Solution Manual for Microelectronic Circuits 3rd Edition by Rashid

Chapter 1 1.1

From Eq. (1.1)

=

V LSB 1.2

=

28

= 136.7 mV

V FS 2

5

=

n

216

= 76 mV

From Eq. (1.1)

=

V LSB 1.4

2

3.5

=

n

From Eq. (1.1)

V LSB 1.3

V FS

V FS

3.5

=

2n

= 53.4 µ V

216

From Eq. (1.2)

= (1x 2−1 + 0 x2−2 + 1x2 −3 ) x 5 = (0.5 + 0.125) x 5 = 0.625 x 5 V vo = 3.125V vo

1.5

From Eq. (1.2)

= (1 x 2−1 + 1 x 2−2 + 0 x 2−3 + 1 x 2−4 ) x 5V vo = (0.5 + 0.25 + 0.125) x 5V = 4.375 V

vo

1.6

200π t = 6.5 + 5 x 10−3 Sin 20 V DC = 6.5 V v AB

I DC =

=

R L

=

v AB

vab

1K Ω

= 6.5 mA

R L

=

5 x 10−3 Sin 20 200 π t 1 K Ω

200 π t = 5 x 10−3 Sin 20

= 6.5 + 5 x 10−3 Sin 20 200π t V

Vab = 6.5 I a

6.5

= 5 x 10−3 Sin 20 200 π t V

vab ia

V DC

=

6 .5

2

2

+(

+(

5 x 10−3 2

5 x 10−3 2

)2

≃

mA

6.5 V

)2 mA

≃

6.5 mA

� � ��

Full file at https://testbanku.eu/Solution-Manua https://testbanku.eu/Solution-Manual-for-Microelect l-for-Microelectronic-Circuitsronic-Circuits-3rd-Edition-by-Ras 3rd-Edition-by-Rashi hi

1.7

Given

v DC = 7.5 + 10 x 10−3 Sin 20 2000π t V

V DC = 7.5 V I DC =

R L

7.5

=

1 K Ω

= 7.5 mA

= 10 x 10−3 Sin 20 2000 π t

vab ia

V DC

=

10 x 10−3 Sin 20 2000π t 1 K

= 10 x 10−3 Sin 20 2000 π t mA

−3

Vab

7.5V

≃

i A

= 7.5 mA + 10 x 10−3 Sin 2000π t

I a

10 x 10−3 2 (7.5) + ( ) mA 2

I a 1.8

=

10 x 10 2 (7.5) + ( ) 2 2

= ≃

mA

2

7.5 mA

v s

= 1.5 + 12 x 10−3 Sin ω t V

vo

= 7.5 + 2.5 Si Sin ω t V

From Eq. (1.9)

AV =

7.5 1.5

=5

From Eq. (1.10)

Av

=

2.5 12 x 10−3

= 208.3

1.9

= 2.5 + 20 x 10−3 Sin ω t vo = 7.5 + 4.5 x 10−3 Sin ω t v s

From Eq. (1.9)

AV =

7.5 2.5

=3

From Eq. (1.10)

Av

=

4.5 20 x 10−3

= 225

1.10 (a)

From Eq. (1.12) Power gain A P =

P L P i

vo 2

=

v s 2

RL Ri

=

1 100 K Ω (40 x 10−3 )2 50 K Ω

= 312.5

� � ��


(b)

= Ri = 50 K Ω

For R L

A P =

P L

=

Pi

vo 2

=

vs 2

1

= 625

(40 x 10−3 )2

From Eq. (1.11)

vo

io

A I =

=

i s

R L

v s

1

=

40 mV

Ri

= 25

1.11 (a)

From Eq. (1.12)

vo 2 A P =

v s

R L

=

2

Ri

1.22

147 K Ω (80 x 10 )2 100 K Ω −3

= 153

(b)

R L

= Ri = 100 K Ω vo

Ai

=

R L

v s

Ri

=

1.2 80 x 10−3

= 15

1.12 (a)

Av =

io =

Ai =

vo(peak) vi(peak) vo RL

=

io(peak) iL(peak)

=

6.5 –3

50 × 10

= 1.30 × 103 = 130 130 or 42.28 dB

6.5 sin1000 sin1000 π t 5000 =

1.3 × 10 –3 –6

1 × 10

= 1.3 sin (1000π t ) mA

= 1300 or 62.28 dB

A p = Av ⋅ Ai = 130 × 1300 = 169 × 103 or 52.28 dB (10 log P log P o/ P P i) Ri = (b)

vi(peak) ii(peak)

=

50 × 10 –3 –6

1 × 10

P dc + V EE dc = V CC CC I CC CC + V EE I EE EE

= 50 k Ω

= 15 (15 + 15) mW = 450 mW

P L =

vo (p (peak ) io (p (peak ) 2

⋅

2

=

6.5 × 1.3 × 10 –3 2

= 4.225 mW P i =

vi (p (peak ) ii (p (peak ) 2

⋅

2

=

50 × 10 –3

× 1 × 10 –6 2

= 25 nW

� � ��


= η = (c)

P L

+ P i

Pdc

4.225 × 10 –3

=

450 × 10 –3

+ 25 × 10−9

≃

0.938%

Av vi(max) = V o(max) o(max) = V CC CC vi(max) =

15

= 115.4 mV

130

1.13

V o = 5.3 V at at V I = 21 mV, V o = 5.8 V at V I = 27 mV

∆V o = (a)

Av =

(b)

Adc =

(c)

Vo – V (min) Av – 5.5 + 2 83.3

5.8 – 5.3 = 0.5 V,

∆V o ∆V I V o

=

–3

6 × 10 5.5

=

V I

0.5

V F – 24 mV ≤

≤

V I – 24 mV ≤

= 83.3 83.3 or 38.4 dB

= 229 or 47.2 dB

24 × 10 –3

≤

∆V I I = 27 – 21 = 6 mV

Vo(max) – V o Av 11 – 5.5 83.3

– 42 mV ≤ V I – 24 mV ≤ 66 mV – 18 mV ≤ V I ≤ 90 mV 1.14 (a)

io = ii =

Ai = Av =

vo RL vi Ri io ii vo vi

= =

= =

2V 10 k Ω

= 0.2 mA

1 mV 100 k Ω

= 10 nA

0.2 × 10 –3 –9

10 × 10 2

–3

1 × 10

= 2 × 104 or 86 dB

= 2 × 103

A p = Av Ai = 2 × 103 × 2 × 104 = 4 × 107 or 152 dB (b)

vi = ii Ri = 1 × 10 –3 × 100 = 10 –1 V vo = io RL = 100 × 10 –3 × 103 = 100 V Av =

Ai =

vo vi io ii

=

=

100 0.1

= 1000 or 60 dB

100 × 10 –3 10 –3

= 100 or 40 dB

A p = Av Ai = 1000 × 100 = 10 5 or 50 dB

� � ��


1.15 (a) From Eq. (1.23)

Av =

=

Avo

(1 + Rs Ri ) (1 + Ro RL ) 150 1800) (1 + 50 4700) (1 + 200 18

= 133.6

From Eq. (1.24) Ai =

Avo Ri

+ Ro

RL

150 × 1800

=

4700 + 50

= 56.84

From Eq. (1.25) A p = Av Ai = 133.6 × 56.84 = 7593.82 (b) Problem 1.15 Amplifier VS

1

0

AC

100MV

RS

1

2

200

RI

2

0

1.8K

E1

3

0

2

R0

3

4

50

RL

4

0

4.7K

. TF

V(4)

0

150

VS

. End

1.16

For maximum power transfer Ro = RL = 50 Ω P L = vo io = ( A Av vi) ( A Ai ii) =

=

=

=

=

∴

P L(max) L(max) =

Avo ⋅ vi

⋅

Avo Ri

(1 + RS Ri ) (1 + Ro RL ) RL + Ro Av2o Ri RL vi

+ Rs )( RL + Ro )

( Ri

⋅

Av2o Ri RL ( RL

+ Ro )

2

( Ri

( RL + Ro ) vi2 2

Av2o Ri RL ( RL

Avo ⋅ vi

⋅

+ Rs )

⋅

+ Ro )2 ( Ri + Rs )

ii

Ri vs

2

( Ri + Ro ) 2

Av2o RL Ri3 vS2 ( RL

+ Ro )2 ( Ri + Rs )3

1502

× 50 × 18003 (100 × 10 –3 ) 2 100 2 (1800 + 200)3

= 820 mW

� � ��


1.17

∆ vo

=

vo

∆ vo

R o

+ Ro

RL

= 0.15 =

vo

Ro =

R o 1.5 k + R o

1.5 × 103

× 0.15

0.85

= 264.7 Ω

1.18

vo = Avo vi

(a)

=

RL RL

+ Ro

RL

= A = Avo

RL

+ Ro

⋅

R i ⋅ V s Ri

+ Rs

200 × 22 × 105

× 50 × 10 –3 = 5.16 V ( 22 + 20) (105 + 1500)

(b) From Problem 1.16

P L =

=

Av2o

RL Ri 3 V s 2

+ R o )2 ( R i + R S )3

( RL

× 22 × 1015 × (50 × 10 –3 ) 2 (22 + 20) 2 ⋅ (105 + 1500) 3

2002

= 1.19 W (c) Av =

(d) Ai =

vo vs io is

=

5.16 50 × 10 –3

= 103.2

5.16

, io =

= 122.86 mA

22 + 20 is =

Ai =

vs R s

+ Ri

=

50 × 10 –3 5

1500 + 10

122.86 × 10 –3 –7

4.93 × 10

= 4.93 × 10 –7 A

= 249 × 103

(e) A p = A = Av Ai = 103.2 × 249 × 103

= 25.7 × 106 1.19

is ≤ 1 µA =

vs Rs

+ Ri

10 ×10 –3

=

2.5k + Ri

Ri > 7500 Ω From Eq. (1.27)

∆ vo vo

=

R o RL

+ Ro

, RL ranging from 2 k Ω to 10 k Ω

� � ��


For

∆ vo vo

≤ 0.5% 5 1000

=

Av =

R o RL

+ Ro

5V 10 mV

, Ro ≤ 10 Ω

= 500 or 53.98 dB

From Eq. (1.23) 500 =

∴

A vo

(1 + Rs

R i ) (1 + Ro RL )

=

A vo

(1 + 2.5 k 7.5 k) (1 + 10 2000)

Avo = 669.8

1.20

From Eq. (1.23) Av =

Avo

(1 + Rs Ri ) (1 + R o RL )

Variation in A in Av will be contributed by Avo, RS, and R and RL. Assume equal contribution by each. Hence the value of R of R o that will keep the variation in gain within 0.5% for variation in R L from 5 k Ω to 20 k Ω can be found from 5 k 5k + Ro

=

Ro =

20 k 20 k + Ro

× 0.995

20k (1 – 0.9 0.995) 95) 4 × 0.995 – 1

> 33.5 Ω

1.21 (a) By Kirchoff’s Kirchoff’s current current law at node A R s

R

A i s

i f

+

V i i i

R o

R i

–

Avo V i

R x

is = ii + if = =

= vi

Rx =

vi is

vi Ri

+

vi – Avo vi R + Ro

 1 1 – Avo   R + R + R   i o =

1 1 Ri

+ (1 − Avo )

( R + Ro )

� � ��


(b) For R For Ri = 50 k, Ro = 75 Ω, Avo = 2, R = R = 10 k

Rx =

1

+ (1 − 2) = – 12.62 k Ω

is =

1 50 50 k V s

Rs

+ Rx

∴

20 mV 1.5 k Ω – 12 12.62 k Ω

= – 1.8 µA

vs

(c) For is = 2.5 µA, Rs + R + Rx =

=

=

(10 k + 75 7 5)

is

20 mV – 2.5 µA

= – 8 k Ω

Rx = – 8 k – 1.5 k = – 9.5 k Ω

1.22

Ais = 200, Ri = 150 Ω, Ro = 2.5 k Ω RL = 100 Ω, is = 4 mA, Rs = 47 k Ω (a) From Eq. (1.30)

Ai =

Ais

(1 + Ri Rs ) (1 + RL Ro )

=

200

(1 + 150 47 k ) (1 + 100 2500)

= 191.7 Av = Ai

RL Rs

= 191.7 ×

100 47 k

= 0.4078

A p = Av Ai = 191.7 × 0.407 = 78.17 (b) Problem 1.22a Amplifier VS

1

0

AC

RS

1

2

47K

RI

3

0

150

F1

0

4

VX

VX

2

3

0V

VY

4

5

DC

R0

4

0

2.5K

RL

4

0

100

.

TF

V(s)

1MV

200

0V

VS

END

.

� � ��


Problem 1.22b Amplifier IS

0

1

AC

RS

1

0

47K

RI

2

0

150

F1

0

3

VX

200

VX

1

2

DC

0V

VY

3

5

DC

0V

R0

3

0

2.5K

RL

5

0

100

.

TF

I(VY)

1MA

IS

END

.

1.23

From Eqs. (1.28) and (1.29) io =

=

A is R o R o

Rs

⋅

+ RL

Rs

+ Ri

10 × 22 × 103 22 × 10

3

+ 150

×

⋅ is

100 × 103 10

5

× 50 × 10 –3

+ 50

= 4.96 A 1.24 i i

+ R s

i s

V i

AI isi

R i

R o

–

(a) Output resistance =

=

open-cir open-circuit cuit voltage voltage short-circuit short-circuit current 12 100 × 10 –3

= 120 Ω

Ais ii = short circuit current = 100 × 10 –3 A ii = is

Ais =

Av =

Rs Ri

+ Rs

=

100 × 10 –3 –6

5 × 10 vo vs

= A = Ais iL

5 × 10 –6

× 100 k

50 + 100 k

≃

5 × 10 –6 A

= 20 × 103 Ro Ro

+ RL

⋅

RL is Rs

� � ��


= 100 × 10 –3 ×

120 120 + 2700

×

2700 5 × 10 –6

× 105

= 22.98 (b) Ai =

20 × 103 100 k) (1 + 2.7 k 12 1 20) (1 + 50 10

= 850.6

(c) A p = A = Av Ai = 22.98 × 850.6

= 19,547 1.25

Following Example 1.4 we have 0.99

Ro Ro

+ 20

=

Ro Ro

+ 500

Ro (1 – 0.99) 0.99) = 500 × 0.99 – 20 Ro = 47.5 k Ω For R For Ri 0.99 ×

100 k 100 k × Ri

=

10 k 1 0 k + R + Ri

Ri = 111.23 Ω Ais =

20 × 10 –3 100 × 10 –6

= 200 A/A

1.26

Assume A Assume Ai varies equally due to contribution from Ais Rs, and R and Ro. Ro

+ RL

Ro

= 0.995, Ro =

Ro Ro

+ 100

100 × 0.995 1 – 0.9 0.995

= 0.995 > 19.9 k Ω

Similarly Rs Rs

+ Ri

= 0.995,

100 k 100 k + Ri

= 0.995

Ri ≤ 503 Ω Ai = 50 =

∴

Ais

≃

Ais

(1 + Ri Rs ) ( RL Ro ) Ais 100 k ) (1 + 100 19 1 9.9 k ) (1 + 503 10

≃

Ais

50

��


1.27 i i

+ R i i s

–

v i

R s

Ai Li

v e

+

R o

+

–

+

R L

v f

R

v o

–

– R x =

v i Li

(a) R allows R allows a voltage that is proportional to lead current io to be fed back to the input side.

Converting current source to voltage source R o

i i

i s

R s

+ Ais i i R o –

R i

+ v i –

I

R x =

R L

+ v o –

R

v i i i

Applying KVL to loop II Ais Ro ii = Ro io = R = R((ii – io) + R + RL io = ( R Ro + R + RL + R + R)) io – Ri – Rii io =

+R + RL + R

Ais Ro Ro

ii

(1)

Applying KVL to loop I vi = R = Ri ii + R + R((ii – io) = ( R Ri + R + R)) ii – Ri – Rio

(2)

Substituting i Substituting io from Eq. (1) into Eq. (2) and simplifying vi = ( Ri + R + R)) ii – R ⋅ vi ii

= R = Ri + R + R – – R R

Rx =

vi ii

+R ii + RL + R

Ais Ro Ro

+R + RL + R

Ais Ro Ro

= Ri + R – R

( Ais + R / Ro ) 1 + ( RL + R) Ro

��


(b) Assume ideal current amplifier with R with Ri = 0 and R and Ro = ∞, we have the reduced figure as i o

i s

+ Ais i L –

R s

R L

+ v o –

R

R x =

v i Li

Rx = R = R – – RA RAis = R = R(1 (1 – A – Ais) For A For Ais > 1, R 1, Rx is negative, and if Ais = 2 Rx = – R – R For R For Rx = – 10 k Ω we need R need R = = 10 k Ω. Thus an ideal current amplifier with Ais = 2 and R and R = = 10 k Ω will simulate a negative resistance. 1.28

Problem 1.27 for A for Ais = 2 and substituting R substituting R by by impedances Z impedances Z ( s) s) (a) Using the result of Problem Z x =

Vi (s) Ii (s)

 

= – Z Z ( s) s) = –  R +

− R ( R + 1 sC )  − R + R + 1 sC 

= – [ R – R – sCR sCR2 – R] R] = sCR = sCR2 Z x = sL = sL e where the effective inductance L e is given by CR 2 . Thus, the circuit simulates an inductance. (b) To simulate L simulate Le = 10 mH, let C let C = 0.01 µF,

10 × 100 –3 = 0.01 × 10 –6 × R2 R2 = 106, R = R = 1 k Ω (c) Problem 1.28 Inductance Simulation Simulation IS

0

4

AC

1MA

V1

4

1

DC

0V

F1

1

0

V1

2

R1

1

2

1K

R2

2

3

1K

C1

3

0

0.01UF

V2

2

5

DC

R3

5

0

1K

F2

5

0

V2

2

. AC LIN

10

1K

10K

. PRINT

AC

VM(1)VP(1)

0V

. END

��


1.29

vo = – Gm1 vi( s) s)

(a)

⋅

1 sC

ii( s) s) = – Gm2 vo( s) s) = Gm1 Gm2 vi ( s)

Z i =

ii (s )

=

vi ( s) sC

SC Gm1Gm2

Z i( j jω ) = j = jω

C Gm1Gm2

(b) Problem 1.29 Amplifier VS

1

0

AC

1V

RS

1

2

100

G1

0

2

3 0

3M

G2

3

0

2 0

3M

C1

3

0

0.1UF

R1

3

0

100M

. TRAN15US1.5 MS . PROBE . END

1.30

Assuming that the discharging time constant τ = = CR i is related to the input frequency by = τ =

10 f , CRi =

10 60 × 103

(a) Let C = = 0.1 µF, then R then Ri =

10 –6

0.1 × 10

× 60 × 103

= 1.67 k Ω (b) Variation in Gm, according to Eq. (1.36), will be contributed by G ms and R and RL. Assume equal

contributions, R contributions, Rs = 0. The gain parameter Gms =

io vs

=

20 cm 170 V

×

5 mA 1 cm

= 0.588 mA/V ± 1% The value of R of Ro that will keep gain variation within 1% for variation in RL from 20 Ω to 500

Ω can be found from 0.99

Ro Ro

+ 20

=

Ro Ro

+ 500

Ro > 47.5 k Ω

��


1.31 R s

Assume 1% for for R i and 1% for R o

Ri Ri

+ Rs

i o

+

+ v i

–

–

V s

R i

Gm v i

R o

R L

> .99, Ri (1 – 0.99) > 0.99 R 0.99 Rs

For R For Rs = 1 k Ω, Ri >

0.99 × 1 k = 99 k Ω 0.01

Similarly Ro Ro

> 0.99,

+ RL

Ro >

0.99 × 200 0.01

Ro > 19.8 k Ω 1.32

For R For Rm = 20 Ω to 100 Ω 0.99 Ro

+ 20

=

1 Ro

+ 100

Ro = 7.9 k Ω Ω o

Gm V i

R o

R m

For R For Rs varying from 2 k Ω to 5 k Ω 0.99 Ri

+ 2k

=

1 Ri

+ 5k

, Ri (1 – 0.99) = 5 k × 0.99 – 2 k

Ri = 295 k Ω For

vs = 10 V, I o = 100 mA Gm =

I o vs

=

100 100 mA 10 V

= 10 mA/V

Transconductance amplifier

��


1.33

Using Eq. (1.41) Z mo

Z m =

(1 + Ro RL ) (1 + Ri Rs )

=

0.5 k 1 0 k ) (1 + 4.7 k 4.7 k ) (1 + 1.5 k 10

= 217.39 V/A From Eq. (1.42) Z mo RL

Av =

( Rs + Ri ) ( RL + Ro )

0.5 × 103

× 4.7 × 103 (10 × 103 + 1.5 × 10 3 ) ( 4.7 × 103 + 4.7 × 103 )

=

= 21.739 × 10 –3 ii =

Rs

⋅ is =

+ Ri

Rs

× 50 × 10−3 10 4 + 1.5 × 103

10 4

=

500 11.5 × 103

mA

From Eq. (1.39) vo = Z mo mo

ii RL

+ R0

RL

= 0.5 k ×

500 11.5 × 103

×

4.7 × 10 1 03

( 4.7 × 103 + 4.7 × 103 )

= 10.87 V 10.87

io =

3

4.7 × 10 io

Ai =

=

is

= 2.31 × 10 –3 A

2.3 × 10 –3 50 × 10 –3

= 0.046

1.34

Since the output variation should be kept within ±2%, variation of effective transimpedance Z m should be kept to ±2%. According to Eq. (1.41) the variation in Z in Z m will be contributed by Z mo and Ro. Assume equal contribution to the variation. mo and R Let R Let Ri = 10 Ω. Then Z mo mo =

10 V 2 cm cm

×

20 cm 100 mA mA

= 1000 V/A ± 1%

The value of R of Ro that will keep gain variations within 1% for variation of RL from 2 k Ω to 10 k Ω is 0.99 × 10 k 2 k = 10 k + Ro 2 k + Ro Ro ≅ 25 Ω Design specifications are Z are Z mo and Ri ≤ 10 Ω. mo = 1000 V/A, Ro ≤ 25 Ω, and R 1.35

Let R Let Ri = 10 k << k << R Rs ii =

Rs Rs

Rs Ri

+ Rs

+ Ri

� is =

105 10

5

× 0 .5 = 454.5 mA + 10 4

= 0.99, Ri × 0.99 + 0.99 Rs = Rs

��


Rs (1 − 0.99)

Ri =

0.99

=

× 0.01

100 k

0.99

= 101 k Ω Rm

+ Ro

Rm

= 0.99, Ro = =

Z mo mo =

V o ii

Rm (1 − 0.99) 0.99 20 k

0.99

5

=

× 0.01

454 �5 � 5 × 10 −3

= 202

= 11 V A ± 1%

1.36 R s

R s

+ v s –

+ Avov i

R i –

R L

+ v o –

Voltage amplifier

For ideal voltage amplifier Ro = 0 and R and Ri = ∞ vo = Avo vi

(1)

For ideal current amplifier Ro =

∞, Ri = 0

io = Ais is

(2)

∴

vo = Avo vi = io Ro = A = Ais is Ro = Ais ii Ro

Using

vi = ii Ri Avo ii Ri = Ais ii Ro Avo = Ais

Ro Ri

(3)

For ideal transconductance amplifier amplifier io = Gms vs for R for Ro = ∞, Ri = ∞ vo = Avo vi = io Ro = Gms vi Ro

∴ Avo =

Gms Ro

(4)

For ideal transimpedance amplifier amplifier Ri = 0, Ro = 0 vo = Z mo mo is = Z mo ⋅ ii vo = Avo vi = Z mo mo

∴

Avo =

Z mo Ri

vo Ri

(5)

��


Using

Ro

Avo = Ais

Ri

Ais = Avo

Ri Ro

= 250 ×

Avo = Gms Ro, Gms =

50 × 10

3

= 12,500 A/A

1 × 103

Avo Ro

=

250 1000

= 0.25 A/V 3 Z mo mo = Avo ⋅ Ri = 250 × 50 × 10

= 12.5 MΩ Equivalent amplifiers are:

i s

R s

Ais i i

R i

R o

R L

Equivalent current amplifier R s

i o

+

+

v i

v s

+ Gms v i

R o

R L

–

–

–

Equivalent transconductance amplifier

R o

+ i s

R s

R i

+

v i

R L

Zmo i i

–

v o

–

Equivalent transimpedance amplifier

1.37 R s

+ v s –

+ v i –

i o R i

Gms v i

R o

+

v o –

R L

Transconductance Transconduc tance amplifier

Gms = 20 mA/V

��


Avo = Gms Ro = 20 × 10 –3 × 2 × 103 = 40 V/V Avo = Ais

Ro Ri

or A or Ais = Avo �

Ri

=

Ro

40 × 100 × 103 2 × 103

= 40 × 50 = 2000 A/A Z mo � Ri = 40 × 100 × 103 mo = Avo � R = 4 MΩ Equivalent circuits R s

R o

+ v i –

+ R 1

v s –

+ Gvo v i –

R L

Voltage amplifier

i i R s

i s

i o + v i –

R i

R o

Ais i i

+ R L

v o

–

Current amplifier v o

i i

R s

i s

+

+

R i

–

Zmi i i

R L

–

Transimpedance amplifier

1.38 (a) From Eq. (1.21)

Av1 = Av2 = A = Avo

Ri2 Ri 2

+ Ro1

=

50 × 2.5 × 103 2.5 × 103

+ 100

= 48 From Eq. (1.45), the overall open-circuit open-circuit voltage gain Avo =

V o V i

= A = Av1 ⋅ Av2 ⋅ Av3

= 48 × 48 × 50 = 115,200 or 101.22 dB

��


Total effective voltage gain using Eq. (1.23) Av =

=

V o

Avo Ri RL

=

V s

( Ri / Rs ) ( RL / Ro )

115200 × 2.5 × 103

× 2.5 × 103 ( 2.5 × 103 + 200 × 103 ) ( 2.5 × 103 + 100 )

= 1367.5 or 62 dB Overall current gain Ai =

io ii

=

Avo ⋅ Ri1 RL

+ Ro3

=

115200 × 2500 2500 + 100

= 110,769 or 100.9 100.9 dB A p =

P L P i

= A = Av ⋅ Ai = 1367.5 × 110769

= 1.5 × 108 or 81.7 dB (b) Problem 1.38 Cascaded Amplifier VS

1

0

AC 100 MV

RS

1

2

200K

RI1

2

0

2.5K

E1

3

0

4

R01

3

4

100

RI2

4

0

2.5K

E2

5

0

40

R02

5

6

100

RI3

6

0

2.5K

E3

7

0

60

R03

7

8

100

RL

8

0

2.5K

. TF

V(8)

50

50

50

VS

. END

1.39 (a) Using Eq. (1.21), A (1.21), Av of stage 1 and 2 is given by

Av1 = Av2 =

Avo1 Ri 2 Ri 2

+ Ro1

=

80 × 2500 2500 + 100

= 76.9 From Eq. (1.45), the overall open-circuit open-circuit voltage gain Avo = Av1 ⋅ Av2 ⋅ Avo3 = 76.92 × 80 = 473,088 or 113.5 dB From Eq. (1.23) Av =

Avo ⋅ Ri RL

( Ri + Rs ) ( RL + Ro )

=

473, 08 088 × 2500 × 1500

( 2500 + 200 × 103 ) (1.5 × 103 + 300 )

= 4867 or 73.7 dB

��


Ai =

Avo RLi

+ Ro3

RL

=

473, 08 088 × 2500 1.5 × 103

+ 300

= 657,066.7 or 116.6 dB

A p = Av Ai = 4867 × 657,066.7 = 3.19 × 109 or 95 dB (b) Problem 1.39 Cascaded Amplifier VS

1

0

AC100MV

RS

1

2

200K

RI1

2

0

2.5K

E1

3

0

2

R01

3

4

100

RI2

4

0

2.5K

E2

5

0

4

R02

5

6

100

RI3

6

0

2.5K

E3

7

0

6080

R03

7

8

300

RL

8

0

1.5K

. TF

V(8)

0

80

0

80

VS

. END

1.40 (a)

io =

=

=

Ai =

=

Ais3 ii3 Ro3 Ro3

+ RL

=

Ais3 Ro3 Ais2 ii 2 Ro 2 Ro3

+ RL

Ais3 Ais2 Ro3 Ro 2

⋅

Ro 2

+ Ri3

Ais1ii1Ro1

( Ro3 + RL ) ( Ro 2 + Ri3 ) ( Ro1 + Ri 2 ) Ais3 Ais2 Ais1 Ro3 Ro2 Ro1

( Ro3 + RL ) ( Ro 2 + Ri3 ) ( Ro1 + Ri2 ) io Ls

=

⋅

Rs ii Rs

Ais3 Ais2 Ais1 Ro3 Ro2 o 2 Ro1

+ Ri1 ⋅

Rs

( Ro3 + RL ) ( Ro 2 + Ri3 ) ( Ro1 + Ri2 ) ( Rs + Ri1) 1003 (4.7 × 103 )3

× 20 × 103 ( 4700 + 100) ( 4700 + 100) ( 4700 + 100) ( 20 × 103 + 100 )

= 934,122 or 119.4 dB Av =

Av =

=

V o V s

=

io Ri ii1 R ⋅ i1

on substitution for A for Av from above

Ais3 Ais2 Ais1 Ro3 Ro 2 Ro1 RL

( Ro3 + RL ) ( Ro 2 + Ri3 ) ( Ro1 + Ri2 ) Ri1 106

× 109 × 4.73 × 102 483 × 106 × 10 2

= 42,498.5 or 92.56 dB AP = Av Ai = 42,498.5 × 934,122 = 3.97 × 1010 or 105 dB

��


(b) Problem 1.40 Cascaded Cascaded Current Amplifier IS

1

0

AC 100MA

RS

1

0

20K

V1

1

2

DC 0V

RI1

2

0

100

F1

0

3

V1 100

R01

3

0

4.7K

V2

3

4

DC 0V

RI2

4

0

100

R02

5

0

4.7K

V3

5

6

DC 0V

RI3

6

0

100

F3

7

0

V3 100

R03

7

0

4.7K

VX

7

8

DC 0V

RL

8

0

100

. TF I(VX)IS . END

1.41

io =

(a)

Ai =

Av =

Gms Vi 2 Ro 2 Ro2 io ii

=

V o V i

+ RL

=

−

Gms Ro 2 Ro 2

+ RL

⋅

Z mo ii Ri 2 Ro1 + Ri 2

− 20 × 10 –3 × 105 10 4 + 106 × 5 3 6 10 + 10 200 + 10 + io

=

RL

ii Ri1

=–

198 × 103 50 × 103

= – 198

= – 3.96

A p = Ai Av = – 198 × – 3.96 = 784 (b) Problem 1.41 Transconductance Amplifier VS

1

0

AC 10MV

RS

1

2

5K

VX

2

3

DC 0V

RI1

3

0

50K

H1

4

0

VX 10K

R01

4

5

200

RI2

5

0

1MEG

G1

6

0

5

R02

6

0

100K

RL

6

0

1K

. TF

V(6)

0

20M

VS

. END

��


1.42

From Eq. (1.54d)

f L

1

=

1

=

2 π R1C1

2π x 147 K Ω x 0.1 nF

= 10.8 KHz

1.43

From Eq. (1.54d)

f L

1

=

2π R1C 1 1

R1 =

2π f LC1

1

=

2π x 4 x 103 x 0.01 x 10−6

= 3.979 K Ω

1.44

From Eq. (1.59d)

f H =

1 2 π x 147 x 103 x 0.1 x 10−9

= 10.8 KHz

1.45

From Eq. (1.59d)

f H = R2

=

1 2 π R2C 2 1

=

2 π f H C2

1 2 π x 25 x 103 x 0.01 x 10−6

= 636.6 K Ω

1.46

Av( j jω ) =

| Av( j jω ) | =

200 1 + jω / 100

100 =

100 + jω

2104 2

ω

(a)

=

2 × 104

+ 10 4

2 × 104 2

ω

H

+ 10 4

, ω H =

3 × 104

= 173.2 rad/s f H =

(b)

50 =

173.2 2π

2 × 104 2

ω

H

ω H =

f H =

= 27.56 Hz = BW

+ 10

4

, ω 2H × 104 = 16 × 104

387.29 rad/s 387.29 2π

= 61.64 Hz = BW

��


1.47 R s

+

+

v s –

(a)

+

v i

Gms v i

R i

R o

C

R L

–

–

 

vo = – Gms vi  Ro ||

1 sC

 

|| RL 

 

= – 20 × 10 –3 vi  980 980 || vi =

5 × 105 vs 3

505 × 10

  sC  1

(i)

= 0.99 vs then

 

vo = – 20 × 10 –3 × (0.99 vs) ×  980 980 || Av =

v o

  sC  1

− 20 × 103 × 0.99 × 980 – 19.404 = − 1 + 980 jω C 1 + jω 980 × 0.1 × 10 6

Av(mid) = 19.404 ω H =

f H =

106 980 × 0.1 10204 2π

= 10,204

= 1624 Hz

f bw = Av(mid) × f H = 19.404 × 1624 = 31,512 (b)

RL = 10 k Ω, Ro || R || RL = 10 k || 50 k = 8333 Ω Av =

ω H =

165 − 20 × 10 –3 × 0.99 × 8333 = − 1 + jω 8333 × 0.1 × 10 –6 1 + jω × 833.3 × 10 –6 106 833.3

, f H = 190 Hz

f bw = 165 × 190 = 31,350 1.48

 

vo = – g m vi  RL || = vi =

 = − vi g m RL sC L  1 + sRL C 2 1

− vi g m C2 ( s + 1 RL C2 ) vs Ri Rs

+ Ri + 1 sC i

=

(1) vs Ri s ( Rs

+ Ri ) [ s + 1 Ci ( Rs + Ri ) ]

(2)

��


From (1) and (2) vo =

ω H

=

or f H = ω L

−

vs g m Ri s C2 ( Rs 1

RL C 2 107

=

f L =

2π

+ Ri ) ( s + 1 RL C2 ) [ s + 1 Ci ( Rs + Ri )] 1

=

10 × 10

3

× 10 × 10

− 12

= 107 rad/s

= 1.59 MHz 1

C1 ( Rs 100 3 × 2π

+ Ri )

=

1 2 0 × 10

−6

× 1500

=

100 3

rad/s

= 5.3 Hz

For A Av(mid) R s

+ V s –

R i

vo = – g m vi Ro = –

=

+ gm v i –

v i

gm Ri � Ro Rs

+ Ri

− 15 × 10−3 × 10 4 × 103 1500

R o

vs

= – 100

Problem 1.48 VS

1

0

AC

RS

1

2

500

C1

2

3

20UF

RI

3

0

1K

G1

4

0

3

RL

4

0

10K

C2

4

0

10PF

. AC

DEC

100

1

AC

VM(4)

.

PRINT

10MV

0

15M

10 MEG

. PROBE . END

��


1.49 R s

C 1

+

+ v i –

+ v s –

gm v i

R i

R o

R L

v o –

Low-frequency equivalent circuit vo = – g m vi ( R Ro|| R || RL) = – 15 × 10 –3 × 5 × 103 vi = – 7.5 vi vi =

=

Ri vs Rs

=

+ Ri + 1 sC 1

(i)

Ri sC1 vs 1 + ( Rs

+ R1 ) sC 1

× 10 × 10 −6 1 + s (25 × 103 + 1 × 103) 10 × 10 − 6 vs s 25 × 103

=

vs s 0.25 1 + 26 × 10−2 s

(ii)

From (i) and (ii) vo =

f L =

75 × 0.255

−

1 + 26 × 10 2 s 1 26 × 10−2

× 2π

(iii)

= 0.612 Hz

For high-frequency equivalent circuit R s

+

+

+

v s

R i

–

v i

gm v i

C i

R o

C o

 

vo = – g m vi  Ro || RL || = – 15 × 10

vi = =

=

v o

 sC o  1

5 × 103 vi

3

=

R L

–

–

1 + 5 × 103

× 10 × 10 − 12 s

− 75 vi 1 + 5 × 10 − 8 s || 1 sCi ) vs ( R1 ||1 Rs + Ri ||1 || 1 sC i Ri vs Ri

+ Rs + sRi Rs C i 25 × 103 vs

3

25 × 10

+ 103 + s × 25 × 103 × 103 × 20 × 10 − 12 ��

� ��


=

Av(s) = = f H = Av(mid) =

25 vs

25 vs = 26 + s 500 × 10 −9 26 + s × 5 × 10 1 0− 7 25 − 75 × − 8 v s (1 + 5 × 10 s) 26 + 5 × 10 − 7 s − 75 × 25 −8 26 (1 + 5 × 10 s) (1 + 5 × 10 − 7 s 26 )

vo

=

1 2π × 5 × 10 10

− 75 × 25 26

−8

= 3.18 MHz

= – 72.1

Problem 1.49 VS

1

0

AC

RS

1

2

1K

C1

2

3

10UF

RI

3

0

25K

G1

4

0

3

RL

4

0

10K

R0

4

0

10K

C0

4

0

10PF

. AC

DEC 100

10MV

0

15M

1

10MEG

. PRINT AC VM(4) . PROBE . END

1.50

R m = 10 k × 400 R 1 Take R Take R1

+ 10 k ≃

1V I m

=

1V 100 100 µA

R 1

= 10 10 k Ω

= 1 V, R1 = 10 k (400 – 1) = 3990 k Ω

400 V

+ –

R m

+ 1V

4 MΩ ± 1%.

1.51

RL = 1.2 × 24 R 1

+ 1.2

6V 5A

R 1

= 1.2 Ω

= 6 V, R1 =

1.2 × 24 6

– 1.2 = 3.6 Ω

+ 24 V –

+ R L

Take R Take R1 = 3.6 Ω ± 5%.

��

Full file at https://testbanku. https://testbanku.eu/Solution-Manualeu/Solution-Manual-for-Microelectron for-Microelectronic-Circuits-3r ic-Circuits-3rd-Edition-by-Rashi d-Edition-by-Rashi

1.52

V 2 RL

=

RL = RL

× 120

R1 + RL 41.667 × 120 50

R 1

502

= 60 W

RL

+ 120 V –

502

+ 50 V

R L

= 41.667 Ω

60

= 50 V = R1 + 41.667, R1 = 58.334 Ω

Take R Take R 62 Ω ± 5%. ≃

1.53

I o = 1 mA i(t ) = V s R

=

V s R 24 R

R

e – t t/ τ V s

24 V

C

= 1 mA, R = R = 24 k Ω ± 5%

1.54

| Av | =

V O V I

=

400 5

= 80, Q = 60°

��

Full file at https://testbanku. https://testbanku.eu/Solution-Manualeu/Solution-Manual-for-Microelectron for-Microelectronic-Circuits-3r ic-Circuits-3rd-Edition-by-Rashi d-Edition-by-Rashi

��


Solution Manual for Microelectronic Circuits 3rd Edition by Rashid

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