CHAPTER 14athena.ecs.csus.edu/.../Spring2006/EEE109/ch14_solutions.pdf14-443 CHAPTER 14 14.1 (a)...

14-443

CHAPTER 14

14.1 (a) Common-collector Amplifier (emitter-follower)

RE

Q1

R1 R2

R3

+

-

v o

RI

v i

14.1 (b) Not a useful circuit because the signal is injected into the drain of the transistor.

R3

+

-

M1

R D

v i

RI

voR1

14.1 (c) Common-emitter Amplifier

1 k Ω

R C+

-

R3

100 k Ω

Q1

R1 R 2

RI

v i

vo

14.1 (d) Common-source Amplifier

14-444

+

-

R3

1 k Ω

RD

470 k Ω

M1

R2R

1v i

RI

vo

14.1 (e) Common-gate Amplifier

R1

R3RD

+

-

M1vi

RI

vo

14.1 (f) Common-collector Amplifier (emitter-follower)

R1

Q1

RER

2 R 3

+

-

vo

RI

v i

14.1 (g) Common-source Amplifier

RI

vi R

G

R 3

R D

+

-

v o

RS

J1

14.1 (h) Common-base Amplifier

14-445

RC

R E

Q1

R3

+

-

RI

vi

vo

14-446

14.1 (i) Not a useful circuit since the signal is being taken out of the base terminal.

R E

Q1

RBR3

+

-

RI

vi

vo

14.1 (j) Common-source Amplifier

1 k Ω

RD

+

-

R3

470 k Ω

M1

R2

R1

RI

vi

vo

14.1 (k) Common-gate Amplifier

R3

RD

+

-

RI

vi

voRS

M1

14.1 (l) Not a useful circuit because the signal is injected into the drain of the transistor.

R3+

-

vo

M1

R D

v i

RI

RS

14-447

14.1 (m) Common-source Amplifier

J 1

RG

R3

R D

+

-vi

RI

vo

14.1 (n) Common-emitter Amplifier

RB

Q1

R3

+

-

v i

RIvo

14.1 (o) Common-drain Amplifier (Source-follower)

RG

M

R 3

+

-

v i

RI

vo

14.2

vo

v i

RIRS

RGRD

CB

CC2

1k Ω

RL

100 k Ω-VSS

VDD

CC1

+

-

14-448

14.3

RI

v i

RS

CC1

CC2

RG

RD

CB

VDD

-V SS

RL1kΩ

100 k Ω

vo

+

-

14.4

RI

v i

RS

-VSS

RG

RD

CB

RL

100 k Ω

CC1

CC2

VDD

1k Ω

vo

+

-

14.5

+15 V

220 kΩ

15 kΩ

-15 V

150 kΩ

22 kΩ

+

-

vO

v I

RI

3.3 k Ω

RE

RB

RC

RLCC1

CC2

CB

Q1

14-449

14.6

(a)

+15 V

220 k Ω

15 kΩ

-15 V150 kΩ22 k Ω

+

-

vO

v I

RI

3.3 k Ω

RE

RB

RC

RL

CC1

CC2

CB

Q1

(b)

+15 V

220 kΩ

-15 V150 kΩ22 kΩ

+

-

vO

v I

RI

3.3 k Ω

RE

RB

RL

CC1

CC2

Q1

14.7

(a)

+15 V

220 kΩ

15 kΩ

-15 V

3.3 k Ω

150 kΩ

22 kΩ

+

-

vO

RE

RB

RC

RL

CC1

CC2

CB

vI

RI

Q1

(b)

+15 V

15 kΩ

-15 V

3.3 kΩ

150 k Ω

22 kΩ

+

-

vO

RE

RC

RL

CC1

CC2

v I

RI

Q1

14.8

a( ) Neglecting Rout : Avt = −gmRL

1+ gmRS= −

0.005S( ) 2000Ω( )1+ 0.005S( ) 200Ω( )

= −5.00

Av = AvtRG

RI + RG= -5

1MΩ75kΩ+ 1MΩ

= -4.65 | Rin = RG = 1MΩ

Rout = ro 1+ gmRS( )=10kΩ 1+ 0.005S( ) 200Ω( )[ ]= 20.0 kΩ >> 2kΩ

Ai = −RGgm

1+ gmRS= −1MΩ

0.005S1+ 0.005S( ) 200Ω( )

= −2500

b( ) Av = −gm RL ro( ) RGRI + RG

= − 0.005S( ) 2kΩ 10kΩ( ) 1MΩ

75kΩ + 1MΩ

= −7.75

Rin = RG = 1MΩ | Rout = ro = 10.0 kΩ | Ai = −RGgm = −1MΩ 0.005S( )= −5000

14-450

14.9

a( ) gm = 0.02S | rπ =75.02

= 3750Ω

Rin = RB rπ + βo + 1( )RE[ ]=15kΩ 3750Ω + 76 200Ω( )[ ]= 8.37 kΩ

Neglecting Rout : Av = AvtRin

RI + Rin

= −

βoRLrπ + βo +1( )RE

RinRI + Rin

Av = −75 12kΩ( )

3750Ω + 76 200Ω( )8.37kΩ

500Ω+ 8.37 kΩ

= −44.8

Rout ≅ ro 1+βoRE

Rth + rπ + RE

= 100kΩ 1+

75 200Ω( )15kΩ 500Ω( )+ 3750Ω+ 200Ω

= 438 kΩ >> 12kΩ

Ai = −βoRB

RB + rπ + βo +1( )RE= −75

15kΩ15kΩ + 3750Ω + 76 200Ω( )

= −33.1

a( ) gm = 0.02S | rπ =75.02

= 3750Ω

Rin = RB rπ + βo +1( )RE[ ]= 15kΩ 3750Ω+ 76 560Ω( )[ ]= 11.3 kΩ

Neglecting Rout : Av = AvtRin

RI + Rin

= −

βoRLrπ + βo +1( )RE

RinRI + Rin

Av = −75 12kΩ( )

3750Ω + 76 560Ω( )11.3kΩ

500Ω+ 11.3kΩ

= −18.6

Rout ≅ ro 1+βoRE

Rth + rπ + RE

= 100kΩ 1+

75 560Ω( )15kΩ 500Ω( )+ 3750Ω + 560Ω

= 976 kΩ >>12kΩ

Ai = −βoRB

RB + rπ + βo +1( )RE= −75

15kΩ15kΩ + 3750Ω + 76 560Ω( )

= −18.3

14.10

a( ) For large βo : Av ≅ −RLRE

= −8.2kΩ 47kΩ330Ω + 680Ω

= −6.91 | b( ) Place a bypass capacitor in

parallel with the 330 Ω resistor. Then Av ≅ −RLRE

= −8.2kΩ 47kΩ

680Ω= −10.3 | c( ) Place a

bypass capacitor in parallel with the 680 Ω resistor. Then Av ≅ −8.2kΩ 47kΩ

330Ω= −21.2

d( ) Place a bypass capacitor from the emitter to ground. e( ) Av ≅ −10 VCC + VEE( )= −240.

14-451

14.11 Rin = rπ + βo + 1( )RE = 500 kΩ

Avt = −βoRL

rπ + βo + 1( )RE= −

75RL500kΩ

= −10 → RL = 66.7 kΩ

Assuming βo +1( )RE >> rπ , RE ≅500 kΩβo + 1

=500 kΩ

76= 6.58 kΩ

Note : The nearest 5% values would be 68 kΩ and 6.8 kΩ.

14.12

Rin = rπ = 500 kΩ | rπ =βogm

=βoVT

IC | IC =

βoVTrπ

=75 0.025V( )

500kΩ= 3.75 µA

Av = −gmRLrπ

Rth + rπ

= −

βoRLRth + rπ

= −75RL

100Ω + 500kΩ= −10 | RL = 66.7 kΩ

Closest 5% value is RL = 68kΩ.

14.13

R thR5

ro

vd

i x

vgs

gm vgs

+ -

ix − gmvgs+gmvgs

=

go −go−go go + GS

vdvs

| vgs = −vs

ix0

=

go − gm + go( )−go gm + go + GS

vdvs

∆ = goGS | vd = gm + go + GS( )ix∆

=gm + go + GS( )

goGSix

Rout =vdix

= RS 1+gmgo

+GSgo

= RS 1+ µf +

roRS

Rout = ro + 1+ µ f( )RS ≅ ro + µ f RS = ro 1+ gmRS( )

14.14

ix + gm' ve

−gm' ve

=

go −go−go go + GE + G1

vxve

|

ix0

=

go − gm' + go( )

−go gm' + go + GE + G1

vxve

∆ = go GE +G1( ) | vx = gm' + go + GE + G1( )ix∆ =gm

' + go + GE +G1( )( )go GE +G1( )

ix

Rout =vxix

= RE R1( )1+ gm'

go+

GE + G1go

= RE R1( )1+ gm' ro + roRE R1( )

Rout = RE R1( )+ ro + ro gmrπRth + rπRE Rth + rπ( )Rth + rπ + RE

= RE R1( )+ ro 1+ βoRERth + rπ + RE

14-452

14.15 From the results in Table 14.1,

AvtCE = −

gmRL1+ gmRE

= −RL

1gm

+ RE= −

RLre + RE

for re =1gm

AvtCS = −

gmRL1+ gmRS

= −RL

1gm

+ RS= −

RLre + RS

for re =1gm

14.16

VEQ =1262kΩ

20kΩ+ 62kΩ= 9.07V | REQ = 20kΩ 62kΩ = 15.1kΩ

IB =12 − 0.7 − 9.07( )V

15.1kΩ + 75 + 1( )3.9kΩ= 7.16µA | IC = 537 µA | VEC = 12− 3900IE − 8200IC = 5.47 V

Active region is correct. | rπ =75 0.025V( )

537µA= 3.49kΩ | VA not specified, choose ro = ∞

Rin = 15.1kΩ 3.49kΩ = 2.83 kΩ | Rout = ro 8.2kΩ = 8.2 kΩ | gm = 40IC = 21.5 mS

RL = ro 8.2kΩ 100kΩ = 8.2kΩ 100kΩ = 7.58kΩ

Av = −gmRLRin

RI + Rin

= − 21.5mS( ) 7.58kΩ( ) 2.83kΩ

1kΩ+ 2.83kΩ

= −120

Ai =RB

RB + rπ−βo( )

RoutRout + R3

=15.1kΩ

15.1kΩ + 3.49kΩ−75( ) 8.2kΩ

8.2kΩ + 100kΩ= −4.62

vbe = v iRin

RI + Rin= v i

2.83kΩ1kΩ + 2.83kΩ

= 0.739v i | v i =5.00mV0.739

= 6.76 mV

Av ≅ −10VCC = −10 12( )= −120. | The voltage gain is identical to the rule - of - thumb estimate.

14.17

VEQ =18500kΩ

1.4MΩ+ 500kΩ= 4.74V | REQ = 500kΩ 1.4MΩ = 368kΩ

4.74 = VGS + 27000ID = 1+2IDS

250x10−6+ 27000ID → ID =104µA

VDS =18 − ID 75kΩ+ 27kΩ( )= 7.39V | Active region operation is correct.

gm = 2 250x10−6( )104x10−6( )= 0.228mS | Assume λ = 0, ro = ∞.

RL = ro 75kΩ 470kΩ ≅75kΩ 470kΩ = 64.7kΩ

Rin = RG = R1 R2 = 368 kΩ | Rout = ro 75kΩ ≅ 75kΩ

14-453

Av = −gmRLRin

RI + Rin= − 0.228mS( ) 64.7kΩ( ) 368kΩ

1kΩ + 368kΩ

= −14.7

Ai = RG −gm( )RD

RD + R3= 368kΩ −0.228mS( ) 75kΩ

75kΩ + 470 kΩ= −11.5

vgs = viRin

RI + Rin= v i

368 kΩ1kΩ + 368kΩ

= 0.997v i | VGS −VTN =2 104µA( )

250µA /V 2= 0.912V

vgs ≤ 0.2 VGS −VTN( )→ vi ≤ 0.20.912V0.997

= 0.183 V | Av ≅ −VDD

VGS −VTN= −

180.912

= −19.7

The rule - of - thumb estimate assumes VRL =VDD

2. We have VRL =104µA 75kΩ( )= 7.80V = 0.433VDD

The estimate also doesn' t account for the presence of R 3 .

14.18

VEQ =182.2MΩ

2.2MΩ + 2.2MΩ= 9.00V | RG = REQ = 2.2MΩ 2.2MΩ = 1.10MΩ

18 = 22000ID − VGS + 9 | 9 = 22000ID + 1+2ID

400x10−6→ ID = 307µA

VDS = − 18 − ID 22kΩ +18kΩ( )[ ]= −5.72V | Active region operation is correct.gm = 2 400x10

−6( ) 307x10−6( )= 0.496mS | Assume λ = 0, ro = ∞RL = ro 18kΩ 470kΩ ≅18kΩ 470kΩ = 17.3kΩ | Rin =1.10 MΩ | Rout = ro 18kΩ = 18kΩ

Av = −gmRLRin

RI + Rin

= − 0.496mS( )17.3kΩ( ) 1.1MΩ

1kΩ + 1.1MΩ= −8.57

Ai = −gmRinRD

RD + R3= −0.496mS 1.1MΩ( ) 18kΩ

18kΩ+ 470kΩ= −18.3

vgs = v iRin

RI + Rin= v i

1.1MΩ1kΩ +1.1MΩ

= 0.999v i | VGS −VTN =2 307µA( )400µA /V 2

=1.24V

vgs ≤ 0.2 VGS − VTN( ) | v i ≤ 0.21.24V0.999

= 0.248 V

14.19

VGS = − 11kΩ( )ID = − 11kΩ( ) 20mA( ) 1−VGS−4

2

→ VGS = −3.50V , ID = −VGS

11kΩ= 318 µA

VDS = 20− ID 11kΩ + 39kΩ( )= 4.10V | Active region operation is correct.

gm =2−4

20mA 318µA( ) = 1.26mS | Assume λ = 0, ro = ∞. | RL = 39kΩ 500kΩ = 36.2kΩ

Rin = RG =1.00 MΩ | Rout = 39kΩ

14-454

Av = −gmRL

1+ gmRS

RinRI + Rin

= −

1.26mS 36.2kΩ( )1+ 1.26mS 11kΩ( )

1MΩ500Ω+ 1MΩ

= −3.07

Ai = −RGgm

1+ gmRS

RDRD + R3

= −106

1.26mS1+1.26mS 11kΩ( )

39kΩ39kΩ + 500kΩ

= −6.14

vgs =Rin

RI + Rinv i =1.00v i | VGS −VP = −3.5− −4( ) = 0.500V

vgs ≤ 0.2 VGS − VP( ) 1+gmRS( ) | v i ≤ 0.2 0.5( )1+ 1.26mS 11kΩ( )[ ]=1.49 V

14.20 VGS = 0 → ID = IDSS = 5.00mA | VDS = 16 −1800ID = 7.00V | Active region operation is correct.

gm =2

−55mA 5mA( ) = 2.00mS | Assume λ = 0, ro = ∞.

Av = -gmRLRG

RI + RG

= − 2.00mS( ) 1.8kΩ 36kΩ( ) 10MΩ

10MΩ +5kΩ

= −3.43

Rin = 10.0 MΩ | Rout = RD ro =1.80 kΩ

Ai = −gmRGRD

RD + R3

= −2.00mS 10MΩ( ) 1.8kΩ

1.8kΩ+ 36kΩ

= −952

vgs = v i10MΩ

10MΩ +5kΩ

≤ 0.2VGS − VP → v i ≤ 1 V

14.21

IB =10 − 0.7( )V

20kΩ+ 80+ 1( )9.1kΩ=12.3µA | IC = 983 µA | VCE = 20− 9100IE =11.0 V


983µA= 2.04kΩ | ro =

100 + 11.0( )V983µA

= 113kΩ

RL = ro 1MΩ =113kΩ 1MΩ = 102kΩ | Rin = RB rπ = 20kΩ 2.04kΩ = 1.85 kΩ | Rout = ro =113 kΩ

Av = -gmRLRin

RI + Rin

= −40 983µA( )102kΩ( ) 1.85kΩ

250Ω + 1.85kΩ= −3530

Ai = −βoRB

RB + rπ

roro + R3

= −80

20kΩ20kΩ + 2.04kΩ

113kΩ113kΩ +1MΩ

= −7.37

vbe = v iRin

RI + Rin

= v i

20kΩ250Ω + 20kΩ

= 0.988v i | v i ≤5.00mV0.988

= 5.06 mV

14.22 From Table 14.3,

AvtCC ≅ Avt

CD =gmRL

1+ gmRL=

RL1

gm+ RL

=RL

re + RL for re =

1gm

14-455

14.23

rπ =80

0.4S= 200Ω | Assume VA = ∞, ro = ∞.

Rin = RB rπ + βo +1( )RL[ ]= 47kΩ 200Ω + 811kΩ( )[ ]= 29.8 kΩ

Rout =Rth + rπβo + 1

=47kΩ 10kΩ( )+ 200Ω

81=104 Ω

Av = +βo + 1( )RL

rπ + βo + 1( )RLRin

RI + Rin

=

811kΩ( )200Ω + 81 1kΩ( )

29.8kΩ10kΩ + 29.8kΩ

= 0.747

Ai = + βo + 1( )RB

RB + rπ + βo + 1( )RL

roro + RL

= 81

47kΩ47kΩ + 200Ω + 811kΩ( )

= 29.7

14.24

Assume λ = 0, ro = ∞. | Rin = RG = 2 MΩ | Rout =1gm

= 100 Ω

Av = +gmRL

1+ gmRL

RinRI + Rin

=

0.01 1kΩ( )1+ 0.01 1kΩ( )

2MΩ100kΩ+ 2MΩ

= 0.866

Ai = +gmRGro

ro + RL

= 0.01 2MΩ( )= 2x104

14.25 Defining v1 as the source node:

a( ) 2kΩ 100kΩ = 1.96kΩv i − v1( )106

+ 3.54 x10−3 v i − v1( )=v1

19603.541x10−3v i = 4.051x10

−3v1v1 = 0.874v i | Av = 0.874

Rin =v iii

=v i

10−6 v i − v1( )= 7.94 MΩ

Driving the output with current source ix :

Rout : ix =v1

106+

v12000

+ 3.54 x10−3v1

Rout =v1ix

= 247 Ω | b( ) Rin = ∞

g mvgs

2 k Ω 100 k Ω

vgs

+

-+

-

1 M Ω

v1v i

RG

14-456

14.26

IB =5 − 0.7( )V

1MΩ + 100 + 1( )430kΩ= 96.8nA | IC = 9.68µA | VCE =10 − 430000IE = 5.80V


9.68µA= 258kΩ | ro =

60 + 5.80( )V9.68µA

= 6.80MΩ - neglected

In the ac model, R1 appears in parallel with rπ . The circuit appears to be using a transistor with

rπ' = 500kΩ rπ =170kΩ and βo

' = gmrπ' = 40 9.68µA( )170kΩ = 65.8

RL = 500kΩ 430kΩ 500kΩ = 158kΩ | Rin = rπ' + βo

' +1( )RL = 170kΩ + 66.8 158kΩ( )= 10.7 MΩ

Av =βo

' + 1( )RLrπ

' + βo' + 1( )RL

RinRI + Rin

=

66.8 158kΩ( )170kΩ + 66.8 158kΩ( )

10.7MΩ500Ω+ 10.7MΩ

= +0.984

Rout = RE R2RI + rπ

'

βo' +1

= 430kΩ 500kΩ500Ω +170kΩ

66.8= 2.52 kΩ

vbe = v irπ

'

RI + rπ' + βo

' + 1( )RL= v i

170kΩ500Ω+ 170kΩ+ 66.8 158kΩ( )

= 1.58x10−2v i

v i ≤0.005V

1.58x10−2= 0.315 V

14.27

VEQ =1851kΩ

51kΩ + 100kΩ= 6.08V | REQ = 51kΩ 100kΩ = 33.8kΩ

IB =6.08 − 0.7 + 18( )V

33.8kΩ + 126( ) 4.7kΩ( )= 37.3µA | IC = 4.67 mA | VCE = 36 − 2000IC − 4700IE = 4.54 V


4.67mA= 669Ω | ro =

50 + 4.54( )V4.67mA

= 11.7kΩ

RB = R1 R2 = 51kΩ 100kΩ = 33.8kΩ | RL = R3 RE ro = 24kΩ 4.7kΩ 11.7kΩ = 2.94kΩ

Rin = RB rπ + βo +1( )RL[ ]= 33.8kΩ 669Ω + 126( )2.94kΩ[ ]= 31.0 kΩ

Av = +βo + 1( )RL

rπ + βo + 1( )RLRin

RI + Rin

=

126 2.94kΩ( )0.669kΩ + 126 2.94kΩ( )

31.0kΩ500Ω + 31.0 kΩ

= 0.982

vbe = v iRin

RI + Rin

rπrπ + βo +1( )RL

=

31.0kΩ500Ω+ 31.0 kΩ

0.669kΩ0.669kΩ+ 126 2.94kΩ( )

= 1.77x103v i

v i ≤0.005V

1.77x10−3= 2.82 V | Rout = RE

RB RI( )+ rπβo +1

= 4.7kΩ33.8kΩ 500Ω( )+ 669Ω

126= 9.22 Ω

14-457

14.28

VGS = 5V | ID =4x10−4

25−1( )2 = 3.2mA | VDS = 5 − −5( )=10V - Pinchoff region

operation is correct. | gm = 2 4x10−4( ) 3.2mA( )1+ 0.02 10( )[ ]= 1.75mS

ro =

10.02

+ 10

3.2VmA

=18.8kΩ − Cannot neglect! | RL =18.8kΩ 100kΩ =15.8kΩ

Rin = RG = 1 MΩ | Rout =1gm

ro = 555 Ω

Av = +Rin

RI + Rin

gmRL1+ gmRL

= +

1MΩ10kΩ + 1MΩ

1.75mS 15.8kΩ( )1+1.75mS 15.8kΩ( )

= 0.956

vgs = v iRin

RI + Rin

11+ gmRL

= v i

106Ω104 Ω+ 106Ω

11+1.75mS 15.8kΩ( )

= 0.0346v i

v i ≤0.2 5 −1( )0.0346

= 23.2 V But, vDS must exceed vGS − VTN ≅ VGS −VTN = 4V for pinchoff.

VDS =10 − vo =10 − 0.956v i ≥ 4 → v i ≤ 6.28 V − Limited by the Q - point voltages

14.29 βo = gmrπ = 3.54 mS 1MΩ( )= 3540 | RL = 2kΩ 100kΩ = 1.96kΩ

Av =βo + 1( )RL

rπ + βo + 1( )RL=

3540 + 1( ) 1.96kΩ( )1MΩ + 3540 +1( )1.96kΩ( )

= 0.874

Rin = rπ + βo + 1( )RL = 1MΩ + 3540 + 1( ) 1.96kΩ( )= 7.94 MΩ

Rout = 2kΩrπ

βo + 1( )= 2kΩ

106

3541( )= 247 Ω

14.30 v i ≤ 0.005 1+ gmRL( ) | RL = RE R7 ≅ RE

v i ≤ 0.005 1+ gmRL( )= 0.005 1+ gmRE( ) = 0.005 1+IC REVT

v i ≤ 0.005 1+ αFIEREVT

≅ 0.005 1+

IEREVT

v i ≤ 0.005 1+VREVT

= 0.005 1+

VRE0.025

= 0.005+ 0.2VRE

14-458

14.31

a( ) vbe = v i − vo | 0.005 ≤ 5 − vo → Av =vov i

≥4.995

5= 0.999

b( ) Av =βo + 1( )RE

rπ + βo + 1( )RE=

1

1 + rπβo +1( )RE

=1

1+ βoβo +1( )

rπβoRE

=1

1+ αogmRE

=1

1+ VTIERE

1

1+ VTIERE

≥ 0.999 →VT

IERE≤ 0.001 → IE RE ≥

0.025V0.001

= 25.0 V

14.32

vbe = v i − vo = 1− Av( )v i | 0.005 ≤ 1− Av( )7.5 → Av =vov i

≥7.5 − 0.005

7.5= 0.999333

From Prob. 14.30, Av =1

1+VT

IERL

| RL = RE 500Ω =500RE

500+ RE | Av =

1

1+VT

IERE

500+ RE500

1

1+VT

IERE

500+ RE500

≥ 0.999333→VT

IERE

500+ RE500

≤ 6.67x10−4

500IE RE500 + RE

≥0.025V

6.67x10−4= 37.5V | VCC ≥ IERE +0.7 +7.5

Some design possibilities are listed in the table below.

RE IE VCC VCC IE

100 Ω 450 mA 53 V 24 W

250 Ω 225 mA 64 V 16 W

360 Ω 179mA 73V 13 W

500 Ω 150 mA 83 V 12 W

750 Ω 125 mA 102 V 13 W

1000 Ω 113mA 120 V 14 W

2000 Ω 93.8 mA 196 V 18 W

Using a result near the minimum-power case in the table: RE = 510 Ω? ?? E?= 149 mA and VCC = 85 V.

Assuming βF = 50 : IB ≅149mA

51= 2.92 mA | Set IR1 = 5IB = 14.6mA ≅ 15mA

R1 =VE + VBE

IR1=

149mA 510Ω( )+ 0.715mA

= 5.07kΩ → 5.1 kΩ | IR2 = IR1 + IB ≅18mA

R2 =85 −VBE − VBE

IR2=

8.3V18mA

= 462Ω → 470 Ω

It is very difficult to achieve the required level of linearity!

14-459

14.33

a( ) Rin = R41gm

= 3kΩ1

0.5mS=1.20 kΩ | Rout = ∞ (assume λ = 0)

Av =gmRL

1+ gm RI R4( )R4

RI + R4

=

0.5mS 100kΩ( )1+ 0.5mS 50Ω 3kΩ( )

3kΩ50Ω + 3kΩ

= 48.0

Ai = 1R4

R4 +1gm

=3kΩ

3kΩ+ 2kΩ= 0.600

b( ) Av =0.5mS 100kΩ( )

1+ 0.5mS 5kΩ 3kΩ( )3kΩ

5kΩ + 3kΩ

= 9.68 | Rin = 5kΩ

10.5mS

= 1.43 kΩ

Rout = ∞ | Ai =5kΩ

5kΩ + 2kΩ= 0.714

14.34

a( ) rπ =100 0.025V( )

12.5µA= 200kΩ | gm = 40 12.5µA( )= 0.5mS

Rin = R4rπ

βo + 1= 100kΩ

200kΩ101

= 1.94 kΩ

Av =gmRL

1+ gm RI R4( )R4

RI + R4

=

0.5mS 100kΩ( )1+ 0.5mS 50Ω 100kΩ( )

100kΩ50Ω +100kΩ

= 48.7

Rout = ro 1+gm RI R4( )[ ]= 60V12.5µA 1+ 0.5mS 50Ω( )[ ]= 4.92 MΩ

Ai = αoR4

RI + R4

= 0.990

100kΩ50Ω +100kΩ

= 0.990

b( ) Av =0.5mS 100kΩ( )

1+ 0.5mS 2.2kΩ 100kΩ( )100kΩ

2.2kΩ +100kΩ

= 23.6 | Rin = 1.94kΩ - no change

Rout = ro 1+gm RI R4( )[ ]= 60V12.5µA 1+ 0.5mS 2.2Ω( )[ ]=10.1 MΩ

14.35 The voltage gain is approximately 0. The signal is injected into the collector and taken out of the emitter. This is not a useful amplifier circuit.

14.36 For R4 >> RI in Table 14.5,

AvtCB = Avt

CG = +gmRL

1+ gmRth= +

RL1

gm+ Rth

= +RL

re + Rth for re =

1gm

14-460

14.37

ID =2x10−4( )

2VGS + 1( )

2 | 15 + VGS

68kΩ= 10−4 VGS + 1( )

2 → VGS = −2.363V

ID =15 + VGS

68kΩ = 186µA | VDS = − 30 − 68kΩ+ 43kΩ( )ID[ ]= −9.35V |

Pinchoff region is correct. | gm =2 186µA( )2.36 −1

= 0.274mS | Rin = 68kΩ1

gm= 3.46kΩ

Rout = RD = 43kΩ | RL = 43kΩ 200kΩ = 35.4kΩ

Av =Rin

RI + RingmRL =

3.46kΩ0.250kΩ + 3.46kΩ

0.274mS( ) 35.4kΩ( )= 9.05

Ai = AvRI + Rin

R3= 9.05

3.71kΩ200kΩ

= 0.168 | vgs = v i

3.46kΩ0.250kΩ+ 3.46kΩ

≤ 0.2 VSG −1( )

v i3.46kΩ

0.250kΩ + 3.46kΩ≤ 0.2 2.36 −1( )→ v i ≤ 0.292 V

14.38

VGS = −12+ 33kΩ( )ID | VGS = −12 +3.3x104( )2x10−4( )

2VGS + 1( )

2

VGS = −2.68V & ID =2x10−4( )

2VGS + 1( )

2 = 282µA

VDS = − 24 − ID 33kΩ + 24kΩ( )[ ]= −7.93 V - Active region operation is correct.gm = 2 2x10

−4( ) 2.82x10−4( )= 3.36x10−4 S | R I RS = 0.5kΩ 33kΩ = 493ΩAssume λ = 0, ro = ∞ | RL = RD R3 = 24kΩ 100kΩ = 19.4kΩ

Av =gmRL

1+ gm RI RS( )RS

RI + RS

=

0.336mS 19.4kΩ( )1+ 0.336mS 493Ω( )

33kΩ500Ω+ 33kΩ

= 5.51

Ai = 1RS

RS +1gm

RDRD + R3

=

33kΩ33kΩ + 2.98kΩ

24kΩ24kΩ+ 100kΩ

= 0.178

Rin = RS1gm

= 2.73 kΩ | Rout = RD = 24kΩ

vgs = viRIN

RI + RIN≤ 0.2VGS +1 | v i

2.73kΩ0.5kΩ + 2.73kΩ

≤ 0.2 1.68( )→ vs ≤ 0.398 V

14-461

14.39

IB =9− 0.7( )V

100kΩ + 50 +1( )82kΩ=1.94µA | IC = 96.9 µA

VCE =18 − 82000IE − 39000IC = 6.12 V | Active region operation is correct.

gm = 40IC = 3.88mS | rπ =βogm

= 12.9kΩ | ro =50 + 6.12( )V

96.9µA= 579kΩ - neglected

RI RE = 0.5kΩ 82kΩ = 497Ω | RL = RC R3 = 39kΩ 100kΩ = 28.1kΩ

Av =gmRL

1+ gm RI RE( )RE

RI + RE

=

3.88mS 28.1kΩ( )1+ 3.88mS 497Ω( )

82kΩ500Ω+ 82kΩ

= 37.0

Rin = 82kΩrπ

βo +1= 252Ω | Ai = Av

RI + RinR3

= 37.0500Ω+ 252Ω

100kΩ= 0.278

Rout = RC = 39.0 kΩ | veb = v iRin

RI + Rin≤ 5.00mV | 0.335v i ≤ 5.00mV | v i ≤ 14.9 mV

14.40

IB =9− 0.7( )V

1000kΩ + 50 +1( )820kΩ=194nA | IC = 9.69 µA

VCE =18 − 820000IE − 390000IC = 6.12 V | Active region is correct.


=129kΩ | ro =50+ 6.12( )V

9.69µA= 5.79MΩ - neglected

RI RE = 5kΩ 820kΩ = 4.97kΩ | RL = RC R3 = 390kΩ 1MΩ = 281kΩ

Av =gmRL

1+ gm RI RE( )RE

RI + RE

=

0.388mS 281kΩ( )1+ 0.388mS 4.97kΩ( )

820kΩ5kΩ+ 820kΩ

= 37.0

Rin = 820kΩrπ

βo +1= 2.52 kΩ | Ai = Av

RI + RinR3

= 37.05kΩ + 2.52kΩ

1MΩ= 0.278

Rout = RC = 390 kΩ | veb = v iRin

RI + Rin≤ 5.00mV | 0.335v i ≤ 5.00mV | v i ≤ 14.9 mV

14.41

VGS = −3900ID = −39005x10−4( )

2VGS + 2( )

2 | VGS = −0.975 VGS + 2( )2 → VGS = −0.9915V

ID =5x10−4( )

2VGS + 2( )

2 = 254µA | VDS =15 − 23.9kΩID = 8.92V - Pinched off.

14-462

gm =2 254µA( )2 − 0.992

= 0.504mS | Rin = 3.9kΩ1gm

= 1.32kΩ | Rout = RD = 20kΩ

RL = 20kΩ 51kΩ =14.4kΩ

Av =gmRL

1+ gm RI RS( )RS

RI + RS

=

0.504mS 14.4kΩ( )1+ 0.504mS 0.796kΩ( )

3.9kΩ1kΩ+ 3.9kΩ

= 4.12

Ai = AvRI + Rin

R3= 4.121kΩ+ 1.32kΩ

51kΩ= 0.187

vgs = v i1.32kΩ

1kΩ + 1.32kΩ≤ 0.2 VGS + 2( ) | v i

1.32kΩ1kΩ +1.32kΩ

≤ 0.2 −0.992 + 2( )→ v i ≤ 0.354 V

14.42

For R th >1gm

, Av ≅RLRth

For large R th, all of the Thevenin equivalent source

current, v thRth

, goes into the transistor source terminal.

14.43

Rin =rπ + 1.5kΩ

βo +1 | rπ =

75 0.025V( )1mA

= 1.88kΩ | Rin =1.88kΩ + 1.5kΩ

76= 44.5 Ω

14.44

gm =2−2

1mA 5mA( ) = 2.24mS | Rin =1gm

= 447 Ω

14.45

gm = 2 1.25mA( ) 1mA( ) = 1.58mS | Rin =1gm

= 633 Ω

14.46

a( ) Rout = ro 1+βoRE

rπ + RE

| IE =

15V − 0.7V143kΩ

= 100µA | For βF =100, IC = 99.0µA

rπ =100 0.025V( )

99.0µA= 25.3kΩ | ro ≅

50V99.0µA

= 505kΩ

Rout = 505kΩ 1+100 143kΩ( )

25.3kΩ + 143kΩ

= 43.4 MΩ b( ) 0 V

14-463

c( ) IE =15V − 0.7V

15kΩ= 953µA | For βF =100, IC = 944µA | rπ =

100 0.025V( )944µA

= 2.65kΩ

ro ≅50V

944µA= 53.0kΩ | Rout = 53.0kΩ 1+

100 15kΩ( )15kΩ + 2.65kΩ

= 4.56 MΩ | VCB ≥ 0 V

14.47

Rout = βo + 1( )ro = βo +1( )VA + VCE

IC

= 126

50 +10.749.6µA

= 154 MΩ

14.48

Rin = 0.5MΩ | Av =104620 = 200 | Fairly large Rin, large gain

A common - emitter amplifier operating at a low current can achieve both alarge gain and input resistance. Av ≅ 20VCC → VCC = 10VAchieving this gain with an FET is much more difficult :

Av ≅VDD

VGS −V TN=

VDD0.25V

→ VDD ≅ 50V which is unreasonably large.

14.49

Rin = 10MΩ | Av = 102620 = 20 | Large Rin, moderate gain

These requirements are readily met by a common - source amplifier.

For example, Av ≅VDD

VGS −V TN=

15V0.5V

= 30.

A common - emitter stage operating at a low collector current with

an unbypassed emitter resistor RE ≅10MΩ

100=100kΩ

is a second possibility,

but the circuit will require careful design.

14.50 An inverting amplifier with a gain of 40 dB is most easily achieved with a common - emitterstage : Av ≅ 10VCC → VCC =10 V . The input resistance can be achieved by shunting the

input with a 5 - Ω resistor. Setting rπ = 5 Ω would require IC ≅100 0.025V( )

5Ω= 0.5A and would

waste a large amount of power to achieve the required input resistance.

14.51 0 - dB gain corresponds to a follower ( Av = 1).

For an emitter - follower, R in ≅ βo + 1( )RL ≅101 20kΩ( )= 2.02 MΩ. So an BJTcannot meet the input resistance requirement. A source follower provides a gain of approximately 1 and can easily achieve the required input resistance.

14.52

14-464

A non - inverting amplifier with a gain of 20 and an input resistance of 5 k Ω shouldbe readily achievable with either a common - base or common - gate amplifier with proper choice of operating point. The gain of 10 is easily achieved with either the

FET or BJT design estimate : Av ≅VDD

VGS − VTN or Av ≅ 10VCC . Rin ≅

1gm

= 5 kΩ is within

easy reach of either device. The gain and input resistance can also be easily metwith either a common - emitter, or common - source stage with a resistor shunt at the input.

14.53

A gain of 0.97 and an input resistance of 400k Ω should be achievable with

either a source - follower or emitter - follower. For the FET, Av ≅gmRL

1+ gmRL= 0.97

requires gmRL = 33.3 : 2IDRL

VGS − VTN= 33.3 → IDRL = 8.3V for a design with VGS − VTN = 0.5V .

The BJT can achieve the required gain with a much lower power supply and

can still meet the Rin requirement : Rin ≅ βoRL ≅100 5kΩ( )= 500kΩ.

Av ≅gmRL

1+ gmRL= 0.97 | gmRL = 33.3→ IC RE = 33.3 0.025( )= 0.833 V .

The requirements can be met with careful bias circuit design and specificationof a BJT with minimum current gain of at least 100.

14.54

Av = 106620 = 2,000. This value of voltage gain approaches the amplification factor

of the BJTs : Av ≤ µ f = 40VA = 40 75( )= 3000. Such a large gainrequirement cannot be met with single - transistor BJT amplifiers using theresistive loaded amplifiers in this chapter (Remember the 10 - VCC limit). FETstypically have much lower values of µf and are at an even worse disadvantage.None of the single - transistor amplifier configurations can meet the gain requirements.

14.55

Rout =Rth + rπβo + 1

| Assuming Rth ≅ RI and rπ = 0, Rout ≥RI

βo +1=

250151

=1.66Ω

14-465

14.56 Rin = rπ + βo + 1( )RE ≅ rπ + βoRE = rπ 1+ gmRE( ) | rπ' = rπ 1+ gmRE( )

gm' =

icv i

=βo

rπ + βo +1( )RE≅

βorπ + βoRE

=βo

rπ 1+ gmRE( )=

gm1+ gmRE

ro' =

icvc v i =0

= ro 1+βoRE

rπ + RE

≅ ro 1+

βoRErπ

= ro 1+ gmRE( ) for rπ >> RE

βo' = gm

' rπ' =

gm1+ gmRE

rπ 1+ gmRE( )= βo | µ f' = gm' ro' =

gm1+ gmRE

ro 1+ gmRE( )= µ f

14.57 *Problem 14.57 - Common-Emitter Amplifier 5mV VCC 6 0 DC 9 VEE 4 0 DC -9 VS 1 0 SIN(0 0.005 1K) C1 1 2 1U RB 2 0 10K RC 6 5 3.6K RE 3 4 2K C2 3 0 50U C3 5 7 1U R3 7 0 10K Q1 5 2 3 NBJT .OP .TRAN 1U 5M .FOUR 1KHZ V(7) .MODEL NBJT NPN IS=1E-16 BF=100 VA=70 .PROBE V(7) .END *Problem 14.57 - Common-Emitter Amplifier 10mV VCC 6 0 DC 9 VEE 4 0 DC -9 VS 1 0 SIN(0 0.01 1K) C1 1 2 1U RB 2 0 10K RC 6 5 3.6K RE 3 4 2K C2 3 0 50U C3 5 7 1U R3 7 0 10K Q1 5 2 3 NBJT .OP .TRAN 1U 5M .FOUR 1KHZ V(7) .MODEL NBJT NPN IS=1E-16 BF=100 VA=70 .PROBE V(7) .END *Problem 14.57 - Common-Emitter Amplifier 15mV VCC 6 0 DC 9 VEE 4 0 DC -9 VS 1 0 SIN(0 0.015 1K) C1 1 2 1U RB 2 0 10K

14-466

RC 6 5 3.6K RE 3 4 2K C2 3 0 50U C3 5 7 1U R3 7 0 10K Q1 5 2 3 NBJT .OP .TRAN 1U 5M .FOUR 1KHZ V(7) .MODEL NBJT NPN IS=1E-16 BF=100 VA=70 .PROBE V(7) .END

Results: 1 kHz 2 kHz 3 kHz THD

5 mV 5.8 mV 0.335 mV (5.7%) 0.043 mV (0.74% 5.9% 10 mV 12.4 mV 1.54 mV (12.5%) 0.258 mV (2.1%) 12.8% 15 mV 20.6 mV 4.32 mV (21%) 1.18 mV (5.4%) 22%

14.58

*Problem 14.58 - Output Resistance VCC 2 0 DC 10 IB1 0 1 DC 10U Q1 2 1 0 NBJT IB2 0 3 DC 10U RE 4 0 10K Q2 2 3 4 NBJT .OP .DC VCC 10 20 .025 .MODEL NBJT NPN IS=1E-16 BF=60 VA=20 .PRINT DC IC(Q1) IC(Q2) .PROBE IC(Q1) IC(Q2) .END

Results: A small value of Early voltage has been used deliberately to accentuate the results. Note that the transistors have significantly different values of βF because of the collector-emitter voltage differences and low value of VA.

NAME Q1 Q2 MODEL NBJT NBJT IB 1.00E-05 1.00E-05 IC 8.77E-04 6.72E-04 VBE 7.61E-01 7.61E-01 VBC -9.24E+00 -2.41E+00 VCE 1.00E+01 3.18E+00 BETADC 8.77E+01 6.72E+01 GM 3.39E-02 2.60E-02 RPI 2.59E+03 2.59E+03 RO 3.33E+04 3.33E+04

14-467

From SPICE : Rout1 =20−10( )

1.17 − 0.877( )VmA

= 34.1 kΩ | Rout2 =20−10( )

903− 673( )VµA

= 43.5 kΩ

For circuit 1 : Rout1 = ro1 = 33.3 kΩ

For circuit 2 : Rout2 = ro 2 1+βoRE

Rth + rπ + RE

+ Rth + rπ( ) RE (See Eq. 14.28)

But Rth = ∞ → Rout2 = ro2 + RE = 33.3kΩ + 10kΩ = 43.3 kΩ

14.59

v th = GmRth = −gm

1 + gmRSro 1+ gmRS( )[ ]v i = −µ f v i = − 0.5mS( ) 250kΩ( )v i = −125v i

Rth = ro 1+ gmRS( ) = ro + µ f RS = 250kΩ + 125 18kΩ( )= 2.50 MΩ

14.60

v th = v irπ

RI + rπ1 + gmro( )=

rπRI + rπ

1+ µ f( )v i = 50kΩ270Ω + 50kΩ 1 + 2mS 250kΩ( )[ ]v i = 498v i

Rth ≅ ro 1+βoRI

RI + rπ

= 250kΩ 1+

100 270Ω( )270Ω + 50kΩ

= 384 kΩ

14.61

v th = v iβo + 1( )ro

RI + rπ + βo +1( )ro= v i

1RI

βo +1( )ro+

rπβo + 1( )ro

+ 1= v i

1gmRI

βo +1( )µ f+

βoβo +1( )µ f

+1≅ v i

Rth ≅RI + rπβo +1

ro ≅RI + rπβo + 1

14.62

a( ) y21 =i2v1 v2 =0

| i2 = −gmv1 | y21 = −gm | y12 =i1v2 v1 = 0

| i1 = 0 | y12 = 0 | y12y21

= 0

b( ) y21 = −400 µS | y12 = 0 | y21 >> y12

14.63

a( ) z21 =v2i1 i2 = 0

| v2 = i1RB( )βo + 1( )RE

RB + rπ + βo + 1( )RE | z21 =

βo + 1( )RERBRB + rπ + βo + 1( )RE

z12 =v1i2 i1 =0

=RE

RE +RB + rπβo + 1( )

RBβo +1( )

=RE RB

RB + rπ + βo + 1( )RE | z21 = βo + 1( )z12

b( ) z21 =101( ) 2.4kΩ( )150kΩ( )

2.4kΩ+ 10kΩ+ 101( )150kΩ= 2.40 kΩ | z12 =

z21101

= 23.7 Ω

14.64

14-468

a( ) h21 =i2i1 v2= 0

=−αoRE

RE +rπ

βo + 1

=−βoRE

rπ + βo + 1( )RE≅ −1

h12 =v1v2 i1= 0

=go

gm + gπ + go + gE=

1

µ f +µ fβo

+ 1+µ f

gmRE

≅1

µ f | h21 = −µ f h12

b( ) h21 =−100 3.9kΩ( )

33.3kΩ + 101( )3.9kΩ= −0.922 | h12 =

1

2400 +2400100

+ 1+240011.7

= 3.8x10−4

14.65

a( ) g21 =v2v1 i2= 0

= +gmRD | g12 =i1i2 v1 = 0

= −RD

RD + ro | g21 = −gm RD + ro( )g12 ≅ −µ f g12

b( ) g21 = +0.5mS 100kΩ( )= 50 | g12 = −100kΩ

100kΩ + 500kΩ= −0.167

14.66

a( ) y21 =i 2v1 v 2= 0

| i 2 =βo

rπ + βo + 1( )REv1 | y21 =

βorπ + βo + 1( )RE

y12 =i1v2 v 1= 0

| i1 = −i2RE

RE + rπ= − v2

Rout

RERE + rπ

≅ − v2

ro 1+βoRE

RE + rπ

RERE + rπ

y12 ≅ −1

βo +1( )ro for βo +1( )RE >> rπ

y12y21

= −1

βo + 1( )rorπ + βo + 1( )RE

βo≅

REβoro

for βo +1( )RE >> rπ | REβoro

=gmREβoµ f

14-469

At the output node vo : gmvx = go + GL( )vo − govx | vo =gm + gogo + GL

vx

ix = gmvx + go vx − vo( ) | ixvx

= gm + go 1−gm + gogo + GL

= gm + go

GL − gmgo + GL

ixvx

= gm + goGL − gmgo + GL

= GL

gm + gogo + GL

| Rin =vxix

=1gm

1+RLro

1+1µf

≅1gm

1+RLro

14.69

IC =1005 − 0.7

104 +101 103( )= 3.87mA | gm = 40IC = 0.155S | rπ =

100gm

= 645Ω

RL = 1kΩ 20kΩ = 952Ω | RE = 1kΩ 20kΩ = 952Ω

Av1 = −βoRL

rπ + βo + 1( )RE= −

100 952Ω( )645Ω +101 952Ω( )

= −0.984

Av2 =βo + 1( )RE

rπ + βo + 1( )RE=

101 952Ω( )645Ω +101 952Ω( )

= 0.993

The small - signal requirement limits the output signal to :

vbe = v i − vo2 = vi 1− 0.993( )= 0.007v i | v i ≤0.0050.007

= 0.714V

vo1 ≤ 0.984 0.714V( )= 0.703VWe also need to check VCB : VC = 5 − 3.87mA 1kΩ( )= 1.13V and VB = −104 IB = −0.387V .The total collector - base voltage of the transistor is therefore : VCB = 1.52V − 0.984v i − v i .We require VCB ≥ 0 for forward - active region operation. Therefore : v i ≤ 0.766 V .The small - signal limit is the most restrictive.

14.70

*Problem 14.70 - Common-Collector Amplifier 14.48(a) VCC 6 0 DC 18 VEE 7 0 DC -18 VI 1 0 AC 1 *For Output Resistance *VI 1 0 AC 0 *VO 8 0 AC 1 RI 1 2 500 C1 2 3 10UF R1 3 0 51K R2 6 3 100K RE 4 7 4.7K RC 6 5 2K C3 5 0 10UF C2 4 8 47UF R3 8 0 24K Q1 5 3 4 NBJT

14-470

.OP

.AC LIN 1 10KHZ 10KHZ

.MODEL NBJT NPN IS=1E-16 BF=125 VA=50

.PRINT AC VM(8) VP(8) VM(3) VP(3) IM(VI) IP(VI) IM(C2) IP(C2)

.END

Results: Q-point: (4.67 mA, 4.57 V), Av = 0.982, Rin = 31.1 kΩ???Rout?= 9.12 Ω

rπ =125 0.025V( )

4.67mA= 669Ω | ro =

50+ 4.57( )V4.67mA

= 11.7kΩ


Rin = RB rπ + βo + 1( )RL[ ]= 33.8kΩ 669Ω + 126( )2.94kΩ[ ]= 31.0 kΩ

Av = +βo +1( )RL

rπ + βo +1( )RLRin

RI + Rin

=

126 2.94kΩ( )0.669kΩ +126 2.94kΩ( )

31.0kΩ500Ω+ 31.0 kΩ

= 0.982

Rout = RERB RI( )+ rπ

βo + 1= 4.7kΩ

33.8kΩ 500Ω( )+ 669Ω126

= 9.22 Ω

14.71

*Problem 14.71 - Common-Emitter Amplifier - 14.48(c) VCC 6 0 DC 12 VI 1 0 AC 1 *VI 1 0 AC 0 *VO 7 0 AC 1 RI 1 2 1K C1 2 3 2.2UF R1 6 3 20K R2 3 0 62K RE 6 4 3.9K C2 6 4 47UF RC 5 0 8.2K C3 5 7 10UF R3 7 0 100K Q1 5 3 4 PBJT .OP .AC LIN 1 5KHZ 5KHZ .MODEL PBJT PNP IS=1E-16 BF=75 VA=60 .PRINT AC VM(7) VP(7) VM(3) VP(3) IM(VI) IP(VI) IM(C3) IP(C3) .END

Results: Q-point: (525 µA, 5.62V), Av = -110, Rin = 3.16kΩ? ?Rout?= 7.69 kΩ

ro =60 + 5.62525x10−6

= 125kΩ | rπ =75 0.025( )525x10−6

= 3.57kΩ

RL = ro 8.2kΩ 100kΩ =125kΩ 8.2kΩ 100kΩ = 7.15kΩ

Rin = 15.1kΩ 3.57kΩ = 2.88 kΩ | Rout = ro 8.2kΩ = 7.70 kΩ | gm = 40IC = 21.0 mS

Av = −gmRLRin

RI + Rin

= − 21.0mS( ) 7.15kΩ( ) 2.88kΩ

1kΩ+ 2.88kΩ

= −111

14.72

14-471

*Problem 14.72 - Common-Source Amplifier - 14.48(d) VDD 4 0 DC 18 VI 1 0 AC 1 *For output resistance *VI 1 0 AC 0 *VO 7 0 AC 1 RI 1 2 1K C1 2 3 2.2U R1 3 0 500K R2 4 3 1.4MEG R4 6 0 27K C2 6 0 47U RD 4 5 75K C3 5 7 10U R3 7 0 470K M1 5 3 6 6 NMOSFET .OP .AC LIN 1 5KHZ 5KHZ .MODEL NMOSFET NMOS VTO=1 KP=250U LAMBDA=0.02 .PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VI) IP(VI) IM(C3) IP(C3) .END

Results: Q-point: (106µA, 7.14V), Av = -14.2, Rin = 369kΩ, Rout = 65.8 kΩ

ro =50 + 7.14106x10−6

= 539kΩ | gm = 2 250x10−6( )106x10−6( )1+ 0.02 7.14( )[ ]= 246µS

RL = ro RD 470kΩ = 539kΩ 75kΩ 470kΩ = 57.7kΩ

Rin = R1 R2 = 500kΩ 1.4MΩ = 368kΩ | Rout = ro 75kΩ = 65.8 kΩ

Av = −gmRLRin

RI + Rin

= − 0.246mS( ) 57.7kΩ( ) 368kΩ

1kΩ + 368kΩ

= −14.2

14.73 *Problem 14.73 - Common-Gate Amplifier - 14.48(e) VSS 5 0 DC 12 VDD 6 0 DC -12 VI 1 0 AC 1 *For Output Resistance *VI 1 0 AC 0 *VO 7 0 AC 1 RI 1 2 500 C1 2 3 10U R1 5 3 33K RD 6 4 24K C2 4 7 47U R3 7 0 100K M1 4 0 3 3 PFET .OP

14-472


.MODEL PFET PMOS KP=200U VTO=-1 LAMBDA=0.02


.END Results: Q-point: (286 µA, 7.72V), Av = +5.50, Rin = 2.73 kΩ, Rout = 21.8 kΩ

gm = 2 2x10−4( ) 2.86x10−4( )1+ 0.02 7.72( )[ ]= 3.63x10−4 S | R I RS = 0.5kΩ 33kΩ = 493Ω

ro =50+ 7.722.86x10−4

= 202kΩ | RL = ro RD R3 = 202kΩ 24kΩ 100kΩ = 17.7kΩ | Rin = RS1gm

= 2.54 kΩ

Rout = ro RD = 21.4 kΩ | Av =gmRL

1+ gm RI RS( )RS

RI + RS

=

0.363mS 17.7kΩ( )1+ 0.363mS 0.493kΩ( )

33kΩ500Ω+ 33kΩ

= 5.37

14.74

IC =1005 − 0.7

500kΩ+ 500kΩ + 101( )430kΩ= 9.68µA | VCE = 5 − IC 430kΩ( )− −5( )= 5.80V

rπ =100

40 9.68µA( )= 258kΩ | ro =

60 +5.89.68x10−6

= 6.80MΩ | RL = 500kΩ 430kΩ 500kΩ ro = 154kΩ

Absorb R1 inro the transistor : rπ' = rπ R1 =170kΩ | βo

' = gmrπ' = 65.8

Rin = rπ' + βo

' + 1( )RL =170kΩ + 66.8 154kΩ( )=10.5 MΩ

Rout = 430kΩ 500kΩ roRI + rπ

'

βo' +1

= 2.53 kΩ

Av = +βo

' + 1( )RLRI + rπ

' + βo' +1( )RL

=66.8 154 kΩ( )

0.5kΩ+ 170kΩ+ 66.8 154kΩ( )= 0.984

*Problem 14.74 - Common-Collector Amplifier - 14.1(f) VCC 7 0 DC 5 VEE 8 0 DC -5 VI 1 0 AC 1 *For Output Resistance *VI 1 0 AC 0 *VO 6 0 AC 1 RI 1 2 500 C1 2 3 10UF R1 3 5 500K R2 5 0 500K RE 4 8 430K C2 4 5 47UF C3 4 6 10UF R3 6 0 500K Q1 7 3 4 NBJT .OP

14-473


.MODEL NBJT NPN IS=1E-16 BF=100 VA=60


.END

Results: Q-point: (9.81 µA, 5.74V), Av = 0.983, Rin = 11.0 MΩ???Rout?= 2.58 kΩ

14.75 *Problem 14.75 - Common-Source Amplifier - 14.48(g) VDD 6 0 DC 20 VI 1 0 AC 1 *For output resistance *VI 1 0 AC 0 *VO 7 0 AC 1 RI 1 2 500 C1 2 3 2.2UF RG 3 0 1MEG R4 4 0 11K C2 5 7 47UF RD 6 5 39K R3 7 0 500K J1 5 3 4 NJFET .OP .AC LIN 1 4KHZ 4KHZ .MODEL NJFET NJF BETA=1.25M VTO=-4 LAMBDA=0.02 .PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VI) IP(VI) IM(C2) IP(C2) .END

Results: Q-point: (319 µA, 4.03V), Av = -3.02, Rin = 1.00 MΩ, Rout = 38.4 kΩ

VGS = − 11kΩ( )ID = − 11kΩ( ) 20mA( ) 1−VGS−4

2

→ VGS = −3.50V , ID = −VGS

11kΩ= 318 µA

VDS = 20− ID 11kΩ + 39kΩ( )= 4.10V | Active region operation is correct.

gm =2−4

20mA 318µA( ) = 1.26mS | Assume λ = 0, ro = ∞. | RL = 39kΩ 500kΩ = 36.2kΩ

14.76

*Problem 14.76 - Common-Base Amplifier - 14.48(h) VCC 7 0 DC -9 VEE 4 0 DC 9 VI 1 0 AC 1 *For output resistance *VI 1 0 AC 0 *VO 8 0 AC 1 RI 1 2 500 C2 2 3 47UF RB 5 0 100K RE 3 4 82K C1 5 0 4.7UF RC 7 6 39K C3 6 8 10UF

14-474

R3 8 0 100K Q1 6 5 3 PBJT .OP .AC LIN 1 12KHZ 12KHZ .MODEL PBJT PNP IS=1E-16 BF=50 VA=50 .PRINT AC VM(3) VP(3) VM(8) VP(8) IM(VI) IP(VI) IM(C3) IP(C3) .END

Results: Q-point: (97.2 µA, 6.10V), Av = 35.5, Rin = 273 Ω??Rout?= 38.1 kΩ


= 12.9kΩ | ro =50 + 6.10( )V

97.2µA= 577kΩ

Rth = RI RE = 0.5kΩ 82kΩ = 497Ω | RL = ro 1+ gmRth( ) RC R3 = 1.69MΩ 39kΩ 100kΩ = 27.6kΩ

Av =gmRL

1+ gm RI RE( )RE

RI + RE

=

3.89mS 27.6kΩ( )1+ 3.89mS 497Ω( )

82kΩ500Ω+ 82kΩ

= 36.4

Rin = 82kΩrπ

βo +1= 252Ω | Rout = ro 1+ gmRth( ) RC = 38.1 kΩ

14-475

14.77 *Problem 14.77 - Common-Source Amplifier 14.48(j) VDD 6 0 DC 18 VI 1 0 AC 1 *For output resistance *VI 1 0 AC 0 *VO 7 0 AC 1 RI 1 2 1K C1 2 3 2.2UF R2 6 3 2.2MEG R1 3 0 2.2MEG RS 6 4 22K C2 6 4 47UF RD 5 0 18K C3 5 7 10UF R3 7 0 470K M1 5 3 4 4 PFET .OP .AC LIN 1 7.5KHZ 7.5KHZ .MODEL PFET PMOS KP=400U VTO=-1 LAMBDA=0.02 .PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VI) IP(VI) IM(C3) IP(C3) .END

Results: Q-point: (268µA, 8.60V), Av = -8.29, Rin = 1.10 MΩ??Rout?= 16.4 kΩ

ro =50 + 8.60268x10−6

= 219kΩ | gm = 2 400x10−6( )268x10−6( )1+ 0.02 8.60( )[ ]= 501µS

RL = ro RD 470kΩ = 219kΩ 18kΩ 470kΩ =16.1kΩ

Rin = R1 R2 = 2.2MΩ 2.2MΩ = 1.1 MΩ | Rout = ro 18kΩ = 16.6 kΩ

Av = −gmRLRin

RI + Rin

= − 0.501mS( )16.1kΩ( ) 1.1MΩ

1kΩ+ 1.1MΩ

= −8.06

14.78

*Problem 14.78 - Common-Gate Amplifier - 14.48(k) VDD 5 0 DC 15 VI 1 0 AC 1 *For output resistance *VI 1 0 AC 0 *VO 6 0 AC 1 RI 1 2 1K C1 2 3 2.2U R1 3 0 3.9K RD 4 5 20K C2 4 6 47U RS 6 0 51K M1 4 0 3 3 NMOSFET .OP .AC LIN 1 20KHZ 20KHZ .MODEL NMOSFET NMOS VTO=-2 KP=500U LAMBDA=0.02 .PRINT AC VM(3) VP(3) VM(6) VP(6) IM(VI) IP(VI) IM(C2) IP(C2) .END

Results: Q-point: (268µA, 8.60V), Av = 4.26, Rin = 1.27 kΩ???Rout?= 18.8 kΩ

14-476

gm = 2 5x10−4( ) 268x10−6( )1+ 0.02 8.60( )[ ]= 0.560mS | Rin = 3.9kΩ 1gm

= 1.22kΩ

ro =50 + 8.60268x10−6

= 219kΩ | Rth = RI RS = 981Ω

Rout = ro 1+ gmRth( ) RD = 18.9 kΩ | RL = Rout 51kΩ = 13.8kΩ

Av =gmRL

1+ gm RI RS( )RS

RI + RS

=

0.560mS 13.8kΩ( )1+ 0.560mS 0.981kΩ( )

3.9kΩ1kΩ+ 3.9kΩ

= 3.97

14.79 *Problem 14.79 - JFET Common-Source Amplifier - 14.48(m) VDD 5 0 DC -16 VI 1 0 AC 1 *For output resistance *VI 1 0 AC 0 *VO 6 0 AC 1 RI 1 2 5K C1 2 3 2.2U R1 3 0 10MEG RD 4 5 1.8K C2 4 6 10U R3 6 0 36K J1 4 3 0 PJFET .OP .AC LIN 1 3KHZ 3KHZ .MODEL PJFET PJF BETA=200U VTO=-5 LAMBDA=0.02 .PRINT AC VM(3) VP(3) VM(6) VP(6) IM(VI) IP(VI) IM(C2) IP(C2) .END

Results: Q-point: (5.59 ?A, 5.93 V), Av = -3.27, Rin = 10.0 MΩ?? Rout?= 1.53 kΩ

gm =2

−55mA 5.59mA( ) = 2.11mS | ro =

50 + 5.935.59x10−3

=10.0kΩ

Av = -gmRLRG

RI + RG

= − 2.11mS( ) 10.0kΩ 1.8kΩ 36kΩ( ) 10MΩ

10MΩ + 5kΩ

= −3.09

Rin = 10.0 MΩ | Rout = RD ro = 1.53 kΩ

14.80 *Problem 14.69 - Common-Emitter Amplifier - 14.48(n) VCC 7 0 DC 10 VEE 6 0 DC -10 VI 1 0 AC 1 *For output resistance *VI 1 0 AC 0 *VO 8 0 AC 1 RI 1 2 250 C1 2 3 4.7UF RB 3 0 20K R4 4 6 9.1K C2 4 0 100UF L 7 5 1H C3 5 8 1UF

14-477

R3 8 0 1MEG Q1 5 3 4 NBJT .OP .AC LIN 1 500KHZ 500KHZ .MODEL NBJT NPN IS=1E-16 BF=80 VA=100 .PRINT AC VM(3) VP(3) VM(8) VP(8) IM(VI) IP(VI) IM(C3) IP(C3) .END

Results: Q-point: (979 µA, 11.0 V), Av = -3420, Rin = 2.09 kΩ???Rout?= 113 kΩ

rπ =80 0.025V( )

979µA= 2.04kΩ | ro =

100 +11.0( )V979µA

=113kΩ | Rout = ro =113 kΩ

RL = ro 1MΩ =113kΩ 1MΩ =102kΩ | Rin = RB rπ = 20kΩ 2.04kΩ = 1.85 kΩ

Av = -gmRLRin

RI + Rin

= −40 979µA( ) 102kΩ( ) 1.85kΩ

250Ω+ 1.85kΩ= −3520

14.81 *Problem 14.81 - Source Follower - 14.48(o) VDD 5 0 DC 5 VSS 6 0 DC -5 VI 1 0 AC 1 *For output resistance *VI 1 0 AC 0 *VO 7 0 AC 1 RI 1 2 10K C1 2 3 2.2U R1 3 0 1MEG L 4 6 100mH C3 4 7 4.7U R3 7 0 100K M1 5 3 4 4 NMOSFET .OP .AC LIN 1 100KHZ 100KHZ .MODEL NMOSFET NMOS VTO=1 KP=400U LAMBDA=0.02 .PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VI) IP(VI) IM(C3) IP(C3) .END

Results: Q-point: (3.84 mA, 10.0 V), Av = 0.953, Rin = 1.00 MΩ?? Rout?= 504 Ω

gm = 2 4x10−4( ) 3.84x10−3( )1+ 0.02 10( )[ ]=1.92mS

ro =

10.02

+ 10

3.84VmA

= 15.6kΩ − Cannot neglect! | RL =15.6kΩ 100kΩ =13.5kΩ

Rin = RG =1 MΩ | Rout =1gm

ro = 504 Ω

Av = +Rin

RI + Rin

gmRL1+ gmRL

= +

1MΩ10kΩ + 1MΩ

1.92mS 13.5kΩ( )1+ 1.92mS 13.5kΩ( )

= 0.953

14-478

14.82

a( ) VEQ =18500 kΩ

1.4MΩ + 500kΩ= 4.74V | REQ = 500kΩ 1.4 MΩ = 368kΩ

4.74 =VGS + 27000 ID =1+2IDS

250 x10−6+ 27000 ID → ID =104µA

VDS =18 − ID 75kΩ+ 27kΩ( )= 7.39V | Active region operation is correct.

gm = 2 250 x10−6( )104 x10−6( )= 0.228mS | ro = 50 + 7.39104 µA = 552kΩ | RG = R1 R2 = 368 kΩ

C1 ≥101

2πf RI + RG( )= 10

2π 400Hz( ) 1kΩ + 368kΩ( ) | C1 ≥ 0.0108 µF → 0.01 µF

C2 ≥101

2πf RS1

gm

= 10

2π 400Hz( ) 27kΩ 10.228 mS

| C2 ≥ 1.05 µF →1.0 µF

C3 ≥101

2πf RD ro( )+ R3[ ]= 10

2π 400Hz( ) 66.0kΩ + 470kΩ( ) | C3 ≥ 7.42 nF → 8200 pF

b( ) C2 =1

2πf RS1

gm

=1

2π 2000 Hz( ) 27kΩ 10.228mS

= 0.0211 µF → 0.022 µF

14.83

a( ) VEQ = 1262kΩ

20kΩ + 62kΩ= 9.07V | REQ = 20kΩ 62kΩ = 15.1kΩ

IB =12 − 0.7 − 9.07( )V

15.1kΩ + 75 + 1( )3.9kΩ= 7.16µA | IC = 537 µA | VEC = 12− 3900IE − 8200IC = 5.47 V

Active region is correct.

rπ =75 0.025V( )

537µA= 3.49kΩ | ro =

60+ 5.47537µA

= 122kΩ | gm = 40IC = 21.5 mS

Rin = R1 R2 rπ = 15.1kΩ 3.49kΩ = 2.83 kΩ | Rout = ro RC = 122kΩ 8.2kΩ = 7.68 kΩ

C1 ≥ 101

2πf RI + Rin( )= 10

2π 100Hz( )1kΩ+ 2.83kΩ( ) | C1 ≥ 4.16 µF → 0.01 µF

C2 ≥ 101

2πfRI REQ + rπ

βo +1

=10

2π 100Hz( )1kΩ 15.1kΩ + 3.49kΩ

76

| C2 ≥ 273 µF → 270 µF

C3 ≥ 101

2πf Rout + R3[ ]=

102π 100Hz( ) 7.68kΩ +100kΩ( )

| C3 ≥ 0.148 µF → 0.15 µF

b( ) C2 =1

2πfRI REQ + rπ

βo +1

=1

2π 1000Hz( )1kΩ 15.1kΩ + 3.49kΩ

76

= 2.73 µF → 2.7 µF

14.84

14-479

IC =1005− 0.7

104 +101 103( )= 3.87mA | gm = 40IC = 0.155S | rπ =

100gm

= 645Ω | ro = ∞

The capacitors will be negligible at a frequency 10 times the individual break frequencies :

Req1 =10kΩ rπ + βo + 1( ) 1kΩ 20kΩ( )[ ]= 9.06kΩ | f1 = 102π 9.06kΩ( )2µF = 87.8 Hz

Req2 = 20kΩ + 1kΩrπ

βo + 1( )

= 20.0kΩ | f 2 =

102π 20.0kΩ( )10µF

= 7.96 Hz

Req3 = Rout + 20kΩ = 21.0kΩ | f 3 =10

2π 21.0kΩ( )10µF= 7.56 Hz

14.85

a( ) IC = αF IE =8081

5 − 0.713.3x103

= 319µA | gm = 40IC = 12.8mS | rπ =

80gm

= 6.26kΩ | ro = ∞

Req1 = 75Ω + 13.3kΩrπ

βo + 1( )

= 152Ω | C1 =

102π 50kHz( )152Ω

= 0.209 µF → 0.20 µF

Req2 = Rout +100kΩ = 8.25kΩ +100kΩ = 108kΩ | C2 =10

2π 50kHz( )108kΩ= 295 pF → 270 pF

b( ) C1 = 0.209µF50kHz100Hz

=105 µF →100 µF | C2 = 295pF

50kHz100Hz

= 0.148 µF → 0.15 µF

14.86

VEQ =1851kΩ

51kΩ + 100kΩ= 6.08V | REQ = 51kΩ 100kΩ = 33.8kΩ

IB =6.08 − 0.7 + 18( )V

33.8kΩ + 126( ) 4.7kΩ( )= 37.3µA | IC = 4.67 mA | VCE = 36 − 2000IC − 4700IE = 4.54 V


4.67mA= 669Ω | ro =

50 + 4.54( )V4.67mA

= 11.7kΩ


Rin = RB rπ + βo +1( )RL[ ]= 33.8kΩ 669Ω + 126( )2.94kΩ[ ]= 31.0 kΩ

Rout = RERB RI( )+ rπ

βo +1= 4.7kΩ

33.8kΩ 500Ω( )+ 669Ω126

= 9.22 Ω

C1 =10

2πf RI + Rin( )= 10

2π 100Hz( ) 500Ω + 31.0kΩ( )= 0.505µF → 0.50 µF

C2 =10

2πf R3 + Rout( )=

102π 100Hz( ) 24kΩ + 9.22Ω( )

= 0.663µF → 0.68 µF

14-480

14.87 Note : RS ≡ R1 = 3.9kΩ

VGS = −3900ID = −39005x10−4( )

2VGS + 2( )

2 | VGS = −0.975 VGS + 2( )

2→ VGS = −0.9915V

ID =5x10−4( )

2VGS + 2( )

2 = 254µA | VDS =15 − 23.9kΩID = 8.92V - Pinched off.

gm =2 254µA( )2 − 0.992

= 0.504mS | ro =50 + 8.92254 x10−6

= 232kΩ | Rin = 3.9kΩ1

gm= 1.32kΩ

Rout = RD ro 1 + gm RS RI( )[ ]= 20kΩ 232kΩ 1 + 0.504mS 3.9kΩ 1kΩ( )[ ]=18.8kΩC1 =

102πf RI + Rin( )

= 102π 400Hz( )1kΩ+ 1.32kΩ( )

=1.72µF →1.8 µF

C2 =10

2πf R3 + Rout( )=

102π 400Hz( ) 51kΩ +18.8Ω( )

= 0.057µF → 0.056 µF

14.88

a( ) Use C2 to set the lower cutoff frequency to 1 kHz. C1 and C3 remain negligible at 1 kHz.C2 = 0.056 µF , C1 = 1800 pF, C3 = 0.015 µF

(b) SPICE Results: fL = 925 Hz

14.89

a( ) Use C3 to set the lower cutoff frequency to 2 kHz. C1 remains negligible at 2 kHz.C1 = 8200 pF, C3 = 820 pF

b( ) Use C1 to set the lower cutoff frequency to 1 kHz. C2 and C3 remain negligible at 1 kHz.C1 = 0.042 µF , C2 = 1800 pF, C3 = 0.015 µF

(c) SPICE Results: For (a) fL = 1.96 kHz. For (b) fL = 1.02 kHz

14.90

Av =gmRL

1+ gmRL≥ 0.95 → gmRL ≥ 19 | gm = 2KnID | VGS − VTN =

2IDKn

= 0.5V

ID =0.5( )2 0.03( )

2= 3.75mA | RL ≥

192 0.03( ) 0.00375( )

=1.27kΩ

RL = RS 3kΩ → RS ≥ 2.19kΩ | VSS = VGS + 3.75mA( )RS | Possible designs :2.4kΩ, 11.5V ; 2.7kΩ, 12.6V ; 3.0kΩ, 13.75V - Making a choice which uses a nearly minimum value of supply voltage gives : VSS = 12 V ,RS = 2.4 kΩ.

14-481

14.91 For a common-emitter amplifier with RE = 0,

Rin ≅ rπ =βoVT

IC | IC =

100 0.025V( )75Ω

= 33.3 mA

14.92

Using Eqn. 14.102 : 50 =RE

1+ 40 4.3( )→ RE = 8.65kΩ | IC ≅

4.3VRE

= 497µA

50 = gmRLRin

RI + Rin | Assuming RI = 50Ω, 50 =

gmRL2

→ RL =2 50( )

40 497µA( )= 5.03kΩ

RL = RC 100kΩ → RC = 5.30kΩ | VEC = 5 + 0.7 − IC RC = 3.07V - Active region is ok.

C1 >>1

2π 500kHz( ) 50Ω + 50Ω( )= 3.18nF → C1 = 0.033 µF

C2 >>1

2π 500kHz( )105kΩ( )= 3.02pF → C2 = 33 pF

14.93 The base voltage should remain half way between the positive and negative power supply voltages. If VEE = +10V and VCC = 0V, then VB should = 5 V which can be obtained using a resistive voltage divider from the +10V supply. We now have the standard four-resistor bias circuit. The base current is 327 µA/80 = 4 µA.

vS 100 k Ω

+

-

vO

75 Ω

C1

+ 10 V

13 k Ω 8.2 k Ω

R1 R2

C2

CB

R ER C

120 k Ω 110 k Ω

Setting the current in R 1 to 10IB = 40µA, R1 =5V

40µA= 125kΩ →120kΩ. The

current in R 2 =11IB = 44µA, and R 2 =5V

44µA= 114kΩ → 110kΩ.

Note that the base terminal must now be bypassed with a capacitor.

14-482

14.94

Using Eqn. 14.102 : 75 =RE

1+ 40 8.3( )→ RE = 25.0kΩ | IC ≅

8.3VRE

= 332µA

50 = gmRLRin

RI + Rin | 50 =

gmRL2

→ RL =2 50( )

40 332µA( )= 7.52kΩ | RL = RC 100kΩ → RC = 8.13kΩ

VEC = 9 + 0.7 − IC RC = 6.98V - Active region is ok.

C1 >>1

2π 500kHz( ) 75Ω + 75Ω( )= 2.12nF → C2 = 0.022 µF

C2 >>1

2π 500kHz( )108kΩ( )= 2.95 pF → C2 = 30 pF

14.95

Rin ≅1gm

→ gm = 2KnID = 0.1S | ID =0.012Kn

| a( ) ID =0.01

2 0.005( )=1 A

b( ) ID =0.01

2 0.5( )= 10.0 mA | The second FET achieves the desired input resistance

at much lower current and hence much lower power for a given supply voltage.

14.96 1

gm=

VTIC

=kTqIC

∝ T

At − 40oC = 233K , 1

gm= 50Ω

233k300K

= 38.8 Ω. | At + 50oC = 323K,

1gm

= 50Ω323K300K

= 53.8 Ω.

Another approach : At 27oC = 300K, IC =300k50q

=6kq

.

At − 40oC = 233K , 1

gm=

2336

= 38.8 Ω. | At + 50oC = 323K, 1

gm=

3236

= 53.8 Ω.

14.97

This analysis assumes that the source and load resistors are fixed, and that only

the amplifier parameters are changing. Av = gmRLRin

RI + Rin

Since RE >> 75Ω, RE RI ≅ 75Ω and Rin ≅ RE1

gm≅

1gm

| Av ≅gmRL

1+ gmRITo achieve Av

max , RL → RLmax ,gm → gm

max which requires

IC → ICmax = 0.988

5.25 − 0.7( )V13kΩ 0.95( )

= 364µA | RLmax = 8.2kΩ 1.05( )100kΩ = 7.93kΩ

14-483

gm = 40 364µA( ) =14.6mS | Avmax =14.6mS 7.93kΩ( )1+ 0.0146 75( )

= 55.3

To achieve A vmin , RL → RL

min , gm → gmmin which requires

IC → ICmin = 0.988

4.75 − 0.7( )V13kΩ 1.05( )

= 293µA | RLmin = 8.2kΩ 0.95( )100kΩ = 7.23kΩ

gm = 40 293µA( ) =11.7mS | Avmin =11.7mS 7.23kΩ( )1+ 0.0117 75( )

= 45.1 | 45.1≤ AV ≤ 55.3

The range is only slightly larger than that observed in the Monte Carlo analysis in Table 14.16.

14.98 *Problem 14.76 - Common-Base Amplifier - Monte Carlo Analysis *Generate Voltage Sources with 5% Tolerances IEE 0 8 DC 5 REE 8 0 RTOL 1 EEE 6 0 8 0 1 * ICC 0 9 DC 5 RCC 9 0 RTOL 1 ECC 7 0 9 0 -1 * VS 1 0 AC 1 RS 1 2 75 C1 2 3 47U RE 3 6 RTOL 13K Q1 4 0 3 PBJT RC 4 7 RTOL 8.2K C2 4 5 4.7U R3 5 0 100K .OP .AC LIN 1 10KHZ 10KHZ .PRINT AC VM(5) VP(5) .MODEL PBJT PNP (BF=80 DEV 25%) (VA = 60 DEV 33.33%) .MODEL RTOL RES (R=1 DEV 5%) .MC 1000 AC VM(5) YMAX .END

Results: Mean value Av = 47.5; 3σ limits: 42.5 = Av = 52.5. However, the worst-case values observed in the

analysis are Avmin = 43.2 and Av

max = 51.9. The mean is 5% lower than the design value. The width of the

distribution is approximately the same as that in Table 14.16.

14-484

14.99 (a)

This analysis assumes that the source and load resistors are fixed, and that only

the amplifier parameters are changing. Av =gmRL

1+ gmRth

RERI + RE

Rth = RE RI and RE

RI + RE=

1

1+RIRE

where RE =13.3kΩ and RI = 75Ω

Since R E >> RI , Rth and RE

RI + REare essentially constant. To achieve Av

max , RL → RLmax ,gm → gm

max

which requires IC → ICmax = 0.988

5.10 − 0.7( )V13.3kΩ 0.99( )

= 330µA | RLmax = 8.25kΩ 1.01( )100kΩ = 7.69kΩ

gm = 40 330µA( )=13.2mS | Avmax =13.2mS 7.69kΩ( )

1+ 0.0132 75( )13.3 0.99( )

75 +13.3 0.99( )

= 50.7

To achieve Avmin , RL → RL

min ,gm → gmmin which requires

IC → ICmin = 0.988

4.90 − 0.7( )V13.3kΩ 1.01( )

= 309µA | RLmin = 8.25kΩ 0.99( )100kΩ = 7.55kΩ

gm = 40 309µA( )=12.4mS | Avmax =12.4mS 7.55kΩ( )1 + 0.0124 75( )

13.3 1.01( )75 + 13.3 1.01( )

= 48.2

(b) Using a Spreadsheet similar to Table 14.16: Mean value Av = 49.6; 3σ limits: 48.2 = Av = 50.9. The worst-case

values observed in the analysis are Avmin = 48.4 and Av

max = 50.8.

14.100 *Problem 14.100 - Common-Base Amplifier - Monte Carlo Analysis *Generate Voltage Sources with 2% Tolerances IEE 0 8 DC 5 REE 8 0 RTOL 1 EEE 6 0 8 0 1 * ICC 0 9 DC 5 RCC 9 0 RTOL 1 ECC 7 0 9 0 -1 * VS 1 0 AC 1 RS 1 2 75 C1 2 3 100U RE 3 6 RR 13.3K Q1 4 0 3 PBJT RC 4 7 RR 8.25K C2 4 5 1U R3 5 0 100K .OP .AC LIN 1 10KHZ 10KHZ .PRINT AC VM(5) VP(5) IM(VS) IP(VS) .MODEL PBJT PNP (BF=80 DEV 25%) (VA = 60 DEV 33.33%) .MODEL RTOL RES (R=1 DEV 2%)

14-485

.MODEL RR RES (R=1 DEV 1%)

.MC 1000 AC VM(5) YMAX *.MC 1000 AC IM(VS) YMAX .END

Results: Mean value: Av = 47.2; 3σ limits: 45.7 = Av = 48.5 Mean value: Rin = 83.4 Ω; 3σ limits: 79.5 Ω = Av = 87.6 Ω

14.101

Rin = RE1

gm=

RE1+ gmRE

=RE

1 + 40IC RE=

RE

1+ 408081

IE RE=

RE1+ 39.5IERE

IE RE = 2.5 − 0.7 = 1.8V | 75 =RE

1 + 39.5 1.8( )→ RE = 5.41kΩ

IC =8081

1.8V5.41kΩ

= 329µA | Rth = 75Ω 5.41kΩ = 74.0Ω | gm = 40 329µA( )= 13.2mS

Av =gmRL

1 + gmRth

RERI + RE

→ RL = 7.59kΩ

7.59kΩ = RC 100kΩ → RC = 8.21kΩ | VC = −2.5V + IC RC = +0.201V | Oops! We are violating

our definition of the forward - active region. If we use the nearest 5% values, RE = 5.6kΩ andRC = 8.2kΩ, IC = 318µA and VC = +0.108V . The transistor is just entering saturation.

14.102 Using a Spreadsheet similar to Table 14.14:

Mean value: Av = 0.960; 3σ limits: 0.942 = Av = 0.979

Mean value: ID = 4.91 mA; 3σ limits: 4.27 mA = ID = 5.55 mA

Mean value: VDS = 7.03 V; 3σ limits: 4.52 V = VDS = 9.54 V

*Problem 14.102 - Common-Drain Amplifier - Fig. 14.34 *Generate Voltage Sources with 5% Tolerances IDD 0 7 DC 5 RDD 7 0 RTOL 1 EDD 5 0 7 0 1 * ISS 0 8 DC 20 RSS 8 0 RTOL 1 ESS 6 0 8 0 -1 * VGG 1 0 DC 0 AC 1 C1 1 2 4.7U RG 2 0 RTOL 22MEG RS 3 6 RTOL 3.6K C2 3 4 68U R3 4 0 3K M1 5 2 3 3 NMOSFET .OP .AC LIN 1 10KHZ 10KHZ .DC VGG 0 0 1 .MODEL NMOSFET NMOS (VTO=1.5 DEV 33.33%) (KP=20M DEV 50%) LAMBDA=0.02

14-486

.MODEL RTOL RES (R=1 DEV 5%)

.PRINT AC VM(4) VP(4) IM(VGG) IP(VGG)

.MC 1000 DC ID(M1) YMAX *.MC 1000 DC VDS(M1) YMAX *.MC 1000 AC IM(VGG) YMAX *.MC 1000 AC VM(4) YMAX .END

Results: Mean value: ID = 4.97 mA; 3σ limits: 4.32 mA = ID = 5.62 mA

Mean value: VDS = 7.19 V; 3σ limits: 6.18 V = VDS = 8.20 V

Mean value: Rin = 22.0 MΩ; 3σ limits: 20.3 Ω = Rin = 24.0 Ω

Mean value: Av = 0.956; 3σ limits: 0.936 = Av = 0.976

14.103

a( ) Avt =gmRL

1+ gm 1+ η( )RL=

0.01S 1kΩ( )1+ 0.01S 1+ 0.5( )1kΩ( )

= 0.625 | Rin = ∞

Rout =1

gm 1+ η( )=

10.01S 1+ 0.5( )

= 66.7Ω

Ai =gmRG

1+ gm 1+ η( )RL=

0.01S 1.5MΩ( )1+ 0.01S 1+ 0.5( )1kΩ( )

= 14.8kΩ

b( ) Avt and Ai are worse; Rout is improved.

14.104

a( ) Avt = −gmRL

1+ gm 1+ η( )R5= −

0.01S 2kΩ( )1+ 0.01S 1+ 0.75( ) 200Ω( )

= −4.44 | Rin = RG

Rout = ro 1+ gm 1+ η( )R5[ ]= 10kΩ 1+ 0.01S 1.75( ) 200Ω( )[ ]= 45.0 kΩ

Ai = −gmRG

1+ gm 1+ η( )R5= −

0.01S 1MΩ( )1+ 0.01S 1+ 0.75( ) 200Ω( )

= 2.22kΩ

b( ) Avt and Ai are reduced; Rout is increased.

14.105

a( ) Avt =gm 1+ η( )RL

1+ gm 1+ η( )Rth=

0.5mS 1+ 1( )100kΩ( )1+ 0.5mS 1+1( ) 50Ω( )

= 95.2

Rin = RS1

gm 1+ η( )= 100kΩ

10.5mS 1+1( )

= 990 Ω

Ai = gm 1+ η( )Rin = 0.990Rout = ro 1+ gm 1+ η( )Rth[ ]= ro 1+ 0.5mS 1+ 1( ) 50Ω( )[ ]= 1.05rob( ) Avt , Ai and Rout are larger; Rin is smaller.

CHAPTER 14athena.ecs.csus.edu/.../Spring2006/EEE109/ch14_solutions.pdf14-443 CHAPTER 14 14.1 (a)...

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Transcript of CHAPTER 14athena.ecs.csus.edu/.../Spring2006/EEE109/ch14_solutions.pdf14-443 CHAPTER 14 14.1 (a)...