Microelectronic Circuit Design Third Edition - Part III Solutions to Exercises doc

€ Values taken from op - amp specification sheets www.jaegerblalock.com or www.analog.com Page 631 € Values taken from op - amp specification sheets www.jaegerblalock.com or www.analog

Microelectronic Circuit Design

Third Edition - Part III Solutions to Exercises

CHAPTER 10

Page 509

I o

I i =

1.58 A 0.167µA = 9.48x10

€

i

( ) The constant slope region spanning a maximum input range is between 0.4 ≤ vI ≤ 0.65,

and the bias voltage V I should be centered in this range : V I =0.4 + 0.65

2 V = 0.525 V.

v i ≤ 0.65 − 0.525 = 0.125 V and vi ≤ 0.525 − 0.40 = 0.125 V For vI = 0.8V, the slope is 0 Av = 0.

(ii) v O = V O + v o For v i = 0, v I = V I = 0.6, V O = 14V and A v = +40 V o = A v V i = 40 0.01V( )= 4V

v O= 14.0 + 4.00sin1000( πt) volts V O = 14 V

A v(j0.95) = 20

12+( )0.12 0.952

CHAPTER 11

Page 545

R1+ R2 ≥ 100kΩ 501R1≥ 100kΩ → R1≥ 200 Ω There are many possibilities

(R1= 200 Ω, R2= 100 kΩ), but ( R1= 220 Ω, R2= 110 kΩ) is a better solution since

resistor tolerances could cause i O to exceed 0.1 mA in the first case

Page 554

R3 =

−6V 3kΩ = −2mA i O= −2sin1000( πt − 2sin2000πt) mA

Page 559

V+= V2 R4

R3+ R4 = 5

100kΩ 10kΩ +100kΩ = 4.545V I O =20.0 − 4.545

36kΩ = −92.1 µA

Page 560

€

v O = −RC dv S

dt = − 20kΩ( ) (0.02µF) (2.50V) (2000π) (cos2000πt)= −6.28cos2000πt V

ii

( ) A v = A vA A vB A vC = −25.2( )3= −16,000 R in = R inA = 2.7 kΩ R out = R outC = 0

Page 570

dC1 =

C1Q

dQ

dR2 R1= R2→ Q =

12

C1

C2 → S R2

Q

= 0 Page 577

R th = 2kΩ 2kΩ = 1kΩ f o= 1

2π 1kΩ 82kΩ( ) (0.02µF) (0.02µF) = 879 Hz Q =

12

82kΩ 1kΩ = 4.53

( ) (0.02µF)

0.02µF + 0.02µF = 4.53The values of the resistors are unchanged C1= C2 =0.02µF

4 = 0.005 µF

2π 1kΩ( ) (82kΩ) (0.005µF) (0.005µF) = 3520 Hz Q =

82kΩ 1kΩ

0.005µF

( ) (0.005µF)

0.005µF + 0.005µF = 4.53

Page 585

€

The diode will conduct and pull the output up to v O = v S = 1.0 V v1= v O + v D = 1.0 + 0.6 = 1.6 V For a negative input, there is no path for current through R, so v O = 0 V The op - amp

sees a -1V input so the output will limit at the negative power supply : v O = −10 V.

The diode has a 10 - V reverse bias across it, so V Z > 10 V.

Page 587

CHAPTER 12

Page 612

( ) Values taken from OP - 27 specification sheet (www.jaegerblalock.com or www.analog.com)

(iii) Values taken from OP - 27 specification sheet

Page 613

€

Values taken from OP - 77 specification sheet (www.jaegerblalock.com or www.analog.com)

Page 616

( ) (0.1Ω)= 20,000

Page 617

( ) (0.1Ω) = 200,000 A dB = 20 log 2x10

5

( )= 106 dB

Page 619

€

Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 620

€

Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 626

€

Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 631

€

Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 634

€

Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 637

Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

A few possibilities : 27 kΩ and 3 kΩ, 270 kΩ and 30 kΩ, 180 kΩ and 20 kΩ, etc.

Page 640

s +105π

s + 316π

f H ≅βf T = 5MHz

1+10

50 20

€

A v( )0 = 50 25( )= 1250 A v( )ωH =1250

2 = 8841+ ωH

CHAPTER 13

Page 671

= 2.7V with v GS − V TN = 4 −1 = 3V , so the transistor has entered the triode region.

(d) Choose two points on the i - v characteristics For example,

V CE = 12 − 4300I C −1500I E = 12 − 4300 1.45mA( )−1500 101

R B = 20kΩ 62kΩ = 15.1 kΩ R L = 8.2kΩ 100kΩ = 7.58 kΩ

€

A vt = −g m R L = −9.80mS 18kΩ( )= −176

( ) The slope of the output characteristics is zero, so λ = 0 and r o = ∞

For the positive change in vgs, g m= Δi D

Δv GS

≅

2.1V 3.3kΩ 0.5V = 1.3 mS

€

V DS ≥ V GS − V TH = 1.981V −1V = 0.981 V

Page 717

€

a

( ) V M ≤ min I[ C R C , V( CE − V BE) ]= min 239µA 10kΩ[ ( ), 3.67 − 0.7( )V]= 2.39 V

Limited by the voltage drop across R C

b

( ) V M ≤ min I[ D R D , V( DS − V DSSAT) ]= min 250µA 30kΩ[ ( ), 4.50 −1( )V]= 3.50 V

Limited by the value of V DS

CHAPTER 14

Page 754

FET : v i ≤ 0.2 V( GS − V TN) (1+ g m R I)R I + R6

R6 = 0.2 0.982( ) [1+ 0.491mS 2kΩ( ) ]2kΩ +12kΩ 12kΩ = 454 mV Neglecting R6, v i ≤ 0.2 V( GS − V TN) (1+ g m R I)= 0.2 0.982( ) [1+ 0.491mS 2kΩ( ) ]= 389 mV

Page 786

= A 3000Ω69.1Ω + 3000Ω= 0.959

Page 805

i

( ) Reverse the direction of the arrow on the ermitter of the transistor as well

as the values of VEE and VCC

= V CCmin − I CmaxR Cmax

= 5V 0.95( )− 368µA 8.2kΩ( ) (1.05)= 1.58 V 1.58 ≥ 0, so active region is ok 10% tolerances I Cmax

= V CCmin − I CmaxR Cmax

= 5V 0.90( )− 410µA 8.2kΩ( ) ( )1.1 = 0.802 V 0.802 ≥ 0, so active region is ok.



 = 49.9 Page 811



 = −50.0

( ) R out = 3300 1

0.0796S +

399090.1

CHAPTER 15

Page 842

V CE1= 15 − 74.3µA 10kΩ( )− −0.7( )= 15.0 V V CE 2 = 15 − 74.3µA − 7.5µA( ) (10kΩ)− −0.7( )= 15.0 V

V EB 3= 74.3µA − 7.5µA( ) (10kΩ)= 0.668 V I S 3= 750µA

exp 0.668V0.025

500µA 50µA

5mA 50µA

R in = 2rπ = 2 50

40 50( µA)= 50 kΩ R out ≅

90V 5mA = 18.0 kΩ

Page 871

( ) Since the device parameters are the same, V BE1 = V EB 2=0.5mA 2.4kΩ( )

= 0.490 MR = 0.5

1+15601+ 0.7

60 +

1.575

= 0.606

MR = 5 A

2 A = 2.50 MR =

2.501+ 3.575

= 2.39 MR = 2.5

1+15601+ 0.7

60 +

3.575

= 2.95

= 7.46 µA IO3= 5 7.46µA( )= 37.3 µA IO4 = 10 7.46µA( )= 74.6 µA

I O2= 10µA 1+

10501+0.7

50 +

1750

= 8.86 µA IO 3= 50µA

1+10501+0.7

50 +

1750

= 44.3 µA

I O4 = 100µA

1+10501+ 0.7

50 +

1750

= 10.1 µA R out 2=50V + 10V

10.1µA = 5.94 MΩ

50V + 10V 529µA = 113 kΩ

Page 906

50V + 10V 117µA = 513 kΩ

Page 907

R out ≅βo r o

2 =

1502

50V + 15V 50µA

67V + 14.3V 50µA

( ) V DS 4 ≥ V GS 4 − V TN = 0.2 V V D 4 ≥ V S 4 + 0.2 V D4 ≥ 0.95 + 0.2 = 1.15 V

iii

( ) I O = 50µA ± 0.1% ΔI O ≤ 50nA R out ≥ 20V

50nA = 400 MΩ Choose R out = 1 GΩ.

Page 917

( ) Since the transistors have the same parameters, V GS1=V DD − −V( SS)

A v = G m R o = 78.5

Page 928

1+ 0.7V

60V +

250

= 1.08 mA I2 = 0.25 1.09mA( ) 1+

21.3V 60V

1+ 0.7V

60V +

250

733µA 18.4µA1

R = 2r 1.3r = 2 60 +13

   1.3 60 +1.3

   = 20.1MΩ 11.1MΩ = 7.15 MΩ

CHAPTER 16

Page 988

2π 100

2+ 10002− 2 500( )2− 2 0( )2 = 114 Hz

Page 989

ω2+ 1002 →ω ≥ 205 rad/s

g m

2πCµ =

0.064S 2π 1pF( )= 10.2 GHz

A mid= −135 is not affected by the value of fT

2π R L(Cµ+ C L)=

12π 4120Ω( ) (0.5 + 5)pF = 7.02 MHz

Page 1043

CHAPTER 17

Page 1078

1041+104(0.0990) = 10.1

R in = R id(1+ Aβ)= 25kΩ 1+ 10[ 4(0.0990) ]= 24.8 MΩ R out = R o

1+ Aβ =

1031+104(0.0990)= 1.01 Ω

ii

( ) A = 4730 1kΩ + 25kΩ + 9.01kΩ

25kΩ + 9.01kΩ = 4870 A v =

48701+ 4870 0.0990( ) = 10.1

R in = R id(1+ Aβ)= 34.0kΩ 1+ 4870 0.0990[ ( ) ]= 16.4 MΩ R out = R o

1+ Aβ =

662Ω1+ 4870 0.0990( ) = 1.37 Ω

Page 1086

€

R E is now connected between the emitter and collector of Q2 in the upper half of

Fig 17.36 However, when the A - circuit is constructed, y22F is connected in parallel

with R E, so the resistance at the output of the A - circuit is still 901 Ω The value of

y11F is unchanged, so overall the A - circuit is unchanged Likewise, β is unchanged,

and so A tr is the same

Page 1102

R out = 2r o3

1+µf

21+ 0 ≅µf r o3 Page 1116

From the Bode plot, the phase shift at the unity gain frequency (10 MHz) is -130o

The phase margin is 180 o - 130 o = 50o

s = 15.4

V

µs ii

= 3.00 V

Page 1149