Chapter 5 Problem Solutions 5.1

(a)

( ) ( )( )

510 856

85 0.98841 861 86 6 516

C

B

E B E

ii

i i i A

β β

βα αββ μ

= = ⇒ =

= = ⇒ =+

= + = ⇒ =

(b)

( ) ( )( )

2.65 530.05053 0.9815541 54 0.050 2.70 E B Ei i i mA

β β

α α

β

= ⇒ =

= ⇒ =

= + = ⇒ =

5.2

(a) For 110:β = 110 0.990991 111

βαβ

= = =+

For 180:β = 180 0.99448181

α = =

0.99099 0.99448α≤ ≤ (b) ( )110 50 A 5.50 C B CI I I mAβ μ= = ⇒ =

or ( )180 50 A 9.00 C CI I mAμ= ⇒ = so 5.50 9.0 CI mA≤ ≤ 5.3

(a)

( )

1.12 9.33 A120

1211.12 1.13 mA120

120 0.9917121

B

E

i

i

μ

α

= ⇒

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

(b)

( )

50 2.5 mA20

21 50 52.5 mA20

20 0.952421

B

E

i

i

α

= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

5.4 (a)

α 1

αβα

=−

0.9 9 0.95 19 0.98 49 0.99 99 0.995 199 0.999 999 (b)

β 1

βαβ

=+

20 0.9524 50 0.9804 100 0.9901 150 0.9934 220 0.9955 400 0.9975 5.5

(a)

( )

( )( )

1.2 14.8 A81

801.2 1.185 mA81

80 0.9877815 1.185 2 2.63 V

B

C

C

I

I

V

μ

α

= ⇒

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

= − =

(b)

( )

( )( )

0.80 9.88 A81

800.80 0.790 mA81

80 0.9877815 0.790 2 3.42 V

B

C

C

I

I

V

μ

α

= ⇒

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

= − =

(c) Yes, C BV V> so B-C junction is reverse biased in both areas. 5.6

For 50, 2.5 mA2C CV I= = =

2.5 2.546 mA0.982

CE E

II I

α= = ⇒ =

5.7

(a)

( )

( )( )

0.75 12.3 A61

600.75 0.738 mA61

60 0.983661

10 0.738 5 106.31 V

B

C

C C C

C

I

I

V I RV

μ

α

= ⇒

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

= − = −= −

(b)

( )

( )( )

1.5 24.6 A61

601.5 1.475 mA61

60 0.983661

1.475 5 10 2.625 V

B

C

C C

I

I

V V

μ

α

= ⇒

⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= − ⇒ = −

(c) Yes, 0CV < in both cases so that B-C junction is reverse biased.

5.8 ( )10 10 1.2 1.76 mA

51.76 1.774 mA0.992

CC

C

CE E

VI

RII Iα

− − −= = =

= = ⇒ =

5.9

( )

/

13 0.68510 exp 27.67 mA0.026

1 91 27.6790

27.98 mA27.67 0.307 mA

90

RE TV VC S

C

E C

D

CB B

I I e

I

I I

III I

ββ

β

−

=

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

⎛ ⎞+ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=

= = ⇒ =

5.10 Device 1: / 3 0.650 / 0.026

1 10.5 10EB Tv VE Eo Eoi I e I e−= ⇒ × =

So that 15

1 6.94 10 EOI A−= ×

Device 2: 3 0.650 / 0.026212.2 10 EoI e−× =

Or 13

2 1.69 10 EoI A−= × 13

215

1

1.69 10Ratio of areas Ratio 24.46.94 10

Eo

Eo

II

−

−

×= = ⇒ =×

5.11

(a) 250 250 1

Ao o

C

Vr r kI

= = ⇒ = Ω

(b) 250 2.50 0.1

Ao o

C

Vr r MI

= = ⇒ = Ω

5.12

00 33

0

60100

12.9 V

C BC E

C E

BVBV

BVβ

= =

=

5.13

00 3

33

220 22056 3.9356

60.6

C BC E

BVBV

β

ββ

β

=

= ⇒ = =

=

5.14

( ) ( )

00 3

330 0

0

50 50

184 V

C BC E

C B C E

C B

BVBV

BV BV

BV

β

β

=

= =

=

5.15

(a) ( )

( )

0.7 101.86 mA

5751.86 1.836 mA76

0.7 4 3.3 V10 3.3 3.65 K1.836

E

C

C

C C

I

I

V

R R

− − −= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

= − + =−= ⇒ =

(b)

( )( )

( )

( )

0.5 0.00658 mA76

0.00658 25 0.164 V750.5 0.493 mA76

1 58.11 K

0.493

B

B B B B

C

C C

I

V I R V

I

R R

= =

= = ⇒ =

⎛ ⎞= =⎜ ⎟⎝ ⎠

− − −= ⇒ =

(c) ( ) ( )( )

( )

( )( ) ( )( )

10 0.7 4 876

7.3 4 0.132 1.767 mA751.767 1.744 mA76

8 1.744 4 1.767 4 8

16 6.972 7.068 1.96 V

EE

E E

C

CE

CE

IO I

I I

I

V

V

= + + −

= + ⇒ =

⎛ ⎞= =⎜ ⎟⎝ ⎠

= − − −⎡ ⎤⎣ ⎦= − − ⇒ =

(d) ( ) ( ) ( ) ( )

( )( )

5 10 20 0.7 2 10 0.263 2 0.776

0.3506 mA 4.61 A 5 0.3506 10

1.49 V

EE E E

E B C

C

II I I

I I V

V

μ

⎛ ⎞= + + + = + + +⎜ ⎟⎝ ⎠

= ⇒ = = −

=

5.16 For Fig. 5.15 (a) 5 5% 5.25 KER = + =

( )

( )

0.7 101.77 mA

5.251.75 mA10 3.3 3.83 K

1.755 5% 4.75 K

0.7 101.96 mA

4.751.93 mA10 3.3 3.47 K

1.93

E

C

C

E

E

C

C

I

I

R

R

I

I

R

− − −= =

=−= =

= − =− − −

= =

=−= =

So 1.75 1.93 mA 3.47 3.83 KC CI R≤ ≤ ≤ ≤ For Fig. 5.15(c) 4 5% 4.2 KER = + =

( )( )

( )( ) ( )( )

( )( )

( )( ) ( )( )

8 0.7 0.0222 mA 1.66 mA10 76 4.21.69 mA16 1.66 4 1.69 4.216 6.64 7.098 2.26 V

4 5% 3.8 K8 0.7 0.0244 1.83 mA

10 76 3.81.86 mA16 1.83 4 1.86 3.816 7.32 7.0681.61 V

B C

E

CE

CE

E

B C

E

CE

CE

I I

IV

V

R

I I

IV

V

−= = =+

== − −= − − ⇒ =

= − =−= = =

+== − −= − −=

So 1.66 1.83 mA 1.61 2.26 VC CEI V≤ ≤ ≤ ≤ 5.17

( )( )

2.5 0.7 120 0.015

70 15 1.05 5 2.5 2.38 1.05

BB EBB B

B

CQ

CC ECQC C

CQ

V VR R kI

I A mAV V

R R kI

μ

− −= = ⇒ = Ω

= ⇒− −= = ⇒ = Ω

5.18 (a)

( )

( )

( )

( )

1500

2.0 A1 0.7 1.7 V

3 1.7 3 0.2708 mA4.8

0.27081 135.4 134.40.002

0.99261

0.269 mA

3 3 1.7 4.7 V

BB B B B

B

B

E

EE

E

E

B

C B C

CE E CE

VV I R IR

IV

VI

RII

I I I

V V V

μ

β β

βα αβ

β

− −−= − ⇒ = =

== − − = −

− − − += = =

= + = = ⇒ =

= ⇒ =+

= ⇒ =

= − = − − ⇒ =

(b)

( )

( ) ( )( )

( )

5 4 0.5 mA2

4 0.7 5

4 0.7 100 0.5 8 50.50.043 1 11.63

0.043

10.63, 0.91401

E E

B B B C C

B C E

B C E

B

EB

B

I I

I R I I RI I II I I

IIII

β

ββ α αβ

−= ⇒ =

= + + + −+ =+ =

= + + −

= ⇒ = + = =

= = ⇒ =+

5.19

( )

( )

[ ]

0.7 5 4.33 3

4.350 5051 3 51

4.3 505 5 103 51

Now , so 1 3.27 5 14.1 9.052.12 V

2.12 4.3 0.727 mA3

B BE

BC E

BC C C

B C B

B

E E

V VI

VI I

VV I R

V V VV

I I

− − − += =

+⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

+⎛ ⎞ ⎛ ⎞= − = − ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= + = − = −

= −

− += ⇒ =

5.20

10 10 2 0.80 mA10 10

0.7 2 0.7 1.3 V1.3 0.026 mA50

0.80 0.026 0.774 mA

EE E

B E

BB B

B

C E B C

VI I

V VV

I IR

I I I I

− −= = ⇒ =

= − = − =

= = ⇒ =

= − = − ⇒ =

( )( )( )

0.774 29.770.026

29.77 0.96751 30.77

102 0.774 10 10

4.26 V

C

B

EC E C E C C

EC

II

V V V V I R

V

β β

βα αβ

= = ⇒ =

= = ⇒ =+

= − = − −= − −⎡ ⎤⎣ ⎦=

Load line developed assuming the VB voltage can change and the RB resistor is removed.

1 mA

0.774

4.26 VEC20

IC

Q-point

5.21

( )( )( )( )

( )

5 0.7 17.2 A250

120 0.0172 2.064 mA2.064 1.5 5 1.90 V

5 1.90 6.90 V

B

C

C

EC EC

I

IV

V V

μ−= ⇒

= == − = −= − − ⇒ =

VEC(V)

Q-point

IC(mA)

6.67

2.06

6.9 10 5.22

( )

( )( )

( ) ( )( )

50 1 0.98 51

9 0.98 4.7 9 4.39

1 0.0196 51

0.0196 50 0.7 1.68

C

C C C C

B

E B B EB E

I mA

V I R or V V

I mA

V I R V on or V V

⎛ ⎞= =⎜ ⎟⎝ ⎠

= − = − = −

= =

= + = + =

5.23

( )

( ) ( )( )( )( )

( )( )( ) ( )( )

50 0.50.5 0.49 , 0.0098 51 51

0.0098 50 0.7 or 1.19 9 0.49 4.7 9 6.70

Then 1.19 6.7 7.89 0.49 7.89 0.0098 0.7 or 3.87

C B

E B B EB E

C C C

EC E C

Q C EC B EB Q

I mA I mA

V I R V on V VV I R V

V V V VP I V I V P mW

⎛ ⎞= = = =⎜ ⎟⎝ ⎠

= + = + == − = − = −

= − = − − === + = + =

( ) ( )( )Power Dissipated 9 0.5 9 1.19S Q EP I V= = − = − Or 3.91 SP mW= 5.24

( )( )

1 2 1 2

1 2

1 2 1 2

0.5 mA20.5 mA5 0.5 4 3 V

E E E E

C C

C C C C

II I I I

I IV V V V

= = ⇒ = =

= ≈= = − ⇒ = =

5.25

(a)

( )

2 0.7 1.30

1.3 5 280 0.8 3.75 K

130 K

E BB B

C CB C

B

R IR R

I RR R

R

−= = =

⎛ ⎞ −= = = ⇒ =⎜ ⎟⎝ ⎠

=

(b)

( )( ) ( )( )( ) ( )( )

0.8 811 K 0.010 mA 0.8 0.81 mA80 80

2 0.010 0.7 0.81 1 49 K

5 0.8 2 0.81 1 2.74 K

E B E

B B

C C

R I I

R R

R R

⎛ ⎞= = = = =⎜ ⎟⎝ ⎠

= + + ⇒ =

= + + ⇒ =

(c) For part (a) 2 0.7 0.01 mA130BI −= =

( )( )( )( )

120 0.01 1.20 mA

5 1.2 3.75 0.5 VC C

CE CE

I I

V V

= ⇒ =

= − ⇒ =

For part (b) ( ) ( ) ( )2 49 0.7 121 1B BI I= + +

( )( ) ( )( )0.00765 mA, 0.925 mA, 0.918 mA5 0.918 2.74 0.925 1 1.56 V

B E C

CE CE

I I IV V

= = == − − ⇒ =

Including ER result in smaller changes in Q-point values. 5.26

a. ( )on

2 0.0333 mA60

24 0.7 699 k0.0333

24 12 6 k2

CC BEBQ

B

CQBQ

B B

CC CEQCQ C C

C

V VI

RI

I

R R

V VI R R

R

β

−=

= = =

−= ⇒ = Ω

− −= ⇒ = ⇒ = Ω

b. ( )

( )( )( )

( )( )

on 24 0.7699

0.0333 mA Unchanged100 0.0333 3.33 mA

24 3.33 6 4.02 V

CC BEBQ

B

CQ BQ CQ

CEQ CC CQ C CEQ

V VI

R

I I I

V V I R V

β

− −= =

== = ⇒ =

= − = − ⇒ =

(c) ( )24 6CE CC C C CV V I R I= − = − IC (mA)

4

3.33

4.02 12 24

2

VCE

Q-pt (� � 100)

Q-pt (� � 60)

5.27 a. 0 Cutoff 0, 6 VB E CV I V= ⇒ ⇒ = =

b.

( ) ( )

1 0.71 V, 0.3 mA1

6 0.3 10 3 V

B E E

C E C C

V I I

I I V V

−= = ⇒ =

≈ ⇒ = − ⇒ =

c. 2 V.BV = Assume active-mode 2 0.7 1.3 mA

1E E CI I I−= = = ≈

( )( )6 1.3 10 7 V!CV = − = −

( )( )

Transistor in saturation 2 0.7 1.3 mA

11.3 V, sat 0.2 V

sat 1.3 0.2 1.5 V

E E

E CE

C E CE C

I I

V VV V V V

−= ⇒ =

= == + = + ⇒ =

5.28 a.

( )0

0

0.

10Cutoff 510 5

3.33 V

BB

LCC

C L

V

RV VR R

V

=

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=

b.

( ) ( )0 0

0 0

1 V1 0.7 6 μA

5075 6 0.45 mA

55 10

1 11 0.45 1.83 V5 10

BB

B

C B C

C

V

I

I I IV VI

V V

β

=−= ⇒

= = ⇒ =−

= +

⎛ ⎞− = + ⇒ =⎜ ⎟⎝ ⎠

c. Transistor in saturation ( )0 sat 0.2 VCEV V= =

5.29 (a) 100β =

(i) ( )

( )( )

1000.1 0.1 0.0990 101

5 0.099 5 4.505

Q C

O O

I mA I mA

V V V

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= − ⇒ =

(ii) ( )

( )( )

1000.5 0.5 0.495 101

5 0.495 5 2.525

Q C

O O

I mA I mA

V V V

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= − ⇒ =

(iii) ( ) ( )

2 mA Transistor is in saturation0.7 0.2 0.5 V

Q

O BE CE O

IV V sat V sat V

== − + = − + ⇒ = −

(b) 150β =

(i) ( )

( )( )

1500.1 0.1 0.09934 151

5 0.09934 5 4.503

4.503 4.505% change 100% 0.044%4.503

Q C

O O

I mA I mA

V V V

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= − ⇒ =

−= × = −

(ii) ( )

( )( )

1500.5 0.5 0.4967 151

5 0.4967 5 2.517

2.517 2.525% change 100% 0.32%2.525

Q C

O O

I mA I mA

V V V

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= − ⇒ =

−= × = −

(iii) 2 Transistor in saturation8.5 V No change

Q

o

I mAV

== −

5.30

( )

5 0.50.5 0.5 , 0.90 5

101 0.90 0.909 100

CB O C

Q Q

V V V V I mA

I I mA

−= ⇒ = = =

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

5.31 For 0,QI = then 0QP =

For ( )500.5 , 0.5 0.49 51Q CI mA I mA⎛ ⎞= = =⎜ ⎟

⎝ ⎠

( )( )( )( )

0.5 0.0098 , 0.490 , 1.19 510.49 4.7 9 6.70 7.89

0.49 7.89 3.87

B B E

C EC

C EC

I mA V V V V

V V V VP I V P mW

= = = =

= − = − ⇒ =≅ = ⇒ =

For 1.0 ,QI mA= Using the same calculations as above, we find 5.95 P mW= For 1.5 , 6.26 QI mA P mW= = For 2 , 4.80 QI mA P mW= = For 2.5 , 1.57 QI mA P mW= = For 3 ,QI mA= Transistor is in saturation.

( ) ( )

( ) ( ) ( )

0.7 50 0.2 4.7 93

Then, 0.7 3 50 0.2 4.7 9

B C

E Q B C B C

C C

I II I I I I I

I I

+ = + −= = + ⇒ = −

+ − = + −

Which yields 2.916 and 0.084 C BI mA I mA= =

( )( ) ( )( )0.084 0.7 2.916 0.2B EB C ECP I V I V= + = + or 0.642 P mW= 5.32

( )

( ) ( )

( ) ( )

on 9 0.7 2.075 mA4

0.9920 2.075 2.06 mA

9 2.06 2.2 4.47 V

EE EBE E

E

C E C

BC C C CC

BC BC

V VI I

RI I I

V I R VV V

α

− −= = ⇒ =

= = ⇒ =

+ == − ⇒ =

5.33

( )

( )( )

2

1 2

1 1 1 1

12 6 2.73 mA2.2

2.73 0.091 mA30

0.7 120.127 mA

1000.127 0.091 0.218 mA

0.7 0.218 15 0.7 3.97 V

CC CEQCQ

C

CQBQ BQ

R

R R BQ

R

V VI

RI

I I

I

I I IV I R V

β

− −= = =

= = ⇒ =

− −= =

= + = + == + = + ⇒ =

5.34

( )

( ) ( ) ( )

2

1 2

1 1 1

For 4.55 4.5 0.5 mA

10.5 0.02 mA250.7 5

0.057 mA100

0.057 0.02 0.077 mAon 0.077 15 0.7 1.86 V

CE

CQ

BQ

R

R R BQ

R BE

V

I

I

I

I I IV I R V

=−= =

= =

− −= =

= + = + == + = + =

( )( )

2

1 2

1

For 1.05 1 4 mA

14 0.16 mA250.057 mA

0.057 0.16 0.217 mA0.217 15 0.7 3.96 V

CE

CQ

BQ

R

R R BQ

V

I

I

II I IV

=−= =

= =

== + = + =

= + ⇒

So 11.86 3.96 VV≤ ≤

Range ofQ-pt values

IC

5

4

0.5

0 1 4.5 5 5.35

(a) 5 2.5 5 K 0.5

0.5 0.00417 mA1205 0.7 1032 K

0.00417

C

B

B

R

I

R

−= =

= =

−= =

IC(mA)

VCE(V)

Q-point

1.0

0.5

2.5 5 (b) Choose 5.1 K

1 MC

B

RR

== Ω

For 1 M 10% 1.1 M, 5.1 k 10% 5.61 K5 0.7 3.91 A 0.469 mA

1.12.37 V

B C

BQ CQ

CEQ

R R

I I

V

μ

= Ω + = = + =−= = ⇒ =

=

1 M 10% 1.1M, 5.1 10% 4.59 K3.91 A 0.469 mA2.85 V

B C

BQ CQ

CEQ

R R KI I

Vμ

= Ω + = = − == ⇒ ==

1 M 10% 0.90 M 5.1 k 10% 5.61 K5 0.7 4.78 A 0.573 mA0.90

1.78 V

B C

BQ C

CEQ

R R

I I

V

μ

= Ω − = Ω = + =−= = ⇒ =

=

1 M 10% 0.90 M 5.1 k 10% 4.59 K4.78 A 0.573 mA2.37 V

B C

BQ C

CEQ

R RI I

Vμ

= Ω − = Ω = − == ⇒ ==

IC(mA)

VCE(V)

1.09

0.891

0.5730.469

2.851.78 2.37 5 5.36

( ) ( )2 2 1 1

2 1 2 1

1 2

5 55 5

E BE E BE

O E E BE BE

O BE BE

V V V VV V V V VV V V

= − = −= − = − − −= −

11

22

We have ln

ln

EBE E

EO

EBE T

EO

IV VI

IV VI

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

( )

1 2

1

2

ln ln

10ln ln

ln 10

E EO T

EO EO

E IO T T

E

O

I IV VI I

IV V VI I

kTVe

⎡ ⎤⎛ ⎞ ⎛ ⎞= −⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

=

5.37 (a) 0ER =

0.7200

IB

VI

−= C BI Iβ= ( ) ( )( ) ( )120 4

5 4 5 0.7200O C IV I V= − = − −

When 0.2, 0.2 5 2.4 1.68 2.7O I IV V V= = − + ⇒ =

VO(V)

5

0.2

0.7 2.7 5 VI (V) (b)

( )( )( )( ) ( )

1 K0.7 0.7

200 121 1 321

120 45 0.7

321When 0.2 5 1.495 1.047

3.91 V

E

I IB C IB

O I

O I

I

RV VI I

V V

V VV

β

=− −

= = =+

= − −

= = − +=

VO(V)

5

0.2

0.7 3.91 5 VI (V) 5.38 For 4.3 5IV≤ ≤ Q is cutoff 0CI =

0OV = If Q reaches saturation, 4.8

4.8 1.2 mA4

5 0.71.2 0.015 1.680 180

So 1.6, 4.8

O

C

IB I

I O

V

I

VI V

V V

=

= =

− −= = = ⇒ =

≤ =

VO(V)

4.8

1.6 4.3 5 VI (V) 5.39 (a) For 4.3,IV ≥ Q is off and 0OV =

When transistor enters saturation, ( ) ( )1015 1 0.2 4 0.958 mA100 C C CI I I⎛ ⎞= + + ⇒ =⎜ ⎟⎝ ⎠

( )( ) ( )( )

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

3.832 V0.00958 mA1015 0.958 1 0.7 0.00958 1801005 0.7 0.9676 1.7244 1.61 V

For 0, transistor in saturation5 1 0.2 4 5 1 1 0.2 45 1 0.7 180 5 1 1 0.7 180

O

B

I

I I

I

E C C B C

E B C B B

E C B

VI

V

V VVI I I I II I I I I

I I I

==

⎛ ⎞= + + +⎜ ⎟⎝ ⎠

= − − − ⇒ ==

= + + ⇒ = + + += + + = + + +

= +( )

( )( )

4.8 5 14.3 1 181

4.8 54.3 181 4.8 5904 864.5

0.956 mA3.825 V

C B

C B

B C

C C

C

C

O

I II I

I II I

IIV

= += +

= −= + −

===

VO(V)

3.832

3.825

1.61 4.3 5 VI(V) 5.40

( )

( )( ) ( )( )

( )( ) ( )( )

1 2

2

1 2

33 10 7.67 k

10 18 4.186 V10 33

on 4.186 0.71 7.67 51 1

0.0594 mA2.97 mA

3.03 mA

18 2.97 2.2 3.03 1 8.44 V

TH

TH CC

TH BEBQ

TH E

BQ

CQ BQ CQ

EQ

CEQ CC CQ C EQ E

CEQ

R R R

RV V

R R

V VI

R RII I I

IV V I R I R

V

β

β

= = = Ω

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠− −= =

+ + +== ⇒ =

== − −= − − ⇒ =

5.41

( ) ( )

( ) ( )( )2

1 2 1 1

1.2 , 9 , 50 1.2Also 0.015 80

1

1 1 50 18

CQ CEQ TH

B

TH BQ TH BE BQ E

TH CC TH CC

I mA V V R k

I mA

V I R V on I R

RV V R VR R R R

β

= = = Ω

= =

= + + +

⎛ ⎞= = ⋅ ⋅ =⎜ ⎟+⎝ ⎠

( )( ) ( )( ) ( )( )( ) 11

22

2

1Then 50 18 0.015 50 0.7 81 0.015 1 338 .

338Then 50 58.7 338

or R kR

R R kR

= + + = Ω

= ⇒ = Ω+

( )

( ) ( )( )

81 1.2 1.215 80

1818 1.2 9 1.215 1 6.49

EQ

CQ C CEQ EQ E

C C

I mA

I R V I RR R k

⎛ ⎞= =⎜ ⎟⎝ ⎠

= + += + + ⇒ = Ω

5.42

( ) ( )

( )

( )

1 2

2

1 2

20 15 8.57

15 10 4.29 15 20

18.5710 1 0.7 4.29101

10 0.7 4.29 5.01Then 4.62 8.57 1.0851101

4.1

TH

TH CC

EQCC EQ E EB TH TH

EQ EQ

EQ EQ

EQB TH TH

R R R k

RV V VR R

IV I R V on R V

I I

I I mA

IV R V

β

β

= = = Ω

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

= + + ⋅ ++

⎛ ⎞= + + +⎜ ⎟⎝ ⎠

− −= = ⇒ =+

= ⋅ + =+

( )62 8.57 4.29 or 4.68 101 BV V⎛ ⎞ + =⎜ ⎟

⎝ ⎠

5.43 (a)

( )

( )( )

42 58 24.36 K42 24 10.08 V

10010.08 0.7 9.38 7.30 A

24.36 126 10 1284.360.913 mA 0.920214.8 V

TH

TH

BQ

CQ EQ

CEQ

R

V

I

I IV

μ

= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

−= = ⇒+

= ==

IC(mA)

2.38

0.913

14.8 24 VCE(V)

Q-point

(b)

( )

1 25% 60.9, 5% 44.1 25.58 K 10.0810.08 0.7 9.38 7.30 A

25.58 126 10 1285.580.912 mA 0.91914.81

TH TH

BQ

CQ EQ

CEQ

R R R V

I

I IV

μ

+ = + = = =−= = ⇒

+= ==

( )( )

1 25% 60.9, 5% 39.90 24.11 K 9.509.50 0.7 8.8 6.85 A

24.11 126 10 1284.110.857 mA 0.8635 mA15.37 V

TH TH

BQ

CQ EQ

CEQ

R R R V

I

I IV

μ

+ = − = = =−= = =

+= ==

1 25% 55.1 K 5% 44.1 K 24.50 K 10.67 V10.67 0.7 9.97 7.76 A

24.50 1260 1284.50.970 mA 0.978 mA14.22 V

TH TH

BQ

CQ EQ

CEQ

R R R V

I

I IV

μ

− = + = = =−= = =

+= ==

1 25% 55.1 K 5% 39.90 23.14 K 10.0810.08 0.7 9.38 7.31 A

23.14 1260 1283.140.914 mA 0.9211 mA14.79 V

TH TH

BQ

CQ EQ

CEQ

R R R V

I

I IV

μ

− = − = = =−= = =

+= ==

So we have 0.857 0.970 mA14.22 15.37 V

CQ

CEQ

IV

≤ ≤≤ ≤

5.44 a.

( )

( ) ( )( )

( )( ) ( )( )

1 2

2

1 2

25 8 6.06 k

8 248 25

5.82 V(on) 5.82 0.7

1 6.06 76 10.0624 mA, 4.68 mA

4.74

24 4.68 3 4.74 15.22 V

TH

TH CC

TH BEBQ

TH E

BQ CQ

EQ

CEQ CC CQ C EQ E

CEQ

R R R

RV VR R

V VIR R

I I

I

V V I R I R

V

β

= = = Ω

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=

− −= =+ + +

= =

=

= − −= − −=

b.

( )( )

( )( ) ( )( )

5.82 0.7 0.0326 mA6.06 151 14.89 mA

4.9224 4.89 3 4.92 14.41 V

BQ BQ

CQ

EQ

CEQ

CEQ

I I

I

IVV

−= ⇒ =+

=

== − −=

5.45 (a)

( )( )

( ) ( ) ( )

( )

( ) ( )

1 2

2

1 2

2 21 2

1 2

2 22

0.4 3 37.5 ; 7.5

0.4 0.49 112.5

0.2 0.4

1

112.5 0.4, 0.004 112.5 100

112.59 0.004112.5 112.5

CQ EQ

C C E E

TH CC BQ TH BE BQ E

TH BQ

I I mA

R R k R R k

R R k

RV V I R V on I RR R

R RR RR I mAR R

R RR

β

≅ =

= ⇒ = Ω = ⇒ = Ω

+ ≅ = Ω

⎛ ⎞= = + + +⎜ ⎟+⎝ ⎠

−= = = =

+

−⎡⎛ ⎞ =⎜ ⎟⎝ ⎠ ⎣

( )( )( )0.7 101 0.004 7.5⎤

+ +⎢ ⎥⎦

We obtain ( ) 5 22 2 20.08 0.004 3.56 10 3.73R R R−= − × +

From this quadratic, we find 2 148 64.5 R k R k= Ω ⇒ = Ω (b) Standard resistor values:

1 2

1 2

Set 7.5 and 62 , 47 Now 62 47 26.7

E C

TH

R R k R k R kR R R k

= = Ω = Ω = Ω= = = Ω

( ) ( )

( ) ( )

2

1 2

47 9 3.88 47 621

TH CC

TH BQ TH BE BQ E

RV V VR R

V I R V on I Rβ

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= + + +

( )( )

( )( )

3.88 0.7So 0.00406 26.7 101 7.5

Then 0.406 0.406 7.5 3.05

BQ

CQ

RC RE

I mA

I mAV V V

−= =+

== = =

5.46 (a)

( ) ( )

1 2

2

1 2

12 2 1.714 K

210 5 10 5 3.571 V14

TH

TH TH

R R R

RV VR R

= = =

⎛ ⎞ ⎛ ⎞= − = − ⇒ = −⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠

(b) ( ) ( )

( ) ( ) ( )

( )( )

( )( ) ( )( )

on 1 53.57 1.714 0.7 101 0.5 5

5 0.7 3.571 0.729 13.96 A1.714 101 0.5 52.211.396 mA, 1.410 mA10 1.396 5 1.41 0.5 2.32 V

TH BQ TH BE BQ E

BQ BQ

BQ

CQ EQ

CEQ CEQ

V I R V I RI I

I

I IV V

β

μ

= + + + −− = + + −

− −= = ⇒+

= == − − ⇒ =

(d)

( )( )

( )( ) ( )( )

0.5 5% 0.525 K 5 5% 5.25 K0.729 13.32 A

1.714 101 0.5251.332 mA 1.345 mA

10 1.332 5.25 1.345 0.52510 6.993 0.7061 2.30 V

0.5 5% 0.525 K 5 5% 4.75 K1.332 mA 1.345 mA

E C

BQ

CQ EQ

CEQ

CEQ

E C

CQ EQ

C

R R

I

I I

VV

R RI I

V

μ

= + = = + =

= ⇒+

= =

= − −= − − ⇒ =

= + = = − == =

( )( ) ( )( )

( )( )

( )( ) ( )( )

10 1.332 4.75 1.345 0.52510 6.327 0.7061 2.97 V

0.5 5% 0.475 K 5 5% 5.25 K0.729 14.67 A

1.714 101 0.4751.467 mA 1.482 mA

10 1.467 5.25 1.482 0.47510 7.70175 0.70395 1.59 V

EQ

CEQ

E C

BQ

CQ EQ

CEQ

CEQ

V

R R

I

I I

VV

μ

= − −= − − ⇒ =

= − = = + =

= ⇒+

= =

= − −= − − ⇒ =

( )( ) ( )( )

0.5 5% 0.475 K 5 5% 4.75 K1.467 mA 1.482 mA

10 1.467 4.75 1.482 0.47510 6.96825 0.70395 2.33 V

E C

CQ EQ

CEQ

CEQ

R RI I

VV

= − = = − == =

= − −= − − ⇒ =

5.47

( ) ( )

( )

1 2

2

1 2

9 1 0.91 k

112 12 1.2 V1 9

on 0

TH

TH

EQ E EB BQ TH TH

R R R

RVR R

I R V I R V

= = = Ω

⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠+ + + =

( ) ( )( )(on) 1.2 0.7

1 0.90 76 0.10.0588, 4.41 mA

4.47 mA

TH EBBQ

TH E

BQ CQ

EQ

V VIR R

I I

I

β− − −= =

+ + += =

=

( ) ( ) ( )

Center of load line 6 V

12 04.47 0.1 6 4.41 12 1.26 k

ECQ

EQ E ECQ CQ C

C C

V

I R V I RR R

⇒ =

+ + − =+ + = ⇒ = Ω

5.48 (a)

( )

( )( )

( )( ) ( )( )

36 68 23.5 K36 10 3.46 V

36 683.46 0.7 0.00178 mA

23.5 51 300.0888 mA 0.0906 mA

10 0.0888 42 0.0906 3010 3.73 2.72 3.55 V

TH

TH

BQ

CQ EQ

CE

CE

R

V

I

I I

VV

= =

⎛ ⎞= =⎜ ⎟+⎝ ⎠−= =

+= =

= − −= − − ⇒ =

(b)

( )( )

( )( ) ( )( )

1 222.7, 12 K, 14 K, 10 K7.85 k 3.46

3.46 0.7 0.00533 mA7.85 51 100.266 mA 0.272 mA10 0.266 14 0.272 103.56 V

C E

TH TH

BQ

CQ EQ

CE

CE

R R R RR V

I

I IVV

= = = == =

−= =+

= == − −=

5.49 (a)

( ) ( )

( ) ( ) ( )

2

1 2

6836 68 23.5 K 10 5 10 5 1.54 V36 68

5 51 30 0.7 23.5 1.54

TH TH

BQ B

RR VR R

I I

⎛ ⎞ ⎛ ⎞= = = − = − =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= + + +

( )( ) ( )( )

2.76 1.78 μA 0.0888 mA1553.5

0.0906 mA

10 0.0906 30 0.0888 4210 2.718 3.7296 3.55 V

BQ CQ

EQ

ECQ

ECQ

I I

I

VV

= = ⇒ =

=

= − −= − − ⇒ =

(b)

( ) ( ) ( )

( )( ) ( )( )

12 22.7 7.85 K1.54 10 K 14 K

5 51 10 0.7 7.85 1.54

2.76 5.33 A 0.266 mA517.85

0.272 mA

10 0.272 10 0.266 1410 2.72 3.724 3.56 V

TH

TH E C

BQ B

BQ CQ

EQ

ECQ

ECQ

RV R R

I I

I I

I

VV

μ

= == = =

= + + +

= = =

=

= − −= − − ⇒ =

5.50 (a)

( )( ) ( )( )( )

( ) ( )1

0.1 1 0.1 101 0.5 5.05 1 1

0.8 0.008 100

TH E

TH TH CC BQ TH BE BQ E

CQBQ

R R k

V R V I R V on I RRI

I mA

β

β

β

= + = = Ω

= ⋅ ⋅ = + + +

= = =

( )( ) ( )( ) ( )( )( )

( )

1

21 2

2

1Then 5.05 10 0.008 5.05 0.7 101 0.008 0.5

44.1or 44.1 , 5.05 5.70 44.1

101Now 0.8 0.808 100EQ

RRR k R kR

I mA

= + +

= Ω = ⇒ = Ω+

⎛ ⎞= =⎜ ⎟⎝ ⎠

( ) ( )( )10 0.8 5 0.808 0.55.75

CC CQ C CEQ EQ E

C

C

V I R V I RR

R k

= + += + += Ω

(b) For 75 150β≤ ≤

( ) ( )

( ) ( )

2

1 2

5.7 10 1.145 5.7 44.11

TH CC

TH BQ TH BE BQ E

RV V VR R

V I R V on I Rβ

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= + + +

For ( )( )

1.145 0.775, 0.0103 5.05 76 0.5BQI mAβ −= = =

+

Then ( )( )75 0.0103 0.775 CQI mA= =

For ( )( )

1.145 0.7150, 0.00552 5.05 151 0.5BQI mAβ −= = =

+

Then 0.829 CQI mA=

0.829 0.775% Change 100% % Change 6.75%0.80

CQ

CQ

II

Δ −= = × ⇒ =

(c) For 1 ER k= Ω

( )( )( )

( )( ) ( )( ) ( )( )( )1 1

1

0.1 101 1 10.1 1 1 10.1 10 0.008 10.1 0.7 101 0.008 1

which yields 63.6

TH

TH TH CC

R k

V R VR R

R k

= = Ω

= ⋅ ⋅ = = + +

= Ω

And 22

2

63.6 10.1 12.0 63.6

R R kR

= ⇒ = Ω+

Now ( ) ( )2

1 2

12 10 1.587 12 63.6TH CC

RV V VR R

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

For ( )( )

1.587 0.775, 0.0103 10.1 76 1BQI mAβ −= = =

+

So 0.773 CQI mA=

For ( )( )

1.587 0.7150, 0.00551 10.1 151 1BQI mAβ −= = =

+

Then 0.826 CQI mA=

0.826 0.773% Change 100% % Change 6.63%0.8

CQ

CQ

II

Δ −= = × ⇒ =

5.51

( )( )( )10 0.8 5 6.25

CC CQ C E CEQ

C E C E

V I R R VR R R R k

≅ + += + + ⇒ + = Ω

Let 0.875 ER k= Ω Then, for bias stable ( )( )( )0.1 121 0.875 10.6 THR k= = Ω

( )( ) ( )( ) ( )( )( )1

0.8 0.00667 120

1 10.6 10 0.00667 10.6 0.7 121 0.00667 0.875

BQI mA

R

= =

= + +

So 1 71.8 R k= Ω and 22

2

71.8 10.6 12.4 71.8

R R kR

= ⇒ = Ω+

Then 10 0.119 71.8 12.4RI mA≅ =

+

This is close to the design specification. 5.52

( )( )12 2 0.26

CC CQ C ECQ EQ CEQ

CQ

V I R RI I VI

= − +≈ ⇒= − +

2.73 mA, 0.0218 mA

6 VCQ BQ

CEQ

I I

V

= =

=

( ) ( )

( )21 2

1 2

on 1 6

12 6,

TH BQ TH BE BQ E

TH TH

V I R V I R

RV R R R

R R

β= + + + −

⎛ ⎞= − =⎜ ⎟+⎝ ⎠

( )( ) ( )( )( )

( )( )1

Bias stable 0.1 1 0.1 126 0.2 2.52 k

1 12 6

TH E

TH TH

R R

V RR

β⇒ = + = = Ω

⎛ ⎞= −⎜ ⎟⎝ ⎠

( )( ) ( )( ) ( )( )( )

( )1

1

21

2

2

1 2.52 12 6 0.0218 2.52 0.7 126 0.0218 0.2 6

1 30.24 0.7549 0.5494

23.2R23.2 k , 2.5223.2 R

2.83 k

R

R

R

R

− = + + −

= +

= Ω =+

= Ω

5.53

a. ( )

( )( ) ( )( )

811 mA. 1 1.01 mA80

12 1 2 1.01 0.2 9.80 V

1 0.0125 mA80

CQ EQ

CEQ CEQ

BQ

I I

V V

I

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= − − ⇒ =

= =

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

2

1 2 1 1

0.1 1 0.1 81 0.2 1.62 k

1 112 6 12 6 19.44 6

on 1 6

TH E

TH TH

TH BQ TH BE BQ E

R R

RV R

R R R RV I R V I R

β

β

= + + = = Ω

⎛ ⎞= − = − = −⎜ ⎟+⎝ ⎠= + + + −

( ) ( )( ) ( )( )( )

( )1

1

21

2

2

1 19.44 6 0.0125 1.62 0.7 81 0.0125 0.2 6

1 19.44 0.923

21.121.1 k , 1.6221.1

1.75 k

R

RRRR

R

− = + + −

=

= Ω =+

= Ω

b. 1 1

2 2

22.2 k or 20.0 k1.84 k or 1.66 k

R RR R

= Ω = Ω= Ω = Ω

( ) ( )2 1max , minR R

( ) ( )

( )

( )( )( )

1.84 20.0 1.685 k1.84 12 6 4.99 V

1.84 20.04.99 0.7 6 0.31 0.0173 mA

1.685 81 0.2 17.891.39 mA, 1.40 mA

TH

TH

BQ

CQ EQ

R

V

I

I I

= = Ω

⎛ ⎞= − = −⎜ ⎟+⎝ ⎠− − − −

= = =+

= =

( )( ) ( )( )For max, 12 1.39 2 1.40 0.28.94 V

C CE

CE

R VV

⇒ = − −=

( ) ( )2 1min , maxR R

( ) ( )

( )

( )( )( )( ) ( )( )

1.66 22.2 1.545 k1.66 12 6 5.165 V

1.66 22.25.165 0.7 6 0.135 0.00761 mA 0.609 mA, 0.616

1.545 81 0.20 17.74512 0.609 2 0.616 0.2

10.7 V

TH

TH

BQ CQ E

CEQ

CEQ

R

V

I I I

VV V

= = Ω

⎛ ⎞= − = −⎜ ⎟+⎝ ⎠− − += = = ⇒ = =

+= − −=

So 0.609 1.39mA8.94 10.7 V

C

CEQ

IV

≤ ≤≤ ≤

5.54

( )( )5 12 3 2.33

CEQ CC CQ C E

C E C E

V V I R RR R R R k

≅ − += − + ⇒ + = Ω

Let 0.333 ER k= Ω and 2 CR k= Ω Nominal value of 100β =

( )( ) ( )( )( )

( ) ( )( )1 1

0.1 1 0.1 101 0.333 3.36 3 0.03

1001 112 6 3.36 12 6

TH E

BQ

TH TH

R R k

I mA

V RR R

β= + = = Ω

= =

= ⋅ ⋅ − = −

( ) ( )( )( ) ( )( ) ( )( )( )

1

1 2

Then 1 61 3.36 12 6 0.03 3.36 0.7 101 0.03 0.333 6

which yields 22.3 and 3.96

TH BQ TH BE BQ EV I R V on I R

RR k R k

β= + + + −

− = + + −

= Ω = Ω

( ) ( )

( ) ( )

( ) ( )( )

2

1 2

3.96Now 12 6 12 6 or 4.19 3.96 22.3

For 75, 1 66 0.7 4.19 6 0.7 0.0387 2.90

1 3.36 76 0.333

TH TH

TH BQ TH BE BQ E

THBQ C

TH E

RV V VR R

V I R V on I RVI mA I mA

R R

β β

β

⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= = + + + −

+ − − + −= = = ⇒ =+ + +

For ( )( )

4.19 6 0.7150, 0.0207 3.36 151 0.333BQI mAβ − + −= = =

+

Then 3.10 CI mA= Specifications are met. 5.55

( )

1 2

2

1 2

3 12 2.4

12 20 16 12 3

TH

TH CC

R R R k

RV V VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

(a) For 75β = ( ) ( )

( )( )

( )( ) ( )( )

20 120 0.7 16 76 2 2.4

So 0.0214 , 1.60 , 1.62 20 1.6 1 1.62 2 or 15.16

BQ E EB BQ TH TH

BQ

BQ CQ EQ

ECQ ECQ

I R V on I R VI

I mA I mA I EV V V

β= + + + +− − = +⎡ ⎤⎣ ⎦

= = == − − =

(b) For 100,β = we find 0.0161 , 1.61 , 15.13 , 1.63 BQ CQ ECQ EQI mA I mA V V I mA= = = = 5.56

( )( ) ( )( )( ) ( )( )( )

( ) ( )

( )( )1 1

4.8 mA 4.84 mA

6 18 4.8 2 4.84 0.496 k

0.1 1 0.1 121 0.496 6.0 kon 1

0.040 mA1 1 6.0 18

CQ EQ

CEQ CC CQ C EQ E

E E

TH E

TH BQ TH BE BQ E

BQ

TH TH CC

I IV V I R I R

R R

R RV I R V I RI

V RR R

V

ββ

= → == − −

= − − ⇒ = Ω

= + = = Ω= + + +=

= ⋅ =⋅

( )( ) ( )( ) ( )( )( )

( )1

1

21

2

2

1 6.0 18 0.04 6.0 0.70 121 0.04 0.496

1 108 3.34

32.332.3 k , 6.032.3

7.37 k

R

RRRR

R

= + +

=

= Ω =+

= Ω

5.57 For nominal 70β =

( )( ) ( )( )( ) ( )( )( )

( )

2 0.0286 mA 2.03 mA70

10 20 2 4 2.03 0.985 K

0.1 1 0.1 71 0.985 6.99 Kon

BQ EQ

CEQ CC CQ C EQ E

E E

TH E

TH BQ TH BE EQ E

I I

V V I R I RR R

R RV I R V I R

β

= = → =

= − −= − − ⇒ =

= + = == + +

( )

( )( ) ( )( ) ( )( )

( )

1

1

1

21

2

2

1 on

1 6.99 20 0.0286 6.99 0.70 2.03 0.985

1 139.8 2.90

48.248.2 K, 6.9948.2

8.18 K

TH CC BQ TH BE EQ ER V I R V I RR

R

RRRR

R

⋅ ⋅ = + +

= + +

=

= =+

=

Check: For 50β =

( )

( )( ) ( )( )

8.18 20 2.908.18 48.2

on 2.90 0.7 0.0384 mA1 6.99 51 0.985

1.92 mA

TH

TH BEBQ

TH E

CQ

V

V VI

R RI

β

⎛ ⎞= =⎜ ⎟+⎝ ⎠− −= = =

+ + +=

For 90β =

( )( )2.90 0.7 0.0228 mA

6.99 91 0.9852.05 mA

BQ

CQ

I

I

−= =+

=

Design criterion is satisfied.

5.58

( )( ) ( )( )( ) ( )( )( )

( )

( )( ) ( )( ) ( )( )1

1

1

1 mA 1.017 mA

5 15 1 5 1.017 4.92 k

Bias stable: 0.1 1 0.1 61 4.92 30.0 k1 0.0167 mA601 on

1 30.0 15 0.0167 30.0 0.70 1.017 4.92

1 4

CQ EQ

CEQ CC CQ C EQ E

E E

TH E

BQ

TH TH CC BQ TH BE EQ E

I IV V I R I R

R R

R R

I

V R V I R V I RR

R

R

β

= → == − −

= − − ⇒ = Ω

= + = = Ω

= =

= ⋅ ⋅ = + +

= + +

( )

21

2

2

48.5 6.197

72.572.5 k , 30.072.5

51.2 k

RRR

R

=

= Ω =+

= Ω

Check: For 45β =

( )

( )( ) ( )( )

51.2 15 6.21V51.2 72.5

on 6.21 0.7 0.0215 mA1 30 46 4.92

0.967 mA, 3.27%

TH

TH BEBQ

TH E

CCQ

C

V

V VI

R RII

I

β

⎛ ⎞= =⎜ ⎟+⎝ ⎠− −= = =

+ + +Δ

= =

Check: For 75β =

( )( )6.21 0.7 0.0136 mA

30.0 76 4.92

1.023 mA, 2.31%

BQ

CCQ

C

I

II

I

−= =+

Δ= =

Design criterion is satisfied. 5.59 (a)

( )( )( )3 0.1 5 1.4 2.67

10013.3 , 0.833 120

CC CQ C E CEQ

E E E

C BQ

V I R R VR R R k

R k I Aμ

≅ + += + + ⇒ = Ω

= Ω = =

( )( ) ( )( )( )( )( )

( ) ( )( )( ) ( )( )( )

1 1

1 2

0.1 1 0.1 121 2.67 32.31 1 32.3 3

10.000833 32.3 0.7 121 0.000833 2.67

which gives 97.3 , and 48.4

TH E

TH TH CC

BQ TH BE BQ E

R R k

V R VR RI R V on I R

R k R k

β

β

= + = = Ω

= ⋅ ⋅ =

= + + += + +

= Ω = Ω

(b)

( ) ( )( )

1 2

3 3 20.6 97.3 48.4

100

100 20.6 3

or 362

R

CQ

CQ R CC

I AR R

I A

P I I V

P W

μ

μ

μ

≅ = ⇒+ +

=

= + = +

=

5.60

( )( ) ( )( )( )

( ) ( )

( )

( )( ) ( )( ) ( )( )

( )( )

1 2

2

1 2 1

1

11

2

2

5 5 1.67 mA3

|| 0.1 1 0.1 101 3 30.3 k

14 2 4 2

0.0165 mA1

5 on15 1.67 3 0.7 0.0165 30.3 30.3 4 2

10.80 30.3 4 152 k

152 30152

EE

E

TH E

TH TH

EQBQ

EQ E EB B TH TH

VIR

R R R R

RV RR R RI

I

I R V I R V

R

RR

RR

β

β

−= = =

= = + = = Ω

⎛ ⎞= − = ⋅ ⋅ −⎜ ⎟+⎝ ⎠

= =+

= + + +

= + + + −

= ⇒ = Ω

=+ 2.3 37.8 kR⇒ = Ω

5.61 a.

( ) ( )

1 2

2

1 2

10 20 6.67 k

2010 5 10 5 1.67 V20 10

TH TH

TH TH

R R R R

RV VR R

= = ⇒ = Ω

⎛ ⎞ ⎛ ⎞= − = − ⇒ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

b. ( ) ( )

( )( )

( )( )

( )( )

10 1 on10 0.7 1.67 7.63 0.0593 mA6.67 61 2 128.73.56 mA, =3.62 mA

10 10 3.62 22.76 V

10 3.56 2.2 102.17 V

BQ E EB BQ TH TH

BQ BQ

CQ EQ

E EQ E

E

C CQ C

C

I R V I R V

I I

I I

V I RV

V I RV

β= + + + +− −= = ⇒ =

+=

= − = −=

= − = −= −

5.62

( )( )( ) ( )20 0.5 8 24

CQ C E ECQ

C E C E

V V I R R VR R R R k

+ −− ≅ + += + + ⇒ + = Ω

Let 10 ER k= Ω then 14 CR k= Ω Let 60β = from previous problem.

( ) ( ) ( )( )( )0.1 1 0.1 61 10TH ER Rβ= + = Or 61 THR k= Ω

( )2

1 2 1

0.5 0.00833 60

110 5 10 5

BQ

TH TH

I mA

RV RR R R

= =

⎛ ⎞= − = ⋅ ⋅ −⎜ ⎟+⎝ ⎠

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )1

Now 10 1110 61 0.00833 10 0.7 0.00833 61 61 10 5

BQ E EB BQ TH THI R V on I R V

R

β= + + + +

= + + + −

1 2

1 2

Then 70.0 and 474

10 10 18.4 70 474R

R k R k

I AR R

μ

= Ω = Ω

≅ = ⇒+ +

So the 40 Aμ current limit is met. 5.63 a.

( ) ( )

1 2

2

1 2

35 20 12.7 k

207 5 7 5 2.45 V20 35

TH TH

TH TH

R R R R

RV VR R

= = ⇒ = Ω

⎛ ⎞ ⎛ ⎞= − = − ⇒ = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

b. ( ) ( )( )

( )( )

on 101

2.45 0.7 10 0.136 mA12.7 76 0.510.2 mA, 10.4 mA

TH BEBQ

TH E

BQ

CQ EQ

V VI

R R

I

I I

β− − −

=+ +

− − += ⇒ =+

= =

( )( ) ( )( )2020 10.2 0.8 10.4 0.56.64 V

CEQ CQ C EQ E

CEQ

V I R I R

V

= − −= − −=

c.

( )

( )

( )( )( )

2

1

1 2

2

1 2

20 5% 21 k35 5% 33.25 k0.5 5% 0.475 k

21 33.25 12.9 k

7 5

21 7 5 2.29 V21 33.252.29 0.7 10

0.143 mA12.9 76 0.47510.7 mA, 10.9 mA

E

TH

TH

BQ

CQ EQ

RRR

R R R

RV

R R

I

I I

= + = Ω= − = Ω= − = Ω= = = Ω

⎛ ⎞= −⎜ ⎟+⎝ ⎠⎛ ⎞= − = −⎜ ⎟+⎝ ⎠− − − −

= =+

= =

For 0.8 5% 0.84 kCR = + = Ω

( )( ) ( )( )20 10.7 0.84 10.9 0.475 5.83 VCEQ CEQV V= − − ⇒ = For 0.8 5% 0.76 kCR = − = Ω

( )( ) ( )( )

( )

( )( )( )

2

1

1 2

20 10.7 0.76 10.9 0.475 6.69 V20 5% 19 k35 5% 36.75 k0.5 5% 0.525 k

19 36.75 12.5 k19 7 5 2.61 V

19 36.752.61 0.7 10

0.128 mA12.5 76 0.5259.58 mA,

CEQ CEQ

E

TH

TH

BQ

CQ EQ

V VRRRR R R

V

I

I I

= − − ⇒ == − = Ω= + = Ω= + = Ω= = = Ω

⎛ ⎞= − = −⎜ ⎟+⎝ ⎠− − − −

= =+

= = 9.70 mA

For 0.84 kCR = Ω

( )( ) ( )( )20 9.58 0.84 9.70 0.525 6.86 VCEQ CEQV V= − − ⇒ = For 0.76 kCR = Ω

( )( ) ( )( )20 9.58 0.76 9.70 0.525 7.63 VCEQ CEQV V= − − ⇒ = So 9.58 10.7 mACQI≤ ≤ and 5.83 7.63 VCEQV≤ ≤ 5.64 a.

( )

( )

500 k 500 k 70 k 250 k 70 k 54.7 k

55 3500 500 705 3 5 1 1 1 0.0554 0.0183

500 500 70 500 500 703.03 V

TH TH

THTH TH

TH TH

TH

R R

VV V

V V

V

= Ω Ω Ω = Ω Ω ⇒ = Ω

− −− −+ =

⎛ ⎞+ − = + + − =⎜ ⎟⎝ ⎠

= −

b. ( ) ( )

( )

( )( )

on 51

3.03 0.7 554.7 101 50.00227 mA

0.227 mA, 0.229

TH BEBQ

TH E

BQ

CQ EQ

V VI

R R

I

I I

β− − −

=+ +

− − +=+

=

= =

( )( ) ( )( )20 0.227 50 0.229 57.51 V

CEQ

CEQ

VV

= − −=

5.65

30 || 60 || 20 10 k

5 530 60 205 5 1 1 1

30 60 30 60 202.5 V

TH TH

TH TH TH

TH

TH

R R

V V V

V

V

= ⇒ = Ω

− −+ =

⎛ ⎞ ⎛ ⎞+ = + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

For 100β =

( ) ( )( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )( )

on 51

2.5 0.7 510 101 0.20.225 mA22.5 mA, 22.7 mA

15 22.5 0.5 22.7 0.20.79! In saturation 0.2 V

52.5 5 0.7 10 0.2

6.8 10 0

TH BEBQ

TH E

BQ

CQ EQ

CEQ

CEQ CEQ

TH BQ TH BE EQ E

BQ EQ

BQ EQ

V VI

R R

II I

VV V

V I R V I RI I

I I

β− − −

=+ +

− +=+

== =

= − −= − ⇒ =

= + + −+ − = +

= + ( )( ) ( )

.2

14.8 0.5 0.2CQ EQI I= +

( ) ( ) ( )( ) ( )

( ) ( )( ) ( )

( )

Transistor in saturation,

6.8 10 0.2 0.26.8 10.2 0.251 14.8 0.2 0.7754.8 10.2 35.7748 35.5

21.1 mA0.2 V

EQ BQ CQ

BQ BQ CQ

BQ CQ

BQ CQ

BQ CQ

CQ

CQ

CEQ

I I I

I I II I

I II I

IIV

= +

= + += +

× = += +

===

5.66

50 A, 0.625 A, 50.6 ACQ BQ EQI I Iμ μ μ= = = (a)

( ) ( )( )

1 19.8 K0.0506

5 0.050 5 0.0506 19.8 580 K

E

C

C

R

RR

= =

= + + −=

( )( )( )

( ) ( )

1 2

2

1 2 1

Design bias stable circuit.0.1 51 19.8 101 K

110 5 10 5

TH

TH

TH TH

R R RR

RV RR R R

== =

⎛ ⎞= − = ⋅ ⋅ −⎜ ⎟+⎝ ⎠

( ) ( ) ( ) ( ) ( )1

1So 101 10 5 101 0.7 0.0506 19.8 5BQIR

− = + + −

( )1

21

2

2

1 1010 0.0631 0.7 1

573 573 K 101573

123 K

RRRR

R

= + +

= =+

=

(b)

( )( )( )

( )( ) ( )( )

101 K, 3.23 V0.7 121 19.8 5

1.07 101 2395.8 0.429 A0.0514 mA, 0.0519 mA10 0.0514 80 0.0519 19.810 4.11 1.03 4.86 V

TH TH

TH BQ TH BQ

BQ BQ

CQ EQ

CEQ

CEQ

R VV I R I

I II I

VV

μ

= = −= + + −= + ⇒ == == − −= − − ⇒ =

5.67 (a)

( )

20.80 mA, 2.5 K0.8

1212 0.8 7 2 3.75 K

EQ E

CQ C CEQ EQ E

C C

TH BQ TH BE EQ E

I R

I R V I RR R

V I R V I R

= = =

= + += + + ⇒ =

= + +

( ) ( )( )

( )( ) ( )( )

( )1 1

11

22

2

For a bias stable circuit 0.1 12.1 2.5 30.25 K

1 1 30.25 12 0.00667 30.25 0.7 2

1 363 2.90 125 K

125 30.25 39.9 K125

TH

TH CC

R

R VR R

RR

R RR

= =

⋅ ⋅ = = + +

= ⇒ =

= ⇒ =+

(b) Let 2.4 K, 3.9 KE CR R= =

1 120 KR = 2 39 KR = Then

( )1 2 120 39 29.4 K

39 12 2.94 V120 39

TH

TH

R R R

V

= = =

⎛ ⎞= =⎜ ⎟+⎝ ⎠

( )( )2.94 0.7 2.24 7.00 A

29.4 121 2.4 319.80.841 m 0.848 mA

BQ

CQ EQ

I

I I

μ−= = ⇒+

= Α =

( ) ( ) ( )( )12 0.841 3.9 0.848 2.4 12 3.28 2.046.68 V

CEQ

CEQ

VV

= − − = − −=

5.68 (a)

( )( ) ( )

100 A, 1.18 A, 101 A2 19.8 K

0.1019 0.101 19.8 6 0.1 9

100 K

CQ BQ EQ

E E

C

C

I I I

R R

RR

μ μ μ= = =

= ⇒ =

= + + −=

( )( )( )

( ) ( )( )

( )( ) ( )( ) ( )( )

2

1 2 1

1

21

2

2

Design a bias stable circuit. 0.1 86 19.8 170 K

118 9 170 18 9

19 0.101 19.8 0.7 0.00118 170 170 18 9

203203 K 170203

1046 K

TH

TH

B

RVR R R

RRRR

R

= =

⎛ ⎞= − = −⎜ ⎟+⎝ ⎠

= + + + −

= =+

=

(b) 125β =

( ) ( ) ( ) ( )

( )( ) ( )( )EQ

9 126 19.8 0.7 170 15.07 92.23 0.837 A 0.1046 mA

2664.8I 0.1054mA

18 0.1046 100 0.1054 19.85.45 V

BQ BQ

BQ CQ

ECQ

ECQ

I I

I I

VV

μ

= + + + −

= = =

== − −=

5.69 (a)

( )51 320 20.4 mA 0.147 K50 20.4

618 9 3 6 V 0.3 K20

EQ E

RC C

I R

V R

⎛ ⎞≈ = ⇒ = =⎜ ⎟⎝ ⎠

= − − = = =

( )( )( )

( )( ) ( )( )

( )

( )

1

11

22

2

For a bias stable circuit. 0.1 51 0.147 0.750 K

118 3 0.7 0.4 0.75 0.75 18

114 13.5 0.964 K

0.9640.75 3.38 K

0.964

TH

EQ E EB BQ TH TH

RV I R V I R V

R

RR

RR

R

+

= == + + +

= + + +

= ⇒ =

= ⇒ =+

(b) Let 0.15 K, 0.3 KE CR R= =

( )

( )( )

( )( ) ( )( )

1 21.0 K, 3.3 K3.381 13.3 0.767 K, 18 13.8 V

1 3.3818 13.8 0.7 3.5 0.416 mA

0.767 51 0.15 8.41720.8 mA, 21.2 mA18 20.8 0.3 21.2 0.1518 6.24 3.188.58 V

TH TH

BQ

CQ EQ

ECQ

ECQ

R R

R V

I

I IV

V

= =

⎛ ⎞= / = = =⎜ ⎟+⎝ ⎠− −= = =+

= == − −= − −=

5.70

( ) ( )

( )( ) ( ) ( )

1 2

2

1 2

11

1

1 E1

100 40 28.6 k

4010 10 2.86 V40 100

on 2.86 0.71 28.6 121 1

0.0144 mA

1.73 mA, 1.75 mA

TH

TH

TH BEB

TH E

B

C

R R R

RV

R R

V VI

R RI

I I

β

= = = Ω

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠− −= =

+ + +=

= =

( ) ( )

( ) ( )

( ) ( )( )

21 2

22

2 21

2

2 2

103

on 105

10 0.7 103 121 5

10 9.3 1 11.733 605 3 121 5

1.588 0.335 4.74 V

BC B

B BEE

B BC

B

B B

VI I

V VI

V VI

V

V V

−= +

− − −=

− − += +

⎛ ⎞− − = +⎜ ⎟⎜ ⎟

⎝ ⎠= ⇒ =

( )2 2

2

2

4.74 0.7 102.808 mA

50.0232 mA

2.785 mA

E E

B

C

I I

I

I

− − −= ⇒ =

=

=

( ) ( )( )

1 1

2 2

4.74 1.75 1 2.99 V

10 4.74 0.7 5.96 VCEQ CEQ

CEQ CEQ

V V

V V

= − ⇒ =

= − − ⇒ =

5.71

( )

( )

1

1

2

2 2

2

2

0.70.7 5

0.215 mA20

0.7 0.7 1.41.4 5

3.6 mA1

0.0444 mA

3.56 mA

E

R

E

E E

B

C

V

I

V

I I

I

I

= −− − −

= =

= − − = −− − −

= ⇒ =

=

=

1 1 2

1

1

1

0.215 0.04440.259 mA

0.00320 mA

0.256 mA

E R B

E

B

C

I I II

I

I

= + = +=

=

=

5.72 Current through V − source 1 2E EI I= + and ( ) ( )( )1 2 11 51 8.26 AE E BI I Iβ μ= = + =

So total current ( )( )2 51 8.26 A=843 Aμ μ=

( )( )0.843 5 4.22 mWP I V P− − −= ⋅ = ⇒ =

(From V − source) From Example 5.19, 0.413 mAQI =

So ( )050 0.413 0.405 mA51CI

⎛ ⎞= =⎜ ⎟⎝ ⎠

( )( )0.405 5 2.03 mWP I V P+ + += ⋅ = ⇒ =

(From V + source) 5.73

( )

( )

( )

1 2

2

1 2

1 1 1

50 100 33.3 k

10 5

100 10 5 1.67 V100 50

5 on

TH

TH

E E EB B TH TH

R R R

RVR R

I R V I R V

= = = Ω

⎛ ⎞= −⎜ ⎟+⎝ ⎠⎛ ⎞= − =⎜ ⎟+⎝ ⎠

= + + +

( )

( ) ( )( )

1

1

1

1

101 0.8 0.808 mA100

0.008 mA5 0.808 0.7 0.008 33.3 1.67

2.93 k

E

B

E

E

I

IR

R

⎛ ⎞= =⎜ ⎟⎝ ⎠

== + + += Ω

( ) ( )

( )

1

1 1 1

2

2 22

5 0.808 2.93 2.63 V2.63 3.5 0.87 V

0.87 0.70 1.57 V1.57 5

0.808 4.25 k

E

C E ECQ

E

E EE

VV V VV

I RR

= − == − = − = −= − − = −

− − −= = ⇒ = Ω

2 2

2 2

4 1.57 4 2.43 V5 2.43 3.21 k

0.8

CEQ C

C C

V V

R R

= ⇒ = − + =−= ⇒ = Ω

( )1 1 2

1 1

0.8 0.008 0.792 mA0.87 5

5.21 k0.792

RC C B

C C

I I I

R R

= − = − =− − −

= ⇒ = Ω