Chapter 8 Problem Solutions 8.1 a.

b. i. 80 VD DV =

Maximum power at 40 V2

25 0.625 A40

80 40 64 0.625

D DDS

TD

D S

D D

VV

PIV

R R

= =

= = =

−= ⇒ = Ω

ii. 50 VD DV =

Maximum power at 25 V2

25 1 A25

50 25 25 1

D DD S

TD

DS

D D

VV

PIV

R R

= =

= = =

−= ⇒ = Ω

8.2 a.

(max)2

2 (max) 2(20)So 1.67 A24

( / 2) 24 12 7.2 1.67

1.67 20.8 mA80

24 0.7 1.12 k20.8

C CQ C Q

QC Q

CC

CC CCL L

CQ

CQB

B B

VP I

PI

VV VR R

I

II

R R

β

= ⋅

= = =

− −= = ⇒ = Ω

= = ⇒

−= ⇒ = Ω

b. ( )( )

00

1.67 7.2462

0.026(max) 12(max) 12 V 26 mV

462

CQ Lv m L

T

P Pv

I RA g R

VVV V V

A

⋅= = = =

= ⇒ = = ⇒ ≅

8.3

a. For maximum power delivered to the load, set 2C C

C EQ

VV =

( )susSet 25 VC C CEV V= =

( ) ( )( )

,max

D,max

25Then 0.1

250 mA25 12.5 125 mA

0.1

max 0.125 12.52

1.56 W P125 1.25 mA10025 0.7 19.4 k

1.25

CCCm

L

Cm C

CQ

CCQ CQ

BQ

B B

VI

RI I

I

VP I

I

R R

= =

= <−= =

= ⋅ =

= <

= =

−= ⇒ = Ω

b. ( ) ( ) ( ) ( ) ( )221 1max 0.125 100 max 0.781 W rms2 2L CQ L LP I R P= ⋅ ⋅ = ⇒ =

8.4

Point (b): Maximum power delivered to load. Point (a): Will obtain maximum signal current output. Point (c): Will obtain maximum signal voltage output.

8.5 a.

b.

( )( )( )

( )

2

2

2

2

5 V, 0.25 5 4 0.25 A, 37.5 V, 9.375 W

6 V, 0.25 6 4 1.0 A, 30 V, 30 W

7 V, 0.25 7 4 2.25 A, 17.5 V, 39.375 W

8 V, 0.25 2 8 4

402.92

103.

GG D D S

GG D D S

GG D D S

GG D D S D S

D SD S

D

V I V P

V I V P

V I V P

V I V V

VV

I

= = − = = =

= = − = = =

= = − = = =

⎡ ⎤= = − −⎣ ⎦−

= ⇒ =

=

( ) 2

71 A, 10.8 W

9 V, 0.25 2 9 4

401.88 V

103.81 A, 7.16 W

GG D D S D S

D SD S

D

P

V I V V

VV

I P

=

⎡ ⎤= = − −⎣ ⎦−

= ⇒ =

= =

c. Yes, at ,max7 V, 39.375 W > 35 WGG DV P P= = = 8.6 a.

( )

( )

2

2

1 2

1 2

22

1

Set 25 V2

50 25 1.25 A20

1.25 4 6.5 V0.2

Let 100 kΩ

6.5 50 13 kΩ100

87 kΩ

DDDSQ

DQ

DQ n GS TN

GS

GS DD

VV

I

I K V V

V

RV VR R

R RR R

R

= =

−= =

= −

+ = =

⎛ ⎞= ⎜ ⎟+⎝ ⎠

+ =

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

=

b. ( )( )1.25 25 31.25 WD DQ DSQ DP I V P= = ⇒ = c.

,max ,max

,max ,max

,max

2 2.5 A

50 V

31.25 W

D DQ D

DS DD DS

D

I I I

V V V

P

= ⇒ =

= ⇒ =

=

d.

( )( )( )( )( )

( )

0

022

0

2 2 0.2 1.25 1 /

1 20 0.5 10 V

101 1 2.5 W2 2 20

31.25 2.5 28.75 W

m Li

m n DQ

L LL

Q Q

Vg R

V

g K I A V

V

VP PR

P P

=

= = =

= =

= ⋅ = ⋅ ⇒ =

= − ⇒ =

8.7 (a)

(b) ( )( ),max 25D D jP P Slope T= − −

,max ,max60At 0, 25 145 C0.5D j jP T T= = + ⇒ = °

(c) ,max,max

j caseD

dev amb

T TP

θ −

−= or 145 25 2

60dev amb dev amb C/Wθ θ− −−= ⇒ = °

8.8

,max,rated

dev case

,max ambdev case

,rated

or

150 25 2.5 C/W50

θ

θ

−

−

−=

−=

−= = °

j ambD

j

D

T TP

T TP

( )( ) ( )

dev amb dev case case amb

case amb case amb

Then 150 25 2.5 125 2.5

θ θθ θ

− −

− −

− = +− = + ⇒ = +

D

D D

T T PP P

8.9

( )( )( )

( )( )( )

( )( )

dev amb dev case case snk snk amb

dev dev

dev case dev case

case dev case

case sink case snk

sink c

4 5 20 W

25 20 1.75 0.8 3 111 136 C

20 1.75 3535 136 35 101 C

20 0.8 16 C

θ θ θ

θ

θ

− − −

−

−

= ⋅ = =− = + +− = + + = ⇒ = °

− = ⋅ = == − = − ⇒ = °

− = ⋅ = = °=

D D DS

D

D

D

P I VT T PT T

T T PT T T

T T PT T ase sink16 101 16 85 C− = − ⇒ = °T

8.10

( )( )

dev amb dev case case amb

case amb case amb200 25 25 3 4 C/Wθ θ

θ θ− −

− −

− = +− = + ⇒ = °

DT T P

8.11

,max ambdev case

,rated

,max amb

dev case case snk snk amb

175 25 10 C/W15

175 25 10 W10 1 4

θ

θ θ θ

−

− − −

− −= = = °

−=

+ +−= ⇒ =

+ +

j

D

jD

D

T TPT T

P

P

8.12

( )2

12 50%

L

S

S CC Q

CCL P P Q

CC Q

CC Q

PP

P V IV

P V I I

V I

V I

η

η η

=

= ⋅

⎛ ⎞= ⋅ = ⎜ ⎟⎝ ⎠

⋅ ⋅= ⇒ =

⋅

8.13

( )( )

( ) ( )

( )

3 2

S

max 4.8 V0.7 5

4.3 mA1

min0.7 max 4.3 mA

1so 3.6 5.5 V min 4.3 V

o

C C

I o L

I o

v

i i

vv v i

v v

=− − −

= = =

= + = − =

− ≤ ≤ = −

8.14

( ) ( )

( )

( ) ( )( )( )

( )( )

( ) ( ) ( ) ( )

2 33 3

23 3

23

2

3

3 22

3 2

2 2

02 0

0 5K

12 0.5 52 11 2 0

11 11 4 12 22 12

1.072 V

12 1.072 0.5 3.93 mAsat 1.072 0.5 0.572 V

minmin : max 3.93 min 3.93 V

1mi

GSD GS TN

GS GS

GS GS

GS

GS GS

D D

DS GS TN

o

I

VI V V

RV V

V V

V

V V

I IV V V

Vv i V

v

− − −= − =

− = −− − =

± +=

= =

= = − == − = − =

= − = ⇒ = −

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

( )

( )( ) ( )

1

21 1 1

1

n min 3.93 0.5min 3.43 V

max 5 sat 5 0.572max 4.43 V

4.43max 3.93 8.36 mA1

8.36 12 0.5 1.33 Vmax 4.43 1.33 max 5.76 V

o TN

I

o DS

o

D

D GS GS

I o GS I

v Vv

v Vv

I

I V Vv v V v

= + = − += −

= − = −=

= + =

= = − ⇒ == + = + ⇒ =

8.15 a. Neglect base currents.

( )

( )

( ) ( )( )

( )( )

0

1 1

1

max (sat) 10 0.2 9.8 V9.8 9.8(max) 9.8 mA

10 0.7 10

949 9.8

max 2 max 19.6 mA

min 0

max 9.8 mA

min 9.8 mA

CE

L Q QL

E Q E

E

L Q

L Q

v V V

i I IR

R R

i I i

i

i I

i I

+= − = − =

= = = ⇒ =

− − −= ⇒ = Ω

= ⇒ =

=

= =

= − = −

b.

( )( ) ( ) ( )

( ) ( )( ) ( )

2 21 1max 9.8 1 48.02 mW2 2

0

9.8 20 9.8 10 294 mW

48.02 16.3%294

L L L L

S Q Q

S

L

S

P i R P

P I V V I V

P

PP

η η

+ − −

= = ⇒ =

= − + −

= + ⇒ =

= = ⇒ =

8.16 a.

( ) ( ) ( )

( )

0 max 10min min 100 mA0.1

0 0.7 12113

100

Q QL

vI I

R

R R

= = ⇒ =

− − −= ⇒ = Ω

b.

( )( )( )

1 1 1100 12 1.2 W

(source) 2 12 2.4 WQ Q CE Q

Q

P I V P

P I

= ⋅ = ⇒ =

= =

c. ( )( )

22 101 0.5 W2 2 100

1.2 2.4 3.6 W

0.5 13.9%3.6

PL

L

S

L

S

VP

R

P

PP

η η

= ⋅ = =

= + =

= = ⇒ =

8.17

( ) ( )( )221

1

12 0 1.8

38.9 mAD n GS TN

D

I K V V

I

= − = − −

=

(a)

( )( )

( )

For max 4.8 V

sat 1.8 Vmin 5 1.8 3.2 V

0.7 2.5 5.5 V

L

o

DS GS TN

o

I o I

RvV V Vvv v v

= ∞== − == − + = −

= + ⇒ − ≤ ≤

(b) For ( )500 max 4.8 VL oR v= Ω =

For ( ) 23.20, min 3.2 V 6.4 mA

0.5o

o oL

vv v IR

−′< = − = = = −

2.5 5.5 VIv− ≤ ≤ (c)

( )

( ) ( )

( )

( )

2

2

22

For 2 , max 38.9 mA2min min 51.4

38.921 1 38.9 mW

2 2 51.438.910 38.9 389 mW % 10%389

o

L

oL L

L

L

v V I

R R

vP PR

P

′= − = −−= ⇒ = Ω

−

= ⋅ = ⋅ ⇒ =

= = = =

8.18

( )

( ) ( )

( )

22

2 2

2

1 1 , 2 2

So

100%η η

+

+ −− +

+

= =

= ⋅ + ⋅ = −

=

= ⇒ =

PL

L L

SL L

SL

L

S

VVPR R

V VP V V

R R

VP

R

PP

8.19 (a)

( ) ( ) ( )

( )

As maximum conversion efficiency

, 0.7854

4So max 0.785 5

max 5 V

πη

π

= =

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

P

CC

p

p

VV

V

V

(b) Maximum power dissipation occurs when ( )2 52

3.183 VCCp

VV

π π= = =

(c) ( )

( )

2

2

2

2

max

52 1.27

CC

L

LL

VPR

RR

θ π

π

=

= ⇒ = Ω

8.20

(a) 2

2

12

150 49 V 52 V, 52 V2 24

p

L

pp

VP

RV

V V V+ −

= ⋅

= ⋅ ⇒ = ⇒ = = −

(b) 49 2.04 A24

PP

L

VI

R= = =

(c) 494 4 5274.0%

P

CC

VV

π πη

η

⎛ ⎞= ⋅ = ⎜ ⎟⎝ ⎠

=

8.21 (a)

( )( ) ( )

( ) ( )

( )

( ) ( ) ( )( )( )

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

2

2

2 00

020 0

20 0

2

0

10 max and max

max

max maxSo 10 max

5 0.4

max10 max

2max

100 20 max max2

max 20.5 max 100 0

20.5 20.5 4 100max

2

DS DS GS TN GS

DS o D L n GS

on GS

L

oGS

L n

o oo

L n

V V sat V V V

V V I I K VV

K VR

VV

R K

V VV

R K

VV

VV V

V V

V

≥ = − =

= − = =

=

=⋅

− = =⋅

− =⎡ ⎤⎣ ⎦

− + =

− + =

± −= ( )0 max 8 V

8 1.6 mA5

1.6 2 10 V0.4

L L

LGS I

n

V

i i

iV V VK

⇒ =

= ⇒ =

= = = ⇒ =

b.

( )

( )

281 6.4 mW2 520 1.6

10.2 mW

6.4 62.7%10.2

L

S

L

S

P

P

PP

π

η η

= ⋅ =

= =

= = ⇒ =

8.22

( )2 and O L L L D n GS TNv i R i i K v V= = = − or ( )2 and L n GS GS I Oi K v v v v= = −

( ) ( )

( )( )( )

( ) ( )

( )( )

2 2

0 00

00 0

00

0

Then

or 2

2 2 1

1 4 4

4or

1 4

O n L I O O I O

II I

I II

I

I I

v K R v v v v v

dv dvv v

dv dvdv

v v v vdv

v vdvdv v v

= − = −

⎛ ⎞= − −⎜ ⎟

⎝ ⎠

+ − = −⎡ ⎤⎣ ⎦

−=

+ −

For ( )

( )0 0

0

4 10 810 V, 8 V 0.889

1 4 10 8II I

dv dvv vdv dv

−= = ⇒ = ⇒ =

+ −

At 000, 0 0I

I

dvv vdv

= = ⇒ =

At 001, 0.5 0.667I

I

dvv vdv

= = ⇒ =

8.23 a.

( )

( )( )

3

13

5 10ln 0.026 ln5 10

0.5987 V 1.1973 V2

5 10 50 mW

CBE T

S

BBBE BB

Q C CE Q

iV VI

VV V

P i v P

−

−

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

= = ⇒ =

= ⋅ = ⇒ =

b.

( )

0

3

13

0

13

8 V8 80 mA

0.180 mA

80 10ln 0.026 ln5 10

0.6708 V

0.5987 0.6708 8 8.072 V2

1.1973 0.6708 0.5265 V

exp 5 10 exp

L L

Cp

CpEB T

S

EB

BBI EB I

BE BB EB

BECn S

T

v

i i

i

iv V

Iv

Vv v v v

V V v

vi IV

−

−

−

= −−= ⇒ = −

≈

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠=

= − + = − − ⇒ = −

= − = − =

⎛ ⎞= = ×⎜ ⎟

⎝ ⎠

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

22

0.5265 0.311 mA0.026

80 0.1 640 mW

0.311 10 8 5.60 mW

80 2 160 mW

Cn

L L L L

Qn Cn CE Qn

Qp Cp EC Qp

i

P i R P

P i v P

P i v P

⎛ ⎞ ⇒ =⎜ ⎟⎝ ⎠

= = ⇒ =

= ⋅ = − − ⇒ =

= ⋅ = ⇒ =

8.24 (a) ( )

( )( )

2

0.5 2 2.5 V 5.0 V2 2

0.5 10 5 mW

Dn n GSn TN

BBGSn BB

n n p

i K v V

Vv V

P P P

= −

+ = = = ⇒ =

= ⇒ = =

(b) ( )

( ) ( )( )( )

2 10 maxand

max max2 2

2 1

DS GS TN DS GS DS o

O OLGS TN

n L n

V V V V V V v

v viV VK R K

= − ⇒ = − = −

= + = + = +

so

( ) ( ) ( )0 00

max max10 max 2 2

2 2v v

v− = + − =

so ( )0 max 8 Vv =

8 8 mA1

8 2 4 V2

Dn L Dn L

GS GS

i i i i

V V

= = ⇒ = =

= + ⇒ =

( )

Then 8 4 2.5 9.5 V2

8 9.5 2.52

1 V cutoff 0

BBI o GS I

BBSGp o I

SGp p Dp

Vv v V v

Vv v v

v M i

= + − = + − ⇒ =

⎛ ⎞= − − = − −⎜ ⎟⎝ ⎠

= ⇒ ⇒ =

( ) ( )( )( )

22 8 1 64 mW

8 10 8 16 mW

0

L L L L

Mn Dn DS Mn

Mp Dp SD Mp

P i R P

P i v P

P i v P

= = ⇒ =

= ⋅ = − ⇒ =

= ⋅ ⇒ =

8.25 a.

02424 V 3 A8

3 73.2 mA41

L L N

Bn Bn

v i i i

i i

= ⇒ = ⇒ ≈ =

= ⇒ =

For 125 mA 25 73.2 98.2 mAD Ri i= ⇒ = + =

( ) 12

3ln 0.026 ln6 10

0.7004 V

NBE T

S

iV V

I −

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠=

( )

( )

( )

( )

1 11

3

12

12

30 24 0.7 5.3Then 98.2 53.97 98.2

25 100.026 ln 0.5759 V6 10

2 2 0.5759 0.70040.4514 V

0.4514exp 6 10 exp 0.208 mA0.026

−

−

−

− += ⇒ = ⇒ = Ω

⎛ ⎞×= =⎜ ⎟×⎝ ⎠= − = −

=

⎛ ⎞ ⎛ ⎞= = × ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

D

EB D BE

EBP S P

T

R RR

V

V V V

Vi I iV

b. Neglecting base current

( )1

12

30 0.6 30 0.6 545 mA53.970.5450.026 ln 0.656 V

6 10

D D

D

i iR

V −

− −≈ = ⇒ ≈

⎛ ⎞= =⎜ ⎟×⎝ ⎠

Approximation for Di is okay. Diodes and transistors matched 545 mAN Pi i⇒ = = 8.26 (a)

( )

( )( )

21 1 1

1

23 3 3

2 23 3 4

5 1 2 5

200 2 1 200 /

D GS TN

GS

D GS TN

n p

I K V V

V V

I K V V

K K K A Vμ

= −

= + =

= −

= − ⇒ = =

(b) 4 3 1I SG GS GS Ov V V V v+ + − =

( )

( )( )

21 1 1

11 1

0

00

For large,

So 2 2 10.5 5

32.5

o L n GS TN

oLGS TN TN

n L n

oI

I

v i i K V V

viV V VK R K

vv v

vv v

= = −

= + = +

⎛ ⎞+ + − + =⎜ ⎟⎜ ⎟

⎝ ⎠

= + −

0 0

0

0

0

1 112 2.5

11 12 2.5

I

I I I

I

dv dvdvdv dv dvv

dvdv v

= = + ⋅ ⋅

⎡ ⎤= +⎢ ⎥

⎢ ⎥⎣ ⎦

For 5 :Ov V=

( )( )0 0 011 1 1.1414 0.876

2 2.5 5I I I

dv dv dvdv dv dv

⎡ ⎤⎢ ⎥= + = ⇒ =⎢ ⎥⎣ ⎦

8.27

and 2

DnBBO I GS GS TN

n

IVv v V V VK

= + − = +

For 0, OO Dn DQ L DQ

L

vv I I i IR

≈ = + = +

Then ( )/

2DQ O LBB

O I TNn

I v RVv v VK

+= + − − or 1

2DQ OBB

O I TNn DQ L

I vVv v VK I R

= + − − ⋅ +

For Ov small, 112 2

DQ OBBO I TN

n DQ L

I vVv v VK I R

≅ + − − ⋅ + ⋅

1 112 2

DQ DQBBO I TN

n DQ L n

I IVv v VK I R K

⎡ ⎤+ ⋅ ⋅ = + − −⎢ ⎥

⎢ ⎥⎣ ⎦

Now 1 0.95

1 112

O

I DQ

n DQ L

dvdv I

K I R

= =⎡ ⎤

+ ⋅ ⋅⎢ ⎥⎢ ⎥⎣ ⎦

So 1 1 1 1 0.05262 0.95

DQ

n DQ L

IK I R

⋅ ⋅ = − =

For 10.1 , then 0.01052Ln DQ

R kK I

= Ω =

Or 95.1n DQK I =

We can write 2 190 m n DQg K I mA/V= = This is the required transconductance for the output transistor. This implies a very large transistor. 8.28

v m LA g R= −

So ( )12 2 6 mA/V= CQm m

T

Ig g

V− = − ⇒ =

( )( )6 0.026 0.156 mACQ CQI I= ⇒ =

But for maximum symmetrical swing, set 10 5 mA 122

CCCQ v

L

VI A

R= = = ⇒ >

Maximum power to the load:

( ) ( )( ) ( )

( )( )

22 101 max max 25 mW2 2 2

10 5 50 mWSo 50%

CCL L

L

S CC CQ

VP P

R

P V Iη

= ⋅ = ⇒ =

= ⋅ = ==

( ) ( )

( )( ) ( )( )( )

( )( )

1

1

11

2

5 0.0278 mA180

6 kon 1

Set 20 0.0278 6 0.7 181 0.0278 0.020

0.967 V1

10.967 6 10 62.0 k

6.64 k

CQBQ

TH

TH BQ TH BE BQ E

E

TH

TH

TH TH CC

II

R RV I R V I R

RVV

V R VR

RR

R

β

β

= = =

= = Ω= + + +

= Ω= + +=

= ⋅ ⋅

= ⇒ = Ω

= Ω

8.29

( ) ( )( ) ( )

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

( )( )

22

1 1

1

2

15 15 mA1

15 0.15 mA100

151max max 112.5 mW2 2 1

Let 10 k1

0.15 10 0.7 101 0.15 0.11 13.715 10 15

40.4 k

13.3 k

β

= = =

= =

= ⋅ = ⇒ =

= Ω= + + += + +

= = ⋅ ⋅ = ⋅

= Ω

= Ω

CCCQ

L

BQ

CCL L

L

TH

TH BQ TH BE BQ E

TH TH CC

VIR

I

VP PR

RV I R V I R

V R VR R

R

R

8.30

( )

( ) ( )( )

( ) ( )

2

1 2

1 2

22

1.55 101.55 0.73

6.80 V0.73 1.55 0.496 k

6.80 0.701 0.496 26 0.02

6.0 mA, 150 mA

and 3 8 72

150 5.77 A/V0.026

TH CC

TH

TH BEBQ

TH E

BQ CQ

v m L L L

CQm

T

RV VR R

R R R

V VIR R

I I

A g R R a RI

gV

β

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=

= = = Ω

− −= =+ + +

= =

′ ′= − = = = Ω

= = ⇒

( )( )( )( )

5.77 72 415

415 0.017 7.06 V7.06 2.35 V

37.06 2.35 V

37.06 2.35 V

3

v

o v i

o

o

o

A

V A V

V

V

V

= − = −

′ = ⋅ = =

= =

= =

= =

( )( )0.15 10 1.5 W

0.345 23%1.5

S CQ CC

L

S

P I V

PP

η η

= ⋅ = =

= = ⇒ =

8.31 a. Assuming the maximum power is being delivered, then

( ) rms rms36 9peak 36 V 9 V 6.36 V4 2o oV V V V′ = ⇒ = = ⇒ = ⇒ =

b. 36 25.5 V2o oV V= ⇒ =

c. Secondary rms rmsrms

2 0.314 A6.36

LPI IV

= = ⇒ =

Primary 0.314 78.6 mA4P PI I= ⇒ =

d. ( )( ). 0.15 36 5.4 W

2 37%5.4

η η

= = =

= ⇒ =

S CQ CCP I V

8.32 a.

( )( )

2

1

2

10 0

21

2

1

1

1( )

11

where 1 11

so

e m E m E

E

i e i e

e i e E

EEe e

E Li E i

E

ee

Vv g V R V g R

r r

V Rr

v V v V v v

v v v Rr

RRv r v nR R

v r R v nRr

v nv v vnn

n

ππ π

π π

ππ

π π

π

π

π

π

β

β

ββ

β β

⎛ ⎞ ⎛ ⎞′ ′= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞+ ′= ⎜ ⎟⎝ ⎠

= + ⇒ = −

⎛ ⎞+ ′= − ⎜ ⎟⎝ ⎠

+ ′⋅ ′+ ⎛ ⎞′= = = = ⎜ ⎟+ ′+ + ⎝ ⎠′+ ⋅

⎛ ⎞= − ⎜⎛ ⎞ ⎝ ⎠⎜ ⎟⎝ ⎠

( )( )

0

1

2

11so 1

E

i E

Rvv r Rn

nπ

ββ

⎟

′+= ⋅

′+ +⎛ ⎞⎜ ⎟⎝ ⎠

b.

( )( )

2 2 2

2 22

2 2

1 1 1, , so .2 2 2

.For 50% :

120.5 so

2 0.1 50 5

η

= ⋅ = = =

==

⋅= = = = = ⇒ =

⋅ ⋅

PL P L CQ L CQ L

S CQ CC

CQ LCQ L CC CC CCL

CQ CC CC CQ LS

InP I R a I P a I Rn a

P I V

a I R a I R V V VP a aI V V I RP

c.

( )( )

( )( )0 0

49 0.0260.255

1 1 50 0.1T

CQ

r VR RI

π ββ β

= = = ⇒ = Ω+ +

8.33 a. With a 10:1 transformer ratio, we need a current gain of 8 through the transistor.

( ) 1 2

1 2

1 and e b b iib

R Ri i i i

R R Rβ

⎛ ⎞= + = ⎜ ⎟⎜ ⎟+⎝ ⎠

so we need ( ) 1 2

1 2

8 1e

i ib

R Rii R R R

β⎛ ⎞

= = + ⎜ ⎟⎜ ⎟+⎝ ⎠ where

( ) ( ) ( )( )1 1 101 0.8 80.8ib L LR r R Rπ β β′ ′= + + ≈ + = =

( )

( )( ) ( )( )

1 2

1 2

1 21 2

1 2

1

1 21

Then 8 10180.8

0.0792 or 6.95 k80.8

2 12Set 15 mA2 0.8

15 0.15 mA100

1

1 6.95 12 0.15 6.95 0.7 47.9 k then

⎛ ⎞= ⎜ ⎟⎜ ⎟+⎝ ⎠

= = Ω+

′= ⇒ = = =′

= =

= +

⋅ ⋅ = +

= + ⇒ = Ω

CC CCL CQ

CQ L

BQ

TH BQ TH BE

TH CC BQ TH BE

R RR R

R RR R

R RV VR II R

I

V I R V

R V I R VR

R RR

8.13 k= Ω

b.

( ) ( )

( )( )

2

0.9 13.5 mA 135 mA

1 0.135 8 72.9 mW2

12 15 180 mW

40.5%

Le CQ L

L L

S CC CQ S

L

S

II I I

a

P P

P V I P

PP

η η

= = = ⇒ =

= ⇒ =

= = ⇒ =

= ⇒ =

8.34 a.

( )( )

( )( )

2

2 8 2 5.66 V peak output voltage

5.66 0.708 A peak output current8

Set 0.9 to minimize distortion0.9 18

Then 2.865.66

=

= = =

= = = =

= =

= ⇒ =

P L L

P

PP

L

e CC P

V R P

V

VIR

V V aV

a a

b.

( )( )

1 1 0.708Now 0.275 A0.9 0.9 2.86

Then 18 0.275 4.95 W Power rating of transistor

⎛ ⎞ ⎛ ⎞= = ⇒ =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= = ⇒ =

PCQ CQ

Q CC Q Q

II Ia

P V I P

8.35 a. Need a current gain of 8 through the transistor.

( ) 1 2

1 2

8 1b

i ib

R Rii R R R

β⎛ ⎞

= = + ⎜ ⎟⎜ ⎟+⎝ ⎠ where ( )( )1 0.9 90.9 kΩibR β≈ + =

( )( ) ( ) ( )

1 21 2

1 2

1 21

8 0.0792 or 7.82 k101 90.9

2 12Set 0.9 k 13.3 mA2 0.9

13.3 0.133 mA1001Then 7.82 12 0.133 7.82 0.7 53.9 k and R 9.15 k

CCCQ

CQ

BQ

R RR R

R R

V II

I

RR

⎛ ⎞= = = Ω⎜ ⎟⎜ ⎟+⎝ ⎠

= Ω ⇒ = =

= =

= + ⇒ = Ω = Ω

b.

( )

( ) ( )

( )( )

2

0.9 12 mA 120 mA

1 0.12 8 57.6 mW2

12 13.3 159.6 mW

57.6 36.1%159.6

η η

= = = ⇒ =

= ⇒ =

= = ⇒ =

= = ⇒ =

Le CQ L

L L

S CC CQ S

L

S

II I Ia

P P

P V I P

PP

8.36 a. All transistors are matched.

1 313 mA

61 13 2.90 mA60 60

ββ β

⎛ ⎞+= + = +⎜ ⎟⎝ ⎠

⎛ ⎞= + ⇒ =⎜ ⎟⎝ ⎠

CE B C

C C

ii i i

i i

b. For 6 Vov = , let 200 .LR = Ω

3

3

1

1 1

2 2

2 1

6 0.03 A 30 mA20030 0.492 mA613 0.492 2.508 mA2.508 41.11 A

6133 mA 49.18 A61

49.18 41.11 8.07 A

o E

B

E

B B

E B

I B B I

i i

i

i

i i

i i

i i i i

μ

μ

μ

= = = ≅

= =

= − =

= ⇒ =

≅ ⇒ = ⇒

= − = − ⇒ =

( )

( )

33

6

33

3 13

3

31

1 13

1

Current gain 30 10 3.72 10

8.07 1030 10ln 0.026 ln5 10

0.6453 V

2.508 10ln 0.026 ln5 10

0.5807 V

i i

EBE T

S

BE

EEB T

S

EB

A A

iV V

IV

iV VI

V

−

−

−

−

−

−

×= ⇒ = ××⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

=

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠=

0 3 1

0

6 0.6453 0.58076.0646 V

Voltage gain 6 0.989

6.0646

I BE EB

I

v vI

v v V Vv

vA Av

= + − = + −=

= = ⇒ =

8.37

a. For 0 311 A, 20 mA

50Bi I= ≅ ⇒

We can then write ( )0,max 31

1 1

1010 2 20BEEBv VV

R R

⎡ ⎤− +−= −⎢ ⎥

⎢ ⎥⎣ ⎦

If, for simplicity, we assume 1 3 0.7 V,EB BEV V= = then ,max

1 1

210 40oBE vVR R− = +

If we assume 0,max 4 V,v = then ( )

1 1

2 49.3 40R R

= + which yields 1 2 32.5 R R= = Ω

b. For 0,Iv = 1 1 29.3 0.286 A

32.5E E EI I I= ⇒ = =

Since 3,4 1,210 ,S SI I= then 3 4 2.86 AE EI I= = c.

( )( )

( )( )

13 1

10

3

33

3

11

1

0

0

We can write

112 1

50 0.026Now 0.4545

2.86120 0.026

10.91 0.286

So

10.910.4545 32.51 1212 51

10.9132.5 32.5 0.0902 0.0900121

1Then 2

T

C

T

C

rr RR

VrI

VrI

R

R

ππ

π

π

ββ

β

β

⎧ ⎫+⎪ ⎪+⎪ ⎪= ⎨ ⎬+⎪ ⎪

⎪ ⎪⎩ ⎭

= = = Ω

= = = Ω

⎧ ⎫+⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

= =

= 00.4545 0.0900 or R 0.00534

51+⎧ ⎫ = Ω⎨ ⎬

⎩ ⎭

8.38

( ) ( )( )

( )( )

( ) ( )( )( )

1 1 3

1 3

1 1 1 227.2 mA and 7.2 mA

60 0.026Then 0.217 k

7.21So 0.217 61 2 0.217 61 0.221 0.217 61 2 12.4 or 52.6 k2

π π

π

β β⎡ ⎤= + + + +⎣ ⎦≈ ≈

= = Ω

⎡ ⎤= + +⎣ ⎦

= + ⎡ ⎤ = Ω⎣ ⎦

i L

C C

i

i

R r R r R

i i

r

R

R

8.39 a.

b.

( )

( )

21 1

12

1 21

5 10 2 2.707 V10 2.7075 1.46 k

+ −= + =

= − ⇒ =−= ⇒ = = Ω

SGSG TP

SG SG

V VI K V VR

V V

R RR

c. 100 LR = Ω For a sinusoidal output signal:

( ) ( )

( ) ( )

( )

2 2

3 3

3

1 1

51 1 125 mW2 2 0.1

550 mA

0.1

50 2 4.236 V10

10 4.236 50.523 mA

1.46

oL L

L

oD D

L

GS

D

vP P

Rv

i iR

V

I I

= ⋅ = ⋅ ⇒ =

≈ = ⇒ =

= + =

− += ⇒ =

10.523 2 2.229 V

105 4.236 2.229 7.007 V

SG

I I

V

v v

= + =

= + − ⇒ =

( ) ( ) ( )

( )

( ) ( )( )( )

22

2

2

2

1010 2

1.4617.007

10 4 41.46

14.6 57.4 41.4 0

57.4 57.4 4 14.6 41.42 14.6

I GSD GS

GSGS GS

GS GS

GS

V VI V

VV V

V V

V

− − −= = −

−= − +

− + =

± −=

( )2

22 2

4 2

4 4

2.98 V

10 2.98 2 9.60 mA

7 2.98 4.02 V5 4.02 0.98 V 0

GS

D D

G I GS

SG D

V

I I

V v VV I

=

= − ⇒ =

= − = − == − = ⇒ =

8.40 For 0 0v =

( )

( )

( )

( )( ) ( )

3 2 1

33 2 2

3 2

22 1 1

2 1

3 1

1 1 1

1

1

1

1

1

11

11 1

10 1051 50 5011

Q C C E

n CB E C

n n

C n C

CPB C E

P n

PC n E

P

PC n n E

p

P PQ n n E n E E

P P

E

I I I I

II I I

I I

II I I

I I

I I

I I I I

I

ββ β

ββ

β β

βββ

ββ ββ

β ββ β ββ β

= + +

⎛ ⎞+= = =⎜ ⎟

⎝ ⎠= +

⎛ ⎞= = =⎜ ⎟+⎝ ⎠

⎛ ⎞= ⎜ ⎟+⎝ ⎠

⎛ ⎞= + ⎜ ⎟⎜ ⎟+⎝ ⎠

⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

( ) ( )

( )( ) ( )

1 1

1 1 1

1 1

2 2

3 3

112318.18 45.451.692 A 1.534 A

1050 1.692 76.9 A11

1051 50 1.692 3.92 mA11

E E

Q E E E

E C

C C

C C

I I

I I I II I

I I

I I

μ μ

μ

⎛ ⎞ +⎜ ⎟⎝ ⎠

= + += ⇒ =

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

Because of 1rπ and Z, neglect effect of r0. Then neglecting r01, r02 and r03, we find

3 3 2 2 1 11

XX m m m

VI g V g V g V

r Zπ π ππ

= + + ++

( )( )

[ ]

( )

11 2 1 1 2

1

3 1 1 2 2 3

1 1 2 1 1 2 3

13 1 1 2 2 3

1

1 1 2 33

1

1 1 22 1 2

1 1

Now

,

and

and

Then

X m

m m

m m m

m m m X

X

m X X

rV V V g V rr Z

V g V g V rg V g g V r r

rV g g g r r Vr Z

rV V

r Z

r rV g r V Vr Z r Z

ππ π π π

π

π π π π

π π π π

ππ π π

π

ππ

π

π ππ π

π π

β β β

β

⎛ ⎞= ≅⎜ ⎟+⎝ ⎠

= += +⎡ ⎤⎣ ⎦⎛ ⎞

= + ⋅⎜ ⎟+⎝ ⎠+

= ⋅+

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

( )

( )( )( )

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

1 1 2 3 1 2 1

1 1 1 1

10

1 1 2 1 1 2 3

1

0

0

Then

1

10 0.0260.169 MΩ

1.53425 kΩ

Then169 25

1 10 10 50 10 10 50 50

194 0.00746 kΩ or 7.46 26,011

XX X X X

X

X

o

VI V V Vr Z r Z r Z r Z

r ZVRI

r

Z

R

R R

π π π π

π

π

β β β β β β β

β β β β β β β

+= ⋅ + ⋅ + ⋅ +

+ + + +

+= =

+ + + +

= =

=

+=+ + + +⎡ ⎤⎣ ⎦

= = = Ω

8.41 a Neglect base currents.

( )

Bias

3

13

2 2 ln

5 102 0.026 ln 1.281 V10

BB D TS

BB

IV V VI

V−

−

⎛ ⎞= = ⎜ ⎟

⎝ ⎠⎛ ⎞×= ⇒ =⎜ ⎟⎝ ⎠

1 3

1 3 2

2 3 3

2 2 3

1

1

BE EB BB

E E C

PB C E

P

PC n B n E

P

V V VI I I

I I I

I I I

ββ

ββ ββ

+ == +

⎛ ⎞= = ⎜ ⎟+⎝ ⎠

⎛ ⎞= = ⎜ ⎟+⎝ ⎠

1 3 3

1 3

1

11

PE E n E

P

PE E n

P

I I I

I I

βββ

βββ

⎛ ⎞= + ⎜ ⎟+⎝ ⎠

⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦

1 31 1 1

1n P P

C C nn P P

I Iβ β ββ

β β β⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +

= +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣ ⎦

( ) ( )

1 31 3

1 3

3 3

ln , ln

21 201.01 1 10020 21

21 100 101.0520

C CBE T EB T

S S

C C

C C

I IV V V V

I I

I I

I I

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤= + =⎢ ⎥⎣ ⎦

1 3

3 3

23

2

23

2

3 C3

100.05

100.05ln ln

100.05ln

100.05exp

exp 0.4995 mA100.05

C C

C CT T BB

S S

CT BB

S

C BB

TS

S BBC

T

I I

I IV V V

I I

IV V

I

I VVI

I VI I

V

=

⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞

=⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟

⎝ ⎠

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

( ) ( )

3

1 3 1

2 2

Then 0.5245 mANow 100.05 49.97 mA

20100 0.5245 49.95 mA21

E

C C C

C C

II I I

I I

== = =

⎛ ⎞= = =⎜ ⎟⎝ ⎠

31

1 13

33

3 13

1 3

49.97 10ln 0.026 ln10

0.70037

0.4995 10ln 0.026 ln10

0.58062Note: 0.70037 0.58062 1.28099

CBE T

S

CEB T

S

BE EB

BB

IV V

I

IV V

I

V VV

−

−

−

−

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=+ = + =

=

b.

( )

( )

( )

0 1 1

1

3

13

1 13

3

133

220

1010 V 0.10 A100

100 1 mA100

4 102 0.026 ln 1.2694 V10

0.10.026 ln 0.718410

1.2694 0.7184 0.55099 V0.5509910 exp 0.1598 mA0.026

10100

E C

B

BB

BE

EB

C

L LL

v i i

i

V

V

V

I

VP P

R

−

−

−

−

= ⇒ ≈ = =

= =

⎛ ⎞×= =⎜ ⎟⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= − =

⎛ ⎞= =⎜ ⎟⎝ ⎠

= = ⇒ 1 W=

( )( )( ) [ ]( )

( )( ) ( )( )( ) [ ]( )

1 1 1 1

3 3 3 3

2 3

2 2 2 2

0.1 12 10 0.2 W

0.1598 10 0.7 12 3.40 mW

100 100 0.1598 15.98 mA15.98 10 12 0.352 W

Q C CE Q

Q C EC Q

C C

Q C CE Q

P i v P

P i v P

i iP i v P

= ⋅ = − ⇒ =

= ⋅ = − − ⇒ =

= = == ⋅ = − − ⇒ =

8.42 a.

( )

( )( )

3

12

1 2 3

2 21 3 2

1 2 3

32

3 3

32

12

10 103 0.026 ln 1.74195 V2 10

,

ln ln ln

ln

exp

1.720 20 10 exp

BB BB

BE BE EB BB

C CC C

n n

C C CT T T BB

S S S

CT BB

n S

BBC n S

T

V V

V V V VI II I

I I IV V V VI I I

IV VI

VI IV

β β

β

β

−

−

−

⎛ ⎞×= ⇒ =⎜ ⎟×⎝ ⎠+ + =

≈ ≈

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎡ ⎤

=⎢ ⎥⎣ ⎦

⎛ ⎞= ⎜ ⎟

⎝ ⎠

= ×

( )

( )

( )

3

2 1 3

3

1 112

2 212

3

3 312

41950.026

0.20 A, 10 mA, 0.5 mA

10 100.026 ln 0.58065 V2 10

0.20.026 ln 0.6585 V2 10

0.5 100.026 ln 0.50276 V2 10

C C C

BE BE

BE BE

EB EB

I I I

V V

V V

V V

−

−

−

−

−

⎛ ⎞⎜ ⎟⎝ ⎠

= ≈ ≈

⎛ ⎞×= ⇒ =⎜ ⎟×⎝ ⎠⎛ ⎞= ⇒ =⎜ ⎟×⎝ ⎠⎛ ⎞×= ⇒ =⎜ ⎟×⎝ ⎠

b.

( )2 2

0 00

1 110 W= max 20 V2 2 20L

L

V VP VR

= ⋅ = ⋅ ⇒ =

For ( )0 maxv :

( )

( )( )

220

0

5 4 3

5 45

5 55

5

2020 W

2020max 1 A20

max 1 A

11

1

11 11 1

1 20 1120 21 21

L LL

C C E o

pC n CC

n n n p

pC n C nC

n n n n n p

C

vP P

R

i

i i i i

i ii

i ii

i

βββ β β β

ββ ββ β β β β β

= = ⇒ =

= − = −

+ + = − =

⎛ ⎞+⎛ ⎞+ ⋅ + =⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

⎡ ⎤⎛ ⎞+⎛ ⎞ ⎛ ⎞+ + =⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤⎛ ⎞ ⎛+ +⎜ ⎟⎢ ⎥⎝ ⎠ ⎝⎣ ⎦

1 6 120 5⎡ ⎤⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎢ ⎥⎠ ⎝ ⎠⎣ ⎦

( )5 5

4

3

3

1.05048 1 0.952 A

0.0453 A

0.00272 A50.002726

0.002267 A

C C

C

E

C

i i

i

i

i

= =

=

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

( )3

3 12

1 2

2 2

2.267 100.026 ln 0.54206 V2 10

1.74195 0.54206 1.19989

ln ln 1.19989

EB

BE BE

C CT T

n S S

V

V V

I IV V

I Iβ

−

−

⎛ ⎞×= =⎜ ⎟×⎝ ⎠+ = − =

⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

( )

2

2

1.19989exp0.026

20 18.83 mA93.9 mA

C n S

C

i I

i

β ⎛ ⎞= ⋅ ⎜ ⎟⎝ ⎠

==

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

21

2 2

5

93.9 4.47 mA1 2124 20 0.0939 44 4.13 W

0.952 10 24 13.3 W

C nC

n n

Q C

Q

ii

P I

P

ββ β

⎛ ⎞= = =⎜ ⎟+⎝ ⎠= − − = =

= − − − =