Solution manual for microelectronic circuits 3rd edition by rashid

Assume equal contribution by each... The gain parameter The value of Ro that will keep gain variation within 1% for variation in RL from 20 Ω to 500 Ω can be found from... According to

Chapter 1

8

3.5 136.7

3

6.5 5 10 Sin 200 6.5

6.5 6.5 1

5 10 Sin 200

5 10 Sin 200

5 10 Sin 200 1

AB

DC

DC DC

L

ab

ab a L

1.7 Given

3

3 3

3

7.5 10 10 Sin 2000 7.5

7.5 7.5 1

10 10 Sin 2000

10 10 Sin 2000

10 10 Sin 2000 1

DC

DC

DC DC

7.5 5 1.5

2.5 20 10 Sin 7.5 4.5 10 Sin

V

A = = From Eq (1.10)

312.5 (40 10 )

o L P

v P A

From Eq (1.11)

1 25 40

o

I

s s

s i

s i

v R A

v

R =

6.5 sin10005000

83.3– 42 mV ≤ VI – 24 mV ≤ 66 mV – 18 mV ≤ VI ≤ 90 mV

vo = io RL = 100 × 10–3 × 103 = 100 V

Av = oi

v

v =

1000.1 = 1000 or 60 dB

Ai = oi

i

i =

–3 –3

10

× = 100 or 40 dB

Ap = Av Ai = 1000 × 100 = 105 or 50 dB

1.15 (a) From Eq (1.23)

Av = (1 s i) (vo1 o L)

=

2

vo i L 2

v

=

2

vo i L 2

1.17

o o

v v

v v

∆

o1.5 k

R R

R V R

Ri > 7500 Ω From Eq (1.27)

o

o

v v

For o

o

v v

∆ ≤ 0.5%

A

Variation in Av will be contributed by Avo, RS, and RL Assume equal contribution by each

Hence the value of Ro that will keep the variation in gain within 0.5% for variation in RL from 5 kΩ to 20 kΩ can be found from

v

1

1R + −(1 A ) (R+R )

(c) For is = 2.5 µA, Rs + Rx = s

s

v i

+

50 = (1 503 100 k 1 100 19.9 k)(is )

A

1.27

+ –

Ri ve

+ –

A Li i Ro

RL

+ –

vo

vi

ii

+ –

A iis iRo

Ro

RL

+ –

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

is Rs

A iis L

RL

+ –

R

Rx = R – RAis = R(1 – Ais)

For Ais > 1, Rx is negative, and if Ais = 2

Rx = – R For Rx = – 10 kΩ we need R = 10 kΩ

Thus an ideal current amplifier with Ais = 2 and R = 10 kΩ will simulate a negative resistance

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

Zx = ii

( )( )

( )( )

60×10

(a) Let C = 0.1 µF, then Ri =

100.1 10× ×60×10 = 1.67 kΩ

(b) Variation in Gm, according to Eq (1.36), will be contributed by Gms and RL Assume equal

contributions, Rs = 0 The gain parameter

The value of Ro that will keep gain variation within 1% for variation in RL from 20 Ω to 500

Ω can be found from

+ –

Rs

RL

G vm i Ro

Assume 1% for and 1% for

Ri > 0.99 1 k

0.01

× = 99 kΩ Similarly

R + = o

1100

R + = i

15k

R + , Ri (1 – 0.99) = 5 k × 0.99 – 2 k

Ri = 295 kΩ For vs = 10 V, Io = 100 mA

Gm = os

1.33

Using Eq (1.41)

Zm = (1 o L) (mo1 i s)

Av = ( s imo) ( LL o)

–3 A

Ai = os

i

i =

–3

–32.3 10

×

1.34

Since the output variation should be kept within ±2%, variation of effective transimpedance

Zm should be kept to ±2% According to Eq (1.41) the variation in Zm will be contributed by

Zmo and Ro Assume equal contribution to the variation

Ri = s(1 0.99)

0.99

= 100 k 0.010.99

×

= 101 kΩ m

R

R +R = 0.99, Ro =

m(1 0.99)0.99

= 20 k 0.010.99

× = 202

–

Rs

+ –

Rs

A vvo i

+ –

vo

Voltage amplifier For ideal voltage amplifier

i

Z

Using Avo = Ais o

i

R R

Zmo = Avo⋅Ri = 250 × 50 × 103

= 12.5 MΩ Equivalent amplifiers are:

vi

+ –

+ –

vo

+ –

+ –

–

+ –

Ri vo

Gms = 20 mA/V

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

Avo = Ais o

i

R R

Zmo = Avo · Ri = 40 × 100 × 103

= 4 MΩ Equivalent circuits

vs R1 Gvovi RL

Voltage amplifier

Rs

+ –

+ –

vi

+ –

ii

Ri

io + –

RL

1.38 (a) From Eq (1.21)

Total effective voltage gain using Eq (1.23)

Av = os

Ai = oi

= 110,769 or 100.9 dB

Ap = Li

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

= 76.9 From Eq (1.45), the overall open-circuit voltage gain

Avo = Av1 ⋅ Av2 ⋅ Avo3 = 76.92 × 80 = 473,088 or 113.5 dB From Eq (1.23)

Av = ( i vos)(iL L o)

= ( o3 is3L)(is2 o2is1 o3 o2i3)( o1o1 i2)

(b) Problem 1.40 Cascaded Current Amplifier

| Av(jω) | =

4

21010

1.47

Rs

+ –

vo

+ –

+ –

980×0.1 = 10,204

fH = 10204

2π = 1624 Hz

fbw = Av(mid) × fH = 19.404 × 1624 = 31,512 (b) RL = 10 kΩ, Ro || RL = 10 k || 50 k = 8333 Ω

v g R

sR C

−+

vi

+ –

1.49

+ –

Rs

+ –

C1

RL

+ –

+ –

Rs

+ –

+ –

−

−+ ×

× – 1.2 = 3.6 Ω

60 = 41.667 Ω L

120+

R

× = 50 V