Ω = 2.08 mA

I

C ≅ IE = 2.08 mA

I

D = IC = 2.08 mA

105

(d) IB = 2.08 mA

100

C

I

β

= = 20.8

μ

A

(e) VC = VG − VGS

VGS = 1D

P

DSS

I

VI

⎛⎞

−

⎜⎟

⎝⎠

= 2.08 mA

(6 V)1 6 mA

⎛⎞

−−

⎜⎟

⎝⎠

= −2.47 V

VC = 3.2 − (−2.47 V)

= 5.67 V

VS = VC = 5.67 V

V

D = VDD − IDRD

= 16 V − (2.08 mA)(2.2 kΩ)

= 11.42 V

(f) VCE = VC − VE = 5.67 V − 2.5 V

=

3.17 V

(g) VDS = VD − VS = 11.42 V − 5.67 V

=

5.75 V

24. VGS = 1D

P

DSS

I

VI

⎛⎞

−

⎜⎟

⎝⎠

= (−6 V) 4 mA

18 mA

⎛⎞

−

⎜⎟

⎝⎠

= −1.75 V

VGS = −IDRS: RS = (1.75 V)

4 mA

GS

D

V

I

−−

−= = 0.44 kΩ

R

D = 3RS = 3(0.44 kΩ) = 1.32 kΩ

Standard values: RS = 0.43 kΩ

RD = 1.3 kΩ

25. VGS = 1D

P

DSS

I

VI

⎛⎞

−

⎜⎟

⎝⎠

= (−4 V) 2.5 mA

110 mA

⎛⎞

−

⎜⎟

⎝⎠

= −2 V

VGS = VG − VS

and VS = VG − VGS = 4 V − (−2 V)

= 6 V

RS = 6 V

2.5 mA

S

D

V

I= = 2.4 kΩ (a standard value)

R

D = 2.5RS = 2.5(2.4 kΩ) = 6 kΩ ⇒ use 6.2 kΩ

V

G = 2

12

DD

RV

R

R+ ⇒ 4 V = 2

2

(24 V)

22 M

R

Ω+ ⇒ 88 MΩ + 4R2 = 24R2

20R2 = 88 M

Ω

R2 = 4.4 MΩ

Use R2 = 4.3 MΩ

106

26. ID = k(VGS − VT)2

D

I

k = (VGS − VT)2

D

I

k = VGS − VT

and VGS = VT + D

I

k = 4 V + 32

6mA

0.5 10 A/V

−

× = 7.46 V

RD = 16 V 7.46 V 8.54 V

6 mA 6 mA

D

RDD DS DD GS

DD D

VVV VV

II I

−− −

=== =

= 1.42 kΩ

Standard value: RD = 0.75 kΩ

RG = 10 MΩ

27. (a) ID = IS = 4 V

1 k

S

V

R=Ω = 4 mA

VDS = VDD − ID(RD + RS) = 12 V − (4 mA)(2 kΩ + 1 kΩ)

= 12 V − (4 mA)(3 kΩ)

= 12 V − 12 V

= 0 V

JFET in saturation!

(b) VS = 0 V reveals that the JFET is nonconducting and the JFET is either defective or an

open-circuit exists in the output circuit. VS is at the same potential as the grounded side

of the 1 kΩ resistor.

(c) Typically, the voltage across the 1 MΩ resistor is ≅ 0 V. The fact that the voltage across

the 1 MΩ resistor is equal to VDD suggests that there is a short-circuit connection from

gate to drain with ID = 0 mA. Either the JFET is defective or an improper circuit

connection was made.

28. VG = 75 k (20 V)

75 k 330 k

Ω

Ω+ Ω = 3.7 V (seems correct!)

V

GS = 3.7 V − 6.25 V = −2.55 V (possibly okay)

I

D = IDSS(1 − VGS/VP)2

= 10 mA(1 − (−2.55 V)/(−6 V))2

= 3.3 mA (reasonable)

However, IS = 6.25 V

1 k

S

V

R=Ω = 6.25 mA ≠ 3.3 mA

D

R

V = IDRD = ISRD = (6.25 mA)(2.2 kΩ)

= 13.75 V

and

SD

R

VV+ = 6.25 V + 13.75 V

= 20 V = VDD

∴VDS = 0 V

1. Possible short-circuit from D-S.

2. Actual IDSS and/or VP may be larger in magnitude than specified.

107

29. ID = IS = 6.25 V

1 k

S

V

R=Ω = 6.25 mA

V

DS = VDD − ID(RD + RS)

= 20 V − (6.25 mA)(2.2 kΩ + 1 kΩ)

= 20 V − 20 V

= 0 V (saturation condition)

VG = 2

12

75 k (20 V)

330 k 75 k

DD

RV

RR

Ω

=

+Ω+Ω

= 3.7 V (as it should be)

V

GS = VG − VS = 3.7 V − 6.25 V = −2.55 V

I

D =

2

1GS

DSS

P

V

IV

⎛⎞

−

⎜⎟

⎝⎠

= 10 mA(1 − (−2.55 V)/(6 V))2

= 3.3 mA ≠ 6.25 mA

In all probability, an open-circuit exists between the voltage divider network and the gate

terminal of the JFET with the transistor exhibiting saturation conditions.

30. (a) VGS = 0 V, ID = IDSS = 8 mA

VGS = VP = +4 V, ID = 0 mA

V

GS = 2

P

V = +2 V, ID = 2 mA

V

GS = 0.3VP = 1.2 V, ID = 4 mA

V

GS = IDRS

ID = 4 mA;

V

GS = (4 mA)(0.51 kΩ)

= 2.04 V

Q

D

I = 3 mA, Q

GS

V = 1.55 V

(b) VDS = VDD + ID(RD + RS)

= −18 V + (3 mA)(2.71 kΩ)

= −9.87 V

(c) VD = VDD − IDRD

= −18 V − (3 mA)(2.2 kΩ)

= −11.4 V

108

31. k =

()

(on)

22

2

(on) ( )

4 mA 4 mA

(4 V)

7 V ( 3 V)

D

GS GS Th

I

VV

==

−

−−−

−

= 0.25 × 10−3 A/V2

ID = 0.25 × 10−3(VGS + 3 V)2

VGS ID

−3 V 0 mA

−4 V 0.25 mA

−5 V 1 mA

−6 V 2.25 mA

−7 V 4 mA

−8 V 6.25 mA

VGS = VDS = VDD + IDRD

At ID = 0 mA, VGS = VDD = −16 V

At VGS = 0 V, ID = 16 V

2 k

DD

D

V

R=

Ω

= 8 mA

(b) VDS = VGS = −7.25 V

(c) VD = VDS = −7.25 V

or VDS = VDD + IDRD

= −16 V + (4.4 mA)(2 kΩ)

= −16 V + 8.8 V

VDS = −7.2 V = VD

32. 1.5 V

4 V

GS

P

V

−

= = −0.375

Find −0.375 on the horizontal axis.

Then move vertically to the ID = IDSS(1 − VGS/VP)2 curve.

Finally, move horizontally from the intersection with the curve to the left to the ID/IDSS axis.

D

DSS

I

I = 0.39

and ID = 0.39(12 mA) = 4.68 mA vs. 4.69 mA (#1)

Q

DS

V = VDD − IDRD = 12 V − (4.68 mA)(1.2 kΩ)

=

6.38 V vs. 6.37 V (#1)

109

33. m = 4 V

(10 mA)(0.75 k )

P

DSS S

V

IR

=

Ω

= 0.533

M = m0.533(0)

4 V

GG

P

V

V=

= 0

Draw a straight line from M = 0 through m = 0.533 until it crosses the normalized curve of ID

=

2

1GS

DSS

P

V

IV

⎛⎞

−

⎜⎟

⎝⎠

. At the intersection with the curve drop a line down to determine

GS

P

V

V = −0.49

so that

Q

GS

V = −0.49VP = −0.49(4 V)

= −1.96 V (vs. −1.9 V #6)

If a horizontal line is drawn from the intersection to the left vertical axis we find

D

DSS

I

I = 0.27

and ID = 0.27(IDSS) = 0.27(10 mA) = 2.7 mA

(vs. 2.7 mA from #6)

(a)

Q

GS

V = −1.96 V, Q

D

I = 2.7 mA

(b) −

(c) −

(d) VDS = VDD − ID(RD + RS) = 11.93 V (like #6)

VD = VDD − IDRD = 13.95 V (like #6)

V

G = 0 V, VS = IDRS = 2.03 V (like #6)

34. VGG = 2

12

110 k (20 V)

110 k 910 k

DD

RV

RR

Ω

=

+Ω+Ω

= 2.16 V

m = 3.5 V

(10 mA)(1.1 k )

P

DSS S

V

IR

=

Ω

= 0.318

M = m × (2.16 V)

0.318 3.5

GG

P

V

V= = 0.196

Find 0.196 on the vertical axis labeled M and mark the location. Move horizontally to the

vertical axis labeled m and then add m = 0.318 to the vertical height (≅ 1.318 in total)—mark

the spot. Draw a straight line through the two points located above, as shown below.

110

Continue the line until it intersects the ID = IDSS(1 − VGS/VP)2 curve. At the intersection move

horizontally to obtain the ID/IDSS ratio and move down vertically to obtain the /

GS p

VV ratio.

D

DSS

I

I = 0.33 and Q

D

I = 0.33(10 mA) = 3.3 mA

GS

P

V

V = −0.425 and Q

GS

V = −0.425(3.5 V)

vs. 1.5 V (#12)

35. m = 6 V

(6 mA)(2.2 k )

P

DSS S

V

IR

=Ω

= 0.4545

M = (4 V)

0.4545 (6 V)

GG

P

V

mV=

= 0.303

Find 0.303 on the vertical M axis.

Draw a horizontal line from M = 0.303 to the vertical m axis.

Add 0.4545 to the vertical location on the m axis defined by the horizontal line.

Draw a straight line between M = 0.303 and the point on the m axis resulting from the

addition of m = 0.4545.

Continue the straight line as shown below until it crosses the normalized

ID = IDSS(1 − VGS/VP)2 curve:

vs. 3.3 mA (#12)

=

−

1.49 V

111

At the intersection drop a vertical line to determine

GS

P

V

V = −0.34

and

Q

GS

V = −0.34(6 V)

= −2.04 V (vs. −2 V from problem 15)

At the intersection draw a horizontal line to the ID/IDSS axis to determine

D

DSS

I

I = 0.46

and

Q

D

I = 0.46(6 mA)

= 2.76 mA (vs. 2.7 mA from problem 15)

(a)

Q

D

I = 2.76 mA, Q

GS

V = −2.04 V

(b) VDS = VDD + VSS − ID(RD + RS)

= 16 V + 4 V − (2.76 mA)(4.4 kΩ)

=

7.86 V (vs. 8.12 V from problem 15)

VS = −VSS + IDRS = −4 V + (2.76 mA)(2.2 kΩ)

= −4 V + 6.07 V

= 2.07 V (vs. 1.94 V from problem 15)

112

Chapter 8

1. gm0 = 22(15 mA)

5 V

DSS

P

I

V=− = 6 mS

2. gm0 =

0

22

2(12 mA)

10 mS

DSS DSS

P

Pm

II

V

Vg

⇒= = = 2.4 V

VP = −2.4 V

3. gm0 = 2DSS

P

I

V ⇒ IDSS = ()()

5 mS(3.5 V)

22

mO P

gV

= = 8.75 mA

4. gm = 0

2(12 mA) 1 V

11

3 V 3 V

Q

GS

m

P

V

gV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟

−−

⎝⎠

= 5.3 mS

5. gm = 21Q

GS

DSS

PP

V

I

VV

⎛⎞

−

⎜⎟

⎝⎠

6 mS = 21 V

1

2.5 V 2.5 V

DSS

I−

⎛⎞

−

⎜⎟

−

⎝⎠

IDSS = 12.5 mA

6. gm = 0

2/4

2(10 mA) 1

5 V 4

DSS DSS

D

m

DSS P DSS

II

I

gIVI

==

= 20 mA 1

5 V 2

⎛⎞

⎜⎟

⎝⎠

= 2 mS

7. gm0 = 22(8 mA)

5 V

DSS

P

I

V= = 3.2 mS

g

m = 0

/4

13.2mS1

Q

GS P

m

PP

VV

gVV

⎛⎞ ⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ ⎝⎠

⎝⎠

= 3.2 mS 13

13.2 mS

44

⎛⎞ ⎛⎞

−=

⎜⎟ ⎜⎟

⎝⎠ ⎝⎠

= 2.4 mS

8. (a) gm = yfs = 4.5 mS

(b) rd = 11

25 S

os

y

μ

= = 40 kΩ

9. gm = yfs = 4.5 mS

r

d = 11

25 S

os

y

μ

= = 40 kΩ

Z

o = rd = 40 kΩ

A

v(FET) = −gmrd = −(4.5 mS)(40 kΩ) = −180

113

10. Av = −gmrd ⇒ gm = (200)

(100 k )

v

d

A

r

−

=−

Ω

= 2 mS

11. (a) gm0 = 22(10 mA)

5 V

DSS

P

I

V= = 4 mS

(b) gm = 6.4 mA 3.6 mA

2 V 1 V

D

GS

I

V

Δ−

=

Δ−

= 2.8 mS

(c) Eq. 8.6: gm = 0

1.5 V

14 mS1

5 V

Q

GS

m

P

V

gV

⎛⎞ −

⎛⎞

−= −

⎜⎟

⎜⎟ −

⎝⎠

= 2.8 mS

(d) gm = 3.6 mA 1.6 mA

3 V 2 V

D

GS

I

V

Δ−

=

Δ−

= 2 mS

(e) gm = 0

2.5 V

14 mS1

5 V

Q

GS

m

P

V

gV

⎛⎞ −

⎛⎞

−= −

⎜⎟

⎜⎟ −

⎝⎠

= 2 mS

12. (a) rd =

constant

(15 V 5 V) 10 V

(9.1 mA 8.8 mA) 0.3 mA

GS

DS

DV

V

I

Δ−

==

Δ−

= 33.33 kΩ

(b) At VDS = 10 V, ID = 9 mA on VGS = 0 V curve

∴gm0 = 22(9 mA)

4 V

DSS

P

I

V= = 4.5 mS

13. From 2N4220 data:

gm = yfs = 750

μ

S = 0.75 mS

r

d = 11

10 S

os

y

μ

= = 100 kΩ

14. (a) gm (@ VGS = −6 V) = 0, gm (@ VGS = 0 V) = gm0 = 22(8 mA)

6 V

DSS

P

I

V= = 2.67 mS

(b) gm (@ ID = 0 mA) = 0, gm (@ ID = IDSS = 8 mA) = gm0 = 2.67 mS

15. gm = yfs = 5.6 mS, rd = 11

15 S

os

y

μ

= = 66.67 kΩ

16. gm = 22(10 mA) 2 V

11

4 V 4 V

Q

GS

DSS

PP

V

I

VV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 2.5 mS

r

d = 11

25 S

os

y

μ

= = 40 kΩ

114

17. Graphically,

Q

GS

V = −1.5 V

gm = 22(10 mA) 1.5 V

11

4 V 4 V

Q

GS

DSS

PP

V

I

VV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 3.125 mS

Z

i = RG = 1 MΩ

Z

o = RD || rd = 1.8 kΩ || 40 kΩ = 1.72 kΩ

A

v = −gm(RD || rd) = −(3.125 mS)(1.72 kΩ)

= −5.375

18. Q

GS

V = −1.5 V

gm = 22(12 mA) 1.5 V

11

6 V 6 V

Q

GS

DSS

PP

V

I

VV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 3 mS

Z

i = RG = 1 MΩ

Z

o = RD || rd, rd = 11

40 S

os

y

μ

= = 25 kΩ

= 1.8 kΩ || 25 kΩ

= 1.68 kΩ

Av = −gm(RD || rd) = −(3 mS)(1.68 kΩ) = −5.04

19. gm = yfs = 3000

μ

S = 3 mS

r

d = 11

50 S

os

y

μ

= = 20 kΩ

Z

i = RG = 10 MΩ

Z

o = rd || RD = 20 kΩ|| 3.3 kΩ = 2.83 kΩ

A

v = −gm(rd || RD)

= −(3 mS)(2.83 kΩ)

= −8.49

20. Q

GS

V = 0 V, gm = gm0 = 22(6 mA)

6 V

DSS

P

I

V= = 2 mS, rd = 11

40 S

os

y

μ

= = 25 kΩ

Zi = 1 MΩ

Zo = rd || RD = 25 kΩ || 2 kΩ = 1.852 kΩ

Av = −gm(rd || RD) = −(2 mS)(1.852 kΩ) ≅ −3.7

115

21. gm = 3 mS, rd = 20 kΩ

Z

i = 10 MΩ

Z

o = 3.3 k

3.3 k 1.1 k

1 (3 mS)(1.1 k )

120 k

D

DS

mS

d

R

RR

gR r

Ω

=

+Ω+Ω

+Ω+

++ Ω

= 3.3 k 3.3 k

1 3.3 0.22 4.52

ΩΩ

=

++ = 730 Ω

Av = (3 mS)(3.3 k )

3.3 k 1.1 k

1 (3 mS)(1.1 k )

120 k

mD

DS

mS

d

gR

RR

gR r

−

Ω

=

+Ω+Ω

+Ω+

++ Ω

= 9.9 9.9

1 3.3 0.22 4.52

−=−

++ = −2.19

22. gm = yfs = 3000

μ

S = 3 mS

r

d = 11

10 S

os

y

μ

= = 100 kΩ

Z

i = RG = 10 MΩ (the same)

Z

o = rd || RD = 100 kΩ || 3.3 kΩ = 3.195 kΩ (higher)

A

v = −gm(rd || RD)

= −(3 mS)(3.195 kΩ)

= −9.59 (higher)

23. Q

GS

V = −0.95 V

gm = 21Q

GS

DSS

PP

V

I

VV

⎛⎞

−

⎜⎟

⎝⎠

= 2(12 mA) 0.95 V

1

3 V 3 V

−

⎛⎞

−

⎜⎟

−

⎝⎠

= 5.47 mS

Zi = 82 MΩ || 11 MΩ = 9.7 MΩ

Z

o = rd || RD = 100 kΩ || 2 kΩ = 1.96 kΩ

A

v = −gm(rd || RD) = −(5.47 mS)(1.96 kΩ) = −10.72

V

o = AvVi = (−10.72)(20 mV) = −214.4 mV

24. Q

GS

V = −0.95 V (as before), gm = 5.47 mS (as before)

Zi = 9.7 MΩ as before

Zo =

1

D

DS

mS

d

R

gR r

+

++

but rd ≥ 10(RD + RS)

116

∴Zo = 2 k 2 k 2 k

1 1 (5.47 mS)(0.61 k ) 1 3.337 4.337

D

mS

R

gR

ΩΩΩ

===

++ Ω+

= 461.1 Ω

Av = 1

mD

mS

g

R

g

R

−

+ since rd ≥ 10(RD + RS)

= (5.47 mS)(2 k ) 10.94

4.337 (from above) 4.337

−Ω

=− = −2.52 (a big reduction)

Vo = AvVi = (−2.52)(20 mV) = −50.40 mV (compared to −214.4 mV earlier)

25. Q

GS

V = −0.95 V, gm (problem 23) = 5.47 mS

Zi (the same) = 9.7 MΩ

Z

o (reduced) = rd || RD = 20 kΩ || 2 kΩ = 1.82 kΩ

A

v (reduced) = −gm(rd || RD) = −(5.47 mS)(1.82 kΩ) = −9.94

V

o (reduced) = AvVi = (−9.94)(20 mV) = −198.8 mV

26. Q

GS

V = −0.95 V (as before), gm = 5.47 mS (as before)

Zi = 9.7 MΩ as before

Z

o =

1

D

DS

mS

d

R

gR r

+

++

since rd < 10(RD + RS)

= 2 k

2 k 0.61 k

1 (5.47 mS)(0.61 k ) 20 k

Ω

+Ω

+Ω+

Ω

= 2 k 2 k

1 3.33 0.13 4.46

ΩΩ

=

++

= 448.4 Ω (slightly less than 461.1 Ω obtained in problem 24)

Av =

1

mD

DS

mS

d

gR

R

gR r

−

+

++

= (5.47 mS)(2 k )

2 k 0.61 k

1 (5.47 mS)(0.61 k ) 20 k

−Ω

Ω

+Ω

+Ω+

Ω

= 10.94 10.94

1 3.33 0.13 4.46

−−

=

++ = −2.45 slightly less than −2.52 obtained in problem 24)

27. Q

GS

V = −2.85 V, gm = 22(9 mA) 2.85 V

11

4.5 V 4.5 V

Q

GS

DSS

PP

V

I

VV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 1.47 mS

Zi = RG = 10 MΩ

Z

o = rd || RS || 1/gm = 40 kΩ || 2.2 kΩ || 1/1.47 mS = 512.9 Ω

680.27 Ω

A

v = ( ) (1.47 mS)(40 k 2.2 k ) 3.065

1 ( ) 1 (1.47 mS)(40 k 2.2 k ) 1 3.065

md S

grR

ΩΩ

==

++ ΩΩ+

= 0.754

117

28. Q

GS

V = −2.85 V, gm = 1.47 mS

Zi = 10 MΩ (as in problem 27)

Z

o = d

r || RS || 1/gm = 20 kΩ || 2.2 kΩ || 680.27 Ω = 506.4 Ω < 512.9 Ω (#27)

1.982 kΩ

Av = ( ) 1.47 mS( 20 k 2.2 k ) 2.914

1 ( ) 1 1.47 mS(20 k 2.2 k ) 1 2.914

md S

grR

ΩΩ

==

++ΩΩ+

= 0.745 < 0.754 (#27)

29. Q

GS

V = −3.8 V

gm = 22(6 mA) 3.8 V

11

6 V 6 V

Q

GS

DSS

PP

V

I

VV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 0.733 mS

The network now has the format examined in the text and

Zi = RG = 10 MΩ d

r′ = rd + RD = 30 kΩ + 3.3 kΩ = 33.3 kΩ

Z

o = d

r′ || RS || 1/ m

g′ = (0.733 mS)(30 k ) 21.99

30 k 3.3 k 33.3 k

md

m

dD

gr

grR

Ω

′== =

+

Ω+ Ω Ω = 0.66 mS

= 33.3 kΩ || 3.3 kΩ || 1/0.66 mS

= 3 kΩ || 1.52 kΩ

≅ 1 kΩ

Av = ()0.66 mS(3 k ) 1.98 1.98

1 ( ) 1 0.66 mS(3 k ) 1 1.98 2.98

md S

grR

′′ Ω

===

′′

++Ω+

= 0.66

30. Q

GS

V = −1.75 V, gm = 2.14 mS

rd ≥ 10RD, ∴Zi ≅ RS || 1/gm = 1.5 kΩ || 1/2.14 mS

= 1.5 kΩ || 467.29 Ω

= 356.3 Ω

rd ≥ 10RD, ∴Zo ≅ RD = 3.3 kΩ

r

d ≥ 10RD, ∴ Av ≅ gmRD = (2.14 mS)(3.3 kΩ) = 7.06

V

o = AvVi = (7.06)(0.1 mV) = 0.706 mV

118

31. Q

GS

V = −1.75 V, gm = 22(8 mA) 1.75 V

11

2.8 V 2.8 V

Q

GS

DSS

PP

V

I

VV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 2.14 mS

Zi = RS || 25 k 3.3 k 28.3 k

1.5 k 1.5 k

1 1 (2.14 mS)(25 k ) 54.5

dD

md

rR

gr

⎡⎤ ⎡⎤

+

Ω

+Ω Ω

=Ω =Ω

⎢⎥ ⎢⎥

++Ω

⎣⎦

= 1.5 kΩ || 0.52 kΩ = 386.1 Ω

Zo = RD || rd = 3.3 kΩ || 25 kΩ = 2.92 kΩ

A

v = /(2.14 mS)(3.3 k ) 3.3 k / 25 k

1 / 1 3.3 k / 25 k

mD D d

Dd

gR R r

Rr

+

Ω

+Ω Ω

=

++ΩΩ

= 7.062 0.132 7.194

1 0.132 1.132

+=

+ = 6.36

Vo = AvVi = (6.36)(0.1 mV) = 0.636 mV

32. Q

GS

V ≅ −1.2 V, gm = 2.63 mS

rd ≥ 10RD, ∴ Zi ≅ RS || 1/gm = 1 kΩ || 1/2.63 mS = 1 kΩ || 380.2 Ω = 275.5 Ω

Z

o ≅ RD = 2.2 kΩ

A

v ≅ gmRD = (2.63 mS)(2.2 kΩ) = 5.79

33. rd = 11

20 S

os

y

μ

= = 50 kΩ, Q

GS

V = 0 V

g

m = gm0 = 22(8 mA)

3

DSS

P

I

V= = 5.33 mS

A

v = −gmRD = −(5.33 mS)(1.1 kΩ) = −5.863

V

o = AvVi = (−5.863)(2 mV) = 11.73 mV

34. Q

GS

V = −0.75 V, gm = 5.4 mS

Zi = 10 MΩ

r

o ≥ 10RD, ∴ Zo ≅ RD = 1.8 kΩ

r

o ≥ 10RD, ∴ Av ≅ −gmRD = −(5.4 mS)(1.8 kΩ)

= −9.72

35. Zi = 10 MΩ

Z

o = rd || RD = 25 kΩ || 1.8 kΩ = 1.68 kΩ

A

v = −gm(rd || RD)

g

m = 22(12 mA) 0.75 V

11

3.5 V 3.5 V

Q

GS

DSS

PP

V

I

VV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 5.4 mS

A

v = −(5.4 mS)(1.68 kΩ)

= −9.07

119

36. gm = yfs = 6000

μ

S = 6 mS

r

d = 11

35 S

os

y

μ

= = 28.57 kΩ

r

d ≤ 10RD, ∴Av = −gm(rd || RD)

= −(6 mS)(28.57 kΩ || 6.8 kΩ)

5.49 kΩ

= −32.94

Vo = AvVi = (−32.94)(4 mV)

= −131.76 mV

37. Zi = 10 MΩ || 91 MΩ ≅ 9 MΩ

g

m = 22(12 mA) 1.45 V

11

3 V 3 V

Q

GS

DSS

PP

V

I

VV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 4.13 mS

Z

o = rd || RS || 1/gm = 45 kΩ || 1.1 kΩ || 1/4.13 mS

= 1.074 kΩ || 242.1 Ω

= 197.6 Ω

Av = ( ) (4.13 mS)( 45 k 1.1 k )

1 ( ) 1 (4.13 mS)(45 k 1.1 k )

md S

grR

Ω

=

++ ΩΩ

(4.13 mS)(1.074 k ) 4.436

1 (4.13 mS)(1.074 k ) 1 4.436

Ω

==

+Ω+

=0.816

38. gm =

()

2Q

GS GS Th

kV V−

= 2(0.3 × 10−3)(8 V − 3 V)

= 3 mS

39. Q

GS

V = 6.7 V

gm =

(

)

2Q

GS T

kV V− = 2(0.3 × 10−3)(6.7 V − 3 V) = 2.22 mS

Z

i = 10 M 100 k 2.2 k

1 ( ) 1 (2.22 mS)(100 k 2.2 k )

FdD

md D

RrR

grR

+Ω+ΩΩ

=

++ ΩΩ

= 10 M 2.15 k

1 2.22 mS(2.15 k )

Ω+ Ω

+Ω

≅ 1.73 MΩ

Zo = RF || rd || RD = 10 MΩ || 100 kΩ || 2.2 kΩ = 2.15 kΩ

A

v = −gm(RF || rd || RD) = −2.22 mS(2.15 kΩ) = −4.77

120

40. gm =

(

)

2Q

GS T

kV V− = 2(0.2 × 10−3)(6.7 V − 3 V)

= 1.48 mS

Zi = 10 M 100 k 2.2 k

1 ( ) 1 (1.48 mS)(100 k 2.2 k )

FdD

md D

RrR

grR

+Ω+ΩΩ

=

++ ΩΩ

= 10 M 2.15 k

1 (1.48 mS)(2.15 k )

Ω+ Ω

+Ω

= 2.39 MΩ > 1.73 MΩ (#39)

Zo = RF || rd || RD = 2.15 kΩ = 2.15 kΩ (#39)

A

v = −gm(RF || rd || RD) = −(1.48 mS)(2.15 kΩ)

= −3.182 < −4.77 (#39)

41. Q

GS

V = 5.7 V, gm =

(

)

2Q

GS T

kV V− = 2(0.3 × 10−3)(5.7 V − 3.5 V)

= 1.32 mS

rd = 1

30 S

μ

= 33.33 kΩ

A

v = −gm(RF || rd || RD) = −1.32 mS(22 MΩ || 33.33 kΩ || 10 kΩ)

= −10.15

Vo = AvVi = (−10.15)(20 mV) = −203 mV

42. ID = k(VGS − VT)2

∴k = (on)

2

(on)

()

D

GS T

I

VV− = 2

4 mA

(7 V 4 V)

− = 0.444 × 10−3

gm = ()

2( )

Q

GS GS Th

kV V− = 2(0.444 × 10−3)(7 V − 4 V)

= 2.66 mS

Av = −gm(RF || rd || RD) = −(2.66 mS)(22 MΩ || 50 kΩ || 10 kΩ) = −22.16

8.33 k

Ω

≅ 8.33 kΩ

Vo = AvVi = (−22.16)(4 mV) = −88.64 mV

43. Q

GS

V = 4.8 V, gm =

()

2QTh

GS GS

kV V− = 2(0.4 × 10−3)(4.8 V − 3 V) = 1.44 mS

Av = −gm(rd || RD) = −(1.44 mS)(40 kΩ || 3.3 kΩ) = −4.39

V

o = AvVi = (−4.39)(0.8 mV) = −3.51 mV

121

44. rd = 11

25 S

os

y

μ

= = 40 kΩ

Q

GS

V = 0 V, ∴gm = gm0 = 22(8 mA)

2.5 V

DSS

P

I

V= = 6.4 mS

v

A = gm(rd || RD)

8 = (6.4 mS)(40 kΩ || RD)

8

6.4 mS = 1.25 kΩ = 40 k

40 k

D

R

Ω

⋅

Ω+

and

RD = 1.29 kΩ

Use RD = 1.3 kΩ

45. 11

(3 V)

33

Q

GS P

VV==− = −1 V

22

1 V

112 mA1

3 V

Q

GS

DDSS

P

V

II V

⎛⎞ −

⎛⎞

=− = −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 5.33 mA

RS = 1 V

5.33 mA

Q

S

D

V

I= = 187.62 Ω ∴Use RS = 180 Ω

g

m = 22(12 mA) 1 V

11

3 V 3 V

Q

GS

DSS

PP

V

I

VV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

⎜⎟ −

⎝⎠

= 5.33 mS

A

v = −gm(RD || rd) = −10

or RD || 40 kΩ = 10

5.33 mS

− = 1.876 kΩ

40 k

D

R

⋅Ω

+Ω

= 1.876 kΩ

40 kΩRD = 1.876 kΩRD + 75.04 kΩ2

38.124RD = 75.04 kΩ

RD = 1.97 kΩ ⇒ RD = 2 kΩ

122

Chapter 9

1. (a) 3, 1.699, −1.151

(b) 6.908, 3.912, −0.347

(c) results differ by magnitude of 2.3

2. (a) log10 2.2 × 103 = 3.3424

(b) loge (2.2 × 103) = 2.3 log10(2.2 × 103) = 7.6962

(c) loge (2.2 × 103) = 7.6962

3. (a) same 13.98

(b) same −13.01

(c) same 0.699

4. (a) dB = 10 log10 o

i

P

P = 10 log10

100 W

5 W = 10 log10 20 = 10(1.301)

= 13.01 dB

(b) dB = 10 log10

100 mW

5 mW = 10 log1020 = 10(1.301)

= 13.01 dB

(c) dB = 10 log10 100 W

20 W

μ

= 10 log105 = 10(0.6987)

=

6.9897 dB

5. dBm

G = 10 log10 2

10

600

25 W

10 log

1 mW 1 mW

P

Ω

=

43.98 dBm

6. GdB = 20 log10 2

1

V

V = 20 log10

100 V

25 V = 20 log104 = 20(0.6021)

= 12.04 dB

7. GdB = 20 log10 2

10

1

25 V

20 log 10 mV

V

V= = 20 log10 2500

= 20(3.398) = 67.96 dB

8. (a) Gain of stage 1 = A dB

Gain of stage 2 = 2 A dB

Gain of stage 3 = 2.7 A dB

A + 2A +2.7A = 120

A = 21.05 dB

123

(b) Stage 1: 1

v

A = 21.05 dB = 20 log10 1

1

o

i

V

21.05

20 = 1.0526 = log10 1

1

o

i

V

101.0526 = 1

1

o

i

V

and 1

1

o

i

V

V = 11.288

Stage 2: 2

v

A = 42.1 dB = 20 log10 2

2

o

i

V

2.105 = log10 2

2

o

i

V

102.105 = 2

2

o

i

V

and 2

2

o

i

V

V = 127.35

Stage 3: : 3

v

A = 56.835 dB = 20 log10 3

3

o

i

V

2.8418 = log10 3

3

o

i

V

102.8418 = 3

3

o

i

V

and 3

3

o

i

V

V = 694.624

12 3T

vvvV

AAAA=⋅⋅ = (11.288)(127.35)(694.624) = 99,8541.1

AT = 120 dB

?

= 20 log1099,8541.1

120 dB ≅ 119.99 dB (difference due to level of accuracy carried through calculations)

9. (a) GdB = 20 log10 2

10

1

48 W

10 log 5 W

P

μ

= = 69.83 dB

(b)

Gv = 20 log10 10

10

20 log (48 W)(40 k )

20 log 100 mV

oo

o

ii

PR

V

VV

Ω

==

= 82.83 dB

124

(c) Ri =

22

(100 mV)

5 W

i

V

P

μ

= = 2 kΩ

(d) Po =

2

o

V

R

⇒ Vo = (48 W)(40 k )

oo

PR =Ω = 1385.64 V

10. (a) Same shape except Av = 190 is now level of 1. In fact, all levels of Av are divided by 190

to obtain normalized plot.

0.707(190) =

134.33 defining cutoff frequencies

at low end f1 ≅ 230 Hz (remember this is a log scale)

at high end f2 ≅ 160 kHz

(b)

11. (a) 2

1

1( /)

o

v

i

V

AV

f

==

+ f1 = 11

2 2 (1.2 k )(0.068 F)RC

π

πμ

=Ω

= 1950.43 Hz

2

1

1950.43 Hz

1

v

A

f

=

⎛⎞

+⎜⎟

⎝⎠

(b)

dB

V

A

100 Hz: v

A = 0.051 −25.8

1 kHz: v

A = 0.456 −6.81

2 kHz:

v

A = 0.716 −2.90

5 kHz: v

A = 0.932 −0.615

10 kHz:

v

A = 0.982 −0.162

(c) f1 ≅ 1950 Hz

(d)(e)

125

12. (a) f1 = 1

2

R

C

π

= 1.95 kHz

θ

= tan−11

f

= tan−11.95 kHz

f

(b)

f

θ

= tan−11.95 kHz

f

100 Hz 87.06°

1 kHz 62.85°

2 kHz 44.27°

5 kHz 21.3°

10 kHz 11.03°

(c)

f1 = 1

2

R

C

π

= 1.95 kHz

(d) First find

θ

= 45° at f1 = 1.95 kHz. Then sketch an approach to 90° at low frequencies and

0° at high frequencies. Use an expected shape for the curve noting that the greatest

change in

θ

occurs near f1. The resulting curve should be quite close to that plotted above.

13. (a)

10 kHz

(b)

1 kHz

(c) 20 kHz → 10 kHz → 5 kHz

(d) 1 kHz → 10 kHz → 100 kHz

126

14. From example 9.9, re = 15.76 Ω

Av = 4 k2.2 k40 k

15.76

CLo

e

RRr

r

−−Ω Ω Ω

=Ω

 

= −86.97 (vs. −90 for Ex. 9.9)

s

L

f

: ro does not affect Ri ∴

s

L

f

= 1

2( )

s

iS

R

RC

π

+ the same ≅ 6.86 Hz

C

L

f

= 11

2( ) 2( )

oLC CoLC

R

RC R r RC

ππ

=

++



RC || ro = 4 kΩ || 40 kΩ = 5.636 kΩ

1

2(5.636 k 2 k )(1 F)

C

L

f

π

μ

=Ω+ Ω

=

28.23 Hz (vs. 25.68 Hz for Ex. 9.9)

E

L

f

: Re not affected by ro, therefore, 1

2

E

L

eE

f

R

C

π

= ≅ 327 Hz is the same.

In total, the effect of ro on the frequency response was to slightly reduce the mid-band

gain.

15. (a)

β

RE ≥ 10R2

(120)(1.2 kΩ) ≥ 10(10 kΩ)

144 kΩ ≥ 100 kΩ (checks!)

VB = 10 k (14 V)

10 k 68 k

Ω

Ω+ Ω = 1.795 V

V

E = VB − VBE = 1.795 V − 0.7 V

= 1.095 V

IE = 1.095 V

1.2 k

E

V

R=Ω = 0.913 mA

r

e = 26 mV 26 mV

0.913 mA

E

I= = 28.48 Ω

(b)

mid

( ) (3.3 k 5.6 k )

28.48

LC

V

e

RR

Ar

−Ω Ω

=− = Ω

= −72.91

(c) Zi = R1 || R2 ||

β

re

= 68 kΩ || 10 kΩ || (120)(28.48 Ω)

3.418 kΩ

= 2.455 kΩ

127

(d)

s

ooi

v

s

is

VVV

AVVV

==⋅

2.455 k

2.455 k 0.82 k

ii

sis

VZ

VZR

Ω

==

+Ω+Ω

= 0.75

s

v

A = (−72.91)(0.75)

= −54.68

(e) 11

2 ( ) 2 (0.82 k 2.455 k )(0.47 F)

S

L

sis

fRRC

π

πμ

==

+Ω+Ω

= 103.4 Hz

C

L

f

= 11

2 ( ) 2 (5.6 k 3.3 k )(0.47 F)

oLC

RRC

π

πμ

=

+Ω+Ω

= 38.05 Hz

E

L

f

= 1

2eE

R

C

π

: Re = RE || s

e

Rr

β

′

⎛⎞

+

⎜⎟

⎝⎠

s

R

′ = Rs || R1 || R2 = 0.82 kΩ || 68 kΩ || 10 kΩ

= 749.51 Ω

Re = 1.2 kΩ || 749.51 28.48

120

Ω

⎛⎞

+

Ω

⎜⎟

⎝⎠

= 1.2 kΩ || 34.73 Ω

= 33.75 Ω

11

2 2 (33.75 )(20 F)

E

L

eE

fRC

π

πμ

== Ω

= 235.79 Hz

(f) f1 ≅

E

L

f = 235.79 Hz

(g)(h)

128

0.91 k

0.6 k 470 k

0.91 k 6.93

110

910 12.38

12.21

s

eE e

sB

e

R

RR r

β

′

⎛⎞

=+

⎜⎟

⎝⎠

⎛⎞

=Ω +

⎜⎟

⎝⎠

ΩΩ

⎛⎞

=

Ω+Ω

⎜⎟

⎝⎠

=Ω Ω

=Ω



16. (a) IB = 20 V 0.7 V 19.3 V

( 1) 470 k (111)(0.91 k ) 470 k 101.01 k

CC BE

BE

VV

RR

β

−

==

++ Ω+ Ω Ω+ Ω

= 33.8

μ

A

IE = (

β

+ 1)IB = (111)(33.8

μ

A)

= 3.752 mA

re = 26 mV

3.752 mA = 6.93 Ω

(b)

mid

()

(3 k 4.7 k ) 1.831 k

6.93 6.93

oCL

v

ie

VRR

AVr

−−Ω Ω− Ω

== = =

Ω



= −264.24

(c) Zi = RB ||

β

re = 470 kΩ || (110)(6.93 Ω) = 470 kΩ || 762.3 Ω

= 761.07 Ω

(d)

( mid) mid

761.07 (264.24)

761.07 0.6 k

s

i

vv

is

Z

AA

ZR

Ω

== −

+Ω+Ω

= −147.76

(e)

S

L

f

= 11

2 ( ) 2 (600 761.07 )(1 F)

SiS

RZC

π

πμ

=

+Ω+Ω

=

116.93 Hz

C

L

f

= 11

2 ( ) 2 (3 k 4.7 k )(1 F)

oLC

RRC

π

πμ

=

+Ω+Ω

=

20.67 Hz

E

L

f

= 1

2eE

R

C

π

= 1

2 (12.21 )(6.8 F)

π

μ

Ω

=

1.917 kHz

(f) f1 ≅ E

L

f=1.917 kHz

(g, h)

129

12

1 k 120 k 30 k

0.96 k

ss

RRRR

′

=

ΩΩΩ

=Ω



17. (a)

β

RE ≥ 10R2

(100)(2.2 kΩ) ≥ 10(30 kΩ)

220 kΩ ≥ 300 kΩ (No!)

RTh = R1 || R2 = 120 kΩ || 30 kΩ = 24 kΩ

E

Th = 30 k (14 V)

30 k 120 k

Ω

Ω+ Ω = 2.8 V

I

B = 2.8 V 0.7 V

( 1) 24 k 222.2 k

Th BE

Th E

EV

RR

β

−

=

++ Ω+ Ω

= 8.53

μ

A

IE = (

β

+ 1)IB = (101)(8.53

μ

A)

= 0.86 mA

re = 6 mV 26 mV

0.86 mA

E

I

2= = 30.23 Ω

(b)

mid

v

A = EL

eEL

RR

rRR+

= 2.2 k 8.2 k

30.23 2.2 k 8.2 k

ΩΩ

Ω+ Ω Ω

= 0.983

(c) Zi = R1 || R2 ||

β

(re +

E

R

′

)

E

R

′

= RE || RL = 2.2 kΩ || 8.2 kΩ = 1.735 kΩ

= 120 kΩ || 30 kΩ || (100)(30.23 Ω + 1.735 kΩ)

= 21.13 kΩ

(d)

s

v

A = ooi

s

is

VVV

=⋅ 21.13 k

21.13 k 1 k

ii

sis

VZ

VZR

Ω

==

+

Ω+ Ω = 0.955

(e) 1

2( )

S

L

s

is

f

R

RC

π

=+

= 1

2 (1 k 21.13 k )(0.1 F)

π

μ

Ω+ Ω

= 71.92 Hz

1

2( )

C

L

oLC

f

R

RC

π

=+

Ro = RE || s

e

Rr

β

′

⎛⎞

+

⎜⎟

⎝⎠

= (2.2 kΩ) || 0.96 k 30.23

100

Ω

⎛⎞

+

Ω

⎜⎟

⎝⎠

= 39.12 Ω

1

2 (39.12 8.2 k )(0.1 F)

C

L

f

π

μ

=Ω+ Ω

= 193.16 Hz

130

(f)

low

1

f

≅ 193.16 Hz

(g)(h)

18. (a) IE = 4 V 0.7 V

1.2 k

EE EB

E

VV

R

−−

=Ω = 2.75 mA

re = 26 mV 26 mV

2.75 mA

E

I= = 9.45 Ω

(b)

mid

v

A = 3.3 k 4.7 k

9.45

CL

e

RR

r

ΩΩ

=Ω



= 205.1

(c) Zi = RE || re = 1.2 kΩ || 9.45 Ω

= 9.38 Ω

(d)

(mid)s

v

A = mid

9.38 (205.1)

9.38 100

i

v

is

ZA

ZR

Ω

=

+Ω+Ω

= 17.59

(e)

s

L

f

= 11

2 ( ) 2 (100 9.38 )(10 F)

sis

RZC

π

πμ

=

+Ω+Ω

= 145.5 Hz

C

L

f

= 11

2 ( ) 2 (3.3 k 4.7 k )(10 F)

oLE

RRC

π

πμ

=

+Ω+Ω

=

1.989 Hz

(f) f =

s

L

f

≅ 145.5 Hz

(g, h)

131

Q

GS

D

V

I

≅

2.45 V

2.1 mA

−

19. (a) VGS = −IDRS

ID =

2

1GS

DSS

P

V

IV

⎛⎞

−

⎜⎟

⎝⎠

(b) gm0 = 22(6 mA)

6 V

DSS

P

I

V= = 2 mS

g

m = 0

(2.45 V)

12 mS1

(6 V)

Q

GS

m

P

V

gV

⎛⎞

−

−= −

⎜⎟

⎜⎟ −

⎝⎠

= 1.18 mS

(c)

mid

v

A = −gm(RD || RL)

= −1.18 mS(3 kΩ || 3.9 kΩ) = −1.18 mS(1.6956 kΩ)

= −2

(d) Zi = RG = 1 MΩ

(e)

s

v

A = Av = −2

(f)

sig

11

2 ( ) 2 (1 k 1 M )(0.1 F)

G

L

iG

fRRC

π

πμ

==

+Ω+Ω

=

1.59 Hz

1

2( )

C

L

oLC

f

R

RC

π

=+

= 1

2 (3 k 3.9 k )(4.7 F)

π

μ

Ω+ Ω

=

4.91 Hz

eq

1

2

S

L

S

f

R

C

π

= eq

11

1.2 k 1.18 mS

S

m

RR

g

==Ω = 1.2 kΩ || 847.46 Ω

= 1

2 (496.69 )(10 F)

π

μ

Ω = 496.69 Ω

= 32.04 Hz

(g) f1 ≅ S

L

f

≅ 32 Hz

(h, i)

132

Q

GS

D

V

I

≅

2.55 V

3.3 mA

−

20. (a) same as problem 19

Q

GS

V ≅ −2.45 V, Q

D

I ≅ 2.1 mA

(b) gm0 = 2 mS, gm = 1.18 mS (rd has no effect!)

(c)

mid

v

A = −gm (RD || RL || rd)

= −1.18 mS(3 kΩ || 3.9 kΩ || 100 kΩ)

= −1.18 mS (1.67 kΩ)

= −1.971 (vs. −2 for problem 19)

(d) Zi = RG = 1 MΩ (the same)

(e) (mid)s

v

A = mid

sig

1 M

()

1 M 1 k

i

v

i

ZA

ZR

Ω

=

+Ω+Ω

(−1.971)

= −1.969 vs. −2 for problem 19

(f)

G

L

f

= 1.59 Hz (no effect)

:

C

L

f

Ro = RD || rd = 3 kΩ || 100 kΩ = 2.91 kΩ

C

L

f

= 11

2 ( ) 2 (2.91 k 3.9 k )(4.7 F)

oLC

RRC

π

πμ

=

+Ω+Ω

= 4.97 Hz vs. 4.91 Hz for problem 19

S

L

f

: Req = 1(1 )/(( ))

S

SmddDL

R

gr r R R++ +

= 1.2 k

1 (1.2 k )(1 (1.18 mS)(100 k )) /(100 k 3 k 3.9 k )

Ω

+Ω+ Ω Ω+Ω Ω



=

1.2 k

11.404

Ω

+

≅ 499.2 Ω

S

L

f

: =

eq

11

22(499.2 )(10 F)

S

RC

π

πμ

=Ω

=

31.88 Hz vs. 32.04 for problem 19.

Effect of rd = 100 kΩ insignificant!

21. (a) VG = 68 k (20 V)

68 k 220 k

Ω

Ω+ Ω = 4.72 V

VGS = VG − IDRS

VGS = 4.72 V − ID(2.2 kΩ)

I

D = IDSS(1 − VGS/VP)2

(b) gm0 = 22(10 mA)

6 V

DSS

P

I

V= = 3.33 mS

g

m = 0

(2.55 V)

13.33 mS1

6 V

GS

m

P

V

gV

⎛⎞ −

⎛⎞

−= −

⎜⎟ ⎜⎟

−

⎝⎠

= 1.91 mS

133

(c)

mid

v

A = −gm(RD || RL)

= −(1.91 mS)(3.9 kΩ || 5.6 kΩ)

= −4.39

(d) Zi = 68 kΩ || 220 kΩ = 51.94 kΩ

(e)

(mid)s

v

A = oi

is

VV

⋅

i

s

V

V = 51.94 k

51.94 k 1.5 k

i

is

Z

ZR

Ω

=

+Ω+Ω

= 0.972

(mid)s

v

A = (−4.39)(0.972) = −4.27

(f)

sig

11

2 ( ) 2 (1.5 k 51.94 k )(1 F)

G

L

iG

fRRC

π

πμ

==

+Ω+Ω

= 2.98 Hz

11

2 ( ) 2 (3.9 k 5.6 k )(6.8 F)

C

L

oLC

fRRC

π

πμ

==

+Ω+Ω

= 2.46 Hz

eq

1

2

S

L

S

f

R

C

π

=

= 1

2 (388.1 )(10 F)

π

μ

Ω

= 41 Hz

Req = RS || 1

m

g

= 1.5 kΩ || 1

1.91 mS

= 1.5 kΩ || 523.56 Ω

= 388.1 Ω

(g) f1 ≅ S

L

f

= 41 Hz

(h, i)

134

22. (a)

i

H

f

=

1

2Th i

R

C

π

= 1

2 (614.56 )(931.92 pF)

π

Ω

= 277.89 kHz

112

0.82 k 68 k 10 k 3.418 k

Th s i

RRRRR=

=

ΩΩΩ Ω



= 0.81 kΩ 2.547 kΩ

= 614.56 Ω

Ci = 1(1 )

Wbebc v

CCC A

+

+−

= 5 pF + 40 pF + 12 pF(1 − (−72.91))

= 931.92 pF ↑Prob. 15

o

H

f

=

2

1

2Th o

R

C

π

= 1

2 (2.08 k )(28 pF)

π

Ω

= 2.73 MHz

2

Th

R

= RC || RL = 5.6 kΩ || 3.3 kΩ

= 2.08 kΩ

Co = oo

WceM

CCC

+

= 8 pF + 8 pF + 12 pF

= 28 pF

(b) f

β

≅

mid

1

2( )

ebe bc

rC C

πβ

+ = 1

2 (120)(28.48 )(40 pF + 12 pF)

π

Ω

= 895.56 kHz ↑ Prob. 15

fT =

β

f

β

= (120)(895.56 kHz)

= 107.47 MHz

(c)

23. (a)

1

2

i

H

Th i

f

R

C

π

=

1

Th

R

= Rs || RB || Ri

Ri: IB = 20 V 0.7 V

(1) 470 k (111)(0.91 k)

CC BE

BE

VV

RR

β

−−

=

+

+Ω+Ω

= 33.8

μ

A

IE = (

β

+ 1)IB = (110 + 1)(33.8

μ

A)

= 3.75 mA

re = 26 mV 26 mV

3.75 mA

E

I= = 6.93 Ω

R

i =

β

re = (110)(6.93 Ω)

= 762.3 Ω

1

Th

R

= Rs || RB || Ri = 0.6 kΩ || 470 kΩ || 762.3 Ω

= 335.50 Ω

135

i

H

f

= 1

2 (335.50 )( )

i

C

π

Ω

Ci: Ci = i

Wbe

CC+ + (1 − Av)Cbc

Av: mid

v

A = ()

(4.7 k 3 k )

6.93

LC

e

RR

r

−

ΩΩ

=Ω



= −264.2

Ci = 7 pF + 20 pF + (1 − (−264.2)6 pF

= 1.62 nF

i

H

f

= 1

2 (335.50 )(1.62 nF)

π

Ω

≅ 293 kHz

o

H

f

=

2

1

2Th o

R

C

π

2

Th

R

= RC || RL = 3 kΩ || 4.7 kΩ = 1.831 kΩ

Co = oo

WceM

CCC++

≅ Cf = Cbc

= 11 pF + 10 pF + 6 pF

= 27 pF

o

H

f

= 1

2 (1.831 k )(27 pF)

π

Ω

= 3.22 MHz

(b) f

β

=

mid

1

2( )

ebe bc

rC C

πβ

+

= 1

2 (110)(6.93 )(20 pF + 6 pF)

π

Ω

= 8.03 MHz

fT =

β

midf

β

= (110)(8.03 MHz)

= 883.3 MHz

(c)

136

24. (a)

i

H

f

=

1

2Th i

R

C

π

= 1

2 (955 )(58 pF)

π

Ω

= 2.87 MHz

112Th s b

R

RRRZ

=

 

Zb =

β

re + (

β

+ 1)(RE || RL)

= (100)(30.23 Ω) + (101)(2.2 kΩ || 8.2 kΩ)

= 3.023 kΩ + 175.2 kΩ

= 178.2 kΩ

1

Th

R

= 1 kΩ || 120 kΩ || 30 kΩ || 178.2 kΩ

= 955 Ω

Ci = i

Wbebc

CCC

+

+ (No Miller effect)

= 8 pF + 30 pF + 20 pF

= 58 pF

24 kΩ

o

H

f

=

2

1

2Th o

R

C

π

= 1

2 (38.94 )(32 pF)

π

Ω

= 127.72 MHz

2

Th

R

= RE || RL || 123

e

R

RR

r

β

⎛⎞

+

⎜⎟

⎝⎠



= 2.2 kΩ || 8.2 kΩ || 24 k 1 k

30.23 100

ΩΩ

⎛⎞

Ω+

⎜⎟

⎝⎠



= 1.735 kΩ || (30.23 Ω + 9.6 Ω)

= 1.735 kΩ || 39.83 Ω

= 38.94 Ω

Co = o

Wce

CC

+

= 10 pF + 12 pF

= 32 pF

(b) f

β

=

mid

1

2( )

ebe bc

rC C

πβ

+

= 1

2 (100)(30.23 )(30 pF + 20 pF)

π

Ω

= 1.05 MHz

fT =

β

mid f

β

= 100(1.05 MHz) = 105 MHz

(c)

137

25. (a) i

H

f

=

1

2Th i

R

C

π

1

Th

R

= Rs || RE || Ri

Ri: IE = 4 V 0.7 V

1.2 k

EE BE

E

VV

R

−−

=Ω = 2.75 mA

r

e = 26 mV 26 mV

2.75 mA

E

I= = 9.45 Ω

R

i = RE || re = 1.2 kΩ || 9.45 Ω

= 9.38 Ω

Ci: Ci = i

Wbe

CC+ (no Miller cap-noninverting!)

= 8 pF + 24 pF

= 32 pF

Ri = 0.1 kΩ || 1.2 kΩ || 9.38 Ω = 8.52 Ω

i

H

f

= 1

2 (8.52 )(32 pF)

π

Ω ≅ 584 MHz

o

H

f

=

2

1

2Th o

R

C

π

2

Th

R

= RC || RL = 3.3 kΩ || 4.7 kΩ = 1.94 kΩ

Co = o

Wbc

CC++ (no Miller)

= 10 pF + 18 pF

= 28 pF

o

H

f

= 1

2 (1.94 k )(28 pF)

π

Ω

= 2.93 MHz

(b) f

β

=

mid

1

2( )

ebe bc

rC C

πβ

+

= 1

2 (80)(9.45 )(24 pF + 18 pF)

π

Ω

= 5.01 MHz

fT =

β

midf

β

= (80)(5.01 MHz)

= 400.8 MHz

(c)

138

26. (a) From problem 19 gm0 = 2 mS, gm = 1.18 mS

(b) From problem 19 mid (mid)s

vv

AA≅ = −2

(c)

1

2

i

H

Th i

f

R

C

π

=

1

2 (999 )(21 pF)

i

H

f

π

=Ω

= 7.59 MHz

1

Th

R

= Rsig || RG

= 1 kΩ || 1 MΩ

= 999 Ω

Ci = ii

WgsM

CCC

+

i

M

C = (1 − Av)Cgd

= (1 − (−2)4 pF

= 12 pF

Ci = 3 pF + 6 pF + 12 pF

= 21 pF

o

H

f

=

2

1

2Th o

R

C

π

= 1

2 (1.696 k )(12 pF)

π

Ω

= 7.82 MHz

2

Th

R

= RD || RL

= 3 kΩ || 3.9 kΩ

= 1.696 kΩ

Co = oo

WdsM

CCC

+

o

M

C = 1

14 pF

2

⎛⎞

−

⎜⎟

−

⎝⎠

= (1.5)(4 pF)

= 6 pF

Co = 5 pF + 1 pF + 6 pF

= 12 pF

(d)

27. (a) gm0 = 22(10 mA)

6 V

DSS

P

I

V= = 3.33 mS

From problem #21 Q

GS

V ≅ −2.55 V, Q

D

I ≅ 3.3 mA

gm = 01Q

GS

m

P

V

gV

⎛⎞

−

⎜⎟

⎝⎠

= 3.33 mS 2.55 V

16 V

−

⎛⎞

−

⎜⎟

−

⎝⎠

= 1.91 mS

139

(b)

mid

v

A = −gm(RD || RL)

= −(1.91 mS)(3.9 kΩ || 5.6 kΩ)

= −4.39

Zi = 68 kΩ || 220 kΩ = 51.94 kΩ

sig

51.94 k

51.94 k 1.5 k

ii

si

VZ

VZR

Ω

==

+Ω+Ω

= 0.972

(mid)s

v

A = (−4.39)(0.972)

= −4.27

(c)

i

H

f

=

1

2Th i

R

C

π

1

Th

R

= Rsig || R1 || R2

= 1.5 kΩ || 51.94 kΩ

= 1.46 kΩ

Ci = (1 )

i

Wgs vgd

CC AC++−

= 4 pF + 12 pF + (1 − (−4.39))8 pF

= 59.12 pF

i

H

f

= 1

2 (1.46 k )(59.12 pF)

π

Ω

= 1.84 MHz

o

H

f

=

2

1

2Th o

R

C

π

2

Th

R

= RD || RL = 3.9 kΩ || 5.6 kΩ

Co = 1

1

o

Wds gd

v

CC C

A

⎛⎞

++−

⎜⎟

⎝⎠

= 6 pF + 3 pF + 1

18 pF

(4.39)

⎛⎞

−

⎜⎟

−

⎝⎠

= 18.82 pF

o

H

f

= 1

2 (2.3 k )(18.82 pF)

π

Ω

= 3.68 MHz

(d)

= 2.3 k

Ω

140

28. T

v

A = 1234

vv v v

AAAA⋅⋅⋅

= Av

4

= (20)4

= 16 × 104

29. 2

f

′ =

(

)

1/

2

21

n

f

−

=

(

)

1/4

21(2.5MHz)−

1.18

= 0.435(2.5 MHz)

=

1.09 MHz

30. 1

11/ 1/4

40 Hz

2121

n

f

f′==

−−

= 40 Hz

0.435

= 91.96 Hz

31. (a)

v = 4

π

Vm

11

sin 2 sin 2 (3 ) sin 2 (5 )

35

ss s

f

tftft

ππ π

⎡++

⎢

⎣

11

sin 2 (7 ) sin 2 (9 )

79

ss

ft ft

ππ

⎤

+++

⎥

⎦

…

= 12.73 × 10−333

1

(sin 2 (100 10 ) sin 2 (300 10 )

3

tt

ππ

×+ ×

333

111

sin 2 (500 10 ) sin 2 (700 10 ) sin 2 (900 10 ) )

579

ttt

πππ

+×+ ×+×

(b)

BW ≅ 0.35

r

t

≅ 0.35

0.7 s

μ

tr ≅ 0.75

μ

s − 0.05

μ

s = 0.7

μ

s

≅ 500 kHz

(c)

P = 90 mV 80 mV

90 mV

VV

V

′

−−

= = 0.111

(0.111)(100 kHz)

o

Ls

P

ff

ππ

== ≅ 3.53 kHz

At 90% or 81 mV, t

≅

0.75

μ

s

At 10% or 9 mV, t ≅ 0.05

μ

s

141

Chapter 10

1. Vo = 1

1

250 k (1.5 V)

20 k

F

RV

R

Ω

−=−

Ω = −18.75 V

2. Av =

1

o

F

i

V

R

VR

=−

For R1 = 10 kΩ:

Av = 500 k

10 k

Ω

−Ω = −50

For R1 = 20 kΩ:

Av = 500 k

20 k

Ω

−Ω = −25

3. Vo = 1

1

1 M

20 k

f

RV

R

Ω

⎛⎞

−=−

⎜⎟

Ω

⎝⎠

V1 = 2 V

V

1 = 2 V

50− = −40 mV

4. Vo = 11

1

200 k

20 k

F

RVV

R

Ω

−=−

Ω = −10 V1

For V1 = 0.1 V:

Vo = −10(0.1 V) = −1 V Vo ranges

For V1 = 0.5 V: from

Vo = −10(0.5 V) = −5 V −1 V to −5 V

5. Vo = 1

1

360 k

1 1 ( 0.3 V)

12 k

F

RV

R

⎛⎞ Ω

⎛⎞

+=+ −

⎜⎟

Ω

⎝⎠

= 31(−0.3 V) = −9.3 V

6. Vo = 11

1

360 k

11

12 k

F

RVV

R

⎛⎞ Ω

⎛⎞

+=+

⎜⎟

Ω

⎝⎠

⎝⎠ = 2.4 V

V

1 = 2.4 V

31 = 77.42 mV

142

7. Vo = 1

1

1F

RV

R

⎛⎞

+

⎜⎟

⎝⎠

For

R1 = 10 kΩ:

Vo = 200 k

1(0.5 V)

10 k

Ω

⎛⎞

+

⎜⎟

Ω

⎝⎠

= 21(0.5 V) = 10.5 V

For R1 = 20 kΩ:

Vo = 200 k

120 k

Ω

⎛⎞

+

⎜⎟

Ω

⎝⎠

(0.5 V) = 11(0.5 V) = 5.5 V

V

o ranges from 5.5 V to 10.5 V.

8. Vo = 123

12 3

ff f

RR R

VVV

RR R

⎡⎤

−++

⎢⎥

⎣⎦

= 330 k 330 k 330 k

(0.2 V) + ( 0.5 V)+ (0.8 V)

33 k 22 k 12 k

ΩΩΩ

⎡⎤

−−

⎢⎥

ΩΩΩ

⎣⎦

= −[10(0.2 V) + 15(−0.5 V) + 27.5(0.8 V)]

= −[2 V + (−7.5 V) + 2.2 V]

= −[24 V − 7.5 V] = −16.5 V

9. Vo = 123

12 3

FF F

RRR

VVV

RR R

⎡⎤

−++

⎢⎥

⎣⎦

= 68 k 68 k 68 k

(0.2 V)+ ( 0.5 V) + ( 0.8 V)

33 k 22 k 12 k

ΩΩ Ω

⎡⎤

−−+

⎢⎥

ΩΩ Ω

⎣⎦

= −[0.41 V − 1.55 V + 4.53 V]

= −3.39 V

10. vo(t) = 1

1()

vtdt

RC

−∫

= 11.5

(200 k )(0.1 F) dt

μ

−Ω∫

= −50(1.5t) = −75t

11. Vo = V1 = +0.5 V

12. Vo = 1

1

100 k

20 k

f

RV

R

Ω

−=−

Ω(1.5 V)

= −5(1.5 V) = −7.5 V

13. V2 = 200 k (0.2 V)

20 k

Ω

⎡⎤

−⎢⎥

Ω

⎣⎦ = −2 V

V

3 = 200 k

1(0.2 V)

10 k

Ω

⎛⎞

+

⎜⎟

Ω

⎝⎠

= +4.2 V

143

14. Vo = 400 k 100 k 100 k

1 (0.1 V) (0.1 V)

20 k 20 k 10 k

Ω−Ω Ω

⎛⎞⎛⎞⎛⎞

+⋅+−

⎜⎟⎜⎟⎜⎟

ΩΩΩ

⎝⎠⎝⎠⎝⎠

= (2.1 V)(−5) + (−10)(0.1 V)

= −10.5 V − 1 V = −11.5 V

15. Vo = −600 k 600 k 300 k

(25 mV) + ( 20 mV)

15 k 30 k 30 k

ΩΩ Ω

⎡⎤⎛⎞

−−

⎜⎟

⎢⎥

ΩΩ Ω

⎣⎦⎝⎠

300 k (20 mV)

15 k

⎡⎤Ω

⎛⎞

+− −

⎜⎟

⎢⎥

Ω

⎝⎠

⎣⎦

= −[40(25 mV) + (20)(−20 mV)](−10) + (−20)(−20 mV)

= −[1 V − 0.4 V](−10) + 0.4 V

= 6 V + 0.4 V = 6.4 V

16. Vo =

1

1f

I

oIof

RVIR

R

⎛⎞

++

⎜⎟

⎝⎠

= 200 k

1 (6 mV) + (120 nA)(200 k )

2 k

Ω

⎛⎞

+Ω

⎜⎟

Ω

⎝⎠

= 101(6 mV) + 24 mV

= 606 mV + 24 mV = 630 mV

17. 4 nA

20 nA +

22

Io

IB IB

I

II

+

+=+= = 22 nA

4 nA

20 nA

22

Io

IB IB

I

II

−

−=−= − = 18 nA

18. f1 = 800 kHz

f

c =

2

1

3

800 kHz

150 10

v

f

A=× = 5.3 Hz

19. ACL = 2.4 V/ s

/ 0.3 V/10 s

i

SR

Vt

μ

=

ΔΔ = 80

20. ACL =

1

200 k

2 k

f

R

Ω

=Ω = 100

K = ACL Vi = 100(50 mV) = 5 V

w

s ≤ 0.4 V / s

5 V

SR

K

μ

= = 80 × 103 rad/s

f

s =

3

80 10

22

s

w

π

×

= = 12.73 kHz

144

21. VIo = 1 mV, typical

IIo = 20 nA, typical

Vo(offset) =

1

1f

I

o

RV

R

⎛⎞

+

⎜⎟

⎝⎠

+ IIoRf

=

200 k

1(1 mV)

20 k

Ω

⎛⎞

+

⎜⎟

Ω

⎝⎠

+ (200 kΩ)(20 nA)

= 101(1 mV) + 4000 × 10−6

= 101 mV + 4 mV = 105 mV

22. Typical characteristics for 741

Ro = 25 Ω, A = 200 K

(a)

ACL =

1

200 k

2 k

f

R

Ω

−=− Ω = −100

(b) Zi = R1 = 2 kΩ

(c)

Zo = 25

1

11 (200,000)

100

o

R

A

β

Ω

=

++

= 25

2001

Ω = 0.0125 Ω

23. Ad = 120 mV

1 mV

o

d

V

V= = 120

A

c = 20 V

1 mV

o

c

V

μ

= = 20 × 10−3

Gain (dB) = 20 log 3

120

20log 20 10

d

c

A

−

=×

= 20 log(6 × 103) = 75.56 dB

145

24. Vd = Vi1 − Vi2 = 200

μ

V − 140

μ

V = 60

μ

V

c = 12

(200 V 140 V)

22

ii

VV

μ

++

= = 170

μ

V

(a) CMRR =

d

c

A

A = 200

Ac = 6000

200 200

d

A= = 30

(b) CMRR = 5

10

d

c

A

A=

Ac = 55

6000

10 10

d

A= = 0.06 = 60 × 10−3

Using Vo = AdVd

1

1c

d

V

CMRR V

⎡⎤

+

⎢⎥

⎣⎦

(a) Vo = 6000(60

μ

V) 1170 V

1200 60 V

μ

⎡⎤

+

⎢⎥

⎣⎦

= 365.1 mV

(b) Vo = 6000(60

μ

V) 5

1 170 V

110 60 V

μ

⎡

⎤

+

⎢

⎥

⎣

⎦ = 360.01 mV

146

Chapter 11

1. Vo = 1

1

180 k (3.5 mV)

3.6 k

F

RV

R

Ω

−=−

Ω = −175 mV

2. Vo = 1

1

750 k

11

36 k

F

RV

R

⎛⎞ Ω

⎛⎞

+=+

⎜⎟

Ω

⎝⎠

⎝⎠ (150 mV, rms)

= 3.275 V, rms ∠0°

3. Vo = 510 k 680 k 750 k

1(20 V)

18 k 22 k 33 k

μ

ΩΩΩ

⎛⎞⎡⎤⎡⎤

+−−

⎜⎟

⎢⎥⎢⎥

ΩΩΩ

⎝⎠⎣⎦⎣⎦

= (29.33)(−30.91)(−22.73)(20

μ

V)

= 412 mV

4.

1

420 k

1R

⎛⎞

Ω

+

⎜⎟

⎝⎠

= +15

2

420 k

R

Ω

− = −22

2

420 k

R

Ω

= −30

R1 = 420 k

14

Ω R2 = 420 k

22

Ω

R3 = 420 k

30

Ω

R

1 = 71.4 kΩ R2 = 19.1 kΩ R3 = 14 kΩ

Vo = (+15)(−22)(−30)V1 = 9000(80

μ

V) = 792 mV

= 0.792 V

5.

Vo1 = 1

11

150 k

F

RVV

RR

Ω

−=−

1

11

150 k

15

o

v

VA

VR

Ω

==−=−

R1 = 150 k

15

Ω = 10 kΩ

Vo2 = 2

11

22

150 k

F

RVV

RR

Ω

−=−

2

12

150 k

30

o

v

VA

VR

Ω

==−=−

R2 = 150 k

30

Ω = 5 kΩ

147

6. Vo = 12

12

470 k 470 k

(40 mV) + (20 mV)

47 k 12 k

FF

RR

VV

RR

⎡⎤

ΩΩ

⎡⎤

−+ =−

⎢⎥

ΩΩ

⎣⎦

= −[400 mV + 783.3 mV] = −1.18 V

7. Vo = 12

10 k 150 k 300 k 300 k

10 k 10 k 150 k 150 k

VV

ΩΩ+ΩΩ

⎛⎞⎛ ⎞

−

⎜⎟⎜ ⎟

Ω+ Ω Ω Ω

⎝⎠⎝ ⎠

= 0.5(3)(1 V) − 2(2 V) = 1.5 V − 4 V = −2.5 V

8. Vo = 330 k 470 k 470 k

(12 mV) (18 mV)

33 k 47 k 47 k

⎧⎫

ΩΩΩ

⎡⎤⎛⎞

−+

⎨⎬

⎜⎟

⎢⎥

ΩΩΩ

⎣⎦⎝⎠

⎩⎭

= −[(−120 mV)(10) + 180 mV] = −[−1.2 V + 0.18 V]

= +1.02 V

9.

10.

11. IL = 1

1

12 V

2 k

V

R=Ω = 6 mA

12. Vo = −I1R1 = −(2.5 mA)(10 kΩ) = −25 V

13.

11

1

oF

s

IR

VRR

⎛⎞

=⎜⎟

⎝⎠

100 k 1 (10 mV)

200 k 10

o

IΩ⎛⎞

=⎜⎟

ΩΩ

⎝⎠ = 0.5 mA

14. Vo = 2

1

p

R

⎛⎞

+

⎜⎟

⎝⎠

[V2 − V1]

= 2(5000)

1[1 V 3 V]

1000

⎛⎞

+−

⎜⎟

⎝⎠ = −22 V

148

15. fOH =

11

2 2 (2.2 k )(0.05 F)RC

π

πμ

=Ω

= 1.45 kHz

16. fOL =

11

2 2 (20 k )(0.02 F)RC

π

πμ

=Ω

= 397.9 Hz

17. fOL =

11

22(10 k)(0.05 F)RC

π

πμ

=Ω = 318.3 Hz

f

OH =

22

11

2 2 (20 k )(0.02 F)RC

π

πμ

=Ω

= 397.9 Hz

149

Chapter 12

1. 18 V 0.7 V

1.2 k

Q

CC BE

B

VV

IR

−−

==

Ω = 14.42 mA

QQ

CB

II

β

= = 40(14.42 mA) = 576.67 mA

Pi = VCCIdc ≅ Q

CC C

VI = (18 V)(576.67 mA)

≅ 10.4 W

IC(rms) =

β

IB(rms)

= 40(5 mA) = 200 mA

Po = 2(rms)

CC

IR = (200 mA)2(16 Ω) = 640 mW

2. Q

B

I = 18 V 0.7 V

1.5 k

CC BE

B

VV

R

−−

=Ω = 11.5 mA

Q

C

I = Q

B

I

β

= 40(11.5 mA) = 460 mA

Pi(dc) = VCCIdc =

(

)

QQ

CC C B

VI I+

= 18 V(460 mA + 11.5 mA)

= 8.5 W

= 18 V(460 mA) = 8.3 W

Q

iCCC

PVI

⎡⎤

≈

⎣⎦

3. From problem 2: Q

C

I = 460 mA, Pi = 8.3 W.

For maximum efficiency of 25%:

%

η

= 100% × o

i

P

P = 8.3 W

o

P × 100% = 25%

Po = 0.25(8.3 W) = 21 W

[If dc bias condition also is considered:

VC = VCC − Q

CC

IR

= 18 V − (460 mA)(16 Ω) = 10.64 V

collector may vary ±7.36 V about Q-point, resulting in maximum output power:

Po =

22

() (7.36 V)

2 2(16)

CE

C

VP

R==1.69 W

150

4. Assuming maximum efficiency of 25%

with Po(max) = 1.5 W

%

η

= o

i

P

P × 100%

Pi = 1.5 W

0.25 = 6 W

Assuming dc bias at mid-point, VC = 9 V

18 V 9 V

16

Q

CC C

C

VV

IR

−−

==

Ω = 0.5625 A

Pi(dc) = Q

CC C

VI = (18 V)(0.5625 A)

= 10.38 W

at this input:

%

η

= o

i

P

P × 100% = 1.5 W

10.38 W × 100% = 14.45%

5. Rp =

2

1

2

s

N

R

N

⎛⎞

⎜⎟

⎝⎠ =

2

25 (4 )

1

⎛⎞ Ω

⎜⎟

⎝⎠ = 2.5 kΩ

6. R2 = a2R1

a

2 = 2

1

8 k

8

R

Ω

=Ω = 1000

a = 1000 = 31.6

7. R2 = a2R1

8 kΩ = a2(4 Ω)

a2 = 8 k

4

Ω

Ω = 2000

a = 2000 = 44.7

8. (a)

Ppri = PL = 2 W

(b)

PL =

2

L

V

R

V

L = (2 W)(16 )

LL

PR =Ω

= 32 = 5.66 V

(c)

R2 = a2R1 = (3.87)2(16 Ω) = 239.6 Ω

P

pri =

2

pri

V

R

= 2 W

2

pri

V = (2 W)(239.6 Ω)

Vpri = 479.2 = 21.89 V

[or,

Vpri = aVL = (3.87)(5.66 V) = 21.9 V]

151

(d) PL = 2

LL

IR

IL = 2 W

16

L

P

R=Ω = 353.55 mA

P

pri = 2 W = 2

pri pri

IR = (239.6 Ω)2

pri

I

Ipri = 2 W

239.6 Ω = 91.36 mA

or, Ipri = 353.55 mA

3.87

L

I

a= = 91.36 mA

9. Idc = Q

C

I = 150 mA

P

i = Q

CC C

VI = (36 V)(150 mA) = 5.4 W

%

η

= 2 W

100% 100%

5.4 W

o

i

P

P×= × = 37%

10.

11.

152

12. (a)

Pi = VCCIdc = (25 V)(1.75 A) = 43.77 W

Where,

Idc = 22 222 V

8

p

L

V

IR

ππ π

==⋅

Ω

= 1.75 A

(b) Po =

22

(22 V)

22(8 )

p

L

V

R=Ω = 30.25 W

(c) %

η

= o

i

P

P × 100% = 30.75 W

43.77 W × 100% = 69%

13. (a) max

Pi = VCCIdc

= VCC ⋅ 2225 V

(25 V) 8

CC

L

V

R

ππ

⎛⎞

⎡

⎤

⋅= ⋅

⎜⎟

⎢

⎥

Ω

⎣

⎦

⎝⎠

= 49.74 W

(b) max Po =

22

(25 V)

22(8 )

CC

L

V

R=Ω = 39.06 W

(c) max %

η

= max 39.06 W

100%

max 49.74 W

o

i

P

P×= × 100%

= 78.5%

14. (a)

(peak)

L

V = 20 V

Pi = VCCIdc = 2L

CC

L

V

R

π

⎡⎤

⋅

⎢⎥

⎣⎦

= 220 V

(22 V) 4

π

⎡⎤

⋅

⎢⎥

Ω

⎣⎦

= 70 W

Po =

22

(20 V)

22(4 )

L

V

R=Ω = 50 W

%

η

= o

i

P

P × 100% = 50 W

70 W × 100% = 71.4%

(b) Pi = 24 V

(22 V) 4

π

⎡

⎤

⋅

⎢

⎥

Ω

⎣

⎦ = 14 W

P

o =

2

(4)

2(4) = 2 W

%

η

= o

i

P

P × 100% = 2 W

14 W × 100% = 14.3%

153

15.

16. (a) max Po(ac) for

p

eak

L

V = 30 V:

max Po(ac) =

22

(30 V)

22(8 )

L

V

R=

Ω

= 56.25 W

(b) max Pi(dc) = VCCIdc = 2230 V

8

o

CC CC

L

V

VV

R

ππ

⎡⎤

⎡

⎤

⋅= ⋅

⎢⎥

⎢

⎥

Ω

⎣

⎦

⎣⎦ = 71.62 W

(c) max %

η

= max

max

o

i

P

P × 100% = 56.25 W

71.62 W × 100%

= 78.54%

(d) max

22

2 2 (30)

8

Q

CC

Z

L

V

PR

ππ

=⋅ =⋅ = 22.8 W

17. (a) Pi(dc) = VCCIdc = VCC ⋅ 2o

L

V

R

π

⎛⎞

⎜⎟

⎝⎠

= 30 V ⋅ 228

8

π

⎡

⎤

⋅

⎢

⎥

⎣

⎦ = 27 W

(b)

Po(ac) =

22

(rms) (8 V)

8

L

V

R=

Ω

= 8 W

(c) %

η

= 8 W

100% 27 W

o

i

P

P×= × 100% = 29.6%

(d)

P2Q = Pi − Po = 27 W − 8 W = 19 W

154

18. (a)

Po(ac) =

22

(rms) (18 V)

8

L

V

R=Ω = 40.5 W

(b)

Pi(dc) = VCCIdc = peak

2L

CC

L

V

VR

π

⎡⎤

⋅

⎢⎥

⎣⎦

= (40 V) 2182 V

8

π

⎡⎤

⋅

⎢⎥

Ω

⎣⎦

= 81 W

(c) %

η

= o

i

P

P × 100% = 40.5 W

81 W × 100% = 50%

(d)

2Q

P = Pi − Po = 81 W − 40.5 W = 40.5 W

19. %D2 = 2

1

A

A × 100% = 0.3 V

2.1 V × 100% ≅ 14.3%

%D3 = 3

1

A

A × 100% = 0.1 V

2.1 V × 100% ≅ 4.8%

%

D4 = 4

1

A

A × 100% = 0.05 V

2.1 V × 100% ≅ 2.4%

20. %THD = 222

234

D

DD++ × 100%

= 222

(0.143) (0.048) (0.024)++ × 100%

= 15.3%

21. D2 =

()

max min

1

2CE CE

CE CE

VV

+

− × 100%

=

1(20 V + 2.4 V) 10 V

2

20 V 2.4 V

−

− × 100%

= 1.2 V

17.6 V × 100% = 6.8%

22. THD = 222 2 2 2

234 (0.15) (0.01) (0.05)DDD++= + +

≅ 0.16

P1 =

22

1(3.3 A) (4 )

22

C

IR Ω

= = 21.8 W

P = (1 + THD2)P1 = [1 + (0.16)2)]21.8 W

= 22.36 W

155

23. PD (150°C) = PD(25°C) − (T150 − T25) (Derating Factor)

= 100 W − (150°C − 25°C)(0.6 W/°C)

= 100 W − 125(0.6) = 100 − 75

= 25 W

24. PD = 200 C 80 C

0.5 C/W + 0.8 C/W + 1.5 C/W

JA

JC CS SA

TT

θθθ

−°− °

=

++ ° ° °

= 120 C

2.8 C/W

°

° = 42.9 W

25. PD =

J

A

JA

TT

θ

−

= 200 C 80 C 120 C

(40C/W) 40C/W

°− ° °

=

°°

= 3 W

156

Chapter 13

1.

2.

3.

4.

5.

157

6.

7. Circuit operates as a window detector.

Output goes

low for input above 9.1 k (12 V)

9.1 k + 6.2 k

Ω+

ΩΩ = 7.1 V

Output goes

low for input below 1 k (12 V)

1 k + 6.2 k

Ω

+

Ω

Ω = 1.7 V

Output is high for input between +1.7 V and +7.1 V.

8.

9. 5

11010 26

(16 V) = (16 V)

232

= 13 V

10. Resolution = 12

10 V 10 V

2 2 4096

REF

n

V== = 2.4 mV/count

11. See section 13.3.

12. Maximum number of count steps = 212 = 4096

13. 212 = 4096 steps at T = 11

20 MHz

f= = 50 ns/count

Period = 4096 counts × ns

50 count = 204.8

μ

s

158

14.

f = 1.44

(2)

AB

R

RC+ = 350 kHz

C = 1.44

7.5 k + 2(7.5 k )(350 kHz)

ΩΩ

≅ 183 pF

15.

T = 1.1 RAC

20

μ

s = 1.1(7.5 kΩ)C

C =

6

3

20 10

1.1(7.5 10 )

−

×

= 2.4 × 10−9

= 2400 × 10−12

= 2400 pF

16. T = 11

10 kHzf= = 100

μ

s

T = 1.1 RAC = 1.1(5.1 kΩ)(5 nF) = 28

μ

s

159

17. fo =

11

2C

VV

RC V

+

⎛⎞

−

⎜⎟

⎝⎠

V

+ = 12 V

V

C = 3

23

11 k

() (12 V)

1.8 k + 11 k

RV

RR

+

Ω

=+

+ΩΩ

= 10.3 V

fo = 2 12 V 10.3 V

(4.7 k )(0.001 F) 12 V

μ

−

⎡⎤

⎢⎥

Ω⎣⎦

= 60.3 × 103 ≅ 60 kHz

18. With potentiometer set at top:

VC = 34

234

5 k 18 k (12 V)

510 5 k 18 k

RR V

RRR

+

+Ω+ Ω

=

++ Ω+ Ω+ Ω = 11.74 V

resulting in a lower cutoff frequency of

fo = 3

11

2 2 12 V 11.74 V

(10 10 )(0.001 F) 12 V

C

VV

RC V

μ

+

⎛⎞

−−

⎛⎞

=

⎜⎟ ⎜⎟

×⎝⎠

⎝⎠

= 4.3 kHz

With potentiometer set at bottom:

VC = 4

234

18 k (12 V)

510 5 k 18 k

RV

RRR

+

Ω

=

++ Ω+ Ω+ Ω

= 9.19 V

resulting in a higher cutoff frequency of

fo =

11

2 2 12 V 9.19 V

(10 k )(0.001 F) 12 V

C

VV

RC V

μ

+

⎛⎞

−−

⎡

⎤

=

⎜⎟

⎢

⎥

Ω

⎣

⎦

⎝⎠

= 61.2 kHz

19. V+ = 12 V

V

C = 3

23

10 k (12 V)

1.5 k + 10 k

RV

RR

+

Ω

=

+ΩΩ

= 10.4 V

f

o =

11 1

2 2 12 V 10.4 V

10 k ( ) 12 V

C

VV

RC V C

+

⎛⎞

−−

⎛⎞

=

⎜⎟ ⎜⎟

Ω⎝⎠

⎝⎠

= 200 kHz

C1 = 2(0.133)

10 k (200 kHz)

Ω

= 133 × 10−12 = 133 pF

20. fo =

11

0.3 0.3

(4.7 k )(0.001 F)RC

μ

=Ω

= 63.8 kHz

21. C1 =

1

0.3 0.3

(10 k )(100 kHz)

Rf =Ω = 300 pF

160

22. fL = ±8o

f

V

= ±

3

8(63.8 10 )

6 V

×

11

0.3 0.3

4.7 k (0.001 F)

o

fRC

μ

⎡⎤

==

⎢⎥

Ω

⎣⎦

= 85.1 kHz = 63.8 kHz

23. For current loop: mark = 20 mA

space = 0 mA

For RS − 232 C: mark = −12 V

space = +12 V

24. A line (or lines) onto which data bits are connected.

25. Open-collector is active-LOW only.

Tri-state is active-HIGH or active-LOW.

161

Chapter 14

1. Af = 2000 2000

1

1201

1 ( 2000)

10

A

β

−−

==

+⎛⎞

+− −

⎜⎟

⎝⎠

= −9.95

2.

()

11

(10%)

11000

20

f

dA dA

AAA

β

==

⎛⎞

−−

⎜⎟

⎝⎠

= 0.2%

3. Af = 300 300

1

121

1 ( 300)

15

A

β

−−

==

+⎛⎞

+− −

⎜⎟

⎝⎠

= −14.3

R

if = (1 +

β

A)Ri = 21(1.5 kΩ) = 31.5 kΩ

R

of = 50 k

121

o

R

A

β

Ω

=

+ = 2.4 kΩ

4. RL = oD

oD

R

R+ = 40 kΩ || 8 kΩ = 6.7 kΩ

A = −gmRL = −(5000 × 10−6)(6.7 × 103) = −33.5

β

= 2

12

200 k

200 k 800 k

R

RR

−−Ω

=

+Ω+Ω

= −0.2

A

f = 33.5 33.5

1 1 ( 0.2)( 33.5) 7.7

A

β

−−

==

++−−

= −4.4

5. DC bias:

IB = 16 V 0.7 V

( 1) 600 k 76(1.2 k )

CC BE

BE

VV

RR

β

−−

=

++ Ω+ Ω

= 15.3 V

691.2 k

Ω

= 22.1

μ

A

IE = (1 +

β

)IB

= 76(22.1

μ

A) = 1.68 mA

[VCE = VCC − IC(RC + RE) = 16 V − 1.68 mA(4.7 kΩ + 1.2 kΩ) ≅ 6.1 V]

re = 26 mV 26 mV

(mA) 1.68 mA

E

I= ≅ 15.5 Ω

h

ie = (1 +

β

)re = 76(15.5 Ω) = 1.18 kΩ = Zi

Zo = RC = 4.7 kΩ

162

Av = 75

1.18 k 1.2 k

fe

ie E

h

hR

−−

=

+Ω+Ω

= −31.5 × 10−3

β

= RE = −1.2 × 103

(1 +

β

A) = 1 + (−1.2 × 103)(−31.5 × 10−3)

= 38.8

Af =

3

31.5 10

138.8

v

A

β

−

−×

=

+ = 811.86 × 10−6

f

v

A = −Af RC = −(811.86 × 10−6)(4.7 × 103) = −3.82

f

i

Z

= (1 +

β

Av)Zi = (38.8)(1.18 kΩ) = 45.8 kΩ

f

o

Z

= (1 +

β

Av)Zo = (38.8)(4.7 kΩ) = 182.4 kΩ

without feedback (RE bypassed):

Av = 4.7 k

15.5

C

e

R

r

−Ω

=− Ω = −303.2

6. C = 33

11

2 6 2 (10 10 )(2.5 10 ) 6Rf

ππ

=××

= 2.6 × 10−9 = 2600 pF = 0.0026

μ

F

7. fo = 11

264 c

RC

R

π

⋅⎛⎞

+⎜⎟

⎝⎠

= 312 33

11

2(610)(150010 ) 64(1810/610)

π

−⋅

×× +× ×

= 4.17 kHz ≅ 4.2 kHz

8. fo = 312

11

2 2(1010)(240010 )RC

ππ

−

=××

= 6.6 kHz

9. Ceq = 12

12

(750 pF)(2000 pF)

750 pF + 2000 pF

CC

=

+ = 577 pF

f

o = 612

eq

11

22 40 10 )(577 10 )

LC

ππ

−−

=××

= 1.05 MHz

10. fo =

eq

1

2LC

π

,

= 1

2 (100 H)(3300 pF)

πμ

= 277 kHz

where Ceq = 12

12

CC

+

= (0.005 F)(0.01 F)

0.005 F + 0.01 F

μ

= 3300 pF

163

11. fo =

eq

1

2LC

π

,

= 312

1

2(410)(25010)

π

−−

××

= 159.2 kHz

Leq = L1 + L2 + 2 M

= 1.5 mH + 1.5 mH + 2(0.5 mH)

= 4 mH

12. fo =

eq

1

2LC

π

,

= 1

2 (1800 H)(150 pF)

πμ

= 306.3 kHz

where Leq = L1 + L2 + 2 M

= 750

μ

H + 750

μ

H + 2(150

μ

H)

= 1800

μ

H

13. See Fig. 14.33a and Fig. 14.34.

14. fo = 1

ln(1/(1 ))

TT

RC

η

−

for

η

= 0.5:

fo ≅ 1.5

TT

R

C

(a) Using RT = 1 kΩ

CT = 1.5 1.5

(1 k )(1 kHz)

To

Rf =Ω = 1.5

μ

F

(b) Using RT = 10 kΩ

CT = 1.5 1.5

(10 k )(150 kHz)

To

Rf =Ω = 1000 pF

164

dc

12 V 5mA

2.4 k

L

V

IR

== =

Ω

Chapter 15

1. ripple factor =

dc

(rms) 2 V/ 2

50 V

r

V

V= = 0.028

2. %VR = NL FL

FL

VV

V

− × 100% = 28 V 25 V

25 V

− × 100% = 12%

3. Vdc = 0.318Vm

Vm = dc 20 V

0.318 0.318

V= = 62.89 V

V

r = 0.385Vm = 0.385(62.89 V) = 24.2 V

4. Vdc = 0.636Vm

V

m = dc 8 V

0.636 0.636

V= = 12.6 V

V

r = 0.308Vm = 0.308(12.6 V) = 3.88 V

5. %r =

dc

(rms)

r

V

V × 100%

Vr(rms) = rVdc = 8.5

100 × 14.5 V = 1.2 V

6. VNL = Vm = 18 V

V

FL = 17 V

%VR = NL FL

FL

VV

V

−

× 100% = 18 V 17 V

17 V

−

× 100%

=

5.88%

7. Vm = 18 V

C = 400

μ

F

I

L = 100 mA

Vr = dc

2.4 2.4(100)

400

I

C= = 0.6 V, rms

Vdc = Vm − dc

4.17I

C

= 18 V − 4.17(100)

400 = 16.96 V

≅ 17 V

8. Vr = dc

2.4 2.4(120)

200

I

C= = 1.44 V

9. C = 100

μ

F

V

dc = 12 V

R

L = 2.4 kΩ

Vr(rms) = dc

2.4 2.4(5)

100

I

C= = 0.12 V

165

10. C = dc

dc

2.4 2.4(150)

(0.15)(24)

I

rV = = 100

μ

F

11. C = 500

μ

F

I

dc = 200 mA

R = 8% = 0.08

Using r = dc

dc

2.4I

CV

Vdc = dc

2.4 2.4(200)

0.08(500)

I

rC = = 12 V

Vm = Vdc + dc

4.17I

C = 12 V + (200)(4.17)

500

= 12 V + 1.7 V = 13.7 V

12. C = dc

2.4 2.4(200)

(0.07)

r

I

V= = 6857

μ

F

13. C = 120

μ

F

I

dc = 80 mA

V

m = 25 V

V

dc = Vm − dc

4.17I

C = 25 V − 4.17(80)

120

= 22.2 V

%r = dc

dc

2.4I

CV × 100% = 2.4(80)

(120)(22.2) × 100%

= 7.2%

14. dc 2(80)

100 100

r

rV

V′

⋅

′== = 1.6 V, rms

15. Vr = 2 V

V

dc = 24 V

R = 33 Ω, C = 120

μ

F

XC = 1.3 1.3

120C= = 10.8 Ω

%r =

dc

r

V

V × 100% = 2 V

24 V × 100%

= 8.3%

10.8 (2 V)

33

C

rr

X

VV

R

′== = 0.65 V

dc

V′ = Vdc − IdcR = 24 V − 33 Ω (100 mA)

= 20.7 V

%r′ =

dc

r

V

′

′ × 100% = 0.65 V

20.7 V × 100% = 3.1%

166

16.

dc dc

L

R

VV

RR

′=+

= 500

50 500+(40 V)

= 36.4 V

Idc = dc 36.4 V

500

L

V

R

′=Ω = 72.8 mA

17.

XC = 1.3 1.3

100C= = 13 Ω

13

100

C

rr

X

VV

R

′==

(2.5 V)

= 0.325 V, rms

18. VNL = 60 V

V

FL = dc

1 k

100 1 k

L

RV

RR

Ω

=

+Ω+Ω

(50 V) = 45.46 V

%VR = NL FL

FL

VV

V

− × 100% = 50 V 45.46 V

45.46 V

− × 100%

= 10 %

19. Vo = VZ − VBE = 8.3 V − 0.7 V = 7.6 V

V

CE = Vi − Vo = 15 V − 7.6 V = 7.4 V

I

R = 15 V 8.3 V

1.8 k

iZ

VV

R

−−

=Ω = 3.7 mA

I

L = 7.6 V

2 k

o

L

V

R=Ω = 3.8 mA

I

B = 3.8 mA

100

C

I

β

= = 38

μ

A

I

Z = IR − IB = 3.7 mA − 38

μ

A = 3.66 mA

167

20. Vo = 2

12

2

()

BEz

RR

VV

R

++

= 33 k 22 k

22 k

Ω+ Ω

Ω(10 V + 0.7 V)

= 26.75 V

21. Vo = 1

2

12 k

1110 V

8.2 k

Z

RV

R

⎛⎞ Ω

⎛⎞

+=+

⎜⎟

Ω

⎝⎠

= 24.6 V

22. Vo = VL = 10 V + 0.7 V = 10.7 V

23.

24. IL = 250 mA

V

m = Vr(rms) ⋅ 2 = 2(20 V) = 28.3 V

p

eak

r

V = 3Vr(rms) = dc

2.4

3I

C

⎛⎞

⎜⎟

⎝⎠

= 2.4(250)

3500

⎛⎞

⎜⎟

⎝⎠

= 2.1 V

Vdc = Vm −

p

eak

r

V = 28.3 V − 2.1 V = 26.2 V

Vi(low) = Vdc −

p

eak

r

V = 26.2 V − 2.1 V = 24.1 V

25. To maintain Vi(min) ≥ 7.3 V (see Table 15.1)

p

eak

r

V ≤ Vm − Vi(min) = 12 V − 7.3 V = 4.7 V

so that

Vr(rms) = peak 4.7 V

1.73

3

r

V= = 2.7 V

The maximum value of load current is then

Idc = (rms) (2.7 V)(200)

2.4 2.4

r

VC

= = 225 mA

168

26. Vo = Vref 2

1

R

⎛⎞

+

⎜⎟

⎝⎠

+ IadjRL

= 1.25 V 1.8 k

1240

Ω

⎛⎞

+

⎜⎟

Ω

⎝⎠

+ 100

μ

A(2.4 kΩ)

= 1.25 V(8.5) + 0.24 V

= 10.87 V

27. Vo = Vref 2

1

R

⎛⎞

+

⎜⎟

⎝⎠

+ Iadj R2

= 1.25 V 1.5 k

1220

Ω

⎛⎞

+

⎜⎟

Ω

⎝⎠

+ 100

μ

A(1.5 kΩ)

= 9.9 V

169

Chapter 16

1. (a) The Schottky Barrier diode is constructed using an n-type semiconductor material and a

metal contact to form the diode junction, while the conventional p-n junction diode uses

both p- and n-type semiconductor materials to form the junction.

(b) −

2. (a) In the forward-biased region the dynamic resistance is about the same as that for a p-n

junction diode. Note that the slope of the curves in the forward-biased region is about

the same at different levels of diode current.

(b) In the reverse-biased region the reverse saturation current is larger in magnitude than

for a p-n junction diode, and the Zener breakdown voltage is lower for the Schottky

diode than for the conventional p-n junction diode.

3. 100 A 0.5 A 1.33 A/ C

75 C

R

I

C

μμ μ

Δ−

==°

Δ° °

(1.33 A/ C) C (1.33 A/ C)(25 C) 33.25 A

R

I

μ

μμ

Δ= °Δ= ° °=

0.5 A+33.25 A

R

I

μ

==33.75 A

μ

4. 11

2 2 (1 MHz)(7 pF)

C

XfC

ππ

== =22.7 kΩ

400 mV

10 mA

F

DC

F

V

RI

== = 40 Ω

5. Temperature on linear scale

T(1/2 power level of 100 mW) ≅ 95°C

6. VF a linear scale VF (25°C) ≅ 380 mV = 0.38 V

At 125°C, VF ≅ 280 mV

100 mV

100 C

F

T

VΔ=

Δ°

= 1 mV/°C

∴ At 100°C VF = 280 mV + (1 mV/°C)(25°C)

= 280 mV + 25 mV

= 305 mV

Increase temperature and VF drops.

170

7. (a) CT(VR) =

()

1/3

(0) 80 pF

4.2 V

110.7 V

n

RT

C

VV

=⎛⎞++

⎜⎟

⎝⎠

= 80 pF

1.912 = 41.85 pF

(b)

k = CT(VT + VR)n

= 41.85 pF(0.7 V + 4.2 V)1/3

1.698

≅ 71 × 10−12

8. (a) At

−3 V, C = 40 pF

At

−12 V, C = 20 pF

ΔC = 40 pF − 20 pF = 20 pF

(b) At

−8 V, 40 pF

20 V

R

C

V

Δ=

Δ = 2 pF/V

At −2 V, 60 pF

9 V

R

C

V

Δ=

Δ = 6.67 pF/V

R

C

V

Δ

Δ increases at less negative values of VR.

9. Ratio =

( 1 V) 92pF

( 8 V) 5.5 pF

t

C

−=

− = 16.73

(1.25 V)

(7 V)

t

C

−

− = 13

10. Ct ≅ 15 pF

11

2 2 (10 MHz)(3 )(15 pF)

vs on chart

st

QfR C

ππ

== Ω

=354.61 350

11. TCC =

10

()

o

C

CT T

Δ

− × 100% ⇒ T1 = 100%

()

Co

C

TC C

Δ

× + To

=

(0.11 pF)(100)

(0.02)(22 pF) + 25

= 50°C

12. VR from −2 V to −8 V

Ct(−2 V) = 60 pF, Ct(−8 V) = 6 pF

Ratio =

( 2 V) 60 pF

( 8 V) 6 pF

t

C

−=

− = 10

171

13. Q(−1 V) = 82, Q(−10 V) = 5000

Ratio =

( 10 V) 5000

Q( 1 V) 82

Q−=

− = 60.98

BW =

6

10 10 Hz

82

o

f

Q

×

= = 121.95 kHz

BW =

6

10 10 Hz

5000

o

f

Q

×

= = 2 kHz

14. High-power diodes have a higher forward voltage drop than low-current devices due to larger

IR drops across the bulk and contact resistances of the diode. The higher voltage drops result

in higher power dissipation levels for the diodes, which in turn may require the use of heat

sinks to draw the heat away from the body of the structure.

15. The primary difference between the standard p-n junction diode and the tunnel diode is that

the tunnel diode is doped at a level from 100 to several thousand times the doping level of a

p-n junction diode, thus producing a diode with a “negative resistance” region in its

characteristic curve.

16. At 1 MHz: XC = 612

11

22(110 Hz)(510 F)fC

ππ

−

=××

= 31.83 kΩ

At 100 MHz: XC = 612

1

2 (100 10 Hz)(5 10 F)

π

−

××

= 318.3 Ω

At 1 MHz: S

L

X

= 2

π

fL = 2

π

(1 × 106 Hz)(6 × 10−9 H)

= 0.0337 Ω

At 100 MHz: S

L

X

= 2

π

(100 × 106 Hz)(6 × 10−9 H)

= 3.769 Ω

Ls effect is negligible!

R and C in parallel:

f = 1 MHz

ZT = (152 180 )(31.83 k 90 )

152 31.83 kj

Ω∠ ° Ω∠ − °

−Ω− Ω

= −152.05 Ω∠0.27° ≅ −152 Ω∠0°

f = 100 MHz

Z

T = (152 180 )(318.3 90 )

152 318.3j

Ω∠ ° ∠ − °

−Ω−

= −137.16 Ω∠25.52 ≠ −152 Ω∠0°

At very high frequencies XC has some impact!

172

17. The heavy doping greatly reduces the width of the depletion region resulting in lower levels

of Zener voltage. Consequently, small levels of reverse voltage can result in a significant

current levels.

18. At VT = 0.1 V,

IF ≅ 5.5 mA

At VT = 0.3 V

IF ≅ 2.3 mA

R = 0.3 V 0.1 V

2.3 mA 5.5 mA

V

I

Δ−

=

Δ−

= 0.2 V

3.2 mA− = −62.5 Ω

19. Isat = 2 V

0.39 k

E

R=Ω ≅ 5.13 mA

From graph: Stable operating points: IT ≅ 5 mA, VT ≅ 60 mV

IT ≅ 2.8 mA, VT = 900 mV

20. Isat = 0.5 V

51

E

R=Ω = 9.8 mA

Draw load line on characteristics.

21. fs =

2

11

2

l

R

C

L

LC

π

⎛⎞

−

⎜⎟

⎝⎠

=

26

3

36

1(10 )(110 F)

1510 H

2 (5 10 H)(1 10 F)

π

−

−−

⎛⎞

Ω×

⎜⎟

−

⎜⎟

×

××

⎝⎠

= (2250.79 Hz)(0.9899)

≅ 2228 Hz

22. W = hf=

34 8

10

(6.624 10 J s)(3 10 m/s)

(5000)(10 m)

v

h

λ

−

×⋅×

=

= 3.97 × 10−19 J

3.97 × 10−19 J 19

1 eV

1.6 10 J

−

⎡⎤

⎢⎥

×

⎣⎦

= 2.48 eV

°

173

23. (a) Visible spectrum: 3750 A → 7500

(b) Silicon, peak relative response ≅ 8400 A

(c)

BW = 10,300 A − 6100 A = 4200 A

24.

92

12

410 W/m

1.609 10

−

×

× = 2,486 fc

From the intersection of VA = 30 V and 2,486 fc we find

Iλ ≅ 440

μ

A

25. (a) Silicon

(b) 1 A = 10−10 m,

7

10

610m

10 m/A

−

× ⇒ 6000 A → orange

26. Note that

V

λ

is given and not V.

At the intersection of V

λ

= 25 V and 3000fc we find Iλ ≅ 500

μ

A and

VR = IλR = (500 × 10−6 A)(100 × 103 Ω) = 50 V

27. (a) Extending the curve:

0.1 kΩ → 1000fc, 1 kΩ → 25fc

3

(1 0.1) 10

(1000 25)

cc

R

f

Δ−×Ω

=

Δ−

= 0.92 Ω/fc ≅ 0.9 Ω/fc

(b) 1 kΩ → 25fc, 10 kΩ → 1.3fc

3

(10 1) 10

(25 1.3)

cc

R

f

Δ−×Ω

=

Δ−

= 379.75 Ω/fc ≅ 380 Ω/fc

(c) 10 kΩ → 1.3fc, 100 kΩ → 0.15fc

3

(100 10) 10

(1.3 0.15)

cc

R

f

Δ−×

=

Δ− = 78,260.87 Ω/fc ≅ 78 × 103 Ω/fc

The greatest rate of change in resistance occurs in the low illumination region.

28. The “dark current” of a photodiode is the diode current level when no light is striking the

diode. It is essentially the reverse saturation leakage current of the diode, comprised mainly

of minority carriers.

29. 10fc → R ≅ 2 kΩ

Vo = 6 V =

3

33

(2 10 )

2 10 5 10

i

V×Ω

×Ω+×Ω

V

i = 21 V

°

° ° °

°

°°

174

30. Except for low illumination levels (0.01fc)

the % conductance curves appear above the

100% level for the range of temperature. In

addition, it is interesting to note that for

other than the low illumination levels the %

conductance is higher above and below

room temperature (25°C). In general, the %

conductance level is not adversely affected

by temperature for the illumination levels

examined.

31. (a) (b)

(c) Increased levels of illumination result in reduced rise and decay times.

32. The highest % sensitivity occurs between 5250A and 5750A. Fig 16.20 reveals that the CdS

unit would be most sensitive to yellow. The % sensitivity of the CdS unit of Fig. 16.30 is at

the 30% level for the range 4800A → 7000A. This range includes green, yellow, and orange

in Fig. 16.20.

33. (a) ≅ 5 mW radiant flux

(b) ≅ 3.5 mW 13

3.5 mW

1.496 10 W/lm

−

× = 2.34 × 1010

lms

°

175

34. (a) Relative radiant intensity ≅ 0.8.

(b)

35. At

IF = 60 mA, Φ ≅ 4.4 mW

At 5°, relative radiant intensity = 0.8

(0.8)(4.4 mW) = 3.52 mW

36. 6, 7, 8

37. −

38. The LED generates a light source in response to the application of an electric voltage. The

LCD depends on ambient light to utilize the change in either reflectivity or transmissivity

caused by the application of an electric voltage.

39. The LCD display has the advantage of using approximately 1000 times less power than the

LED for the same display, since much of the power in the LED is used to produce the light,

while the LCD utilizes ambient light to see the display. The LCD is usually more visible in

daylight than the LED since the sun’s brightness makes the LCD easier to see. The LCD,

however, requires a light source, either internal or external, and the temperature range of the

LCD is limited to temperatures above freezing.

40.

η

% =

2

max

2

cm

( )(100 mW/cm )

P

A × 100%

9% = max

22

(2 cm )(100 mW/cm )

P × 100%

Pmax = 18 mW

41. The greatest rate of increase in power will occur at low illumination levels. At higher

illumination levels, the change in VOC drops to nearly zero, while the current continues to rise

linearly. At low illumination levels the voltage increases logarithmically with the linear

increase in current.

42. (a) Fig. 16.48

⇒ 79 mW/cm2

(b) It is the maximum power density available at sea level.

(c) Fig. 16.48

≅ 12.7 mA

(b)

Relative radiant vs degrees off vertical

>30 and relative radiant intensity essentially zero-

Drops off very sharply after 25 !

θ

°

176

43. (a) (b)

(c) The curve of Io vs Pdensity is quite linear while the curve of Vo vs Pdensity is only linear in

the region near the optimum power locus (Fig 16.48).

44. Since log scales are present, the differentials must be as small as possible.

(7000 1000) 6000

(40 0) 40

R

T

Δ−ΩΩ

==

Δ−°°

= 150 Ω/°C

(3 1) 2

40 40

R

T

Δ−ΩΩ

==

Δ°°

= 0.05 Ω/°C

From the above 150 Ω/°C: 0.05 Ω/°C = 3000:1

Therefore, the highest rate of change occurs at lower temperatures such as 20°C.

45. No. 1 Fenwall Electronics Thermistor material.

Specific resistance

≅ 104 = 10,000 Ω cm

R = A

ρ

A 2x ∴R = 2 × (10,000 Ω) = 20 kΩ

twice

46. (a)

≅ 10−5 A = 10

μ

A

(b) Power

≅ 0.1 mW, R ≅ 107 Ω = 10 MΩ

(c) Log scale

≅ 0.3 mW

47. V = IR + IRunk + Vm

V = I(R + Runk) + 0 V

Runk = V

R

I

−

=

0.2 V

2 mA − 10 Ω

= 100

Ω − 10 Ω

=

90 Ω

177

Chapter 17

1. −

2. −

3. −

4. (a) p-n junction diode

(b) The SCR will not fire once the gate current is reduced to a level that will cause the

forward blocking region to extend beyond the chosen anode-to-cathode voltage. In

general, as IG decreases, the blocking voltage required for conduction increases.

(c) The SCR will fire once the anode-to-cathode voltage is less than the forward blocking

region determined by the gate current chosen.

(d) The holding current increases with decreasing levels of gate current.

5. (a) Yes

(b) No

(c) No. As noted in Fig. 17.8b the minimum gate voltage required to trigger all units is 3 V.

(d) VG = 6 V, IG = 800 mA is a good choice (center of preferred firing area).

VG = 4 V, IG = 1.6 A is less preferable due to higher power dissipation in the gate. Not in

preferred firing area.

6. In the conduction state, the SCR has characteristics very similar to those of a p-n junction

diode (where VT = 0.7 V).

7. The smaller the level of R1, the higher the peak value of the gate current. The higher the peak

value of the gate current the sooner the triggering level will be reached and conduction

initiated.

8. (a) VP = sec (rms) 2

2

V

⎛⎞

⎜⎟

⎝⎠

=

()

117 V 2

2 = 82.78 V

VDC = 0.636(82.78 V)

= 52.65 V

(b) VAK = VDC − VBatt = 52.65 V − 11 V = 41.65 V

178

(c) VR = VZ + VGK

= 11 V + 3 V

= 14 V

At 14 V, SCR2 conducts and stops the charging process.

(d) At least 3 V to turn on SCR2.

(e) V2 ≅ 1

2P

V = 1(82.78 V)

2 = 41.39 V

9. −

10. (a) Charge toward 200 V but will be limited by the development of a negative voltage

VGK

(

)

1

ZC

VV=− that will eventually turn the GTO off.

(b)

τ

= R3C1 = (20 kΩ)(0.1

μ

F)

= 2 ms

5

τ

= 10 ms

(c)

5

τ

′ = 1(5 )

2

τ

= 5 ms = 5RGTO C1

RGTO = 6

1

5 ms 5 ms

5 5(0.1 10 F)

C−

=× = 10 kΩ 1(20 k above)

2

⎛⎞

=Ω−

⎜⎟

⎝⎠

11. (a) ≅ 0.7 mW/cm2

(b) 0°C → 0.82 mW/cm2

100°C → 0.16 mW/cm2

0.82 0.16

0.82

− × 100% ≅ 80.5%

12. VC = VBR + VGK = 6 V + 3 V = 9 V

V

C = 40(1 − e−t/RC) = 9

40 − 40e−t/RC = 9

40e−t/RC = 31

e−t/RC = 31/40 = 0.775

RC = (10 × 103 Ω)(0.2 × 10−6 F) = 2 × 10−3 s

loge(e−t/RC) = loge 0.775

−t/RC = −t/2 × 10−3 = −0.255

and t = 0.255(2 × 10−3) = 0.51 ms

13. −

14. 12

B

RBR

VV= ± 10% 2

B

R

V

= 6.4 V ± 0.64 V ⇒ 5.76 V → 7.04 V

15. −

179

16. P

P

VV

I

− > R1

6

40 V [0.6(40 V) + 0.7 V]

10 10−

−

× = 1.53 MΩ > R1

V

VV

I

− < R1 ⇒ 40 V 1 V

8 mA

− = 4.875 kΩ < R1

∴1.53 MΩ > R1 > 4.875 kΩ

17. (a)

η

= 1

12

0E

B

BB

I

R

RR

=

+ ⇒ 0.65 =

2

2 k

B

R

Ω

Ω+ 2

B

R

= 1.08 kΩ

(b)

RBB =

(

)

12

0E

BB

I

RR

=

+ = 2 kΩ + 1.08 kΩ = 3.08 kΩ

(c)

1

B

R

V =

η

VBB = 0.65(20 V) = 13 V

(d)

VP =

η

VBB + VD = 13 V + 0.7 V = 13.7 V

18. (a)

η

= 1

0

E

B

BB I

R

R=

0.55 =

1

10 k

B

R

Ω

!

B

R

= 5.5 kΩ

RBB = !2

B

R

R+

10 kΩ = 5.5 kΩ + 2

B

R

2

B

R

= 4.5 kΩ

(b)

VP =

η

VBB + VD = (0.55)(20 V) + 0.7 V = 11.7 V

(c)

R1 < 20 V 11.7 V

50 A

P

p

VV

I

μ

−−

= = 166 kΩ

ok: 68 kΩ < 166 kΩ

180

(d) t1 = R1C logeV

P

VV

−

− = (68 × 103)(0.1 × 10−6) loge

18.8

8.3 = 5.56 ms

t2 = 12

()

B

R

R+C loge P

V

V = (0.2 kΩ + 2.2 kΩ)(0.1 × 10−6) loge

11.7

1.2

= 0.546 ms

T = t1 + t2 = 6.106 ms

f = 11

6.106 msT= = 163.77 Hz

(e)

(f)

2

2.2 k (20 V)

2.2 k 10 k

R

BB

RV

VRR

Ω

==

+

Ω+ Ω

= 3.61 V

2

1

2

( 0.7 V)

P

R

B

RV

VRR

−

≅+

= 2.2 k (11.7 V 0.7 V)

2.2 k 0.2 k

Ω−

Ω+ Ω

= 10.08 V

(g)

f ≅

1

11

log (1/(1 )) (6.8 k )(0.1 F)log 2.22

ee

RC

ημ

=

−Ω = 184.16 Hz

difference in frequency levels is partly due to the fact that t2 ≅ 10% of t1.

19. IB = 25

μ

A

I

C = e

f

B

hI = (40)(25

μ

A) = 1 mA

181

20.

21. (a)

DF = I

T

Δ

= 0.95 0 0.95

25 ( 50) 75

−=

−− = 1.26%/°C

(b) Yes, curve flattens after 25°C.

22. (a) At 25°C, ICEO ≅ 2 nA

At 50°C, ICEO ≅ 30 nA

9

(30 2) 10 A 28 nA

(50 25) C 25 C

CEO

I

T

−

Δ−×

==

Δ−°°

= 1.12 nA/°C

ICEO (35°C) = ICEO(25°C) + (1.12 nA/°C)(35°C − 25°C)

= 2 nA + 11.2 nA

= 13.2 nA

From Fig. 17.55 ICEO (35°C) ≅ 4 nA

Derating factors, therefore, cannot be defined for large regions of non-linear curves.

Although the curve of Fig. 17.55 appears to be linear, the fact that the vertical axis is a

log scale reveals that ICEO and T( °C) have a non-linear relationship.

23. 20 mA

45 mA

oC

iF

II

==

≅ = 0.44

Yes, relatively efficient.

182

24. (a) PD = VCEIC = 200 mW

IC =

max

200 mW

30 V

D

CE

P

V= = 6.67 mA @ VCE = 30 V

V

CE = 200 mW

10 mA

D

C

P

I= = 20 V @ IC = 10 mA

I

C = 200 mW

25 V

D

CE

P

V= = 8.0 mA @ VCE = 25 V

Almost the entire area of Fig. 17.57 falls within the power limits.

(b)

β

dc = 4 mA

10 mA

C

F

I

I= = 0.4, Fig. 17.56 C

F

I

I ≅ 4 mA

10 mA = 0.4

The fact that the IF characteristics of Fig. 17.57 are fairly horizontal reveals that the level

of IC is somewhat unaffected by the level of VCE except for very low or high values.

Therefore, a plot of IC vs. IF as shown in Fig. 17.56 can be provided without any

reference to the value of VCE. As noted above, the results are essentially the same.

25. (a)

IC ≥ 3 mA

(b) At

IC = 6 mA; RL = 1 kΩ, t = 8.6

μ

s

RL = 100 Ω; t = 2

μ

s

1 kΩ:100 Ω = 10:1

8.6

μ

s:2

μ

s = 4.3:1

ΔR:Δt ≅ 2.3:1

26.

η

= 2

22

33

34

B

BB

R

RR

=

+ = 0.75, VG =

η

VBB = 0.75(20 V) = 15 V

27. VP = 8.7 V, IP = 100

μ

A ZP = 8.7 V

100 A

P

V

I

μ

= = 87 kΩ (≅ open)

VV = 1 V, IV = 5.5 mA ZV = 1 V

5.5 mA

V

I= = 181.8 Ω (relatively low)

87 kΩ: 181.8 Ω = 478.55:1 ≅ 500:1

183

28. Eq. 17.23:

T = RC loge log ()

BB BB

e

BB P BB BB D

VV

RC

VV V VV

η

⎛⎞ ⎛ ⎞

=

⎜⎟ ⎜ ⎟

−−+

⎝⎠ ⎝ ⎠

Assuming

η

VBB  VD, T = RC loge (1 )

BB

V

η

⎛⎞

⎜⎟

−

⎝⎠

= RC loge(1/1−

η

) = RC loge

1

12

1

1B

BB

R

RR

⎛⎞

⎜⎟

−

⎜⎟

+

⎝⎠

= RC loge12 1

22

log 1

BB B

e

BB

RR R

RC

RR

⎛⎞ ⎛⎞

+=+

⎜⎟ ⎜⎟

⎝⎠ ⎝⎠

Eq. 17.24

29. (a) Minimum VBB:

Rmax =

B

BP

P

VV

I

− ≥ 20 kΩ

()

BB BB D

P

VVV

I

η

−+

= 20 kΩ

VBB −

η

VBB − VD = IP 20 kΩ

VBB(1 −

η

) = IP 20 kΩ + VD

VBB = 20 k

1

PD

IV

η

Ω

+

−

= (100 A)(20 k ) 0.7 V

10.67

μ

Ω

+

−

= 8.18 V

10 V OK

(b) R < 12 V 1 V

5.5 mA

BB V

V

VV

I

−−

= = 2 kΩ

R < 2 kΩ

(c) T ≅ RC loge1

2

1B

B

R

⎛⎞

+

⎜⎟

⎝⎠

2 × 10−3 = R(1 × 10−6)loge

10 k

15 k

Ω

⎛⎞

+

⎜⎟

Ω

⎝⎠

log

e3 = 1.0986

R =

3

6

210

(1 10 )(1.0986)

−

×

R = 1.82 kΩ

Solutions for Laboratory Manual

to accompany

Electronic Devices and Circuit Theory

Tenth Edition

Prepared by

Franz J. Monssen

185

186

EXPERIMENT 1: OSCILLOSCOPE AND FUNCTION GENERATOR OPERATIONS

Part 1: The Oscilloscope

a. it focuses the beam on the screen

b. adjusts the brightness of the beam on the screen

c. allows the moving of trace in either screen direction

d. selects volts/screen division on y-axis

e. selects unit of time/screen division on x-axis

g. allows for ac or dc coupling of signal to scope and at

GND position; establishes ground reference on screen

h. locates the trace if it is off screen

i. provide for the adjustment of scope from external

reference source

k. determines mode of triggering of the sweep voltage

m. the input impedance of many scopes consists of the parallel combination of a 1 Meg

resistance and a 30pf capacitor

n. measuring device which reduces loading of scope on a circuit and effectively

increases input impedance of scope by a factor of 10.

Part 2: The Function Generator

d. T = l/f = 1/1000 Hz = l ms

e. (calculated): 1 ms*[l cm/.2 ms] = 5cm

(measured): 5 cm = same

f. (calculated): l ms*[cm/.5ms] =2 cm

(measured): 2 cm = same

g. (calculated): 1 ms*[cm/1ms] = l cm

(measured): l cm = same

h. .2 ms/cm takes 5 boxes to display total wave

.5 ms/cm takes 2 boxes to display total wave

1 ms/cm takes 1 box to display total wave

i. 1. adjust timebase to obtain one cycle of the wave

2. count the number of cm's occupied by the wave

3. note the timebase setting

4. multiply timebase setting by number of cm's occupied

by wave. This is equal to the period of the wave.

5. obtain its reciprocal; that's the frequency.

187

j. (calculated): 2cm * [2V/cm] = 4Vp-p

k. 8 * [.5V/cm] = 4Vp-p

1. the signal occupied full screen; the

peak amplitude did not change with a

change in the setting of the vertical sensitivity

m. no: there is no voltmeter built into function

generator

Part 3: Exercises

a. chosen sensitivities: Vert. Sens. = l V/cm

Hor. Sens. = 50

μ

s/cm

T(calculated): 4cm*[50

μ

s/cm)= 200

μ

s

Fig 1.1

b. chosen sensitivities: Vert. Sens. = .l V/cm

Hor. Sens. = 1 ms/cm

T(calculated):5 cm*[l ms/cm] = 5 ms

Fig 1.2

188

c. chosen sensitivities: Vert. Sens. = l V/cm

Hor. Sens. = l

μ

s/cm

T(calculated):10 cm*[1

μ

s/cm]=10

μ

s

Fig 1.3

Part 4: Effect of DC Levels

a. V(rms)(calculated) = 4V * 1/2 * .707 = 1.41 Volts

b. V(rms)(measured) = 1.35 Volts

c. [(1.41 − 1.35)/1.41) * 100 = 4.74%

d. no trace on screen

e. signal is restored, adjust zero level

f. no shift observed; the shift is proportional to dc

value of waveform

g. (measured) dc level: 1.45 Volts

h. Fig 1.5

i. Switch AC-GND-DC switch, make copy of waveform above.

The vertical shift of the waveform was equal to the battery voltage.

189

The shape of the sinusoidal waveform was not affected by changing the positions of

the AC-GND-DC coupling switch.

j. The signal shifted downward by an amount equal to

the voltage of the battery.

Fig 1.6

Part 5: Problems

1. b. f = 2000/(2*3.14) = 318Hz

c. T = l/f =1/318 = 3.14ms

d. by inspection: V(peak) = 20V

e. V(peak-peak) = 2*Vpeak = 40V

f. V(rms) =.707 * 20 = 14.1V

g. by inspection: Vdc = 0V

2.

a. f = 2 * 3.14 * 4000/(2 * 3.14*) = 4 KHz

c. T = l/f =1/4 Khz = 250

μ

s

d. by inspection:Vpeak)= 8 mV

e. V(peak-peak) = 2 * V(peak) = 16 mV

f. V(rms) = .707 * 8 mV = 5.66 mV

g. by inspection: Vdc = 0V

3. V(t) = 1.7 sin (2.51 Kt) volts

Part 6: Computer Exercise

PSpice Simulation 1-1

See Probe Plot page 191.

190

191

EXPERIMENT 2: DIODE CHARACTERISTICS

Part 1: Diode Test

diode testing scale

Table 2.1

Test

Forward

Reverse

Si (mV)

535

OL

Ge (mV)

252

OL

Both diodes are in good working order.

Part 2. Forward-bias Diode characteristics

b. Table 2.3

VR(V)

VD(mV)

ID (mA)

.1

453

.1

.2

481

.2

.3

498

.3

.4

512

.4

.5

528

.5

.6

532

.6

.7

539

.7

.8

546

.8

VR(V) .9 1 2 3 4 5 6 7 8 9 10

VD (mV) 551 559 580 610 620 630 640 650 650 660 660

ID(mA) .9 1 2 3 4 5 6 7 8 9 10

d. Table 2.4

VR(V)

VD(mV)

ID(mA)

.1

156

.1

.2

187

.2

.3

206

.3

.4

217

.4

.5

229

.5

.6

239

.6

.7

247

.7

.8

254

.8

VR(V) .9 1 2 3 4 5 6 7 8 9 10

VD (mV) 260 266 300 330 340 360 370 380 390 400 400

ID(mA) .9 1 2 3 4 5 6 7 8 9 10

e. Fig 2.5

192

f. Their shapes are similar, but for a given ID, the potential VD is greater for the silicon diode

compared to the germanium diode. Also, the Si has a higher firing potential than the

germanium diode.

Part 3: Reverse Bias

b. Rm = 9.9 Mohms

V

R(measured) = 9.1 mV

IS(calculated) = 8.21 nA

c. VR(measured) = 5.07 mV

I

S(calculated) = 4.58

μ

A

d. The IS level of the germanium diode is approximately 500 times as large

as that of the silicon diode.

e. RDC (Si) = 2.44*109 ohms

R

DC(Ge) = 3.28 M*106 ohms

These values are effective open-circuits when compared to resistors in the kilohm range.

Part 4: DC Resistance

a. Table 2.5

ID (mA) VD (mV) RDC (ohms)

.2 350 1750

1.0 559 559

5.0 630 126

10.0 660 66

b. Table 2.6

ID (mA) VD (mV) RDC (ohms)

.2 80 400

1.0 180 180

5.0 340 68

10.0 400 40

Part 5: AC Resistance

a. (calculated)rac = 3.4 ohms

b. (calculated)rac = 2.9 ohms

c. (calculated)rac = 27.0 ohms

d. (calculated)rac = 26.0 ohms

Part 6: Firing Potential

VT(silicon) = 540 mV

VT(germanium) = 260 mV

193

Part 7: Temperature Effects

c. For an increase in temperature, the forward diode current will increase while the voltage

VD across the diode will decline. Since RD = VD/ID, therefore, the resistance of a diode

declines with increasing temperature.

d. As the temperature across a diode increases, so does the current. Therefore, relative to the

diode current, the diode has a positive temperature coefficient.

Part 9: Computer Exercises

PSpice Simulation 2-1

1. See Probe plot page 195.

2. RD 600mV = 658 Ω

R

D 700 mV = 105 Ω

4. RD 600 mV = 257 Ω

5. See Probe Plot V(D1) versus I(D1)

7. Silicon

8. See Probe plot page 196.

9. See Probe plot page 196.

10. See Probe plot page 196.

194

195

196

EXPERIMENT 3: SERIES AND PARALLEL DIODE CONFIGURATIONS

Part 1: Threshold Voltage VT

Fig 3.2

Firing voltage: Silicon: 595 mV Germanium: 310 mV

Part 2: Series Configuration

b. VD = .59 V

V

O (calculated) = 5 − .595 = 4.41 V

ID = 4.41/2.2 K = 2 mA

c. VD(measured) = .59 V

V

O(measured) = 4.4 V

ID(from measured) = 2 mA

e. VD = 595 mV

V

O(calculated) = (5 − .595) 1 K/(1 K + 2.2 K) = 1.33 V

ID = 1.36 mA

f. VD = .57 V

V

O = 1.36 V

ID(from measured) = 1.36 V/1 K = 1.36 mA

g. VD(measured) = 5 V h. VD(measured) = 5 V

VO(measured) = 0 V VO(measured) = 0 V

ID(measured) = 0 A ID(measured) = 0 A

j. V1(calculated) = .905 V

VO(calculated) = 4.1 V

ID(calculated) = 1.86 mA

Part 7: Computer Exercise

PSpice Simulation 3-2

1. 638.0 mV

197

EXPERIMENT 4: HALF-WAVE AND FULL-WAVE RECTIFICATION

Part 1: Threshold Voltage

VT = .64 V

Part 2: Half-wave Rectification

b. Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

c. Fig 4.4

d. Both waveforms are in essential agreement.

e. Vdc = (4 − .64)/3.14 = 1.07 V

f. Vdc(measured) = .979 V

% difference = (1.07 − .979)/1.07*100 = 8.5%

g. For an ac voltage with a dc value, shifting the coupling switch from its DC to AC

position will make the waveform shift down in proportion to the dc value of the

waveform.

h. Fig 4.6

198

i. Vdc(calculated) = −1.07 V

V

dc(measured) = −.970 V

Part 3: Half-Wave Rectification (continued)

b.

Fig 4.8

c.

Fig 4.9

The results are in reasonable agreement.

d. The significant difference is in the respective reversal of the two voltage waveforms.

While in the former case the voltage peaked to a positive 3.4 volts, in the latter case, the

voltage peaked negatively to the same voltage.

e. VDC = (.318)*3.4 = 1.08 Volts

f. Difference = [1.08 − .979]/1.08*100 = 9.35%

199

Part 4: Half-Wave Rectification (continued)

b.

Fig 4.11

c. Fig 4.12

There was a computed 2.1% difference between the two waveforms.

d.

Fig 4.13

200

We observe a reversal of the polarities of the two waveforms caused by the reversal of

the diode in the circuit.

Part 5: Full-Wave Rectification (Bridge Configuration)

a. V(secondary)rms = 14 V

This value differs by 1.4 V rms from the rated voltage of the secondary of the

transformer.

b. V(peak) = 1.41*14 = 20 V

c.

Fig 4.15

Vertical sensitivity: 5 V/cm

Horizontal sensitivity: 2 ms/cm

d.

Fig 4.16

201

Again, the difference between expected and actual was very slight.

e. Vdc(calculated) = (.6326)*(20) = 12.7 V

V

dc(measured) = 11.36 V

% Difference = −10.6%

g. Vertical sensitivity = 5 V/cm

Horizontal sensitivity = 2 ms/cm

Fig 4.17

i. Vdc(calculated) = (.636)*(12) = 7.63 V

j V

dc(measured) = 7.05 V

% Difference = −7.6%

k. The effect was a reduction in the dc level of the output voltage.

Part 6: Full-Wave Center-tapped Configuration

a. Vrms(measured) = 6.93 V

V

rms(measured) = 6.97 V

As is shown from the data, the difference for both halves of the center-tapped windings

from the rated voltage is .6 volts.

b. Vertical sensitivity = 5 V/cm

Horizontal sensitivity = 2 ms/cm

202

c.

Fig 4.21

d. Vdc(calculated) = 3.5 V

V

dc(measured) = 3.04 V

Part 7: Computer Exercise

PSpice Simulation 4-2

1. Vp = 8.47 V; relative phase shift is equal to 180°

2. PIV = 2 Vp

3. 180° out of phase

4. See Probe plot page 204.

Its amplitude is 7.89 V

5. Yes

6. Reasonable agreement.

203

204

EXPERIMENT 5: CLIPPING CIRCUITS

Part 1: Threshold Voltage

VT(Si) = .618 V

VT(Ge) = .299 V

Part 2 Parallel Clippers

b. VO(calculated) = 4 V

c. VO(calculated) = −1.5 − .618 = −2.2 V

d.

Fig 5.2

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

e.

Fig 5.3

No measured differences appeared between expected and observed waveforms.

f. VO(calculated) = 4 V

g. V

O(calculated) = .62 V

205

Part 3: Parallel Clippers (continued)

b. VO(calculated) = .61 V

c. VO(calculated) = .34 V

d.

Fig 5.7

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

e.

Fig 5.8

The waveforms agree.

Part 4: Parallel Clippers (Sinusoidal Input)

b. VO(calculated) = 4 V when Vi = 4 V

V

O(calculated) = −2 V when Vi = −4 V

VO(calculated) = 0 V when Vi = 0 V

206

Fig 5.9

c. Waveforms agree within 6.5%.

Part 5: Series Clippers

b. VO(calculated) = 2.5 V when Vi = 4 V

c. VO(calculated) = 0 V when Vi = −4 V

d.

Fig 5.12

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

e. agree within 5.1%

f. VO(calculated) = 5.5 V when Vi = 4 V

g. VO(calculated) = 0 V when Vi = −4 V

207

h.

Fig 5.14

Vertical sensitivity = 2 V/cm

Horizontal sensitivity = .2 ms/cm

i. no major differences

Part 6: Series Clippers (Sinusoidal Input)

b. VO(calculated) = 2 V when Vi = 4 V

VO(calculated) = 0 V when Vi = −4 V

VO(calculated) = 0 V when Vi = 0 V

Fig 5.16

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

208

Part 7: Computer Exercises

PSpice Simulation 5-2

1. See Probe plot page 210.

2. VOUT = 4 V

3. No

4. VOUT = −2.067 V

5. Yes, VOUT(ideal) = −1.5 V

6. Reasonable agreement

7. No significant discrepancies

8. See Probe plot page 211.

PSpice Simulation 5-3

1. See Probe plot page 212.

2. In close agreement

3. No

4. For V1 = 4 V; Vout = V1 − VD1 − 1.5 V = 4 V − .6 − 1.5 V = 1.9 V

For V1 = −4 V; I(D1) = 0 A, ∴ Vout = 0 V

5. See Probe plot page 213.

6. See Probe plot page 213.

7. See Probe plot page 213.

8. See Probe plot page 213.

9. Forward bias voltage of about 600 mV when “ON”.

Reverse diode voltage of diode is −4 V − 1.5 V = −5.5 V

209

210

211

212

213

EXPERIMENT 6: CLAMPING CIRCUITS

Part 1: Threshold Voltage

VT = .62 V

Part 2: Clampers (R, C, Diode Combination)

b. VC(calculated) = 4 − 0.62 = 3.38 V

V

O(calculated) = 0.62 V

c. VO(calculated) = −4 − 3.38 V = −7.38 V

d. Fig 6.2

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

214

e. Fig 6.3

f. VC(calculated) = −3.38 V

V

O(calculated) = −0.62 V

g. VO(calculated) = 7.38 V

h. Fig 6.4

i. Fig 6.5

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

215

Part 3: Clampers with a DC battery

b. VC(calculated) = 1.88 V

V

O(calculated) = 0.62 V + 1.5 V = 2.12 V

c. VO(calculated) = −1.88 V − 4 V = −5.88 V

d. Fig 6.7

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

e. Fig 6.8

f. VC(calculated) = 4.88 V

V

O(calculated) = 1.5 V − 0.62 V = 0.88 V

g. VO(calculated) = 4 V + 4.88 V = 8.88 V

216

h. Fig 6.9

Vertical sensitivity = 2 V/cm

Horizontal sensitivity = .2 ms/cm

Part 4: Clampers (Sinusoidal Input)

b. VO(calculated) = 0 V when Vi = 2 V

VO(calculated) = −2 V when Vi = −3.6 V

VO(calculated) = −1.6 V when Vi = 0 V

Fig 6.11

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

217

Part 5: Clampers (Effect of R)

a. Tau(calculated) = R*C = 103 ms

b. T(calculated) = 1/f = 1 ms

T/2(calculated) = 1 ms/2 = .5 ms

c. 5Tau(calculated) = 5*103 ms = 515 ms

d. otherwise the capacitor voltage will not remain constant

e. 5Tau(calculated) = 5 ms

f. 5 ms/.5 ms = 10

g. Fig 6.13

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

i. 5Tau = .5 ms

j. .5 ms/.5ms = 1

k. Fig 6.14

Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

m. 5Tau = 2.5 T or Tau = 1/2 T

218

Part 6: Computer Exercise

PSpice Simulation 6-2

1. See Probe Plot page 220.

2. They are the same.

3. VO(calculated) is close to V(2) of Probe plot.

4. See Probe plot page 221.

5. V(1, 2) remains at 2 V during the cycle of V(1)

6. It rises exponentially toward its final value of 2 V.

7. See Probe plot page 222.

219

220

221

222

EXPERIMENT 7: LIGHT-EMITTING AND ZENER DIODES

Part 1: LED Characteristics

b. VD(measured) = 1.6 V

VR(measured) = 49.1 mV

ID(calculated) = 49.1 mV/101.4 ohms = 484

μ

A

c. VD(measured) = 1.9 V

VR(measured) = 1.55 V

ID(calculated) = 1.55 V/101.4 ohms = 15.3 mA

d.

E(v) 0 1 2 3 4 5 6

VD (V) 0 1 1.71 1.84 1.93 2.01 2.08

VR (V) 0 0 .34 1.2 2.2 3.1 3.9

ID = VR/R (mA) 0 0 3.3 11.8 21.4 30.6 38.5

e. Fig 7.2

h. The reversed biased Si diode prevents any current from flowing through the circuit,

hence, the LED will not light.

k. VR(V) = 3.48 V, therefore ID(mA) = 1.6 mA and LED is in the “good brightness” region.

Part 2: Zener Diode Characteristics

b. and c. Table 7.2

E (V) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

VZ (V) 0 1 2 3 4 5 6 7 8 9 10 10 10.1 10.2 10.3 10.4

VR (V) 0 0 0 0 0 0 0 0 0 .1 .97 1.9 2.8 3.7 4.6 4.6

IZ (mA) 0 0 0 0 0 0 0 0 0 .99 9.6 18.7 27.6 36.5 45.4 45.4

223

d. Fig 7.5

e. VZ(V) (approximated) = (10.4 + 9)/2 = 9.7 V

f. rav(ohms) = (10.4 − 9)/(.045 − .0099) = 39.9 ohms

g. RZ(ohms) = 39.9 ohms

VZ(V) = 9.7 V

Part 3: Zener Diode Regulation

a. R (meas) = 979 ohms

RL (meas) = 986 ohms

VZ (V) = 10.2 V

b. VL (V) = 986*15/(979 + 986) = 7.53 V

VR (V) = 979*15/(979 + 986) = 7.47 V

IR (mA) = 7.47/979 = 7.64 mA

IL (mA) = 7.53/986 = 7.63 mA

IZ (mA) = IR − IL = 10

μ

A

c. VL (measured) = 7.5 V

V

R (measured) 7.49 V

IR (calculated) = 7.65 mA

IL (calculated) = 7.60 mA

IZ (calculated) = 50

μ

A

d. VL (calculated) = 11.5 V

V

R (calculated) 3.54 V

IR (calculated) = 3.62 mA

IL (calculated) = 3.48 mA

IZ (calculated) = .14 mA

224

e. VL (measured) = 9.82 V

V

R (measured) = 3.54 V

IR (calculated) = 3.54 mA

IL (calculated) = 2.98 mA

IZ (calculated) = .56 mA

The difference is expressed as a percent with calculated value as the standard of

reference.

percent change of: VL = −14.6%

VR = 0.%

IR = −2.21%

IL = −14.4%

IZ = 30.0%

f. Rmin/(Rmin + 979)*15 = 9.82 V

R

L(calculated) = 1.86 Kohms

g. Since 2.2 Kohms > Rmin = 1.86 Kohms, therefore, diode is in the “on” state.

Part 4: LED-Zener diode combination

b. VD = 1.86 V

I

D = 15.8 mA

VZ = 10.07 V

Vab(calculated) = 11.9 V

c. VL (calculated) = 11.9 V

IL (calculated) = 5.41 mA

e. E (calculated) = VR + VL = 6.93 + 11.9 = 18.9 V

f. E (measured) = 19.1 V

The two values are in agreement within 1.06% using E (calculated) as reference.

225

Part 5: Computer Exercise

PSpice Simulation 7-1

1. – 8. See Circuit diagram

9. Yes

226

EXPERIMENT 8: BIPOLAR JUNCTION TRANSISTOR (BJT) CHARACTERISTICS

Part 2: The Collector Characteristics

d., f., g., h. Table 8.3

i.

Fig 8.3

227

Part 3: Variation of Alpha and Beta

b. The variations for Alpha and Beta for the tested transistor are not really significant,

resulting in an almost ideal current source which is independent of the voltage VCE.

c. The highest Beta’s are found for relatively large values of IC and VCE. This is a generally

well known factor.

d. Beta did increase with increasing levels of IC.

e. Beta did increase with increasing levels of VCE.

Part 5: Exercises

1. Beta(average) = 141

The arithmetic average occurred in the center of Fig 8.3.

2. VBE(average) = .678 V

Given that .7 V differs by only 3.14% from .678, and given that resistive circuit component

can vary by as much as 20%, the assumption of a constant .7 V is entirely reasonable.

3. The Beta of the transistor is increasing. Table 8.3 does substantiate that conclusion.

Beta would be a constant anywhere along that line.

Part 6: Computer Exercise

PSpice Simulation 8-1

1. See Circuit diagram.

2. Experimental PSpice

α

.99 .99

β

150 208

228

EXPERIMENT 9: FIXED- AND VOLTAGE-DIVIDER BIAS OF BJTs

Part 1: Determining

β

b. VBE(measured) = .67 V

V

RC(measured) = 4.9 V

c. IB = (VBCC − VBE)/RBB = (20 − .67)/1.108 M = 17.4

μ

A

IC = VRC/RC = 4.9/2.73 K = 1.79 mA

d. Beta IC/IB = 1.79 mA/17.4

μ

A = 105 B

Part 2: Fixed-bias configuration

a. IB(calculated) = 17

μ

A B

I

C(calculated) = 1.79 mA

b. VB(calculated) = VBCC − IB * RBB = .67 V

V

C(calculated) = VCC − IC * RC = 13.4 V

VE(calculated) = 0 V(emitter is at ground)

VCE(calculated) = VC − VE = 13.4 V

c. VB(measured) = .67 V B

V

C(measured) = 13.4 V

VE(measured) = 0 V

VCE(measured) = 13.34 V

The difference between measured and calculated values in every case is less

than 10%. It’s almost too good to be true.

d. VBE(measured) = .68 V

V

RC(measured) = 16.7 V

IB(from measured) = 17.4

μ

A B

IC(from measured) = 6.12 mA

Beta(calculated) = 352

Table 9.1

Transistor Type VCE (V) IC (mA) IB (

μ

A)

β

2N3904 13.34 1.79 17.4 105

2N4401 3.2 6.12 17.4 352

e. Table 9.2

% Δ

β

% Δ IC% Δ VCE % Δ IBB

242 242

−76.0 0

229

Part 3: Voltage-divider configuration

b. Table 9.3

2N3904 VB (V) BVE (V) VC (V) VCE (V)

(calculated) 3.52 2.82 12.47 9.7

(measured) 3.3 2.6 12.9 10.1

2N3904 IE (mA) IC (mA) IB

μ

A) B

(calculated) 4.07 4.05 30

(measured) 3.76 3.87 36.5

c. The agreement between measured and calculated values fall entirely within reasonable

limits.

d. and e.

Table 9.4

Transistor Type VCE (V) IC (mA) IB (

μ

A) Beta

2N3904 10.1 3.87 36.5 103

2N4401 9.6 4.03 17.2 234

f. Table 9.5

% Δ

β

% Δ IC% Δ VCE % Δ IBB

56 41 4.9 53

Part 4: Computer Exercises

PSpice Simulation 9-1

1.-3. See circuit diagram.

230

4. See circuit diagram.

5. 8.24%

6. %ΔIB = 0.05% B

%ΔIC = 8.2%

%ΔIE = 8.15%

7. %ΔVCE = −6.57%

8. S(

β

) = .995

PSpice Simulation 9-2

1.-3. See circuit diagram.

231

4. See circuit diagram.

5. %Δ

β

= 8.24%

6. %ΔIB = −6.47% B

%ΔIC = 1.13%

%ΔIE = 1.10%

7. %ΔVCE = −0.94%

8. S(

β

) = 1.13%

8.24% = 0.13

9. Circuit with Q2N2222.

10. Same as #9.

11. Same as #9.

Part 5: Problems and Exercises

1. a. IC(sat, fixed bias) = 20/2.73 K = 7.33 mA

b.

IC(sat,volt-divider bias) = 20/(1.86 K + 692) = 7.83 mA

c. The saturation currents are not sensitive to the Beta’s in either bias configuration.

2. In the case of the 2N4401 transistor, which had a higher Beta than the 2N3904 transistor, the

Q point of the former shifted higher up the loadline toward saturation. (See data in Table 9.4).

3. a.

Table 9.6

Fixed bias % Δ IC% Δ VCE % Δ IBB

% Δ

β

% Δ

β

% Δ

β

1 .314 0

Volt-divider .732 .087 .94

The ideal circuit has Beta independence when the ratio of %Δ IC/%Δ

β

is equal to 0.

Thus, the smaller the ratio, the more Beta independent is the circuit. Using this as a

criterion of stability, it becomes apparent that the voltage divider bias circuit is the more

stable of the two.

232

4. a.

IC =

β

(VCC − .67)/RB mA B

b.

IC = [R2/(R1 + R2)*VCC − .7]/[(R1 || R2)/

β

+ RE] mA

c. In equation 4a, the Beta factor cannot be eliminated by a judicious choice of circuit

components. In 4b however, if the quantity R1 || R2/

β

is made much smaller than RE, then

IC is no longer dependent upon Beta. In particular:

IC = [R2/(R1 + R2)*VCC − .7]/RE mA

In that case, we have achieved Beta independent biasing.

233

EXPERIMENT 10: EMITTER AND COLLECTOR FEEDBACK BIAS OF BJTs

Part 1: Determination of

β

b.

VB (measured) = 5.04 V B

V

RC (measured) = 4.04 V

c. IB (from measured) = (20 − 5.41)/1.1 M = 13.6

μ

A B

I

C (from measured) = 4.04/2.2 K = 1.84 mA

d.

β

= 1.84 mA/13.6

μ

A = 135

Part 2: Emitter-bias configuration

a. Using KVL:

−20 + IC * (1.01 M/

β

) + .67 V + IC * (2.23 K) = 0 V

therefore: IC = (20 − .67)/9.1 K = 2.1 mA

IB = 2.1 mA/135 = 15

μ

A B

b. and c. Table 10.1

Calculated Values

Transistor type VB (V) VC (V) VE (V) VBE (V) VCE (V)

2N3904 5.4 15.3 4.7 .70 10.6

2N4401 8.2 12.6 7.4 .8 5.2

Transistor type IB (

μ

A) BIC (mA)

2N3904 15.0 2.1

2N4401 11.7 3.3

Measured Values

Transistor type VB (V) VC (V) VE (V) VBE (V) VCE (V)

2N3904 4.75 15.9 4.2 .66 11.8

2N4401 8.0 12.5 7.6 .62 4.8

Transistor type IB (

μ

A) BIC (mA) Beta

2N3904 14.7 2.2 150

2N4401 11.9 3.4 286

d. See Table 10.1.

e. See Table 10.1.

f. In every case, the difference between calculated and measured values were less than 10%

apart.

234

g. Table 10.3

% Δ

β

% Δ IC%Δ VCE % Δ IBB

90.7 54.5

−58.5 −19

Part 3: Collector Feedback Configuration (RE = 0 ohms)

b. Using KVL:

−20 + IC(3.2 K) + IC(395 K)/150 + .7 V = 0 V

from which:

IB = 21

μ

A and IBC = 3.4 mA

Table 10.4

Calculated Values

Transistor type VB (V) VC (V) VCE (V) IB (

μ

A) BIC (mA)

2N3904 .62 9.1 9.1 21.2 3.4

2N4401 .55 6.2 6.2 14.4 4.3

Table 10.5

Measured Values

Transistor type VB (V) VC (V) VCE (V) IB (

μ

A) BIC (mA)

2N3904 .68 9.6 9.6 22.4 3.6

2N4401 .63 5.8 5.8 15.1 4.4

Table 10.6

% Δ

β

% Δ IC% Δ VCE % Δ IBB

83 22.8

−39.9 −33

Part 4: Collector Feedback Configuration (with RE)

a. For 2N3904:

−20 + IC(3.2 K) + IC(395 K/150) + IC(2.2 K) = 0 V

from which:

IB = 15

μ

A and IBC = 2.4 mA

for 2N4401:

−20 + IC(3.2 K) + IC(395 K/286) + IC(2.2 K) = 0 V

from which:

IB = 9.7

μ

A and IBC = 2.8 mA

b. See Table 10.7.

c. See Table 10.8.

d. See Table 10.7.

e. See Table 10.8.

f. Table 10.7

Calculated Values

Transistor VB (V) BVC (V) VE (V) VCE (V) IC (mA) IE (mA) IB (

μ

A) B

2N3904 6.2 12.1 5.4 6.7 2.45 2.5 15

2N4401 6.9 10.8 6.3 4.5 2.8 2.9 9.7

235

Table 10.8

Measured Values

Transistor VB (V) BVC (V) VE (V) VCE (V) IC (mA) IE (mA) IB (

μ

A) B

2N3904 5.9 12.6 5.2 7.4 2.3 2.4 19

2N4401 7.0 10.8 6.5 4.3 2.8 2.9 9.2

Table 10.9

% Δ

β

% Δ IC% Δ VCE % Δ IBB

83.2 23.8

−41.2 −50.3

Part 5: Computer Exercises

PSpice Simulation 10-1

1−6. See Circuit diagram.

7. See Circuit diagram.

236

8. %Δ

β

= 8.87%

9. %ΔIB = −2.18% B

%

ΔIC = 6.50%

%

ΔIE = 6.42%

10. %ΔVCE = −7.43%

11. S(

β

) = .73

12. P(Q2N3904) = 46.41 mW

P(Q2N2222) = 49.40 mW

Yes

13. Yes, see circuit diagram above.

14. Yes, see circuit diagram above.

PSpice Simulation 10-2

1−6. See Circuit diagram.

7. See Circuit diagram.

237

8. %Δ

β

= 8.64%

9. %ΔIB = −5.43% B

%

ΔIC = 2.75%

%

ΔIE = 2.69%

10. %ΔVCE = −4.96%

11. S(

β

) = .32

12-14. See circuit diagrams above.

Part 6: Problems and Exercises

1. a. IC(sat) 20/(2.2 K + 2.2 K) = 4.55 mA

b. IC(sat) = 20/3 K = 6.67 mA

c. IC(sat) = 20/5.2 K = 3.85 mA

d. Beta does not enter into the calculations.

2. The Q point shifts toward saturation along the loadline.

3. a. Table 10.10

Emitter bias % Δ IC % Δ VCE % Δ IBB

% Δ

β

% Δ

β

% Δ

β

b. The smaller that ratio, the better is the Beta stability of a particular circuit. Looking at the

results, which were computed from measured data, it appears that the collector feedback

circuit with RE = 0 ohms is the most stable. This is counter to expectations.

Theoretically, the most stable of the two collector feedback circuits should be the one

with a finite RE. Since the stability figures of both of those circuits are so small, the

apparent greater stability of the collector feedback circuit without RE is probably the

result of measurement variability.

4. Using KVL:

−VCC + IC/

β

*RB + VBBE + IC*RE = 0 V

from this:

IC = (VCC − VBE)/(RB/

β

+ RBE) mA

This division results in:

IC =

β

(VCC − VBE)/(RB +

β

*RBE) mA

If

β

*RE >>RB then IBC = (VCC − VBE)/RE mA

5. Using KVL:

−VCC + IC*RC + IC/

β

*RB + VBBE = 0 V

from this:

IC = (VCC − VBE)/(RC + RB/

β

) B

if

RC>>RB/

β

then IBC = (VCC − VBE)/RC mA

6. Using KVL:

−VCC + IC*RC + IC/

β

*RB + VBBE + IC*RE = 0 V

from this:

IC = (VCC − VBE)/(RC + RE + RB/

β

) mA B

if (RC + RE)>>RB/

β

then IBC = (VCC − VBE)/(RC + RE) mA

238

EXPERIMENT 11: DESIGN OF BJT BIAS CIRCUITS

Part 1: Collector-Feedback Configuration

a.

RC = (15 − 7.5)V/5 mA = 1.5 Kohms

R

C(commercial) = 1.5 ohms

d.

VRC(measured) = 5.14 V

V

CEQ(measured) = 7.7 V

ICQ(from measured) = 3.4 mA

β

(calculated) = 104

e. The most critical values for proper operation of this design is the voltage VCEQ measured

at 7.7 V. It being within 2.7% of the design makes this a workable design.

f.

RB/(

β

*RBC) = 214 K/(104*1.5 K) = 1.37

g. RF1 + RF2 = 189 K

R

B(commercial) + 214 K B

h. No, the value of RB is fixed both by VBCC and VBE, neither of which changed.

i.

VRC(measured) = 5.64 V

V

CEQ(measured) = 9.27 V

ICQ(from measured) = 3.76 mA

β

(calculated) = 3.73 mA/([9.27 − .7)/214 K] = 108

j. The measured voltage VCE is somewhat high due to the measured current IC being below

its design value. In general, the lowest IC which will yield proper VCE is preferable since it

keeps power losses down. For the given specifications, this design, for small signal

operation, will probably work since most likely no clipping will be experienced.

k.

RB/(

β

*RBC)(calculated) = 214 K/(108*1.5 K) = 1.4

R

B/(

β

*RBC)(calculated) = 1.34 (see above)

The parameters of the circuit do not change significantly with a change of transistor.

Thus, the design is relatively stable in regard to any Beta variation.

l.

S(

β

) = 3.76 mA − 3.4 mA)/3.4 mA = .8

Part 2: Emitter-bias Configuration

a.

RC(calculated) = [(VCC − (7.5 + 1.5)]V/5 mA = 1.2 K

R

C(commercial) = 1.2 K

b.

RE(calculated) = 1.5 V/5 mA = 300 ohms

R

E(commercial) = 285 ohms

d.

RB(measured) = RB1 + R2 = 392 K

R

B(commercial) = 394 K B

239

e.

VRC(measured) = 6.04 V

V

CE(measured) = 7.55 V

IC(from measured) = 4.7 mA

β

(calculated) = 144

f. All measured values are well within a 10% tolerance of the design parameters. This is

acceptable.

g.

RB/(

β

*RBE) = 9.6

h.

RB(calculated) = 950 K B

RB(commercial) = 1 M B

i. Yes, it changed from 214 K to a value of 950 K. The increase in Beta was compensated

for by the increase in RB in such a way that IBCQ, and consequently VC, VCEQ and VE

remained constant. Hence, so did RC and RE.

j.

VRC(measured) = 5.2 V

V

CEQ(measured) = 8.6 V

ICQ(calculated) = 4.2 mA

β

(calculated) = 372

k. The important voltage VCEQ was measured at 8.61 V while it was specified at 7.5 V. Thus,

it was larger by about 12%. This is probably the largest deviation to be tolerated. If the

design is used for small signal amplification, it is probably OK; however, should the

design be used for Class A, large signal operation, undesirable cut-off clipping may

result.

l. The magnitude of the Beta of a transistor is a property of the device, not of the circuit. All

the circuit design does is to minimize the effect of a changing Beta in a circuit. That the

Betas differed in this case came as no surprise.

m. (calculated)RB/(

β

*RBE)(2N3904) = 10.4

(calculated)RB/(

β

*RBE)(2N4401) = 9.6

n. S(

β

) = .66

Part 3: Voltage-divider Configuration

a.

RC(calculated) = [25 − (1.5 + 7.5)]V/5 mA = 1.2 K

R

C(commercial) = 1.25 K

b.

RE = 1.5 V/5 mA = 300 ohms

RE(commercial) = 285 ohms

d.

R2(calculated) = 2.94 K

R

2(commercial) = 3.2 K

R1(calculated) = 17.1 K

R1(commercial) = 18.2 VK

240

e.

VRC(measured) = 6.47 V

V

CEQ(measured) = 7.09 V

ICQ(calculated) = 5.2 mA

β

(calculated) = 144

The difference between the calculated and the measured values of ICQ and VCEQ are

insignificant for the operation of this circuit.

f.

R1 || R2/(

β

*RE) = .066

g.

VRC(measured) = 6.98 V

V

CEQ(measured) = 6.47 V

ICQ(calculated) = 5.6 mA

β

(calculated) = 368

h. The measured values of the previous part show that the circuit design is relatively

independent of Beta.

i. The Betas are about the same.

j.

R1 || R2/(

β

*RE)(2N4401) = .026

R

1 || R2(

β

*RE)(2N3904) = .066

k.

S(

β

) = .051

Part 4: Problems and Exercises

1. Table 11.1

Configuration ICQ (mA) VCEQ (V)

Collector-feedback 3.4 7.7

Emitter-bias 4.7 7.5

Voltage-divider 5.2 7.1

The critical parameter for this design is the voltage VCEQ. Given that its variation for the

various designs is less than 10%, the results are satisfying.

2. Table 11.2

Configuration Stability factors

RB/(

β

RBC) S(

β

)

Collector-feedback 1.4 .8

Emitter-bias 0.6 .66

Voltage-divider .06 .051

The data in adjacent columns is consistent.

The voltage-divider bias configuration was the least sensitive to variations in Beta. This is

expected since the resistor R2, while decreasing the current gain of the circuit, stabilized the

circuit in regard to any current changes.

241

3. Using KVL:

−VCC + IC*RC + IC/

β

*RB + VBBE = 0 V

from which:

IC = (VCC − VBE)/(RC + RB)/

β

) mA B

for stable operation, make: RC>>RB/

β

B

4. Using KVL:

−VCC + IC/

β

*RB + IBC * RE + VBE = 0 V

from which:

IC = (VCC − VBE)/(RE + RB)/

β

) mA B

for stable operation, make: RE>>RB/

β

B

5. Using KVL:

−VBB + IC/Beta*R1 || R2 + VBE + IC*RE = 0 V

where:

VBB = R1/(R1 + R2)*VCC

from which:

IC = (VBB − VBE)/(RE + R1 || R2/

β

) mA

for stable operation: make RE>>R1 || R2/

β

Part 5: Computer Exercises

PSpice simulation 11-1

1. See Circuit diagram.

2.

β

= 170.5

3. S = 1.095

4. Yes

5. See Circuit diagram above.

242

PSpice simulation 11-2

1. See Circuit diagram.

2.

β

= 170.96

3. S = 0.08

4. No

5. See Circuit diagram.

6. Yes

7. Not needed

8. See circuit diagram above.

243

EXPERIMENT 12: JFET CHARACTERISTICS

Part 1: Measurement of the Saturation Current IDSS and Pinch-off Voltage VP

c. VR(measured) = .754 V

d. IDSS = 7.44 mA

e.

Vp(measured) = −2.53 V

f.

1. IDSS = 8.3 mA, Vp = −3.1 V

2.

IDSS = 9.1 mA Vp = −3.9 V

It is extremely unlikely that two 2N4416 ever have the same saturation current and pinch-off

voltage.

Fig 12.1

Part 2: Drain-Source Characteristics

a. through d.

Table 12.1

VGS(V) 0 −1.0 −2.0

VDS(V) ID(mA) ID(mA) ID(mA)

0.0 0.0 0.0 0.0

1.0 4.63 2.1 .25

2.0 5.61 2.6 .28

3.0 7.32 3.06 .34

4.0 7.40 3.1 .36

5.0 7.43 3.2 .39

6.0 7.5 3.16 .42

7.0 7.5 3.31 .43

8.0 7.5 3.33 .44

9.0 7.3 3.36 .46

10.0 7.3 3.36 .50

11.0 7.1 3.36 .50

12.0 6.81 3.36 .51

13.0 6.76 3.36 .52

14.0 6.71 3.36 .53

244

Fig 12.3

IDSS (Fig 12.3) = 7.5 mA

IDSS (Part 1) = 7.44 mA

V

P (Fig 12.3) = −3 V

VP(Part 1) = −2.53 V

Part 3: Transfer Characteristics

a. b. Table 12.2

V

DS(V) 3V 6V 9V 12V

VGS(V) ID(mA) ID(mA) ID(mA) ID(mA)

0 7.32 7.5 7.4 6.81

−1 3.06 3.26 3.36 3.36

−2 .34 .42 .46 .51

d. Given that the various variables in the above Table vary by less than 10%, it is reasonable

that the curves can be replaced on an approximate basis by a single curve defined by

Shockley’s equation if the average values of both IDSS and VGS(off) are used.

Part 5: Problems and Exercises

1. Shockley’s equation involves four parameters. Given two of them, such as ID and VGS, an

infinite number of curves can be drawn through their interception all of which can satisfy

Shockley’s equation for particular IDSS and VP.

2.

VG = VP*[1 − (ID/IDSS)1/2] V

3. For: IDSS = 10 mA; VP = −5 V; and ID = 4 mA

VGS(calculated) = (−5)*[(.4)1/2] = −3.16 V

a.

gmO(calculated) = 2*(7.44 mA)/2.53 = 5.88 ms

b. The slope of the Shockley curve is maximum at VGS = 0 V.

c.

gm(calculated) = gmO(1 − VP/VP) = 0 S when VGS = VP.

The slope of the transfer curve at VGS = VP = 0 S

245

d. VGS/VP = 1/4 VGS/VP = 1/2 VGS/VP = 3/4

gm 4.41 mS 2.94 mS 1.47 mS

Note: gm0 = 5.88 mS

e. The slope is a constant value.

f. It is proportional to the derivative of Shockley’s equation.

Part 6: Computer Exercises

PSpice Simulation 12-1

1-4. See Circuit diagram.

PSpice Simulation 12-2

Part A

4. See Probe Plot page 247.

5. IDSS = 16 mA

6. VP = −1.5 V

Part B

4. See Probe plot page 248.

5. IDSS = 18.2 mA

VP = −1.4 V

246

247

248

EXPERIMENT 13: JFET BIAS CIRCUITS

Part 1: Fixed-Bias Network

b. IDSS = 12 mA

c. VP(measured) = −6 V

e. ID = 12−3(1 − 1/6)1/2 = 8.33 mA

f. VRD(measured) = 8 V

IDQ(measured) = 8.2 mA

RD(measured) = 976 ohms

Part 2: Self-Bias Network

b. IDQ = 2.64 mA

V

GSQ = −3.3 V

c. VGS(calculated) = −3.3 V

V

D(calculated) = 12.4 V

VS(calculated) = 3.1 V

VDS(calculated) = 9.3 V

VG(calculated) = 0 V

d. VGS(measured) = −3.4 V

V

D(measured) = 12.2 V

VS(measured) = 2.1 V

VDS(measured) = 9.1 V

VG(measured) = 0 V

The percent differences are determined with the calculated values as the reference.

VGS(calculated %) = 3.1%

VD(calculated %) = −1.6%

VS(calculated %) = −.64%

VDS(calculated %) = −2.3%

VG(calculated %) = 0%

Part 3: Voltage Divider-Bias Network

b. For voltage divider-bias-line see Fig. 13.2

c. IDQ(calculated) = 4.8 mA

V

GS(calculated) = −2.4 V

d. VD(calculated) = 10.3 V

V

S(calculated) = 5.2 V

VDS(calculated) = 5.1 V

e. VGSQ(measured) = −2.3 V

V

D(measured) = 10.4 V

VS(measured) = 5.3 V

VDS(measured) = 5.1 V

249

f. The percent differences are determined with calculated values as the reference.

VGS(calculated %) = −4.2 %

VD(calculated %) = .97%

VS(calculated %) = 1.9%

VDS(calculated %) = 1.2%

g. IDQ(measured) = 4.8 mA

I

DQ(calculated %) = .4%

Part 4: Computer Exercises

PSpice Simulation 13-1

1. 928

μ

A

2. 12.96 V

3. −1.114 V

4. 13.92 mW

5. See Circuit diagram.

6. Negligible due to back bias of gate-source function

7. 12.03 mW

8. No

250

PSpice Simulation 13-2

1-8. See circuit diagram.

9. No

251

EXPERIMENT 14: DESIGN OF JFET BIAS CIRCUITS

Part 1: Determining IDSS and VP

b. IDSS(measured) = 10.8 mA

c. VP(measured) = −6 V

Part 2: Self-bias Circuit Design

a. IDQ(calculated) = 5.4 mA

V

DSQ(calculated) = 15 V

VDD(calculated) = 30 V

d. RS(calculated) = 1/gm = 333 ohms

R

S(commercial) = 330 ohms

e. VRD = VDD − VDSQ − VRS = 30 − 15 − 1.8 = 13.2 V

R

D = 2.4 K

f. VDSQ(measured) = 14.7 V

I

DQ(measured) = 5.6 mA

VDSQ(calculated) = 15 V

IDQ(calculated) = 5.4 mA

g. Agreements between calculated and measured values are within 10% of each other and

thus are within acceptable limits.

h. VDSQ(measured) = 13.7 V

I

DQ(measured) = 6 mA

IDSS(borrowed JFET) = 9.8 mA

VP(borrowed JFET) = −5.1 V

Even though in our case, the variations between JFETs was relatively small, such may

not be the case in general. Thus, the values of the biasing resistors for the same bias

design but employing different JFETs may differ considerably.

Best is not to use the arithmetic but the geometric average for the range of IDSS and VP.

Thus in our case, the geometric averages would be:

IDSS(geometric average) =

[IDSS(min) * IDSS(max)]1/2 = [5 mA*15mA)]1/2 = 8.66 mA

VP (geometric average) = [1*6]1/2 = 2.45 V

Statistically, these values are most likely the ones encountered.

Part 3: Voltage-divider Circuit Design

a. VGS(calculated) = −2.6 V

b. RS = (VGG − VGS)/IDQ = (6 − 2.6)V/4 mA = 850 ohms

R

S(commercial value) = 820 ohms

V

G(calculated) = VGS + ID*RS = 2.6 + 4 mA*820 = 5.85 V

252

c. VRD(calculated) = VDD − VDSQ − VRS = 20 − 8 − 3.28 = 8.72 V

where VRS = IDQ*RS = 4 mA*820 = 3.28 V

RD = [VDD −(VDSQ + VRS)]/ID = [20 − (8 + 3.28)] = 2.18 Kohms

R

D(commercial value) = 2 Kohms

d. R2 = 10*RS = 10*820 = 8.2 Kohms

R

2(commercial value) = 10 Kohms

Solving equation 14.3 for R1 we obtain:

R1 = R2*(VDD − VG)/VG = 10 K*(20 − 5.85)/5.85 = 24.2 Kohms

R

1(commercial value) = 22 Kohms

e. VDSQ(measured) = 7.9 V

IDQ(measured) = 4.2 mA

VDSQ(specified) = 8 V

I

DQ(specified) = 4 mA

f. %IDQ(calculated) = 5%

%VDSQ(calculated) = −1.25%

Such relative small percent deviations are almost too good to be true.

The voltage divider bias line is parallel to the self-bias line. To shift the Q point in either

direction, it is easiest to adjust the bias voltage VG to bring the circuit parameters within

an acceptable range of the circuit design.

g. In the present case, the percent differences for IDQ and VDSQ were well within the 10%

tolerance allowed. If not, the easiest adjustment would be the moving of the voltage-

divider bias line parallel to itself by means of raising or lowering of VG. This could best

be accomplished by a change of the voltage divider R2/(R1 + R2)*VDD. Its value

determines the voltage VG which in turn determines the Q point for the design.

h. VDSQ(measured) = 13.7 V

I

DQ(measured) = 3.68 mA

I

DSS(borrowed JFET) = 9.8 mA

V

P(borrowed JFET) = −5.1 V

Part 4: Problems and Exercises

1. RD(commercial value) = 2.7 Kohms

R

S(commercial value) = 180 ohms

2. RD(commercial value) = 2.4 Kohms

R

S(commercial value) = 680 ohms

R

1(commercial value) = 6.8 Kohms

R

2(commercial value) = 33 Kohms

253

3. In the design, use the geometric mean of both the given ranges on IDSS and VP for a

given type JFET.

Part 5: Computer Exercises

PSpice Simulation 14-1

1-6. See circuit diagram.

PSpice Simulation 14-2

1. See circuit diagram.

254

2. See circuit diagram.

3. See above circuit diagrams.

4. %VDS = 7.37%

5. Yes

255

EXPERIMENT 15: COMPOUND CONFIGURATIONS

Part 1: Determining the BJT(

β

) amd JFET (IDSS and VP) Parameters

a. RB(measured) = 982 Kohms B

R

C(measured) = 2.6 Kohms

IB = (VBCC − VBE)/RBB = (20 − .7)/982 K = 19.7

μ

A

VRC(measured) = 6.45 V

I

C = VRC/RC = 6.45/2.6 K = 2.48 mA

β

(calculated) = 1.48 mA/19.7

μ

A = 126

Part 2: Capacitive-Coupled Multistage System with Voltage-Divider Bias

b. VB1 = 4.7 K/(4.7 K + 15 K)* 20 = 4.8 V

V

E1 = 4.8 − .7 = 4.1 V

I

E1 = IC1 = VE1/RE1 = 4.1/1 K = 4.1 mA

V

C1 = VCC − IC1 * RC1 = 20 − 4.1 mA*2.7 K = 9.2 V

V

B2 = 2.4 K/(2.4 K + 15 K)*20 = 2.8 V

V

E2 = 2.8 − .7 = 2.1 V

I

E2 = IC2 = VE2 /RE2 = 2.1/470 = 4.6 mA

V

C2 = VCC − IC2 * RC2 = 20 − 4.6 mA * 1.2 K = 14.5 V

Table 15.1

VB1(V) VC1(V) VB2(V) VC2(V)

Calculated Values 4.8 9.2 2.8 14.5

Measured values 4.7 9.1 2.7 14.2

% Difference −1.1 −1.1 −1.8 −2.1

As can be seen from the above data, the differences between the calculated and measured

values were much less than 10%.

f. We note that the voltages VC1 and VB2 are not the same as they would be if the voltage

across capacitor CC was 0 Volts, indicating a short circuit across that capacitor.

Part 3: DC-Coupled Multistage Systems

Use the same equations to determine the circuit parameters as in Part 2 except that VB2=VC1.

b. Table 15.2

VB1(V) VC1(V) VB2(V) VC2(V)

Calculated Values 4.8 9.2 9.2 13.0

Measured values 4.7 9.1 9.1 12.9

% Difference −1.7 −1.0 −1.0 −.8

Again, the percent differences between calculated and measured values are less than 10%

in every instance.

256

f. The dc collector voltage of stage 1 determines the dc base voltage of stage 2. Note that no

biasing resistors are needed for stage 2.

Part 4: A BJT-JFET Compound Configuration

b. VB = 4.7 K/(4.7 k + 15 k) * 30 = 7.2 V B

V

E = VB − .7 V = 6.5 V B

I

E = ID = 6.5 V/1.2 K = 5.4 mA

V

D = VDD*RD = 30 − 5.4 mA*985 = 24.7 V

For the JFET used: IDSS = 10.1 mA

VP = −3.2 V

determine VGS:

ID/IDSS = [1 − VGS/VP]1/2 mA = 5.4 mA/10.1 mA = [1 − VGS/3.2]1/2 mA

therefore:

[5.4 mA/10.1 mA]2 = [1 − VGS/3.2]

.286 = [1 − VGS/3.2]

from which: VGS = (1 − .286)*3.2 = −2.28 V

remember: VGS is a negative number:

VC = VB − VBGS = 7.2 − (−2.28) = 9.5 V

Table 15.3

VB(V) BVD(V) VC(V)

Calculated Values 7.2 23.6 9.5

Measured values 7.1 24.4 8.7

% Difference −.56 3.4 −8.4

d. See Table 15.3.

e. Differences were less than 10%.

f. VGS(calculated from measured values) = VB − VBC = 7.1 − 8.7 = −1.6 V

VGS(measured) = −1.7 V

g. VRD = VDD − VD = 30 − 24.7 = 5.3 V

I

D = 5.3 V/985 = 5.4 mA

I

D(measured) = 6.4 mA

The percent difference between the measured and the calculated values of ID was 18.5%,

with the calculated value of ID used as the standard of reference.

VE(calculated) = 7.2 − .7 = 6.5 V

I

C(calculated) = 6.5 V/1.26 K = 5.2 mA

I

C(measured) = 5.06 mA

257

The percent difference between the measured and the calculated values of IC was −2.7%,

with the calculated value of ID used as the standard of reference.

Part 5: Problems and Exercises

1. a. There will be a change of VB and VBC for the two stages if the two voltage divider

configurations are interchanged.

b. The voltage divider configuration should make the circuit Beta independent, if it is

well designed. Thus, there should not be much of a change in the voltage and current

levels if the transistors are interchanged.

2. Again, depending on how good the design of the voltage divider bias circuit is, the

changes in the circuit voltages and currents should be kept to a minimum.

Part 6: Computer Exercises

PSpice Simulation 15-1

1-11. See below.

258

PSpice Simulation 15-2

1-11. See below.

259

260

EXPERIMENT 16: MEASUREMENT TECHNIQUES

Part 1: AC and DC Voltage Amplitude Measurements

DC MEASUREMENT

e. VO(calculated) = 2K/(2K + 3.9 K)*12 = 3.86 V

f. VO(measured) = 3.78 V

%Diff. (calculated) = −2%

g. VO(measured shift) = 3.8 V

The shift was down from the center of the screen.

There is almost complete agreement between the two sets of measurements.

The measurement taken with the DMM is the more accurate of the two, especially for a

DMM, since it reads to 1/100 of a volt.

AC MEASUREMENTS

h. Vi(rms)(calculated) = 8/2*.707 = 2.82 V

i. V

O(rms)(calculated) = [(2 K || 3.9 K + j0)*(2.82 + j0)]/(2.41 K − j1.59 K) = 1.34∠33.4 V

j. VO(measured) = 1.31 V

% diff. (calculated) = −1.51%

k. VO(p − p)(measured) = 3.72 V

l. If we convert the measured rms value of VO to peak value, we obtain 3.78 volts.

Comparing that to the measured peak value of VO which was 3.72 V, we can be satisfied

with the results.

Part 2: Measurements of the Periods and Fundamental Frequencies of Periodic Waveforms

b. Horizontal sensitivity = 100

μ

s/div

c. number of divisions = 5.6

d. Period(T) = 100

μ

s/div*5.6 div = 560

μ

s

e. Frequency(f) = 1/T = 1/560

μ

s = 1800 Hz

f. f(dial setting) = 1750 Hz

g. The dial setting on the signal generator at best can only give an approximate setting of the

frequency.

h. f(counter) = 1810 Hz

i. Indeed it is, the difference between calculated and measured values is only 10 Hz using

the counter, whereas the difference between signal generator setting and calculated values

was 50 Hz. That measurement which is closest to that of the counter is the better

measurement. In our case, the scope measures better than the signal generator.

261

Part 3: Phase-Shift Measurements

b. Vi(rms)(calculated) = 6/2*.707 = 2.12 V

c. VO(rms) = (0 − j1.59 K)*(2.12 + j0)/(1k − j1.59 K) = 1.81∠−31.6 V

VO(p−p)(rms) = 1.81*1.41*2 = 5.1 V

f. A(number of divisions) = .8

g. B

(number of divisions) = 10

h. angle

θ

(calculated) = −31.6 degrees

j. The network is a lag network, i.e., the output voltage VO lags the input voltage by the

angle theta, in our case it lags it by −31.6 degrees.

k. VR(rms)(calculated) = 1.1 V

V

R(p−p)(calculated) = 3.1 V

angle theta = 58.4 degrees

The output voltage VO leads the input voltage by 58.4 degrees. Note that an angle of 58.4

degrees is the complement of an angle of 31.6 degrees.

l. VR(p−p)(measured) = 3 V

angle

θ

= 58 degrees

It’s a lead angle.

Part 4: Loading Effects

c. VO(p−p)(calculated) = 1 K/(1 K + 1 K)*8 = 4 V

d. VO(p−p)(measured) = 3.98 V

f. VO(p−p)(calculated) = 1 M/(1 M + 1 M)*8 = 4 V

VO(p−p)(measured) = 2.7 V

g. The real part of the input impedance of the scope is now in parallel with the R2 resistor

and since for many scopes, that real part is about 1 Mohm, therefore, Rscope || R2 = 500

kohms.

Thus, VO is considerably reduced.

h. R(prime) = 1 M/[Vi/VO − 1] = 1 M/[8/2.7 − 1] = 588 kohms

R

(scope) = −R(prime)*R2/[R(prime) − R2] = 1.43 Megohms

Most general purpose oscilloscopes have an input impedance consisting of a real part of

1 Megohms in parallel with a 30 pf capacitor. The result obtained for the real part of that

impedance is reasonably close to that.

i. VO(p−p)(calculated) = 1 K/(1 K + 1 M)*8 = 8 mV

j. VO(p−p)(measured) = 7.9 mV

k. The results agree within 1.25 percent.

262

Part 5: Problems and Exercises

1. No. for the frequency of operation, the capacitor represents an impedance of 1.59k∠−90

ohms. Therefore, in relationship to the existing resistors in the circuit, it cannot be

neglected without making a serious error.

2. It depends upon the waveform. In case of sinusoidal voltages, the advantage is probably

with the DMM. For more complex waveforms, the nod goes to the oscilloscope.

3. For measuring sinusoidal waves, the DMM gives a direct reading of the rms value of the

measured waveform. However, for non-sinusoidal waves, a true rms DMM must be

employed. The oscilloscope only gives peak-peak values, which, if one wants to obtain

the power in an ac circuit, must be converted to rms.

4. T = 5 div*.1 ms/div = .5 ms

f = 1/T = 1/.5 ms = 2 KHz

5. angle theta = 1.5/8*360 = 67.5 degrees

VO/Vi = R′/(R′ + R1)

therefore: Vi/VO = (R′ + R1)/R′

solving for R′: R′(Vi/VO) = R′ + R1

R′(Vi/VO − 1) = R1

Hence: R′ = R1/(Vi/VO − 1) ohms

Part 6: Computer Exercises

PSpice Simulation 16-1

1. See Probe plot page 264.

2. See Probe plot page 264.

3. See Probe plot page 265.

4. See Probe plot page 265.

5. 33.74°

6. Vout

9. See Probe plot page 266.

10 See Probe plot page 266.

11. Vin(rms) = 2.84 V

Vout(rms) = 1.32 V

12. Yes

13. See Probe plot page 267.

14. See Probe plot page 267.

15. Vout = 32 K

(12 V) = (12 V)

2 3 (212 + 3.9 K)

R

RR+

= 4.067 V

16. Agree

263

264

265

266

267

PSpice Simulation 16-2

1. Using VOM, R2 = 100 kΩ

2. Using DMM, R2 = 1 kΩ

3. For R2 = 1 kΩ

4. Both circuits

5. No

268

EXPERIMENT 17: COMMON-EMITTER TRANSISTOR AMPLIFIERS

Part 1: Common-Emitter DC Bias

b. VBB = R2/(R1 + R2) * VCC = 10 K/(10 K + 33 K) * 10 = 2.33 V

V

E = VBB − .7 = 1.63 V

V

C = VCC − IC * RC = 10 − 1.63 mA * 3 K = 5.1 V

I

E = VE/RE = 1.63/1 K = 1.63 mA

r

e = 26 mV/IE = 26 mV/1.63 mA = 16 ohms

c. VB(measured) = 2.25 V B

VE(measured) = 1.57 V

V

C (measured) = 4.95 V

I

E = VE/RE = 1.57/978 = 1.6 mA

r

e = 26 mV/1.6 mA = 16.2 ohms

The two values for re obtained are within .2 ohms.

This represents a 1.25 percent difference.

Part 2: Common-Emitter AC Voltage Gain

a. AV(no load) = −RC/re = 3.2 K/16 = 198

b. Vsig = 8.3 mV(rms)

VO(no load) = 1.47 V (rms)

A

V(no load) = 177

The two values of AV agree within 10.6 percent of each other.

Part 3: AC Input Impedance, Zi

Zin = R1 || R2 || Beta * re = 10 K || 33 K || (150 * 16) = 1.8 Kohms

Vi (measured) = 12 mV (rms)

Vsig = 20 mV(rms)

Zin = [12 mV/(20 mV − 12 mV)] * 1 K = 1.5 Kohms

The two values of the input impedance were within 18.9% of each other. This relatively large

divergence is in part the result of using an assumed value of Beta for our transistor. For a 2N3904

transistor, the geometric average of Beta is closer to 126.

Part 4: Output Impedance

a. ZO(calculated) = RC = 3.2 Kohms

b. Vsig(rms) = 10 mV(rms)

VO(no load)(rms) = 1.8 V (rms)

V

O(loaded)(rms) = .913 V(rms)

R

L = 3.2 Kohms

Z

O = [(VO − VL)/VL] * RL = [(1.8 − .913)/.913] * 3.2 K = 3.1 K

The two values for ZO are within 3.15% of each other.

269

Part 6: Computer Analysis

PSpice Simulation 17-1

1. See Circuit diagram.

2. re = 6.93 Ω

3. See Probe plot page 271.

4. See Probe plot page 271.

5. 180°

6. As I(B) increases, so does I(C).

As I(C) increases, so does V(RC) and V(RE). Therefore V(C) decreases.

7. Zin (theoretical) = 937.3 Ω

8. See Probe plot page 272.

9. See Probe plot page 272.

10. Zin(PSpice) = 1.1323 k ≈ Zin(theoretical)

Determining output impedance

1. Zout ≈ RC = 3 k

2. See Probe plot page 273.

3. See Probe plot page 273.

4. Zout(PSpice) = 2.6392 k ≈ RC

270

271

272

273

EXPERIMENT 18: COMMON-BASE AND EMITTER-FOLLOWER

(COMMON-COLLECTOR TRANSISTOR AMPLIFIERS

Part 1: Common-Base DC Bias

a. VB(calculated) = 10 K/(10 K + 33 K) * 10 = 2.33 V B

V

E = VB − .7 V = 1.63 V B

I

E = IC = VE/RE = 1.63 V/1 K = 1.63 mA

V

C = 10 − IC * RC = 10 − (1.63 mA) * 3 K = 5.1 V

r

e = 26 mV/IE = 26 mV/1.63 mA = 16 ohms

b. VB(measured) = 2.26 V B

V

E(measured) = 1.57 V

V

C(measured) = 4.95 V

I

E(from measured values) = VE/RE = 1.57 V/978 = 1.6 mA

r

e(from measured values) = 26 mV/IE = 26 mV/1.6 = 16.3 ohms

In every case, the differences between the two sets of values are less than 10% apart.

Such divergence is not excessive given the variability of electronic components.

Part 2: Common-Base AC Voltage Gain

a. AV(calculated) = RC/re = 3.2 K/16.3 = 197

b. Vsig = 50 mV

VO = 2.43 V

A

V = 2.43/Vsig = 2.43/.05 = 122

The two gains differed by −38 percent with the calculated gain used as the standard of

comparison.

Part 3: CB Input Impedance, Zi

a. Zi = re = 16.3 ohms

b. Vsig = 50 mV

Vi = 9.9 mV

R

X = 100 ohms

Zi = [Vi/(Vsig − Vi)] * RX = [9.9 mV/(50 mV − 9.9 mV)] * 100 = 23.7 ohms

The two values of the input impedance differed by 45 percent with the theoretical value

of re (16.3 ohms) used as the standard of comparison.

Part 4: CB Output Impedance, ZO

a. ZO = RC = 3.2 K

b. V

sig = 20 mV

VO(measured, no load) = 2.43 V

V

L(measured, loaded) = 1.22 V

Z

O = [(VO − VL)/VL] * RL = [(2.43 − 1.22)/1.22] * 3 K = 3.18 Kohms

The agreement between the two values of the output impedance is within less than 1

percent.

274

Part 5: Emitter-Follower DC Bias

a. VB(calculated) = 2.33 V B

VE(calculated) = 1.63 V

IE(calculated) = 1.63 V

VC(calculated) = 10 V

re(calculated) = 26 mV/IE = 26 mV/1.63 mA = 16 ohms

b. VB(measured) = 2.26 V B

V

E(measured) = 1.78 V

V

C(measured) = 10.1 V

I

E = VE/RE = 1.78 V/1 K = 1.78 mA

r

e = 26 mV/1.78 mA = 14.3 ohms

Part 6: Emitter-Follower AC Voltage Gain

a. AV = RE/(RE + re) = 1 K/(1 K + 14.3) = .986

b. Vsig = 1 V

VO(measured) = .987 V

A

V = VO/Vsig = .987/1 = .987

The two values of gain are within .1 percent of each other.

Part 7: Emitter Follower (EF) Input Impedance, Zi

a. Zi = R1 || R2 || (Beta * (1 K + re) = 7.31 Kohms

b. Vsig = 2 V

RX = 10 Kohms

f = 1 KHz

V

i(measured) = .85 V

Z

i = [Vi/(Vsig − Vi)] * RX = [.85/(2 − .85)] * 10 K = 7.39 Kohms

The input impedance calculated from measured values is within 1.1 percent of the

theoretically calculated value of Zi.

Part 8: Emitter Follower (EF) Output Impedance, ZO

a. ZO = re = 16 ohms

b. VO(measured) = 19.8 mV

VL(measured) = 11.2 mV

Z

O = [(VO − VL)/VL]*RL = [(19.8 mV − 11.2 mV)/11.2 mV] * 100 = 76.8 ohms

In the theoretical formulation, ZO was equated with re, this is an approximation. A better

expression for the output impedance is: ZO = re + (RG || R1 || R2)/Beta. Thus it can be seen

that the given formulation was actually a minimum value of the output impedance.

275

Part 9: Computer Analysis

PSpice Simulation 18-1

Bias Point Analysis

1. See Circuit diagram.

2. See circuit diagram.

4. re = 16.71 Ω

5. Av = 179.53

6. Zin = re = 16.71 Ω

7. Zout = 3 kΩ

Transient Analysis

1. See Probe plot page 277.

2. 38°

4. Av = 141.59, see Probe plot page 278.

Input Impedance

1. Zin = 20.7 Ω, see Probe plot page 279.

Output Impedance

1. Zout = 2.87 kΩ, see Probe plot page 280.

276

277

278

279

280

PSpice Simulation 18-2

Bias Point Analysis

1. See Circuit diagram.

2. See circuit diagram.

4. re = 16.65 Ω

5. Av = 0.98

6. Zin = 7.31 kΩ

7. Zout r≅e = 16.65 kΩ

Transient Data

1. See Probe plot page 282.

2. 0.0°

4. 0.981

Input Impedance

1. 7.35 kΩ

Output Impedance

1. 58.63

Ω

281

282

EXPERIMENT 19: DESIGN OF COMMON-EMITTER AMPLIFIERS

Part 2. Computer Analysis

PSpice Simulation 19-1

1. See Circuit diagram.

3.

β

= 139.6

4. VCE = 4.78 V

5. Yes

Transient Analysis

1. AV = 147.9

2. Yes

4. Zin = 2.78 kΩ

5. Yes

8. Zout = 3.893 kΩ

9. Yes

Part 3: Build and Test CE Circuit

b. VB(measured) = 1.54 V B

V

E(measured) = .87 V

V

C(measured) = 7.15 V

I

C = IE = VE/RE = .87 V/979 = .89 mA

r

e = 26 mV/IE = 26 mV/.89 mA = 29.3 ohms

c. Vsig = 10 mV

V

L(measured) = .815 V

A

V = (RC || RL)/re = (3.2 K || 10.2 K)/29.3 = 80.7

d. Vsig = 20.5 mV

R

X = 3.17 Kohms

Vi(measured) = 8.8 mV

Z

i = (R1 || R2 || Beta*re) = (100.2 K || 21.6 K || 100 * 29.3 = 2.4 Kohms

e. VO(measured) = 1.08 V

ZO = (VO − VL)/VL*RL = (1.08 − .82)/.82 * 10.2 K = 3.25 Kohms

283

f. Design parameter Measured value

AV 100 min. 80.7

Zi (Kohms) 1 Kmin. 2.38 K

Z

O (Kohms) 10 Kmax. 3.35 K

V

Omax(p−p) 3 Vp−p min. 7.1 Vp−p

The design of the circuit was successful with all parameters, but the gain, meeting and

even exceeding the design specification. The gain is about 20 percent below the expected

value. To increase it, the supply voltage VCC could be increased. This would increase the

quiescent current, lower the dynamic resistance re and consequently increase the gain of

the amplifier.

284

EXPERIMENT 20: COMMON-SOURCE TRANSISTOR AMPLIFIERS

Part 1: Measurement of IDSS and VP

a. IDS = 9.1 mA

b. V

P = −2.9 V

Part 2: DC Bias of Common-Source Circuit

a. VGS = −1.33 V

I

D = 2.55 mA

V

D = VDD

−

ID * RD = 20 − 2.55 mA * 2.2 K = 13.8 V

c. VG(measured) = 0 V

V

S(measured) = 1.46 V

V

D(measured) = 13.8 V

V

GS(measured) = −1.37 V

I

D = VD/RS = 13.8/488 = 2.99 mA

The agreement between calculated and measured values was in most cases within 10

percent of each other, the exception being the 17.3 percent difference between the

calculated and measured value of ID.

Part 3: AC Voltage Gain of Common-Source Amplifier

a. AV = −gmRD

where

gm = IDSS/(2 * | VP |) * (1 − VGS/VP)2 = 2 * 9.1 mA/2.9 * (1 − 1.33/2.9) = 3.4 mS

therefore: AV = −3.4 mS * 2.2 K = 7.48

b. Vsig = 10 mV

f = 1 KHz

V

O(measured) = 758 mV

A

V = VO/Vsig = 758 mV/100 mV = 7.58

The difference between the theoretical gain and the gain calculated from measured values

was only 1.34 percent.

Part 4: Input and Output Impedance Measurements

a. Zi = RG

Zi(expected) = 1 Megohm

b. ZO = RD

ZO(expected) = 2.25 Kohms

c. Vi(measured) = 37.2 mV

Z

i(calculated) = Vi * RX(Vsig − Vi) = 592 Kohms

285

d. VO(measured) = 760 mV

R

L(measured) = 9.9 Kohms

V

L(measured) = 620 mV

Z

O = (VO − VL) * RL/VL = (760 mV − 620 mV) * 9.9 K/620 mV = 2.24 Kohms

The infinite input impedance of the JFET is predicated upon the assumption of the zero

reverse gate current. Such may not be entirely true. Hence, we observe a 41 percent

difference between the theoretical input impedance and the input impedance calculated from

measured values.

The two values of the output impedance are in far better agreement. They differ only by .44

percent.

Part 5: Computer Exercises

PSpice Simulation 20-1

1. See Circuit diagram.

3. gmo = 21.15 mS; gm = 7.48 mS

4. Av = −17.9

6. Av = −19.47

8. Zin = 954.64 kΩ

11. Zout = 2.34 kΩ

13. Av

14.

D

v

A

R

= 4 kΩ = 21.93

286

EXPERIMENT 21: MULTISTAGE AMPLIFERS: RC COUPLING

Part 1: Measurement of IDSS and VP

IDSS = 10.4 mA

V

P = −3.2 V

Part 2: DC Bias of Common-Source Circuit

a. VGS1(calculated) = −1.36 V

ID1(calculated) = 3.1 mA

V

D1(calculated) = VDD − ID1 * RD1 = 20 V − 3.1 mA * 2.2 K = 13.2 V

V

GS2(calculated) = −1.38 V

I

D2(calculated) = 3.54 mA

V

D2(calculated) = VDD − ID2 * RD2 = 20 V − 3.54 mA * 2.2 K = 12.2 V

c. VG1(measured) = 0 V

V

S1(measured) = 1.49

V

D1(measured) = 13.81 V

V

GS1(measured) = −1.04 V

I

D1 = VS1/RS1 = 1.49 V/496 = 3 mA

V

G2(measured) = 0 V

V

S2(measured) = 1.52 V

V

D2(measured) = 11.3 V

V

GS2(measured) = −.8 V

I

D2 = VS2/RS2 = 1.52 V/468 = 3.25 mA

The theoretical and the measured bias values were consistently in close agreement.

Part 3: AC Voltage Gain of Amplifier

a. For stage 2:

AV2 = −gm2(RD2 || RL) = (−3.64 mS)(2.2 K || 10 K) = 6.6

For stage 1:

AV1 = −gm1(RD1 || Zi2) = (−3.51 mS)(2.2 K || 1 M) = 7.72

note: Zi2 = RG2 = 1 Megohm

AV = AV1 * AV2 = 6.6 * 7.72 = 50.7

b. Vsig(measured) = 20 mV

V

L(measured) = 945 mV

A

V = VL/Vsig = 945 mV/20 mV = 47.3

V

O1(measured) = 145 mV

V

sig(measured) = 20 mV

A

V1 = VO1/Vsig = 145 mV/20 mV = 7.25

A

V2 = VL/VO1 = 945 mV/145 mV = 6.52

The voltage gains differed by less than 10 percent from each other.

287

Part 4: Input and Output Impedance Measurements

a. Zi = RG1 = 1 Megohm

b. ZO = RD2 = 2.2 Kohms

c. Vi1(measured) = 7.5 mV

Vsig = 20 mV

R

X = 1 Megohm

Z

i = Vi1 * RX/(Vsig − Vi1) = 7.5 mV * 1 M/(20 mV − 7.5 mV) = 600 Kohms

d. VL(measured) = 330 mV

V

O(measured) = 410 mV

Z

O = (VO − VL) * RL/VL = (410 mV − 330 mV) * 10 K/330 mV = 2.42 Kohms

Again, the input impedance calculated from measured values is about 40 percent below

that which we expected from the assumption that the JFET was ideal and had no reverse

gate current. This seems not to be the case in actuality. There is a reverse leakage current

at the gate which reduces the effective input impedance below that of RG by being in

parallel with it.

The output impedances again are in reasonable agreement, differing by no more than 9

percent from each other.

Part 5: Computer Exercise

Pspice Simulation 21-1

1. See circuit diagram.

3. gmo = 21.15 mS

gmJ1 = 7.48 mS

g

mJ2 = 7.48 mS

4. Av1 = 17.95 Av2 = 7.48

6. Av1 = 19.498

288

7. Av2 = 8.275

10. (Av1)(Av2) = 161.35

11. (Av1)(Av2) = 161.35

14. Yes

16. Interchange J1 with J2

17. Zin = 956.89 kΩ

20. Zout = 989.74 Ω

289

EXPERIMENT 22: CMOS CIRCUITS

Part 1: CMOS Inverter Circuit

Table 22.1 Table 22.2

IN OUT IN OUT

0V 5V 0V 5V

5V .3V 5 V .3 V

Part 2: CMOS Gate

Table 22.3

A B OUTPUT

0 V 0 V 5 V

0 V 5 V 0 V

5 V 0 V 0 V

5 V 5 V 0 V

Part 3: CMOS Input-Output Characteristics

a.

IN (V) 0.0 .2 .4 .6 .8 1.0 1.2 1.4 1.8

OUT (V) 5.0 5.0 5.0 5.0 4.9 4.8 4.8 4.7 4.4

IN (V) 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6

OUT (V) 3.9 3.4 1.6 1.1 .75 .6 .4 .3

IN (V) 3.8 4.0 4.2 4.4 4.6 4.8 5.0

OUT (V) .1 .1 .08 .02 .02 .005 0

Part 4: Computer Exercise

1. See Probe plot page 291.

2. No

VPlot data

1. See Probe plot page 292.

290

291

292

EXPERIMENT 23: DARLINGTON AND CASCODE AMPLIFIER CIRCUITS

Part 1: Darlington Emitter-Follower Circuit

a. VB(calculated) = 2.21 V B

VE(calculated) = 1.01 V

A

V = RE/(RE + re) = 47/(47 + 10) = .83

b. VB(measured) = 5.9 V B

V

E(measured) = 4.94 V

I

B(calculated) = 199

μ

A B

I

E(calculated) = 106 mA

β

(calculated) = 106 mA/199

μ

A = 535

c. Vi(measured) = 350 mV

V

O(measured) = 340 mV

A

V = VO/Vi = 340 mV/350 mV = .97

Part 2: Darlington Input and Output Impedance

a. Zi(calculated) = 20.6 K || (535 * 47) = 11.3 Kohms

Z

O = re + (RG || RB)/

β

= 9 ohms B

b. Vsig = 500 mV

Vi(measured) = 55.6 mV

Z

i = [Vi/(Vsig − Vi) * Rx = [55.6 mV/(500 mV − 55.6 mV)] * 100 K = 12.5 Kohms

c. VO(measured) = 492 mV

V

L(measured) = 476 mV

R

L = 100 ohms

ZO = [(VO − VL)/VL] * RL = [(492 mV − 476 mV)/476 mV] * 100 = 4.2 ohms

The two values of the input impedance differed by about 10.6 percent while the two

values of the output impedance differed by 53 percent. It is to be noted however that with

such small values the difference in just one ohm manifests itself as a large percent

change. Given the tolerances of electronic circuit due to their components and that of the

Darlington chip, the results are quite satisfactory.

Part 3: Cascode Circuit

a. VB1 (calculated) = 5.5 V

V

E1 (calculated) = 4.8 V

V

C1 (calculated) = 11 V

V

B2 (calculated) = 12 V

V

E2 (calculated) = 11.3 V

V

C2 (calculated) = 12.5 V

I

E1 = VE1/RE1 = 4.8 V/1 k = 4.8 mA

I

E2 = 11.3/1.8 K = 6.24 mA

r

e1 = 26 mV/IE1 = 26 mV/4.8 mA = 5.4 ohms

r

e2 = 26 mV/IE2 = 26 mV/6.24 mA = 4.2 ohms

b. VB1 (measured) = 4.69 V

V

E1 (measured) = 4.0 V

293

V

C1 (measured) = 10.7 V

V

B2 (measured) = 12.0 V

V

E2 (measured) = 10.5 V

V

C2 (measured) = 12.3 V

I

E1 (calculated) = VE1/RE1 = 4 V/1 K = 4 mA

I

E2 (calculated) = VE2/RE2 = 10.5/1.8 K = 5.2 mA

r

e1 = 26 mV/IE1 = 26 mV/4 mA = 6 ohms

r

e2 = 26 mV/IE2 = 26 mV/5.2 mV = 5 ohms

c. AV1 = −1 (as per equation 23.5)

A

V2 = RC/re2 = 1.8 K/5 = 360

d. Vsig = 10 mV

V

i (measured) = 8 mV

V

O2 (measured) = 7.91 mV

V

O1 (measured) = 948 mV

AV1 (calculated) = −VO1/Vi = 7.91/8 mV = −.98

A

V2 (calculated) = VO2/VO1 = 948 mV/7.91 mV = 120

A

V = VO2/Vi = −948 mV/8 mV = −119

The voltage gains for stage 1 were within 2 percent of each other, while the overall

theoretical gain of 180 differs from the calculated gain from measured values by 34 percent.

Part 4: Computer Exercises

PSpice Simulation 23-1

1. See circuit diagram.

2. re = 249 Ω

3. See Probe plot page 295.

4. See Probe plot page 295.

5. See Probe plot page 295.

6. AV = 0.787

9. Zin = 47.123 kΩ

Zout = 2.04 kΩ

294

295

PSpice Simulation 23-2

1. See circuit diagram.

2. = 5.63 Ω = 5.6 Ω

1Q

e

r2Q

e

r

5. See Probe plot page 297.

7. See Probe plot page 298.

8. For Q1; AV = 1.8 k

5.63

C

e

R

r=Ω = 319

For Q2; AV = 5.6

5.6

e

r

Ω

=Ω= 1

296

297

298

EXPERIMENT 24: CURRENT SOURCE AND CURRENT MIRROR CIRCUITS

Part 1: JFET Current Source

a. VDS (measured) = 9.64 V

b. IL = (VDD − VDS)/RL = (10 − 9.64)/51.2 = 7.03 mA

c. Table 24.1

RL (ohms) 20 51 82 100 150

V

DS (Volts) 9.88 9.64 9.44 9.34 8.85

I

L (mA) 6.1 7.03 6.83 6.60 7.57

Part 2: BJT Current Source

a. IL = 1.9 mA

b. VE (measured) = −.68 V

VC (measured) = .404 V

c. IE (calculated) = 2.13 mA

I

L (calculated) = 1.88 mA

d.

Table 24.2

RL (kohms) 3.6 4.3 5.1

V

E (Volts) −.68 −.67 −.68

V

C (Volts) 3.03 1.74 .404

I

E (mA) 2.14 2.17 2.13

I

L (mA) 1.94 1.92 1.88

Part 3: Current Mirror

a. IX = .9 mA

b. VB1 = .669 V

VC2 = 2.24 V

I

X = .89 mA

I

L = 1.0 mA

c. IX (calculated) = 1 mA

V

B1 (measured) = .669 V

V

C2 (measured) = 4.1 V

I

X = .9 mA

I

L = 1.5 mA

299

Part 4: Multiple Current Mirrors

a. IX (calculated) = 1 mA

b. VB1 (measured) = .672 V

VC2 (measured) = 1.67 V

V

C3 (measured) = 1.65 V

I

X = 1.01 mA

I

L1 = 1.58 mA

I

L2 = 1.78 mA

c. IX (calculated) = 1 mA

V

B1 (measured) = .672 V

V

C2 (measured) = 3.81 V

V

C3 (measured) = 2.87 V

I

X = 1.02 mA

I

L1 = 1.73 mA

I

L2 = 1.44 mA

Part 5: Computer Exercise

PSpice Simulation 24-1

1. See circuit diagram.

2. I(RX) = 933.6

μ

A

I(RL) = 1.020 mA

3. Yes

300

4. See Circuit diagram.

5. I(RX) = 933.6

μ

A

I(RL) = 991.3

μ

A

6. Yes

8. Yes

10. Yes

11. No

12. Yes

301

EXPERIMENT 25: FREQUENCY RESPONSE OF COMMON-EMITTER AMPLIFIERS

Résumé

fL,1 = 1/(2 * 3.24 * 1.39 K * 10

μ

f) = 11.5 Hz

f

L,2 = 1/(2 * 3.24 * 6.1 K * 1

μ

f) = 26 Hz

fL,E = 1/(2 * 3.14 * 2.2 K * 20

μ

f) = 3.62 Hz

f

H,i = 1/(2 * 3.14 * 1.68 K * 960

μ

f) = 98.7 KHz

f

H,O = 1/(2 * 3.14 * 1.43 K * 45 pf) = 2.43 MHz

Part 1: Low-Frequency Response Calculations

a. Cbe (specified) = 100 pf

C

bc (specified) = 10 pf

Cce (specified) = 15 pf

CW,i (approximated) = 20 pf

CW,o (approximated) = 30 pf

b.

β

(measured) = 126

c. VB (calculated) = 4.08 V B

VE (calculated) = 3.38 V

V

C (calculated) = 14 V

I

E (calculated) = 1.54 mA

r

e = 26 mV/IE = 26 mV/1.54 mA = 16.9 ohms

d. A

V (mid) = (RC || RL)/re = (3.9 K || 2.2 K)/16.9 = 83.2

e. fL,1 (calculated) = 11.5 Hz

fL,2 (calculated) = 26.2 Hz

f

L,E (calculated) = 3.62 Hz

Part 2: Low Frequency Response Measurements

b. Vsig (measured) = 30 mV

V

O (measured) = 2.1 V

A

V (mid) = 70 Table 25.1

f (Hz) 50 100 200 400 600 800 1 K 2 K 3 K 5 K 10 K

VO(p−p) .4 .5 .9 1.6 1.8 1.9 2.0 2.1 2.1 2.1 2.2

Table 25.2

f (Hz) 50 100 200 400 600 800 1 K 2 K 3 K 5 K 10 K

AV 13.2 16.7 30 53.3 60 63.3 66.7 70 70 70 73.3

Part 3: High Frequency Response Calculations

a. fH,I (calculated) = 98.7 KHz

f

H,O (calculated) = 2.47 MHz

302

b. Table 25.3

f (KHz) 10 50 100 300 500 600 700 900 1000 2000

VO(p−p) 2.2 2.2 2.1 1.9 1.6 1.5 1.4 1.3 1.3 .8

Table 25.4

f (KHz) 10 50 100 300 500 600 700 900 1000 2000

AV73 73 70 63 53 20 46 40 40 27

Part 4: Plotting Bode Plot and Frequency Response

Fig 25.2

from plot: f1 = 400 Hz

f2 = 500 Hz

Part 5: Computer Exercise

PSpice Simulation 25-1

1. See circuit diagram.

2. re = 17.2 Ω; AV mid = 81.3

4. See Probe plot 346.

5. Almost identical

6. See Probe plot page 346.

7. See Probe plot page 347.

8. 20 log(78.028) = 37.84 both gains agree

303

304

305

EXPERIMENT 26: CLASS A AND CLASS B POWER AMPLIFIERS

Part 1: Class A Amplifier: DC Bias

a. VB (calculated) = 1.53 V B

V

E (calculated) = .83 V

I

E (calculated) = IC = VE/RE = .83/20 = 41 mA

VC (calculated) = 5.1 V

b. VB (measured) = 1.59 V B

VE (measured) = .88 V

V

C (measured) = 5.3 V

I

E (calculated) = IC = VE/RE = .88/20 = 44 mA

Part 2: Class-A Amplifier: AC Operation

a. Pi (calculated) = 400 mW

V

O (calculated) = 5.3 Vp−p

PO (calculated) = 29.3 mW

% efficiency (calculated) = 7.3 percent

b. Vi (measured) = 65 mV

V

o (measured) = 5 Vp−p

c. Pi = 400 mW

P

O = 26 mW

% efficiency (calculated) = 6.5 percent

While the values for the power and the efficiency are fairly consistent between the

theoretical and those calculated from measured values, the low efficiency of the amplifier

is an undesirable feature. In general, Class A amplifiers operate close to a 25 percent

efficiency. This circuit would need to be redesigned to make it a practical circuit.

d. Vi (measured) = 32.5 mVp−p

VO (measured) = 3 Vp−p

e. Pi (calculated) = 400 mW

P

O (calculated) = 9.38 mW

% efficiency (calculated) = 2.3 percent

f. Pi = 400 mW

P

O = 93 mW

% efficiency = 2.3 percent

As stated previously, while the data is consistent, the values of the efficiency makes this

not a practical circuit.

306

Part 3: Class-B Amplifier Operation

a. for VO = 1 Vpeak

Pi (calculated) = 1.59 W

P

O (calculated) = 50 mW

% efficiency (calculated) = 3.1 percent

for VO = 2 Vpeak

Pi = 1.59 W

P

O = 200 mW

% efficiency (calculated) = 12.6 percent

b. Vi (measured) = 2.9 Vp−p

V

O (measured) = 2.7 Vp−p

P

i = 890 mW

P

O = 91 mW

% efficiency = 10.2%

c. Vi (measured) = 5 Vp−p

V

O (measured) = 4 Vp−p

I

dc (measured) = 159 mA

P

i = 1.27 W

P

O = 637 mW

% efficiency = 50.2%

Note that the efficiency of the Class B amplifier increases with increasing input signal

and consequent increasing output signal. Also observe that the two stages of the Class B

amplifier shown in Figure 26.2 are in the emitter follower configuration. Thus, the

voltage gain for each stage is near unity. This is what the data of the input and the output

voltages show. Note also, that as the output voltage approaches its maximum value that

the efficiency of the device approaches its theoretical efficiency of about 78 percent.

Part 4: Computer Exercises

PSpice Simulation 26-1

1. See Circuit diagram.

307

2. Yes.

3. VCE = 5.709 V − 0.719 V = 4.98 V



1(10 V)

2

4. 442.7 mW

5. Q1

7. See Probe plot page 309.

8. Yes

9. No

10. 180°

11. Po(AC) = 4.59 mW

12. %

η

= 1.04%

13. DC values remain the same

Po(AC) = 1.16 mW

%

η

= 0.26%

See Probe plot page 310.

308

309

310

PSpice Simulation 26-2

1. See circuit diagram.

Pi(DC) = 372.1 mW

2. Q1 and Q2

3. Yes.

4. V(E2) = 4.947 V



1(10 V)

2

5. V(BE)Q1 = 0.77 V

V(BE)Q2 = −0.81 V

6. Maintain proper bias across Q1 and Q2.

7. 0.7 V

9. See Probe plot page 312.

10. V(OUT)p−p = 4.077 V

11. %

η

= 55.8%

14. V(OUT)p−p = 2.187 V

%

η

= 16.1%

311

312

EXPERIMENT 27: DIFFERENTIAL AMPLIFIER CIRCUITS

Part 1: DC Bias of BJT Differential Amplifier

a. VB (calculated) = 0V B

V

E (calculated) = −.7 V

V

C (calculated) = 5.43 V

I

E (calculated) = 457

μ

A

r

e (calculated) = 57 ohms

b. Q1 Q2

VB (measured) −.10 V 0 V B

V

E (measured) −.65 V −.65 V

V

C (measured) 5.10 V 4.9 V

I

E (calculated) 490

μ

A 510

μ

A

r

e 53 ohms 51 ohms

Part 2: AC Operation of BJT Differential Amplifier

a. AV,d (calculated) = 179

A

V,c (calculated) = .5

b. V

O1 (measured) = 1.48 V

V

O2 (measured) = 1.43 V

V

O,d = (VO,1 + VO,2)/2 = (1.48 + 1.43)/2 = 1.46 V

A

V,d = VO,d/Vi = 72.8

c. VO,c (measured) = .55 V

A

V,c = VO,c/Vi = .55

Part 3: DC Bias of BJT Differential Amplifier with Current Source

a. For either Q1 or Q2:

VB (calculated) = 0 V B

V

E (calculated) = −.7 V

V

C (calculated) = 9 V

I

E (calculated) = .5 mA

r

e (calculated) = 52 ohms

For Q3:

VB (calculated) = −5 V B

V

E (calculated) = −5.7 V

V

C (calculated) = −.7 V

I

E (calculated) = 1 mA

r

e (calculated) = 26 ohms

313

b. For Q1, Q2, and Q3:

Q1 Q2 Q3

VB (measured) 47 mV 0 mV −4.69 V B

V

E (measured) −.64 V −.64 V −5.35 V

V

C (measured) 7.91 V 2.97 V −.64 V

I

E (calculated) 110

μ

A 612

μ

A 783

μ

A

r

e (calculated) 236 ohms 42.5 ohms 33.2 ohms

Part 4: AC Operation of Differential Amplifier with Transistor Current Source

a. AV,d = RC/(2 * re) = 10 K/(2 * 57.8) = 173

Part 5: JEFT Differential Amplifier

a. For Q1: IDSS = 7.9 mA

VP = −3.1 V

For Q2: IDSS = 8.1 mA

VP = −3.4 V

For Q3: IDSS = 11.2 mA

VP = −4.2 V

b. VD,1 (calculated) = 9.84 V

V

D,2 (calculated) = 9.84 V

V

S,1 (calculated0 = .845 V

c. VG,1 (measured) = 0 V

V

D,1 (measured) = 9.71 V

V

D,2 (measured) = 9.72 V

V

D,3 (measured) = .84 V

d. AV,d = .184

e. VO,1 (measured) = 50 mV

V

O,2 (measured) = 46 mV

A

V1,d = .5

A

V2,d = 4.6

314

Part 6: Computer Exercises

Pspice Simulations 27-1

1. See circuit diagram.

P(DC)VCC = 9.283 mW

P(DC)VEE = 9.356 mW

2. Practically yes

3. = 6 V

12

() ()

QQ

VCE VCE=

4. Yes.

5. I(Q1) = 464.2

μ

A

I(Q2) = 464.2

μ

A

6. Yes

9. See Probe plot page 316.

10. (VOUT1)p−p = (VOUT2)p−p = 3.23 V

phase shift = 180°

11. See Probe plot page 317.

AV = 114

14. See Probe plot page 318.

15. (VOUT1)p−p = (VOUT2)p−p = 0.98 V

phase shift = 0°

16. See Probe plot page 319.

AV = 0

315

316

317

318

319

Pspice Simulations 27-2

1. See circuit diagram.

P(DC)VCC = 9.853 mW

P(DC)VEE = 14.97 mW

2. = 5.074 V

1

()

Q

VC

= 5.074 V

2

()

Q

VC

Yes

3. I(Q1) = 492.6

μ

A

I(Q2) = 492.6

μ

A

I(Q3( = 993.0

μ

A

4. 12

() ()IQ IQ=

31

()2()2()IQ IQ IQ==

2

320

7. See Probe plot page 322.

8. Both voltages are 1.7602 Vp−p

phase shift = 180°

9. See Probe plot page 323.

AV = 125

12. See Probe plot page 324.

13. 1.6 mVp−p phase shift = 0°

14. See probe plot page 325.

AV = 0

321

322 322

323

324

325

EXPERIMENT 28: OP-AMP CHARACTERISTICS

Part 1: Determining the Slew Rate

f. 5 V p-p

g. 12 us

h. 0.41 V/us

Part 2: Determining the Common Mode Rejection Ratio

g. Vout(rms) = 0.263 V Vin(rms)= 8.7 V

h. A(cm) = V(out)/V(in) = 0.0302

i. A(dif)=R1/R2=1000

j. CMR(dB)= 90.4 dB

k. Published values: 90-95 dB

Part 3: Computer Exercises

PSpice Simulation: Determining the Slew Rate

b. V(Vout) max= 5 V V(Vout)min= 0 V

c. Time interval = 12 us

d. SR= 0.40 us

e. Published values: 0.3-0.7 us

PSpice Simulation: Determining the Common Mode Rejection Ratio

b. A(cm) = V(out)/V(in) = 0.26 V/8.7 V = 0.03

c. A(dif) = R1/R2 = 1000

d. CMR(dB) = 90.4

e. Published values: 90-95 dB

326

EXPERIMENT 29: LINEAR OP-AMP CIRCUITS

Part 1: Inverting Amplifier

a. VO/Vi (calculated) = −RO/Ri = 100 K/20 K = −5

b. VO (measured) = −4.87

AV = −VO/Vi = 4.87/1 = −4.87

c. VO/Vi (calculated) = −RO/Ri = −100 K/100 K = −1

V

O (measured) = 1 V

AV = −VO/Vi = −1.06/1 = −1.06

d. Fig 29.6

Part 2: Noninverting Amplifier

a. AV (calculated) = (1 + RO/Ri) = (1 + 100 K/20 K) = 6

b. VO (measured) = 5.24 V

AV = VO/Vi = 5.25/1 = 5.25

The two gains are within 12.5 percent of agreement.

c. AV (calculated) = (1 + 100 K/100 K) = 2

V

O (measured) = 2.17 V

V

O/Vi = 2.17

The two gains are within 8.5 percent of agreement.

Part 3: Unity-Gain Follower

a. Vi (measured) = 2.06 V

V

O (measured) = 2.05 V

The ratio of the computed gain from measured values is equal to .995, which is

practically identical to the theoretical unity gain.

327

Part 4: Summing Amplifier

a. VO (calculated) = −[100 K/100 K * 1 + 100 K/20 K * 1] = −6 V

b. VO (measured) = −5.02 V

The difference between the two values of VO is equal to 16.3 percent.

c. VO (calculated) = −[100 K/100 K * 1 + 100 K/100 K*1] = −2 V

V

O (measured) = −2.01 V

The difference between the two values of VO is equal to .5 percent.

Part 5: Computer Exercises

PSpice Simulation 29-1

1. See Probe plot page 329.

2. (VOUT)peak = 5 V

(VIN)peak = 1 V

3. AV = out

in in

o

VR

=− = −5

4. VOUT 5 V

VIN 1 V

=− = −5

5. Yes

6. 180°

7. Yes

328

29-1

329

PSpice Simulation 29-2

1. See Probe plot page 331.

2. (VOUT)peak = 6 V

(VIN)peak = 1 V

3. out

in in

100 k

11

20 k

o

VR

⎛⎞ Ω

⎛⎞

=+ =+

⎜⎟

Ω

⎝⎠

⎝⎠ = 6

4. out

in

6 V

V1 V

V= = 6

5. Yes

6. 0°

7. Yes

330

29-2

331

EXPERIMENT 30: ACTIVE FILTER CIRCUITS

Part 1: Low-Pass Active Filter

a. fL (calculated) = 1/(2 * 3.14 * 10 K * .001

μ

F) = 15.9 KHz

b. Table 30.1 Low Pass Filter

f (Hz) 100 500 1 K 2 K 5 K 10 K 15 K 20 K 30 K

VO (V) 1.0 1.0 1.0 .99 .95 .85 .74 .59 .52

c. Fig 30.4

d. fL (from graph) = 15 KHz

Part 2: High-Pass Filter

a. fH = 1/(2 * 3.14 * R2 * C2) = 1/(2 * 3.14 * 10 K*.001

μ

F) = 15.9 KHz

b. Table 30.2 High-Pass Filter

f (Hz) 1 K 2 K 5 K 10 K 20 K 30 K 50 K 100 K 300 K

VO (V) .06 .13 .31 .54 .78 .94 1.0 1.0 1.0

c. Fig. 30.5

332

d. fH (from graph) = 15 KHz

Part 3: Band-Pass Active Filter

c. Table 30.3 Band-Pass Filter

f (Hz) 100 500 1 K 2 K 5 K 10 K 15 K 20 K 30 K

VO (V) .01 .035 .07 .15 .32 .51 .57 .57 .49

f (Hz) 50 K 100 K 200 K 300 K

VO (V) .35 .10

d. Fig 30.6

Part 4: Computer Exercises

PSpice Simulation 30-1

1-2. See Probe plot page 334.

3-4. See Probe plot page 335.

5. Slight variance due to PSpice cursor position.

6. fC (calculated) = 15.923 KHz

fC (numeric gain) = 15.937 KHz

f

C (log. gain) = 15.848 KHz

333

30-1

334

30-1

335

PSpice Simulation 30-2

1-5. See Probe plot page 337.

6-8. See Probe plot page 338.

9. Numeric Logarithmic

fC (low): 6.5151 KHz 6.6408 KHz

fC (high): 38.826 KHz 38.214 KHz

Bandwidth: 32.311 KHz 31.573 KHz

10. See tabulation in #9.

336

30-2

337

30-2

338

EXPERIMENT 31: COMPARATOR CIRCUITS OPERATION

Part 1: Comparator with 74IC Used as a Level Detector

a. R3 = 10 Kohms, Vref = 5 V

R3 = 20 Kohms, Vref = 6.7 V

c. Vref (measured) = 4.97 V

d. Vi (measured) (LED goes on) = 5.01 V

V

i (measured) (LED goes off) = 4.98 V

e. Vref (measured) = 6.63 V

V

i (measured) (LED goes on) = 6.65 V

V

i (mesasured) (LED goes off) = 6.61 V

All values of voltages measured and calculated relative to a particular R3 are in very close

agreement.

Part 2: Comparator IC Used as a Level Detector

a. R3 = 10 Kohms Vref (calculated) = 4.98 V

R3 = 20 Kohms Vref (calculated) = 6.63 V

c. Vref (measured) = 4.97 V (R3 = 10 Kohms)

d. Vi (measured) (LED goes on) = 5.01 V

V

i (measured) (LED goes off) = 4.97 V

e. Replace R1 with 20 Kohm resistor.

Vref (measured) = 6.67 (R3 = 20 Kohms)

V

i (measured) (LED goes on) = 6.69 V

V

i (measured) (LED) goes off) = 6.65 V

f. Vi (measured) (LED goes on) = 6.65 V

V

i (measured) (LED goes off) = 6.67 V

The agreement between calculated and measured values in every case was near perfect.

Part 3: Window Comparator

a. V+(pin5, calculated) = 7.5 V

V

− (pin6, calculated) = 2.5 V

c. Vi (pin1, measured) = 7.6 V

V

+ (pin5, measured) = 7.36 V

V

− (pin6, measured) = 2.3 V

d. Vi (measured) (LED goes on) = 7.6 V

V

i (measured) (LED goes off) = 2.6 V

339

e. Vi (measured) (LED goes on) = 7.46 V

V

i (measured) (LED goes off) = 2.2 V

f. Vi (measured) (LED goes on) = 7.46 V

V

i (measured) (LED goes off) = 5.01 V

Again as in the previous case, the agreement between measured and calculated values

was excellent.

Part 4: Computer Exercises

PSpice Simulation 31-1

1. See circuit diagram.

3

1

-

1

:

2. Vin = 6 V; Vref = 5 V

3. Yes. I(D1) = 9.006 mA

340

4-6. See circuit diagram.

31-1:

8. Vin = 4 V; Vref = 5 V

9. No, I(D1) < 8 mA; I(D1) = 118.8

μ

A

PSpice Simulation 31-2

1-3. See Probe plot page 342.

341

31-1

342

EXPERIMENT 32: OSCILLATOR CIRCUITS 1: THE PHASE-SHIFT OSCILLATOR

Part 1: Determining Vout

d. f(theoretical) = 650 Hz

f. Estimated setting of RPot = 3 kohm

g. Vout (peak-peak) = 29 V

h. Period = 1.54 ms

i. f(experimental) = 649.4 Hz

j. Calculated % difference = 0.15

k. RPot + Rf = 29.5 kohm

l. Open-loop gain = 29.5

m. Calculated % difference = 7.8%

Part 2: PSpice Simulation

b. Vout(peak-peak) = 28.8 V

c. Vout(period) = 1.54 ms

d. Vout(frequency) = 649.4 Hz

e. Vout(peak-peak) = 19.1 V

f. Vout(frequency) = 646.5 Hz

j. P(V(feedback) = -89.9 degrees

P(V(VOUT) = 89.4 degrees

P(V(VOUT) – P(V(feedback) = 180 degrees

343

EXPERIMENT 33: OSCILLATOR CIRCUITS 2

Part 1: Wien Bridge Oscillator

c. T (measured) = 305

μ

s

d. f = 1/T = 1/305

μ

s = 3.28 KHz

e. T (measured, C = 0.01

μ

F) = 3 ms

f (calculated, C = 0.01

μ

F) = 328 Hz

f. f (calculated, C = .001

μ

F) = 3.12 KHz

f (calculated, C = .01

μ

F) = 312 Hz

Again, the agreement between the two sets of values was well within 10 percent.

Part 2: 555 Timer Oscillator

c. T (measured) = 20.1

μ

s

d. f = 1/T = 49.3 KHz

e. T (measured, C = 0.01

μ

F) = 203

μ

s

f = 1/T = 4.93 KHz

f. k = fRC = .48

f = 4.91 KHz

The agreement between the two values differed by only .4 percent.

Part 3: Schmitt-trigger Oscillator

c. T (measured) = 21

μ

s

d. f = 1/T = 46.9 KHz

e. T (measured, C = 0.01

μ

F) = 210

μ

s

f = 1/T = 4.69 KHz

f. f (calculated, C = 0.001

μ

F) = 46 KHz

f (calculated, C = 0.01

μ

F) = 4.6 KHz

The measured and calculated values of the frequency for each capacitor were within 2

percent of each other.

344

33-1

345

PSpice Simulation 33-1

1. See Probe plot page 347.

(VOUT)min = 0 V

(VOUT)max = 10 V

2. Yes.

3. 15.87

μ

s

4. PW = 8.63

μ

s

5. f= 63.2 KHz

6. See Probe plot page 348.

7. Yes

8. No

9. P = 31.115

μ

s

10. PW = 23.993

μ

s

11. f = 41.67 KHz

12. Yes

346

33-1

347

33-1

348

EXPERIMENT 34: VOLTAGE REGULATION—POWER SUPPLIES

Note: The data obtained in this experiment was based on the use of a 10 volt Zener diode.

Part 1: Series Voltage Regulator

a. VL = VZ − VBE = 10 V − .7 V = 9.3 V

b. VO (measured) = 9.3 V Table 34.1

Vi (V) 10 11 12 13 14 15 16

VO (V) 9.25 9.26 9.28 9.30 9.32 9.33 9.35

The voltage regulation of the system was −.54 percent.

Part 2: Improved Series Regulator

a. A = 1 + R1/R2 = 1 + 1 K/2 K = 1.5

V

L = AVZ

VL (calculated) = 15 V

b. Table 34.2

Vi (V) 10 12 13 14 16 18 20 22 24

VL (V) 9.44 9.44 9.60 9.64 14.7 14.8 14.9 14.9 14.9

Upon coming near the nominal voltage level, the regulation of the system was −2 percent.

Part 3: Shunt Voltage Regulator

a. VL = (R1 + R2) * VZ/R1 = 3 K/2 K * 10 V = 15 V

b. VL (measured) = 14.7 V

Table 34.3

Vi (V) 24 26 28 30 32 34 36

VO (V) 14.3 14.4 14.5 14.7 14.7 14.9 15.1

The regulation of this system was 2.7 percent.

Part 4: Computer Exercises

PSpice Simulation 34-1

1. See Probe plot page 350.

2. Vin is swept linearly from 2 V to 8 V in 1 V increments.

3. V(V2) = 4.68 V

V(OUT) = 4 V

4. Approx. at V(VIN)) = 6.5 V

5. 0.68 V

6. Yes

7. VL = 4.68 V − 0.68 V = 4 V

349

34-1

350

PSpice Simulation 34-2

1. See Probe plot page 352.

2. V(IN) increases linearly from 6 V to 16 V in 0.5 V increments.

3. VL = V(OUT) = 1 k 1 k

1 k

Ω+ Ω

Ω(4.68 V) = 9.36 V

4. theor.

(OUT)V = 9.36 V

PSpice

(OUT)V = 8.9197 V

5. V(V2) = 4.68 V

V(VOUT) = 8.9197 V

351

34-2

352

EXPERIMENT 35: ANALYSIS OF AND, NAND, AND INVERTER LOGIC GATES

Part 1: The AND Gate: Computer Simulation

a. Table 35-1

Input terminal 1 Input terminal 2 Output terminal 3

1 1 1

0 1 0

1 0 0

0 0 0

Traces U1A:A and U1A:B are the inputs to the gate.

Trace U1A:Y is the output of the gate.

b. The output is at a logical HIGH if and only if both inputs are HIGH.

c. Over the period investigated, the Off state is the prevalent one.

d.

Terminal 25 ms 125 ms 375 ms

1 1 1 0

2 1 1 0

3 1 0 0

353

Part 2. The AND Gate: Experimental Determination of Logic States

a. Ideally, the same.

b. 10 Hz

c. Should be the same as that for the simulation.

d. The amplitude of the TTL pulses are about 5 volts, that of the Output terminal 3 is

about 3.5 volts.

e. The internal voltage drop of across the gate causes the difference between these voltage

levels.

Part 3: Logic States versus Voltage Levels

b. Example of a calculation: assume: V(V1A:Y) = 3.5 volts, VY = 3.4 volts

3.5V 3.4V

%deviation *100 2.86 percent

3.5V

−

==

c. For this particular example, the calculated percent deviation falls well within the

permissible range.

Part 4: Propagation delay

a. For the current case, the propagation delay at the lagging edge of the applied TTL pulse

should be identical to that at the leading edge of that pulse. Thus, it should measure

about 18 nanoseconds.

b. Ideally, the propagation delays determined by the simulation should be identical to that

determined in the laboratory.

c. From Laboratory data, determine the percent deviation using the same procedure as

before.

354

Part 5: NOT-AND Logic

A. Computer Simulation

a.

Table 35-2

Input1(7408) Input 2(7408) Input1(7404) Output(7404)

1 1 1 0

0 1 0 1

1 0 0 1

0 0 0 1

Traces

U1A: A and U1A:B are the inputs to the 7408 gate, U1A:Y its output trace.

Trace

U2A:Y is the output of the 7404 gate.

b. The Output of the 7404 gate will be HIGH if and only if the input to both terminals of

the 7408 gate are HIGH, otherwise, the output of the 7404 gate will be LOW.

c. The most prevalent state of the Output terminal of the 7404 gate is HIGH.

d. The PSpice cursor was used to determine the logic states at the requested times. The logic

states are indicated at the left margin.

At

t = 25 milliseconds:

A1 =

25.397m,

1

A2 = 0.000, 1

dif = 25.397m

355

At

t = 125 milliseconds

A1 =

125.397m,

1

A2 = 0.000, 1

dif = 125.397m

At

t = 375 milliseconds

A1 =

375.397m,

0

A2 = 0.000, 1

dif = 375.397m

B. Experimental Determination of Logic States

a. They should be relatively close to each other.

b. They are identical.

c. The output of the 7404 gate is the negation of the output of the 7408 gate.

356

Part 6: The 7400 NAND Gate

A. Computer Simulation Table 35-3

a.

Input terminal 1 Input terminal 2 Output terminal 3

1 1 0

0 1 1

1 0 1

0 0 1

b.

B. Experimental Determination of Logic States

Table 35-4

Input terminal 1 Input terminal 2 Output terminal 3

1 1 0

0 1 1

1 0 1

0 0 1

357

EXPERIMENT 36: ANALYSIS OF OR, NOR AND XOR LOGIC GATES

Part 1: The OR Gate: Computer Simulation

a. Table 36-1

Input terminal 1 Input terminal 2 Output terminal 3

1 1 1

0 1 1

1 0 1

0 0 0

Traces U1A:A and U1A:b are the inputs to the gate.

Trace U1A:Y is the output of the gate.

b. The output is a logical LOW if and only if both inputs are LOW, otherwise the output is

HIGH.

c. Over the period investigated, the ON, or HIGH, state is the prevalent one. This differs

from that of the AND gate. Its prevalent state was the OFF or LOW state.

358

d. The PSpice cursor was used to determine the logic states at the requested times. The

logic states are indicated at the left margin.

At t = 25 milliseconds:

A1 =

25.397m,

1

A2 = 0.000, 1

dif = 25.397m

At t = 125 milliseconds

A1 =

125.397m,

1

A2 = 0.000, 1

dif = 125.397m

At

t = 375 milliseconds

A1 =

375.397m,

0

A2 = 0.000, 1

dif = 375.397m

359

Part 2: The OR Gate: Experimental Determination of Logic States

a. The pulse of 100 milliseconds of the TTL pulse is identical to that of the simulation

pulse.

b. The frequency of 10 Hz of the TTL pulse is identical to that of the simulation pulse.

c. They were determined to be the same at the indicated times.

d. The voltage of the TTL pulse was 5 volts. The voltage at the output terminal was 3.5

volts.

e. The difference in these two voltages is caused by the internal voltage drop across the

7432 gate.

Part 3: Logic States versus Voltage Levels

a. The PSpice simulation produced the identical traces as shown on the PROBE plot for

Figure 36-2.

b. Example of a calculation: assume V(V1A:Y) = 3.6 volts, VY = 3.4 volts

3.6V 3.4V

%deviation *100 5.56 percent

3.6V

−

==

a. It is larger by (5.56-2.86) = 2.7 percent.

Part 4: Combining AND with OR Logic

A. Computer Simulation

a.

360

Table 36-2

U1A:A U1A:B U1A:Y U2A:A U2A:B U2A:Y U3A:A U3A:B U3A:Y

1 1 1 1 1 1 1 1 1

0 1 0 0 1 0 0 0 0

1 0 0 1 1 1 0 1 1

0 0 0 0 1 0 0 0 0

c.

At t = 25 milliseconds

A1 =

25.397m,

1

A2 = 0.000, 1

dif = 25.397m

At

t = 125 milliseconds

A1 =

125.397m,

1

A2 = 0.000, 1

dif = 125.397m

361

At

t = 375 milliseconds

A1 =

375.397m,

0

A2 = 0.000, 1

dif = 375.397m

b. The output of the 7432 gate, U3A:Y, is evenly divided between the ON state and the OFF

state during the simulation.

B. Experimental Determination of Logic States

a. The logic states of the simulation and those experimentally determined are identical.

b. The logic state of the output terminal U3A:Y is identical to that of the TTL clock.

c. The logic state of the output terminal U3A:Y is identical to that of the output terminal

U2A:Y of the U2A gate.

Part 5: NOR and XOR Logic combined

A. Computer Simulation

a.

The output trace of the 7402 NOR gate, U1A:Y and the output trace of the XOR gate,

U2A:Y are both shown in the above plot.

362

b. Table 36-3

U1A:A U1A:B U1A:Y U2A:A U2A:B U2A:Y

1 1 0 1 1 0

0 1 0 0 1 1

1 0 0 1 0 1

0 0 1 0 0 0

c. The output of the 7402 gate, U1A:Y is HIGH if and only if both inputs are LOW,

otherwise the output is LOW.

d. This is a logical inversion of the OR gate.

c. The output of the 7486 gate is HIGH if and only if the two inputs U2A:A and U2A:B

are at opposite logic levels.

f. The logic state of the OR gate is HIGH if both inputs are at opposite logic levels and if

both inputs are HIGH.

B. Experimental Determination of Logic States

a. The experimental data is identical to that obtained from the simulation.

b. Refer to the data in Table 36-3.

c. Refer to the data in Table 36-3.

d. Refer to the data in Table 36-3.

e. The output of the 7486 XOR gate is HIGH if and only if its input terminals have

opposite logic levels, otherwise, its output is at a LOW.

f. For an OR gate, its output is HIGH if both, or at least one input terminal, is HIGH.

Its output will be LOW if both inputs are LOW. For an XOR gate, its output is HIGH if

and only if both input terminals are at opposite logic levels, otherwise, the output will be

LOW.

g. The output of an XOR gate will be HIGH when both input terminals are at opposite

logic levels. Otherwise, its output is at a logical LOW.

363

EXPERIMENT 37: ANALYSIS OF INTEGRATED CIRCUITS

Part 1: Positive Edge-Triggered D Flip-Flop

A. Computer Simulation

a. The PROBE data shows the flip flop to be in the SET condition.

b. The flip flop goes to RESET at 200 milliseconds because the D input terminal goes

negative. The flip flop goes to SET at 400 milliseconds because both the CLOCK input

and the D input are positive.

c. The importance to note is that the D input can be negative and positive during the time

that the Q output is low.

d. After the initial SET condition of the flip flop, and after a RESET state of 200

milliseconds, the flip flop returns to its SET condition because at 400 milliseconds, both

the CLOCK and the D inputs are positive.

e. Starting from a SET condition, a transition to RESET will occur when the D input is

negative and the CLOCK pulse goes positive. The flip flop will SET again when the D

input is positive and the CLOCK goes positive.

f. The conditions stated in previous answer define a positive edge triggered flip flop as

defined in the first paragraph of Part 1.

g. See above answers.

h.

364

i. Let us assume that D is high when a positive CLOCK pulse goes high. This will SET the

flip flop. This SET will be stored, or remembered, until D is negative and the CLOCK

triggers positive again. At that time, the flip flop will RESET. This RESET will be

stored, or remembered, until D is positive and the CLOCK triggers positive again. At

that time the flip flop will SET. Events repeat themselves after this.

B. Experimental Determination of Logic States

a. Both input terminals are held at 5 volts during the experiment.

b. The amplitude of the voltage of the TTL pulse is 5 volts.

c. The amplitude of the output voltage at the Q terminal is 3.5 volts.

d. The difference between the input voltages and the output voltage is caused by the voltage

drop through the flip flop.

e. The experimental and the simulation transition states occur at the same times.

365

Part 2: Frequency Division

A. Computer Simulation

Answer all questions below with reference to the following PROBE plot.

a. The frequency at the U1A:Q terminal is 5 Hz.

b. The frequency at the U1A:Q terminal is one-half that of the U1A:CLK terminal.

c. The frequency at the U2A:Q terminal is 2.5 Hz.

d. The frequency of the U2A:Q terminal is one-half that of the U2A:CLK terminal.

e. The overall frequency reduction of the output pulse U2A:Q relative to the input pulse

U1A:CLK is one-fourth.

f. Each flip flop reduced its input frequency by a factor of two.

g. It would take four 74107 flip-flops.

B. Experimental Determination of Logic States.

a. The J and CLR terminals of both flip flops are kept at 5 volts during the experiment.

b. The voltage level of the U1A:CLK terminal is 5 volts. The voltage level of the U2A:CLK

terminal is 3.5 volts. The voltage level of the U2A:Q terminal is 3 volts.

366

c. Refer to the above PROBE plot.

d. Pulse Frequency

U1A:CLK 10.0 Hz

U1A:Q 5.0 Hz

U2A:CLK 5.0 Hz

U2A:Q 2.5 Hz

e. They are identical.

Part 3: An Asynchronous Counter: the 7493A Integrated Circuit

A. Computer Simulation

a.

A1 =

22.152m,

0

A2 = 0.000, 0

dif = 22.152m

b. See PROBE plot above.

d. t = 175 milliseconds. There is one clock pulse to the left of the cursor.

A1 =

174.051m,

1

A2 = 0.000, 0

dif = 174.051m

367

e. t = 375 milliseconds. There are three clock pulses to the left of the cursor.

f. t = 575 milliseconds. There are five clock pulses to the left of the cursor.

g. t = 1.075 seconds. There are ten clock pulses to the left of the cursor.

A1 =

376.582m,

1

A2 = 0.000, 0

dif = 376.582m

A1 =

575.949m,

1

A2 = 0.000, 0

dif = 575.949m

368

le riginal condition.

j. The output terminal QA represents the most significant digit.

k. The indicated propagation delay is about 12.2 nanoseconds.

B. Experimental Determination of Logic States

a. The logic states of the output terminals were equal to the number of the TTL pulses.

b. The experimental data is equal to that obtained from the simulation.

c. The propagation delay measured was about 13 nanoseconds.

d. The difference in the experimentally determined propagation delay was 13 nanoseconds

compared to a propagation delay of 12 nanoseconds as obtained from the simulation data.

h. At t = 1.075 milliseconds, the output terminals, QA, QB, QC and QD have resumed their

initial states.

i. The MOD 10 counts to ten in binary code after whic it recyc s to its oh

A1 =

1.0760,

1

A2 = 0.000, 0

dif = 1.0760

A1 = 1.0000,

A2 = 1.0000,

1.7628

4.9975

dif = 12.200n, −3.2347

369

BoylestadnashelIM Solution Manual Electronic Devices And Circuit Theory 10th Edition

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