1

Microelectronic Circuit DesignThird Edition

Solutions to Exercises

CHAPTER 1

Page 11

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VLSB =5.12V

210 bits=

5.12V1024bits

= 5.00 mV VMSB =5.12V

2= 2.560V

11000100012 = 29 + 28 + 24 + 20 = 78510 VO = 786 5.00mV( ) = 3.925 V

or VO = 2−1 + 2−2 + 2−6 + 2−10( ) 5.12V = 3.912 V

Page 12The dc component is VA = 4V. The signal consists of the remaining portion of vA: va = (5 sin 2000πt + 3 cos 1000 πt) Volts.

Page 19

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iSC = i1 + βi1 =vsR1

+ βvsR1

= β +1( ) vsR1

vOC =β +1( )RS

β +1( )RS + R1vS

Rth =vOCiSC

=β +1( )RS

β +1( )RS + R1R1β +1( )

=1

β +1( )R1

+1RS

→ Rth = RSR1β +1( )

Page 23

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vo = −5cos 2000πt + 25o( ) = − −5sin 2000πt + 25o − 90o( )[ ] = 5sin 2000πt − 65o( )

Vo = 5∠− 65o Vs = 0.001∠0

o Av =5∠− 65o

0.001∠0o= 5000∠− 65o

2

Page 26

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vs = 0.5sin 2000πt( ) + sin 4000πt( ) +1.5sin 6000πt( )[ ]The three spectral components are f1 =1000 Hz f2 = 2000 Hz f3 = 3000 Hz

a( ) The gain of the band - pass filter is zero at both f1 and f3. At f2, Vo =10 1V( ) =10V ,and vO =10.0sin 4000πt( ) volts.b( ) The gain of the low - pass filter is zero at both f2 and f3. At f2, Vo = 6 0.5V( ) = 3V ,

and vO = 3.00sin 2000πt( ) volts.

Page 27

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39kΩ 1− 0.1( ) ≤ R ≤ 39kΩ 1+ 0.1( ) or 35.1 kΩ≤ R ≤ 42.9 kΩ3.6kΩ 1− 0.01( ) ≤ R ≤ 3.6kΩ 1+ 0.01( ) or 3.56 kΩ≤ R ≤ 3.64 kΩ

Page 29

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P =VS

2

R1 + R2 Pnom =

152

54kΩ= 4.17 mW

Pmax =1.1x15( )

2

0.95x54kΩ= 5.31 mW Pmin =

0.9x15( )2

1.05x54kΩ= 3.21 mW

Page 33

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R =10kΩ 1+10−3

oC−55− 25( )oC

= 9.20 kΩ R =10kΩ 1+

10−3

oC85− 25( )oC

=10.6 kΩ

3

CHAPTER 2

Page 47

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ni = 2.31x1030 K−3cm−6( ) 300K( )3 exp −0.66eV

8.62x10−5eV / K( ) 300K( )

= 2.27x1013 cm3

ni = 1.08x1031 K−3cm−6( ) 50K( )3 exp −1.12eV

8.62x10−5eV / K( ) 50K( )

= 4.34x10−39 cm3

ni = 1.08x1031 K−3cm−6( ) 325K( )3 exp −1.12eV

8.62x10−5eV / K( ) 325K( )

= 4.01x1010 cm3

L3 =cm3

4.34x10−3910−2

mcm

3

→ L = 6.13x1010 m

Page 48

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vp = µ pE = 500cm2

V − s10

Vcm

= 5.00x10

3 cms

vn = −µnE =1350cm2

V − s1000

Vcm

=1.35x10

6 cms

E =VL

=1

2x10−4Vcm

= 5.00x103Vcm

Page 48

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µn =vnE

=4.3x105cm / s

100V / cm= 4300

cm2

s µ p =

v pE

=2.1x105cm / s

100V / cm= 2100

cm2

s

µn =vnE

=8.5x105cm / s

100V / cm= 8500

cm2

s

4

Page 50

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ni2 =1.08x1031 400( )

3exp

−1.12

8.62x10−5 400( )

= 5.40x1024 / cm6

ρ =1σ

=1

1.60x10−19 2.32x1012( ) 1350( ) + 2.32x1012( ) 500( )[ ]=1450 Ω− cm

ni2 =1.08x1031 50( )

3exp

−1.12

8.62x10−5 50( )

=1.88x10−77 / cm6

ρ =1σ

=1

1.60x10−19 4.34x10−39( ) 6500( ) + 4.34x10−39( ) 2000( )[ ]=1.69x1053 Ω− cm

Page 54

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ni2 =1.08x1031 400( )

3exp

−1.12

8.62x10−5 400( )

= 5.40x1024 / cm6

p = N A − N D =1016 − 2x1015 = 8x1015

holes

cm3 n =

ni2

p=

5.40x1024

8x1015= 6.75x108

electrons

cm3

n = N D = 2x1016 electrons

cm3 n =

ni2

p=

1020

2x1016= 5.00x103

holes

cm3 n > p→ n - type silicon

Page 55Reading from the graph for NT = 10

16/cm3, 1250 cm2/V-s and 400 cm2/V-s.Reading from the graph for NT = 10

17/cm3, 800 cm2/V-s and 230 cm2/V-s.

Page 57

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σ =1000 =1.60x10−19µnn → unn = 6.25x1021 / cm3 = 6.25x1019( ) 100( )

(a) NT = 2x1016 / cm3 (b) NT = 5x10

16 / cm3

5

Page 58

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p = N A − N D = 4x1018 holes

cm3 n =

ni2

p=

1020

4x1018= 25

electrons

cm3

NT =4x1018

cm3 and mobilities from Fig. 2.8

p = N A − N D = 7x1019 holes

cm3 n =

ni2

p=

1020

7x1019=1.4

electrons

cm3

NT =7x1019

cm3 and mobilities from Fig. 2.8

ρ =1σ

=1

1.60x10−19 1.4( ) 100( ) + 7x1019( ) 50( )[ ]=1.79 mΩ− cm

Page 59

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VT =kTq

=1.38x10−23 50( )

1.602x10−19= 4.31 mV

VT =1.38x10−23 300( )

1.602x10−19= 25.8 mV

VT =1.38x10−23 400( )

1.602x10−19= 34.5mV

Page 59

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Dn =kTq

µn = 25.8mV 1362cm2

V − s

= 35.1

cm2

s Dp =

kTq

µ p = 25.8mV 492cm2

V − s

=12.7

cm2

s

jn = qDndndx

=1.60x10−19C 20cm2

s

1016

cm3 −µm

104µmcm

= 320

A

cm2

jp = −qDpdpdx

= −1.60x10−19C 4cm2

s

1016

cm3 −µm

104µmcm

= −64

A

cm2

6

CHAPTER 3

Page 79

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Emax =1εs

−qN A−x p

0

∫ dx =qN A x pεs

Emax = −1εs

qN D0

xn

∫ dx = qN D xnεs

For the values in Ex.3.2 :

Emax =1.6x10−19C 1017 / cm3( ) 1.13x10−5cm( )

11.7 8.854x10−14 F / cm( )=175

kVcm

Emax =1.6x10−19C 1020 / cm3( ) 1.13x10−8cm( )

11.7 8.854x10−14 F / cm( )=175

kVcm

Emax =2 1.05V( )

2.63x10−6cm= 799

kVcm

xp = 0.0258µm 1+2x1018

1020

−1

= 0.0253 µm xn = 0.0258µm 1+1020

2x1018

−1

= 5.06x10−4 µm

Page 82

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T =1.602x10−19 25.8mV( )

1.03 1.38x10−23( )= 291K

Page 84

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iD = 5x10−15 A exp

−0.040.025

−1

= −3.99 fA iD = 5x10

−15 A exp−2.00.025

−1

= −5.00 fA

Page 86

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VBE = VT ln 1+IDIS

= 0.025V ln 1+

40x10−6 A

2x10−15 A

= 0.593 V

VBE = VT ln 1+IDIS

= 0.025V ln 1+

400x10−6 A

2x10−15 A

= 0.651 V ΔVBE = 57.6 mV

VBE = VT ln 1+IDIS

= 0.0258V ln 1+

40x10−6 A

2x10−15 A

= 0.612 V

VBE = VT ln 1+IDIS

= 0.0258V ln 1+

400x10−6 A

2x10−15 A

= 0.671 V ΔVBE = 59.4 mV

Page 89

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wd = 0.113µm 1+10V

0.979V= 0.378 µm Emax =

2 10.979V( )0.378x10−4cm

= 581kVcm

7

Page 89

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IS =10 fA 1+10V0.8V

= 36.7 fA

Page 92

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Cjo =11.7 8.854x10−14 F / cm( )

0.113x10−4cm= 91.7

nF

cm2 Cj 0V( ) = 91.7

nF

cm210−2cm( ) 1.25x10−2cm( ) =11.5 pF

Cj 5V( ) =11.5 pF

1+5V

0.979V

= 4.64 pF

Page 92

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CD =10−5 A0.025V

10−8 s = 4.00 pF CD =8x10−4 A0.025V

10−8 s = 320 pF CD =5x10−2 A0.025V

10−8 s = 0.02 µF

Page 97

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Two points on the load line : VD = 0, ID =5V5kΩ

=1 mA; ID = 0, VD = 5V

Page 108

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From the answer, the diodes are on,on,off.

I1 = ID1 + ID2 10V - VB

2.5kΩ=

VB − 0.6V − −20V( )10kΩ

+VB − 0.6V − −10V( )

10kΩ= 0→VB =1.87 V

ID1 =1.87 − 0.6 − −20V( )

10kΩ= 2.13 mA ID2 =

1.87 − 0.6 − −10V( )10kΩ

=1.13 mA

VD3 = − 1.87 − 0.6( ) = −1.27 V

Page 110

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Rmin =5kΩ

205−1

=1.67 kΩ VO = 20V1kΩ

5kΩ+1kΩ= 3.33 V (VZ is off) VO = 5 V (VZ is conducting)

Page 111

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VL − 20V1kΩ

+VL − 5V0.1kΩ

+VL

5kΩ= 0→VL = 6.25 V IZ =

6.25V − 5V0.1kΩ

=12.5mA

PZ = 5V 12.5mA( ) +100Ω 12.5mA( )2

= 78.1 mW

Page 112

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Line Regulation =0.1kΩ

0.1kΩ+ 5kΩ=19.6

mVV

Load Regulation = 0.1kΩ 5kΩ = 98.0 Ω

8

Page 117

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Vdc = VP −Von = 6.3 2 −1= 7.91 V Idc =7.91V0.5Ω

=15.8 A Vr =IdcC

T =15.8A0.5F

160

s = 0.527 V

ΔT =1ω

2VrVP

=1

2π 60( )2

0.527V8.91V

= 0.912 ms θc =120π 0.912ms( )180

o

π=19.7o

Page 118

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Vdc = VP −Von =10 2 −1=13.1 V Idc =13.1V

2Ω= 6.57 A C =

IdcVr

T =6.57A0.1F

160

s =1.10 F

ΔT =1ω

2VrVP

=1

2π 60( )2

0.1V14.1V

= 0.316 ms θc =120π 0.316ms( )180

o

π= 6.82o

Page 118

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For VT = 0.025V , VD =kTq

ln 1+IDIS

= 0.025V ln 1+

48.6 A

10−15 A

= 0.961 V

VD =kTq

ln 1+IDIS

=

1.38x10−23 J / K 300K( )1.60x10−19C

ln 1+48.6 A

10−15 A

= 0.994 V

VD =kTq

ln 1+IDIS

=

1.38x10−23 J / K 273K + 50K( )1.60x10−19C

ln 1+48.6 A

10−15 A

=1.07 V

9

CHAPTER 4

Page 151

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(i) Kn' = µn

εoxTox

= 500cm2

V − s

3.9 8.854x1014 F / cm( )25x10−7cm

= 69.1x10−6C

V 2 − s= 69.1

µAV 2

(ii) Kn = Kn' W

L= 50

µAV 2

20µm1µm

=1000

µAV 2

Kn = 50µAV 2

60µm3µm

=1000

µAV 2

Kn = 50µAV 2

10µm0.25µm

= 2000

µAV 2

(iii) For VGS = 0V and 1V , VGS < VTN and the transistor is off. Therefore ID = 0.

VGS - VTN = 2 -1.5 = 0.5V and VDS = 0.1 V → Triode region operation

ID = Kn' W

LVGS −VTN −

VDS2

VDS = 25

µAV 2

10µm1µm

2 −1.5−

0.12

0.1V

2 =11.3 µA

VGS - VTN = 3 -1.5 = 1.5V and VDS = 0.1 V → Triode region operation

ID = Kn' W

LVGS −VTN −

VDS2

VDS = 25

µAV 2

10µm1µm

3−1.5−

0.12

0.1V

2 = 36.3 µA

Kn' = 25

µAV 2

100µm1µm

= 250

µAV 2

Page 152

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Ron =1

Kn' W

LVGS −VTN( )

=1

250µAV 2

2 −1( )= 4.00 kΩ Ron =

1

250µAV 2

4 −1( )=1.00 kΩ

VGS = VTN +1

KnRon=1V +

1

250µAV 2

2000Ω( )= 3.00 V

10

Page 155

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VGS - VTN = 5-1 = 4 V and VDS = 10 V → Saturation region operation

ID =Kn2

VGS −VTN( )2

=12

mA

V 25−1( )

2V 2 = 8.00 mA

Kn = Kn' W

L→

WL

=1mA40µA

=251

W = 25L = 8.75 µm

€

Assuming saturation region operation,

ID =Kn2

VGS −VTN( )2

=12

mA

V 22.5−1( )

2V 2 =1.13 mA

gm = Kn VGS −VTN( ) =1mAV 2 2.5−1( )V =1.50 mS

Checking : gm =2 1.125mA( )

2.5−1( )V=1.50 mS

Page 156

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(i) VGS - VTN = 5-1 = 4 V and VDS = 10 V → Saturation region operation

ID =Kn2

VGS −VTN( )2

1+ λVDS( ) =12

mA

V 25−1( )

2V 2 1+ 0.02 10( )[ ] = 9.60 mA

ID =Kn2

VGS −VTN( )2

1+ λVDS( ) =12

mA

V 25−1( )

2V 2 1+ 0 10( )[ ] = 8.00 mA

(ii) VGS - VTN = 4 -1 = 3 V and VDS = 5 V → Saturation region operation

ID =Kn2

VGS −VTN( )2

1+ λVDS( ) = 252µAV 2

4 −1( )2V 2 1+ 0.01 5( )[ ] =118 µA

ID =Kn2

VGS −VTN( )2

1+ λVDS( ) =252

µAV 2

5−1( )2V 2 1+ 0.01 10( )[ ] = 220 µA

Page 157

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(i) Assuming pinchoff region operation, ID =Kn2

0 −VTN( )2

=502

µAV 2

+2V( )2

=100 µA

VGS = VTN +2IDKn

= 2V +2 100 µA( )50µA /V 2

= 4.00 V

(ii) Assuming pinchoff region operation, ID =Kn2

VGS −VTN( )2

=502

µAV 2

1− −2V( )[ ]2

= 225 µA

11

Page 159

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(i) VTN = VTO + γ VSB + 0.6V − 0.6V( ) =1+ 0.75 0 + 0.6 − 0.6( ) =1 VVTN =1+ 0.75 1.5+ 0.6 − 0.6( ) =1.51 V VTN =1+ 0.75 3+ 0.6 − 0.6( ) =1.84 V

(ii − a) VGS is less than the threshold voltage, so the transistor is cut off and ID = 0.

(b) VGS - VTN = 2 -1 = 1 V and VDS = 0.5 V → Triode region operation

ID = Kn VGS −VTN −VDS2

VDS =1

mA

V 22 −1−

0.52

0.5V

2 = 375 µA

(c) VGS - VTN = 2 -1 = 1 V and VDS = 2 V → Saturation region operation

ID =Kn2

VGS −VTN( )2

1+ λVDS( ) = 0.5mA

V 22 −1( )

2V 2 1+ 0.02 2( )[ ] = 520 µA

Page 160

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a( ) VGS is greater than the threshold voltage, so the transistor is cut off and ID = 0.

b( ) VGS - VTN = -2 +1 = 1 V and VDS = 0.5 V → Triode region operation

ID = Kn VGS −VTN −VDS2

VDS = 0.4

mA

V 2−2 +1−

−0.52

−0.5( )V 2 =150 µA

c( ) VGS - VTN = −2 +1 = 1 V and VDS = 2 V → Saturation region operation

ID =Kn2

VGS −VTN( )2

1+ λVDS( ) =0.42

mA

V 2−2 +1( )

2V 2 1+ 0.02 2( )[ ] = 208 µA

Page 165

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Active area =120Λ2 =120 0.5µm( )2

= 30 µm2 L = 2Λ = 1µm W = 10Λ = 5µm

Gate area = 20Λ2 = 20 0.5µm( )2

= 5 µm2 N =104µm( )

2

14Λ( ) 16Λ( ) 0.5µm( )2

=1.79 x 106 transistors

12

Page 167

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CGC = 200µFm2

5x10

−6 m( ) 0.5x10−6 m( ) = 0.500 fF

Triode region : CGD = CGS =CGC

2+ CGSOW = 0.25 fF + 300

pFm

5x10

−6 m( ) =1.75 fF

Saturation region : CGS =23

CGC + CGSOW = 0.333 fF + 300pFm

5x10

−6 m( ) =1.83 fF

CGD = CGSOW = 300pFm

5x10

−6 m( ) =1.50 fF

Page 169

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KP = Kn =150U LAMBDA = λ = 0.0133 VTO = VTN =1 PHI = 2φF = 0.6

W = W = 1.5U L = L = 0.25U

Page 173

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i( ) Assume saturation region operation and λ = 0. Then ID is indpendent of VDS, and ID = 50 µA.VDS = VDD − ID RD =10 − 50kΩ 50µA( ) = 7.50 V . VDS ≥VGS −VTN , so our assumption was correct.Q - Point = 50.0 µA, 7.50 V( )

ii( ) VEQ =270kΩ

270kΩ+ 750kΩ10V = 2.647 V REQ = 270kΩ 750kΩ Assume saturation region.

ID =25x10−6

2A

V 22.647 −1( )

2V 2 = 33.9 µA VDS = VDD − ID RD =10 −100kΩ 33.9µA( ) = 6.61 V .

VDS ≥VGS −VTN , so our assumption was correct. Q - Point = 33.9 µA, 6.61 V( )

iii( ) VGS does not change : VGS = 3.00 V . ID = 30x10−6

2A

V 23−1( )

2V 2 = 60.0 µA

VDS = VDD − ID RD =10 −100kΩ 60.0µA( ) = 4.00 V . VDS ≥VGS −VTN , so our assumption was correct.Q - Point = 60.0 µA, 4.00 V( )

iv( ) VGS does not change : VGS = 3.00 V . ID =25x10−6

2A

V 23−1.5( )

2V 2 = 28.1 µA

VDS = VDD − ID RD =10 −100kΩ 28.1µA( ) = 7.19 V . VDS ≥VGS −VTN , so our assumption was correct.Q - Point = 28.1 µA, 7.19 V( )

13

Page 174

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VDS =10 −25x10−6 105( )

23−1( )

21+ 0.01VDS( )→VDS =10 − 5 1+ 0.01VDS( )

VDS =10 − 51.05

V = 4.76 V ID =25x10−6

23−1( )

21+ 0.01 4.76( )[ ] = 52.4 µA

Page 176

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For ID = 0, VDS = 10 V . For VDS = 0, ID =10V

66.7kΩ=150µA.

The load line intersects the VGS = 3 - V curve at ID = 50 µA, VDS = 6.7 V .

Page 178

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i( ) 4 = VGS +30x10−6 3.9x104( )

2VGS −1( )

2→VGS

2 − 0.291VGS − 5.838 = 0→VGS = 2.566V , − 2.275V

ID =30x10−6

22.566 −1( )

2= 36.8 µA VDS =10 −114kΩ 36.8µA( ) = 5.81 V

Q - Point : 36.8 µA,5.81 V( )

ii( ) 4 = VGS +25x10−6 3.9x104( )

2VGS −1.5( )

2→VGS

2 − 0.949VGS − 5.955 = 0→VGS = 2.960V , − 2.012V

ID =25x10−6

22.960 −1.5( )

2= 26.7 µA VDS =10 −114kΩ 26.7µA( ) = 6.96 V

Q - Point : 26.7 µA,6.96 V( )

ii( ) 4 = VGS +25x10−6 6.2x104( )

2VGS −1( )

2→VGS

2 − 0.710VGS − 4.161= 0→VGS = 2.426V , −1.716V

ID =25x10−6

22.426 −1( )

2= 25.4 µA VDS =10 −137kΩ 25.4µA( ) = 6.52 V

Q - Point : 25.4 µA,6.52 V( )

14

Page 180

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i( ) Assume saturation. For ID = 99.5µA, VGS = 4V - 99.5µA 1.8kΩ( ) = 3.821V

ID =25µA

23.821−1( )

2= 99.5µA which agrees. VDS =10V − 99.5µA 40.8kΩ( ) = 5.94 V

Saturation region operation is correct.

ii( ) VEQ =1.5MΩ

1.5MΩ+1MΩ10V = 6.00 V REQ =1.5MΩ 1MΩ = 600kΩ Assume saturation region.

6 = VGS +25x10−6 22x103( )

2VGS −1( )

2→VGS

2 +1.636VGS − 20.82 = 0→VGS = 3.818V

ID =25x10−6

23.818 −1( )

2= 99.3 µA VDS =10 − 40kΩ 99.3µA( ) = 6.03 V

Saturation region operation is correct. Q- Point : 99.3 µA, 6.03 V( )

iii( ) R1 + R2 =10V2µA

= 5MΩ R1

R1 + R210V = 6V → R1 = 5MΩ

6V10V

= 3 MΩ→ R2 = 2 MΩ

REQ = 3MΩ 2MΩ =1.2 MΩ

Page 182

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VGS = 6 − 22000ID VSB = 22000ID VTN =1+ 0.75 VSB + 0.6 − 0.6( ) ID = 25µA2 VGS −VTN( )2

Spreadsheet iteration yields ID = 83.2 µA.

Page 183

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Equation 4.55 becomes 6 - 1+ 0.75 VSB + 0.6 − 0.6( ) + 2.83[ ] −VSB = 0.VSB

2 − 6.065VSB + 7.231= 0→VSB =1.63V .

RS =1.63V

10−4 A=16.3 kΩ→16 kΩ RD =

10 − 6 −1.63( )V10−4 A

= 23.7kΩ→ 24 kΩ

Page 185

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i( ) Using Eq. (4.58) VGS = 3.3−2x10−4 10kΩ( )

2VGS −1( )

2→VGS

2 −VGS − 2.3 = 0→VGS = 2.097V

VGS =2x10−4

22.097 −1( )

2=120.3µA VDS = VGS Q - Po int : 120 µA, 2.10 V( )

ii( ) Since there was no voltage drop across RG in (i), the equations and answers are identical tothe first part.

15

Page 186

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4 −VDS1.8x106

= 2.5x10−4 4 −1−VDS2

VDS →VDS = 2.96 mV ID =

4 −VDS1.8x106

= 2.22 µA

Q - Po int : 2.22 µA, 2.96 mV( ) Checking; ID Ron =2.22µA

2.5x10−4 4 −1( )= 2.96 mV

Page 189

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i( ) 10 − 6 =25x10−6 62x104( )

2VGS +1( )

2−VGS →VGS

2 + 0.710VGS − 4.161= 0→VGS = −2.426V , +1.716V

ID =25x10−6

2−2.426 +1( )

2= 25.4 µA VDS = − 10 −137kΩ 25.4µA( )[ ] = −6.52 V

Q - Point : 25.4 µA, - 6.52 V( )

Page 192

€

i( ) Circuits/cm2 ∝α2 = 1µm0.25µm

2

= 16

Power - Delay Product ∝1

α3=

1

43=

164

→ 64 times improvement

ii( ) iD* = µnεox

Tox /αW /αL /α

vGS −VTN −vDS2

vDS =α iD P

* = VDDiD* = VDD α iD( ) =α P

P*

A*=

αPW /α( ) L /α( )

=α3PA

iii( ) fT =1

2π500cm2 /V − s

10−4cm( )2

1V( ) = 7.96 GHz fT =1

2π500cm2 /V − s

0.25x10−4cm( )2

1V( ) =127 GHz

Page 193

€

The field in the first sentence at the top of the page should be 105V / cm.

VL

=104Vcm

→ L = 105Vcm

10

−4cm( ) =10 V L = 105 Vcm

10

−5cm( ) =1 V

16

CHAPTER 5

Page 213

€

βF =αF

1−αF=

0.9701− 0.970

= 32.3 βF =0.993

1− 0.993=142 βF =

0.2501− .250

= 0.333

αF =βF

βF +1=

4041

= 0.976 αF =200201

= 0.995 αF =34

= 0.750

Page 215

€

iC =10−15 A exp

0.7000.025

− exp

−9.300.025

−

10−15 A0.5

exp−9.300.025

−1

=1.45 mA

iE =10−15 A exp

0.7000.025

− exp

−9.300.025

+

10−15 A100

exp0.7000.025

−1

=1.46 mA

iB =10−15 A

100exp

0.7000.025

−1

+

10−15 A0.5

exp−9.300.025

−1

=14.5 µA

Page 217

€

iC =10−16 A exp

0.7500.025

− exp

0.7000.025

−

10−16 A0.4

exp0.7000.025

−1

= 563 µA

iE =10−16 A exp

0.7500.025

− exp

0.7000.025

+

10−16 A75

exp0.7500.025

−1

= 938 µA

iB =10−16 A

75exp

0.7500.025

−1

+

10−16 A0.4

exp0.7000.025

−1

= 376 µA

Page 217

€

iT =10−15 A exp

0.7500.025

− exp

−20.025

=10.7 mA

iT =10−16 A exp

0.7500.025

− exp

−50.025

=1.07 mA

17

Page 220

€

VBE = VT lnICIS

+1

= 0.025V ln

10−4 A

10−16 A+1

= 0.691 V

VBE = VT lnICIS

+1

= 0.025V ln

10−3 A

10−16 A+1

= 0.748 V

Page 221

€

npn :VBE > 0, VBC < 0→ Forward −Active Region pnp :VEB > 0, VCB > 0→ Saturation Region

Page 223

€

βF =αF

1−αF=

0.950.05

=19 βR =αR

1−αR=

0.250.75

=13

VBE = 0, VBC

18

Page 226

€

a( ) The currents do not depend upon VCC as long as the collector - base junction is reversebiased. (Later when Early voltage VA is discussed, one should revisit this problem.)

b( ) Forward - active region : IB =100 µA, IE = βF +1( )IB = 5.10 mA, IC = βF IB = 5.00 mA

VBE = VT lnICIS

+1

= 0.025V ln

5.00mA

10−16 A+1

= 0.789 V Checking : VBC = −5 + 0.789 = −4.21 V

c( ) Forward - active region with VCB ≥ 0 requires VCC ≥VBE or VCC ≥ 0.764 V

Page 228

€

IE =−0.7V − −9V( )

5.6kΩ=1.48 mA IB =

IEβF +1

= 29.1 µA, IC = βF IB =1.45 mA

VCE = VC −VE = 9 − 4300IC( ) − −0.7( ) = 3.47 V Q - Point : 1.45 mA, 3.47 V( )

IE =βF +1βF

IC =5150

100µA =102µA R =−0.7V − −9V( )

102µA=

8.3V102µA

= 81.4 kΩ

Nearest 5% value is 82 kΩ.

Page 229

€

i( ) IE =−0.7V − −9V( )

5.6kΩ=1.48 mA IB =

IEβF +1

= 29.1 µA, IC = βF IB =1.45 mA

ii( ) IE =βF +1βF

IC =5150

IC = 1.02IC VBE = VT lnICIS

+1

= 0.025ln 2x10

15 IC +1( )

VBE + 8200 5.10x10−16 exp

VBE0.025

−1

= 9→VBE = 0.7079 V using a calculator solver or spreadsheet.

IC = 5x10−16 exp

0.70790.025

= 992 µA VCE = 9 − 4300IC − −0.708( ) = 5.44 V

Page 229

€

ISD =ISBJTαF

=2x10−14 A

0.95= 21.0 fA

Page 232

€

−IC =−0.7V − −9V( )

5.6kΩ=1.48 mA IB =

−ICβR +1

= 0.741 mA, - IE = βR IB = 0.741 mA

19

Page 234

€

i( ) VCESAT = 0.025V( ) ln 10.5

1+1mA

2 40µA( )1−

1mA

50 40µA( )

= 99.7 mV

ii( ) VBE = 0.025V( ) ln0.1mA + 1− 0.5( )1mA

10−15 A150

+1− 0.5

= 0.694 mV

VBC = 0.025V( ) ln0.1mA−

1mA50

10−15 A1

0.5

150

+1− 0.5

= 0.627 mV VBE −VBC = 67.7mV

Page 237

€

Dn =kTq

µn = 0.025V 500cm2 /V − s( ) =12.5cm2 / s

IS =qADnni

2

N ABW=

1.6x10−19C 50µm2( ) 10−4cm / µm( ) 12.5cm2 / s( ) 1020 / cm6( )1018 / cm3( ) 1µm( )

=10−18 A

Page 240

€

i( ) VT =1.38x10−23 J / K( ) 373K( )

1.60x10−19C= 32.2mV CD =

ICVTτF =

10A0.0322V

4x10−9 s( ) =1.24 µF

ii( ) fβ =fTβF

=300MHz

125= 2.40 MHz

Page 241

€

(a) IC =10−15 Aexp

0.70.025

1+

1050

=1.74 mA βF = 75 1+

1050

= 90.0 IB =

1.74mA90.0

=19.3 µA

(b) IC =10−15 Aexp

0.70.025

=1.45 mA βF = 75 IB =

1.45mA75

=19.3 µA

Page 243

€

gm = 40 10−4( ) = 4.00 mS gm = 40 10−3( ) = 40.0 mS

CD = 4.00mS 25 ps( ) = 0.100 pF CD = 40.0mS 25 ps( ) =1.00 pF

20

Page 249

€

VEQ =180kΩ

180kΩ+ 360kΩ12V = 4.00V | REQ =180kΩ 360kΩ =120kΩ

IB =4.00 − 0.7

120 + 75+1( )16VkΩ

= 2.470µA | IC = 75IB =185.3µA | IE = 76IB =187.7µA

VCE =12 − 22000IC −16000IE = 4.92V | Q - po int : 185 µA, 4.92 V( )

Page 250

€

i( ) I2 =IC5

=50IB

5=10IB

ii( ) VEQ =18kΩ

18kΩ+ 36kΩ12V = 4.00V | REQ =18kΩ 36kΩ =12kΩ

IB =4.00 − 0.7

12 + 500 +1( )16VkΩ

= 0.4111µA | IC = 500IB = 205.5µA | IE = 76IB = 206.0µA

VCE =12 − 22000IC −16000IE = 4.18V | Q - po int : 206 µA, 4.18 V( )

Page 251

€

The voltages all remain the same, and the currents are reduced by a factor of 10. Hence all

the resistors are just scaled up by a factor of 10.

120 kΩ→1.2 MΩ 82 kΩ→ 820 kΩ 6.8 kΩ→ 68 kΩ

Page 253

€

i( ) VCE = 0.7 V at the edge of saturation. 0.7V = 12V - RC +7675

16kΩ

205µA( )→ RC = 38.9 kΩ

ii( ) VBESAT = 4 −12kΩ 24µA( ) −16kΩ 184µA( ) = 0.768 V VCESAT =12 − 56kΩ 160µA( ) −16kΩ 184µA( ) = 0.096 V

Page 254

€

IB =9 − 0.7

36 + 50 +1( )1VkΩ

= 95.4µA | IC = 50IB = 4.77 mA | IE = 76IB =187.7µA

VCE = 9 −1000 IC + IB( ) = 4.13V | Q - po int : 4.77 mA, 4.13 V( )

21

CHAPTER 6

Page 280

€

NM L = 0.8V − 0.4V = 0.4 V NMH = 3.6V − 2.0V =1.6 V

Page 282

€

V10% = VL + 0.1 ΔV( ) = −2.6V + 0.1 −0.6 − −2.6( )[ ] = −2.4 V orV10% = VH − 0.9 ΔV( ) = −0.6V − 0.9 −0.6 − −2.6( )[ ] = −2.4 V

V90% = VL + 0.9 ΔV( ) = −2.6V + 0.9 −0.6 − −2.6( )[ ] = −0.8 V orV90% = VH − 0.1 ΔV( ) = −0.6V − 0.1 −0.6 − −2.6( )[ ] = −0.8 V

V50% =VH + VL

2=−0.6− 2.6

2= −1.6 V tr = t4 − t3 = 3 ns t f = t2 − t1 = 5 ns

Page 283

€

At P = 1 mW : PDP =1mW 1ns( ) =1 pJ At P = 3 mW : PDP = 3mW 1ns( ) = 3 pJAt P = 20 mW : PDP = 20mW 2ns( ) = 40 pJ

Page 285

€

Z = A + B( ) B + C( ) = AB + AC + BB + BC = AB + BB + AC + BB + BCZ = AB + B + AC + B + BC = B A +1( ) + AC + B C +1( ) = B + AC + BZ = B + B + AC = B + AC

Page 288

€

IDD =P

VDD=

0.4mW2.5V

=160 µA R =VDD −VL

IDD=

2.5V − 0.2V160µA

=14.4 kΩ

1.6x10−4 A =10−4A

V 2WL

S

2.5− 0.6 −0.22

0.2 V

2 →WL

S

=4.44

1

Page 289

€

IDD =VDD −VL

R=

3.3V − 0.1V102kΩ

= 31.4µA

31.4x10−6 A = 6x10−5A

V 2WL

S

3.3− 0.75−0.12

0.1 V

2 →WL

S

=2.09

1

22

Page 290

€

0.15V =Ron

Ron + 28.8kΩ2.5V → Ron =1.84 kΩ

WL

S

=1

10−4 2.5− 0.60 −0.15

2

1.84kΩ( )

→WL

S

=2.98

1

Page 291

€

i( ) Ron =1

6x10−51.03

1

3.3− 0.75−

0.22

= 6.61 kΩ VL =6.61kΩ

6.61kΩ+102kΩ3.3V = 0.201 V

ii( ) 1KnR

=

V 2

A1Ω

= V

Page 293

€

KnR = 6x10−5( ) 1.031

1.02x10

5( ) = 6.30V

NM H = 3.3− 0.75+1

2 6.30( )−1.63

3.36.30

=1.45 V NML = 0.75+1

6.30−

2 3.3( )3 6.30( )

= 0.318 V

Page 297

€

i( ) VH = 2.5− 0.6 + 0.75 VH + 0.6 − 0.6( )[ ]→VH =1.416 Vii( ) 80x10−6 A =100x10−6 A

V 2WL

S

1.55− 0.60 −0.15

2

0.15 V

2 →WL

S

=6.10

1

VTNL = 0.6 + 0.5 .15 + 0.6 − 0.6( ) = 0.646 V 80x10−6 A =

100x10−6

2A

V 2WL

L

2.5− 0.15− 0.646( )2V 2 →

WL

L

=0.551

1=

11.82

(iii ) 80x10−6 A =100x10−6A

V 2WL

S

1.55− 0.60 −0.12

0.1 V

2 →WL

S

=8.89

1

VTNL = 0.6 + 0.5 .1+ 0.6 − 0.6( ) = 0.631 V 80x10−6 A =

100x10−6

2A

V 2WL

L

2.5− 0.1− 0.631( )2V 2 →

WL

L

=0.511

1=

11.96

23

Page 300

€

The high logic level is unchanged : VH = 2.11

60x10−6 A = 50x10−6A

V 2WL

S

2.11− 0.75−0.12

0.1 V

2 →WL

S

=9.16

1

VTNL = 0.75 + 0.5 .1+ 0.6 − 0.6( ) = 0.781 V60x10−6 A =

50x10−6

2A

V 2WL

L

3.3− 0.1− 0.781( )2V 2 →

WL

L

=0.410

1=

12.44

Page 302

€

γ = 0→VTN = 0.6V and VH = 2.5 - 0.6 = 1.9 V IDD = 0 for vO = VH

100x10−6101

1.9 − 0.6 −

VL2

VL =

100x10−6

221

2.5−VL − 0.6( )

2

6VL2 −116.8VL + 3.61 = 0→VL = 0.235V IDD =100x10

−6 101

1.9 − 0.6 −

0.2352

0.235 = 278 µA

Checking : IDD =100x10−6

221

2.5− 0.235− 0.6( )

2= 277 µA

Page 307

€

VTNL = −1.5+ 0.5 0.2 + 0.6 − 0.6( ) = −1.44V60.6x10−6 =

100x10−6

2WL

L

0 −1.44( )2→

WL

L

=0.585

1=

11.71

60.6x10−6 =100x10−6WL

S

3.3− 0.6 −0.22

0.2→

WL

S

=1.17

1

Page 308

€

IDS =100x10−6 2.22

1

2.5− 0.6 −

0.22

0.2 = 79.9 µA which checks.

Page 309

€

The PMOS transistor is still saturated so IDL =144 µA, and VH = 2.5 V .

144x10−6 =100x10−651

2.5− 0.6 −

VL2

VL →VL = 0.158 V

Page 314

€

Place a third transistor with WL

=2.22

1 in parallel with tansistors A and B.

24

Page 315

€

Place a third transistor in series with tansistors A and B.

The new W/L ratios of transistors A, B and C are WL

= 32.22

1=

6.661

.

Page 321

€

M L1 is saturated for all three voltages. IDD =40x10−6

21.11

1

L

−2.5− −0.6( )[ ]2

= 80.1 µA

Note that the voltages in two rows are in error in the table.

11000 132 64.4 0 11111 64.6 31.9 31.9

The voltages can be estimated using the on - resistance method.

For the 11000 case, RonA =132mV − 64.4mV

80.1µA= 844 Ω RonB =

64.4mV80.1µA

= 804 Ω

For the 00101 case, RonE =64.4mV80.1µA

= 804 Ω.

For the 01110 case, RonC =203mV −132mV

80.1µA= 886 Ω RonD =

132mV − 64.4mV80.1µA

= 844 Ω

The voltage across a given conducting device is IDRon. Small variations in Ron are ignored.

ABCDE Y (mV) 2 (mV) 3 (mV) IDD (uA) ABCDE Y (mV) 2 (mV) 3 (mV) IDD (uA)

00000 2.5 V 0 0 0 10000 2.5 V 2.5 V 0 000001 2.5 V 0 0 0 10001 2.5 V 2.5 V 0 000010 2.5 V 0 0 0 10010 2.5 V 2.5 V 2.5 V 000011 2.5 V 0 0 0 10011 200 130 64 80.100100 2.5 V 0 2.5 V 0 10100 2.5 V 2.5 V 2.5 V 000101 130 0 64 80.1 10101 130 130 64 80.100110 2.5 V 2.5 V 2.5 V 0 10110 2.5 V 2.5 V 2.5 V 000111 130 64 64 80.1 10111 100 83 64 80.101000 2.5 V 0 0 0 11000 130 64 0 80.101001 2.5 V 0 0 0 11001 130 64 0 80.101010 2.5 V 0 0 0 11010 130 64 64 80.101011 2.5 V 0 0 0 11011 110 43 22 80.101100 2.5 V 0 2.5 V 0 11100 130 64 64 80.101101 130 0 64 80.1 11101 66 32 32 80.101110 200 64 130 80.1 11110 110 64 87 80.101111 114 21 43 80.1 11111 65 32 32 80.1

Page 322

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Pav =2.5V 80.1 µA( )

2= 0.100 mW

25

Page 323

€

PD = 10-12 F 2.5V( )

232x106 Hz( ) = 2x10−4W = 200 µW or 0.200 mW

Page 324

€

i( ) The inverter in Fig. 6.38(a) was designed for a power dissipation of 0.2 mW.To reduce the power by a factor of two, we must reduce the W/L ratios by a factor of 2.

WL

L

=12

11.68

=

13.36

WL

S

=12

4.711

=

2.361

ii( ) To increase the power by a factor of 4mW0.2mW

, we must increase the W/L ratios by a factor of 20.

WL

L

= 201.81

1

=

36.21

WL

S

= 22.22

1

=

44.41

iii( ) To reduce the power by a factor of three, we must reduce the W/L ratios by a factor of 3.WL

L

=13

1.811

=

0.6031

=1

1.66

WL

A

=13

3.331

=

1.111

WL

BCD

=13

6.661

=

2.221

Page 327

€

tr = 2.2 28.8x103Ω( ) 2x10−13 F( ) =12.7 ns τ PLH = 0.69 28.8x103Ω( ) 2x10−13 F( ) = 3.97 ns

Page 330

€

t f = 3.7 2.37x103Ω( ) 2.5x10−13 F( ) = 2.19 ns τ PHL =1.2 2.37x103Ω( ) 2.5x10−13 F( ) = 0.711 ns

tr = 2.2 28.8x103Ω( ) 2.5x10−13 F( ) =15.8 ns τ PLH = 0.69 28.8x103Ω( ) 2.5x10−13 F( ) = 4.97 ns

τP =0.711 ns + 4.97 ns

2= 2.84 ns

Page 335

€

i( ) The PMOS transistor is saturated for vO = VL. IDD =40x10−6

223.7

1

L

−2.5− −0.6( )[ ]2

=1.71 mA

Pav =2.5V 1.71mA( )

2= 2.14 mW PD = 5x10

−12 F 2.5V − 0.2V( )2 1

2x10−9 s

=13.2 mW

ii( ) We must increase the power by a factor of 20pF5pF

2ns1ns

= 8,

so the W/L ratios must also be increased by a factor of 8.

WL

L

= 823.7

1

=

1901

WL

S

= 847.4

1

=

3791

PD = 20x10−12 F 2.5V − 0.2V( )

2 1

10−9 s

= 106 mW

26

CHAPTER 7

Page 355

€

a( ) K p = 40x10−6201

= 800

µAV 2

Kn =100x10−6 20

1

= 2000

µAV 2

= 2.00 mA

V 2

b( ) VTN = 0.6 + 0.5 2.5 + 0.6 − 0.6( ) =1.09 Vc( ) VTP = −0.6 − 0.75 2.5 + 0.7 − 0.7( ) = −1.31 V

Page 357

€

a( ) For vI =1V , VGSN −VTN =1− 0.6 = 0.4V and VGSP −VTP = −1.5+ 0.6 = −0.9VMN is saturated for vO ≥ 0.4 V . MP is in the triode region for vO ≥1.6 V . ∴ 1.6 V ≤ vO ≤ 2.5 V

b( ) MP is saturated for vO ≤1.6 V . ∴ 0.4 V ≤ vO ≤1.6 Vc( ) MN is in the triode region for vO ≤ 0.4 V . MP is saturated for vO ≤1.6 V . ∴ 0 ≤ vO ≤ 0.4 V

Page 357

€

WL

P

=KnK p

WL

N

= 2.5101

=

251

Page 358

€

i( ) Both transistors are saturated since VGS = VDS .Kn2

VGSN −VTN( )2

=K p2

VGSP −VTP( )2 Kn = K p VTN = VTP

VGSN = −VGSP → vI = VDD − vI → vI =VDD

2

ii( ) 10K p

2VGSN −VTN( )

2=

K p2

VGSP −VTP( )2→ 10 VGSN −VTN( ) = −VGSP + VTP

10 vI − 0.6( ) = 4 − vI − 0.6→ vI =1.273 V

27

Page 360

€

KR =

KnWL

N

K pWL

P

=KnK p

= 2.5

VIH =2KR VDD −VTN + VTP( )

KR −1( ) 1+ 3KR−

VDD − KRVTN + VTP( )KR −1

VIH =2 2.5( ) 2.5− 0.6 − 0.6( )

2.5−1( ) 1+ 3 2.5( )−

2.5− 2.5 0.6( ) − 0.6( )2.5−1

=1.22V

VOL =KR +1( )VIH −VDD − KRVTN −VTP

2KR=

2.5 +1( )1.22 − 2.5− 2.5 0.6( ) + 0.62 2.5( )

= 0.174V

VIL =2 KR VDD −VTN + VTP( )

KR −1( ) KR + 3−

VDD − KRVTN + VTP( )KR −1

VIL =2 2.5 2.5− 0.6 − 0.6( )

2.5−1( ) 2.5 + 3−

2.5− 2.5 0.6( ) − 0.6( )2.5−1

= 0.902V

VOH =KR +1( )VIL + VDD − KRVTN −VTP

2=

2.5+1( )0.902 + 2.5− 2.5 0.6( ) + 0.62

= 2.38V

NM H = VOH −VIH = 2.38 −1.22 =1.16 V NML = VIL −VOL = 0.902 − 0.174 = 0.728 V

Page 362

€

C should be 0.75 pF, and the delay in the inverter in Fig. 7.13(b) should be 1.6 ns.

i( ) Symmetrical Inverter : τ P =1.2RonnC =1.20.75x10−12 F

2 10−4( ) 2.5− 0.6( )Ω = 2.4 ns

ii( ) Symmetrical Inverter : Ronn =τP

1.2C=

10−9 s

1.2 5x10−12 F( )=167Ω

WL

N

=1

RonnKn' VGS −VTN( )

=1

167 10−4( ) 2.5− 0.6( )=

31.51

WL

P

= 2.5WL

N

= 78.8

Page 364

€

The inverters need to be increased in size by a factor of 280ps250ps

=1.12.

WL

N

=1.123.77

1

=

4.221

WL

P

=1.129.43

1

=

10.61

28

Page 364

€

WL

N

=3.77

1

3.3− 0.753.3− 0.5

=

3.431

WL

P

=9.43

1

3.3− 0.753.3− 0.5

=

8.591

Page 364

€

τPHL = 2.4RonnC =2.4C

Kn VGS −VTN( )=

2.4C

Kn 2.5− 0.6( )=1.26

CKn

τPLH = 2.4RonpC =2.4C

K p VGS −VTN( )=

2.4C

K p 2.5− 0.6( )=1.26

CK p

Page 365

€

τPHL = 2.4RonnC =2.4C

Kn VGS −VTN( )=

2.4C

Kn 3.3− 0.75( )= 0.94

CKn

τPLH = 2.4RonpC =2.4C

K p VGS −VTN( )=

2.4C

K p 3.3− 0.75( )= 0.94

CK p

Page 366

€

The inverter in Fig. 7.12 is a symmetrical design, so the maximum current occurs for vO = vI =VDD

2.

Both transistors are saturated. iDN =10-4

221

1.25− 0.6( )

2= 42.3 µA

Checking : iDP =4x10-5

251

1.25− 0.6( )

2= 42.3 µA

Page 367

€

PDP ≅CVDD

2

5=

10−13 F 2.5V( )2

5= 0.13 pJ PDP ≅

CVDD2

5=

10−13 F 3.3V( )2

5= 0.22 pJ

Page 371

€

Remove the NMOS and PMOS transistors connected to input E, and ground the source of

the NMOS transistor connected to input D. The are now 4 NMOS transistors in series, and

WL

N

= 421

=

81

WL

P

=51

29

Page 376

€

There are two NMOS transistors in series in the AB and CD NMOS paths, and three PMOS

transistors in the ACE and BDE PMOS paths. Therefore :

WL

N −ABCD

= 221

=

41

WL

N −E

=21

WL

P

= 351

=

151

Page 380

AB+C

B

A

C

€

P = CVDD2 f = 50x10−12 F( ) 5V( )2 107 Hz( ) =12.5 mW

Page 383

€

β =50 pF50 fF

1

2

= 31.6 τ P = 31.6τ o + 31.6τ o = 63.2τ o | z = eln z | z

1

ln z = e ln z( )1

ln z = e

β =50 pF50 fF

1

7

= 2.683

1, 2.68 , 2.6832 = 7.20, 2.6833 = 19.3, 2.6834 = 51.8, 2.6835 = 139, 2.6836 = 373

A6 = 1+ 3.16 +10 + 31.6 +100 + 316( ) Ao = 462AoA7 = 1+ 2.68 + 7.20 +19.3 + 51.8 +139 + 373( ) Ao = 594 Ao

30

CHAPTER 8

Page 401

€

NS =28 •220

27 •210= 211 = 2048 segments NS =

230

29 •210= 211 = 2048 segments

Page 403

€

i( ) N = 28 •220 = 228 = 268,435,456 IDD =0.05W3.3V

=15.2 mA Current/cell =15.2mA

228cells= 56.4 pA

ii( ) Reverse the direction of the substrate arrows, and connect the substrates of the PMOStransistors to VDD.

Page 406

€

MA1 : At t = 0+, VGS −VTN = 4V and VDS = 2.5V, so transistor MA1 is operating in the triode region.

i1 = 60x10−6 1

1

5−1−

2.52

2.5 = 413 µA

MA2 : At t = 0+, VGS = VDS, so transistor MA2 is operating in the saturation region.

VTN 2 =1+ 0.6 2.5+ 0.6 − 0.6( ) =1.592V i2 = 60x10−6

211

5− 2.5−1.592( )

2= 24.8 µA

Page 408

€

MA1 : At t = 0+, VGS = VDS, so transistor MA1 is operating in the saturation region.

i1 =60x10−6

211

5−1( )

2= 480 µA

MA2 : At t = 0+, VGS = VDS, so transistor MA2 is operating in the saturation region.

i1 =60x10−6

211

5−1( )

2= 480 µA

Page 411

€

i( ) At t = 0+, VGS −VTN = 3 - 0.7 = 2.3 V and VDS =1.9 V, so transistor MA is operating in the triode region.

i1 = 60x10−6 1

1

3− 0.7 −

1.92

1.9 =154 µA t f = 3.6RonC = 3.6

50x10−15 F

60x10−6 3− 0.7( )=1.30 ns

ii( ) VC = VBL −VTN VC = 3− 0.7 + 0.5 VC + 0.6 − 0.6( )[ ]→VC =1.89 V | VC = 3− 0.7 = 2.3Viii( ) n = CV

q=

25x10−15 F 1.89V( )1.60x10−19C

= 2.95 x 105electrons

31

Page 413

€

ΔV =VC −VBLCBLCC

+1=

1.9 − 0.9549CCCC

+1V = 0.019 V ΔV =

VC −VBLCBLCC

+1=

0 − 0.9549CCCC

+1V = −0.019 V

τ = RonCC

CCCBL

+1= 5kΩ

25 fF1

49+1

= 0.123 ns or τ ≅ RonCC = 5kΩ 25 fF( ) = 0.125 ns

Page 414

€

At t = 0+, VGS −VTN = 3 - 0( ) - 0.7 = 2.3 V and VDS =1.5 V , so transistor MA2 is operating in

the triode region. iD = 60x10−6 2

1

3− 0.7 −

1.52

1.5 = 279 µA

Page 416

€

i( ) In setting the drain currents equal, we see that the change in W/L cancels out, andthe voltages remain the same.

∴ iD =12

60x10-6( ) 51

1.33− 0.7( )

2= 59.5 µA PD = 2 59.5µA( ) 3V( ) = 0.357 mW

As a check, the current should scale with W/L : iD =52

23.5µA( ) = 58.8 µA

ii( ) 12

25x10-6( ) 21

2.5−VO − 0.6( )

2=

12

60x10-6( ) 21

VO − 0.6( )

2

1.4VO2 + 0.92VO − 2.746 = 0→VO =1.11V

iD =12

25x10-6( ) 21

2.5−1.11− 0.6( )

2=15.6 µA PD = 2 15.6µA( ) 2.5V( ) = 78.0 µW

Checking : 12

60x10-6( ) 21

1.11− 0.6( )

2=15.6 µA

Page 418

€

Ron =1

60x10−6 3−1.3−1( )= 23.8 kΩ τ = 23.8kΩ 25 fF( ) = 0.595 ns

Page 420

€

For all possible input combinations there will be two inverters and 3 output lines in the low state.

PD = 5 0.2mW( ) =1.0 mW

Page 422

€

WL

L

=2

2.221.81

1

=

1.631

32

Page 424

€

i( ) For a 0 - V input, all transistors will be on and the input nodes will all discharge to 0 V.For the 3 - V input, the nodes will all charge to 3 V as long as VTN ≤ 2V.

VTN = 0.7 + 0.5 3 + 0.6 − 0.6( ) =1.26 V . Thus the nodes will all be a 3 V.2 ≥ 0.7 + γ 3 + 0.6 − 0.6( )→γ ≤1.158ii( ) The output will drop below VDD / 2. For the PMOS device, VGS −VTP = 3−1.9 − 0.7 = 0.4V .

The PMOS transistor will be saturated. For the NMOS device, VGS −VTP =1.9 − 0.7 =1.2V .

Assume linear region operation.

40x10-6

251

−1.1+ 0.7( )

2=100x10-6

21

1.9 − 0.7 −

VO2

VO

VO2 − 2.4VO + 0.16 = 0→VO = 68.6 mV

33

CHAPTER 9

Page 442

€

iC2iC1

= exp0.2V

0.025V

= 2.98 x 10

3 iC2iC1

= exp0.4V

0.025V

= 8.89 x 10

6 2.98x103( )2

= 8.88 x 106

Page 444

€

The current must be reduced by 5 while the voltages remain the same.

IEE =300µA

5= 60 µA RC = 5 2kΩ( ) =10 kΩ

Page 445

€

IB =IE

βF +1 IB3 =

92.9µA21

= 4.42 µA IB 4 =107µA

21= 5.10 µA

IB3RC = 4.42µA 2kΩ( ) = 8.84 mV

34

Page 452

€

a( ) For all inputs low : IEE =−1− 0.7 − −5.2( )

11.7VkΩ

= 299µA

ΔV2

= VH −VREF = −0.7− −1( ) = 0.3 V ΔV = 0.6 V RC2 =0.6V

299µA= 2.00 kΩ

For an inputs high : IEE =−0.7− 0.7 − −5.2( )

11.7VkΩ

= 325µA RC1 =0.6V

325µA=1.85 kΩ

b( ) Based upon analysis above, RC = 0.6V325µA =1.85 kΩ

Page 453

€

For all inputs low : IEE =−1− 0.7 − −5.2( )

11.7VkΩ

= 299µA

ΔV2

= VH −VREF = −0.7− −1( ) = 0.3 V ΔV = 0.6 V RC =0.6V

299µA= 2.00 kΩ

Page 454

€

RE =VE − −VEE( )

0.3mA=

0 − 0.7 − −5.2V( )0.3

VmA

=15.0 kΩ

Page 455

€

a( ) For IE = 0, vO = -5.2 V. b( ) For IE = 0, vO = -5.2V10kΩ

10kΩ+15kΩ= −2.08 V

Page 456

€

i( ) The transistor's power dissipation is P = VCB IC + VBE IE = 5V 2.55mA5051

+ 0.7V 2.55mA( ) =14.3 mW

The total power dissipation in the circuit is P = VCC IC + VEE IE = 5V 2.55mA5051

+ 5V 2.55mA( ) = 25.3 mW

For vO = −3.7V , IE =−3.7 − −5( )

1300−

3.75000

= 260 µA. At the Q - point, IE =−0.7 − −5( )

1300−

0.75000

= 3.17 mA

The transistor's power dissipation is P = VCB IC + VBE IE = 5V 3.17mA5051

+ 0.7V 3.17mA( ) =17.8 mW

ii( ) - 4V = -5.2V 10kΩ10kΩ+ RE

→ RE = 3.00 kΩ

IE =5.2V3kΩ

=1.73 mA IE =−4− −5.2( )

3000−

410000

= 0 IE =4 − −5.2( )

3000+

410000

= 3.47 mA

35

Page 458

€

Increase the value of each resistor by a factor of 10.

Page 461

€

RC =ΔVIEE

=0.6V

0.5mA=1.2kΩ τP = 0.69 1.2kΩ( ) 2 pF( ) =1.66 ns

P = 5.2V 0.5 + 0.1 + 0.1( )mA = 3.64mW PDP = 6.0 pJ

Page 464

€

For vO = VH , IC = 0, and P = 0. P = VDD IDD = 5V 2.43mA( ) =12.1 mAIncrease R by a factor of 10 : R = 10 2kΩ( ) = 20kΩ.

Page 466

€

i( ) Γ = exp 0.10.0258

= 48.2 IB ≥

10A20

1+20

0.1 48.2( )1−

1148.2

= 3.34 A βFOR =10A

3.34A= 3.00

ii( ) αR =0.2

0.2 +1=

16

IB ≥10A20

1+20

0.2 54.6( )1−

654.6

=1.59 A

iii( ) Γ = exp 0.150.025

= 403 IB ≥

10A20

1+20

0.1 403( )1−

11403

= 0.769 A

iv( ) VT =1.38x10−23 273+150( )

1.60x10−19= 36.5 mV VCEMIN = 36.5mV ln

0.05+10.05

=111 mV

Page 467

€

Γ = 54.6 αR =0.25

1+ 0.25=

15

IB ≥10mA

40

1+40

0.25 54.6( )1−

554.6

=1.08 mA βFOR =10mA

1.08mA= 9.24

36

Page 467

€

i( ) 1ns = 6.4ns ln 1mA− IBR2.5mA40.7

− IBR

1.169 =

1mA− IBR0.0614mA− IBR

→ IBR = −5.49 mA

ii( ) iCMAX =VCC −VCE

βF≅

5− 02500

= 2.5mA QXS = 6.4ns 1mA−2.5mA40.7

= 6.01 pC

QF = iFτF = 2.5mA 0.25ns( ) = 0.625 pC QXS >> QF

Page 471

€

i( ) vI = VL = 0.15 V IIL = − 5− 0.954000 = −1.01 mA VBE2 = VL + VCESAT1Using the value of VCESAT in Fig.9.32, VBE2 = 0.15 + 0.04 = 0.19 V

A better estimate is VCESAT1 = 25mV ln2 +1

2

=10.1 mV

VBE2 = 0.15 + 0.010 = 0.16 V

ii( ) vI = VH = 5 V IIH = 25−1.54000

=1.75 mA VBE2 = 0.8 V

Using Eq. (5.29), VBESAT = 0.025V ln

0.875mA + 1−23

2.4mA( )

10−15 A1

40+ 1−

23

= 0.729 V

Page 477

€

i( ) 5V - N (2kΩ)βR IB ≥1.5V IB =5−1.54000

= 0.875mA

βR ≤3.5V

5(2kΩ) 0.875mA( )= 0.4 βR ≤

3.5V

10(2kΩ) 0.875mA( )= 0.2

ii( ) IB2 = 2 +1( ) 5−1.54000 = 2.63 mA 2.43mA + N 1.01mA( ) ≤ 28.3 2.63mA( )→ N ≤ 71

37

Page 479

€

vI = VL and vO = 0 : IB 4 =5−VB 41600

=5− 0 + 0.7 + 0.7( )

1600= 2.25mA IL = 41IB 4 = 92.3 mA

5−1600IL41− 0.7 − 0.7 ≥ 3

5− 3+ 0.7 + 0.7( )1600

= 0.375mA IL = 41IB 4 =15.4 mA

VCE = 5−130ΩIC −VO = 5V −130Ω 15.4mA( ) − 3.7 = −0.702V Oops! - the transistor is not forward active. Assume saturation with VCESAT = 0.15V .

IL = IB + IC =5− 0.8 + 0.7 + 3.0( )

1600+

5− 0.15 + 0.7 + 3( )130

= 9.16 mA

38

CHAPTER 10

Page 509

€

Vo = 2PoRL = 2 20W( ) 16Ω( ) = 25.3 V Av =VoVi

=25.3V0.005V

= 5.06x103

Io =VoRL

=25.3V16Ω

=1.58 A Ii =Vi

RS + Rin=

0.005V10kΩ+ 20kΩ

= 0.167µA Ai =IoIi

=1.58A

0.167µA= 9.48x106

AP =PoPS

=25.3V 1.58A( )

0.005V 0.167µA( )= 4.79x1010 Checking : Ap = 5.06x10

3( ) 9.48x106( ) = 4.80x1010

Page 510

€

i( ) AvdB = 20 log 5060( ) = 74.1 dB AidB = 20 log 9.48x106( ) =140 dB APdB =10 log 4.80x1010( ) =107 dBii( ) AvdB = 20 log 4x104( ) = 92.0 dB AidB = 20 log 2.75x108( ) =169 dB APdB =10 log 1.10x1013( ) =130 dB

Page 511

€

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

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

2V = 0.525 V .

vi ≤ 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) vO = VO + vo For vi = 0, vI = VI = 0.6, VO =14V and Av = +40. Vo = AvVi = 40 0.01V( ) = 4VvO = 14.0 + 4.00sin1000πt( ) volts VO =14 V

Page 519

€

g11 =1

20kΩ+ 76 50kΩ( )= 0.262 µS g21 = 0.262µS 76( ) 50kΩ( ) = 0.995

g22 =1

50kΩ+

120kΩ

+75

20kΩ= 3.82 mS g12 = −

1

g22 20kΩ( )= −

1

3.82mS 20kΩ( )= -0.0131

Rin =1

g11= 3.82 MΩ A = g21 = 0.995 Rout =

1g22

= 262 Ω

39

Page 520

€

i( ) P = Av Ai = Av2RS + Rin

RL

ii( ) Vo = 2 100W( ) 8Ω( ) = 40 V 40 = 0.001 50kΩ5kΩ+ 50kΩ

A

8Ω0.5Ω+ 8Ω

→ A = 46,800

P =Io

2RL2

=0.5Ω

240V8Ω

2

= 6.25 W Ai =40V8Ω

5kΩ+ 50kΩ0.001V

= 2.75 x 10

8

iii( ) 40 = 0.001 5kΩ5kΩ+ 5kΩ

A

8Ω8Ω+ 8Ω

→ A =160,000

P =Io

2RL2

=8Ω2

40V8Ω

2

=100 W! Ai =40V8Ω

5kΩ+ 5kΩ0.001V

= 5.00 x 10

7

Page 521

€

Av s( ) =300s

s + 5000( ) s +100( ) Zeros at s = 0 and s = ∞; Poles at s = -5000 and s = −100.

Page 523

€

Av s( ) = −2π x 106

s + 5000π=

−400

1+s

5000π

→ Amid = −400 fH =5000π

2π= 2.50 kHz

BW = fH − fL = 2.50 kHz − 0 = 2.50 kHz GBW = 400( ) 2.50kHz( ) =1.00 MHz

Page 524

€

i( ) Av j5( ) = 5052 − 4

52 − 2( )2

+ 4 52( )= 41.87 20 log 41.87( ) = 32.4 dB

∠Av j5( ) =∠ 52 − 4( ) − tan−1−2 5( )52 − 2

= 0 − −23.5o( ) = 23.5o

Av j1( ) = 5012 − 4

12 − 2( )2

+ 4 12( )= 67.08 20 log 41.87( ) = 36.5 dB

∠Av j1( ) =∠ 12 − 4( ) − tan−1−2 1( )12 − 2

=180o − −63.43o( ) = 243o = −117o

40

Page 524

€

ii( ) Av jω( ) = 201+ j

0.1ω1−ω2

Av j0.95( ) =20

12 +0.1( )

20.952( )

1− 0.952( )2

=14.3 ∠Av j0.95( ) =∠20 − tan−10.1 0.95( )1− 0.952

= 0 − 44.3o( ) = −44.3o

Av j1( ) =20

12 +0.1( )

212( )

1−12( )2

= 0 ∠Av j1( ) =∠20 − tan−10.1 1( )1−12

= 0 − 90o( ) = −90.0o

Av j1.1( ) =20

12 +0.1( )

21.12( )

1−1.12( )2

=17.7 ∠Av j1.1( ) =∠20 − tan−10.1 1.1( )1−1.12

= 0 − −27.6o( ) = 27.6o

Page 526

€

fH =1

2π1

1kΩ 100kΩ( ) 200 pF( )= 804 kHz

Page 527

€

Av s( ) =250

1+250π

s

Ao = 250 fL =250π2π

=125 Hz fH =∞ BW =∞−125 =∞

Page 528

€

fL =1

2π1

1kΩ 100kΩ( ) 0.1µF( )=15.8 Hz

41

Page 529

€

i( ) Av s( ) = −4001+

100s

1+

s50000

Ao = 400 or 52 dB

fL =1002π

=15.9 Hz fH =50000

2π= 7.96 kHz BW = 7960 −15.9 = 7.94 kHz

ii( ) ∠Av j0( ) = −90− 0 − 0 = −90o

∠Av j100( ) = −90o − tan−1 100100

− tan

−1 10050000

= −90− 45− 0.57 = −136

o

∠Av j50000( ) = −90o − tan−150000100

− tan

−1 5000050000

= −90− 89.9 − 45 = −225

o

∠Av j∞( ) = −90− 90 − 90 = −270o

Page 531

€

The numerator coefficient should be 6 x106.

Av s( ) = 302x105 s

s2 + 2x105 s +1014 Ao = 30

fo =1

2π1014 =1.59 MHz Q =

107

2x105= 50 BW =

1.59MHz50

= 31.8 kHz

Page 533

€

The transfer fucntion should be Av s( ) =6.4x1012π 2s

s + 200π( ) s + 80000π( )2

.

Av s( ) =1000

1+200π

s

1+

s80000π

2 Ao =1000 or 60 dB

fL =200π2π

=100 Hz fH = 0.64480000π

2π

= 25.8 kHz BW = 25800 −100 = 25.7 kHz

42

CHAPTER 11

Page 545

€

vid =10V100

= 0.100V =100 mV vid =10V

104= 0.001 V =1.00 mV vid =

10V

106=1.00x10−5V =10.0 µV

Page 547

€

Av = −360kΩ68kΩ

= −5.29 vO = −5.29 0.5V( ) = −2.65 V iS =0.5V68kΩ

= 7.35 µA iO = −i2 = −iS = −7.35 µA

Page 549

€

IS =2V

4.7kΩ= 426 µA I2 = IS = 426 µA Av = −

24kΩ4.7kΩ

= −5.11 VO = −5.11 2V( ) = −10.2 V

Page 551

€

Av =1+36kΩ2kΩ

= +19.0 vO =19.0 −0.2V( ) = − 3.80 V iO =−3.80V

36kΩ+ 2kΩ= −100 µA

Page 552

€

i( ) Av =1+39kΩ1kΩ

= +40.0 AvdB = 20log 40.0( ) = 32.0 dB Rin =100kΩ ∞ =100kΩ

vO = 40.0 0.25V( ) =10.0 V iO =10.0V

39kΩ+1kΩ= 250 µA

ii( ) Av =1054

20 = 501 1+R2R1

= 501 R2R1

= 500 iO =vO

R2 + R1

10R2 + R1

≤ 0.1 mA

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 iO to exceed 0.1 mA in the first case.

Page 554

€

Inverting Amplifier : Av = −30kΩ1.5kΩ

= −20.0 Rin = R1 =1.5 kΩ

vO = −20.0 0.15V( ) = −3.00 V iO =vOR2

=−3.00V30kΩ

= −100 µA

Non - Inverting Amplifier : Av =1+30kΩ1.5kΩ

= +21.0 Rin =vSiS

=0.15V

0A=∞

vO = 21.0 0.15V( ) = 3.15 V iO =vO

R2 + R1=

3.15V30kΩ+1.5kΩ

=100 µA

43

Page 555

€

Vo1 = 2V −3kΩ1kΩ

= −6V Vo2 = 4V −

3kΩ2kΩ

= −6V vO = −6sin1000πt − 6sin2000πt( ) V

The summing junction is a virtual ground : Rin1 =v1i1

= R1 =1 kΩ Rin2 =v2i2

= R2 = 2 kΩ

Io1 =Vo1R3

=−6V3kΩ

= −2mA Io2 =Vo2R3

=−6V3kΩ

= −2mA iO = −2sin1000πt − 2sin2000πt( ) mA

Page 559

€

i( ) I2 =3V

10kΩ+100kΩ= 27.3 µA

ii( ) Av = −100kΩ10kΩ

= −10.0 VO = −10 3V − 5V( ) = 20.0 V IO =VO −V−100kΩ

=VO −V+100kΩ

V+ = V2R4

R3 + R4= 5

100kΩ10kΩ+100kΩ

= 4.545V IO =20.0 − 4.545

100kΩ=155 µA I2 =

5V10kΩ+100kΩ

= 45.5 µA

iii( ) Av = − 36kΩ2kΩ = −18.0 VO = −18 8V − 8.25V( ) = 4.50 V IO =VO −V−36kΩ

=VO −V+36kΩ

V+ = V2R2

R1 + R2= 8.25

36kΩ2kΩ+ 36kΩ

= 7.816V IO =4.50 − 7.816

36kΩ= −92.1 µA

Page 560

€

I =VA −VB

2R1=

5.001V − 4.999V2kΩ

=1.00 µA

VA = V1 + IR2 = 5.001V +1.00µA 49kΩ( ) = 5.05 VVB = V2 − IR2 = 4.999V −1.00µA 49kΩ( ) = 4.95 V

VO = −R4R3

VA −VB( ) = −

10kΩ10kΩ

5.05− 4.95( ) = −0.100 V

Page 564

€

i( ) Av = −R2R1

= −1026

20 = −20.0 R1 = Rin =10kΩ R2 = 20R1 = 200kΩ

C =1

2π 3kHz( ) 200kΩ( )= 265 pF Closest values : R1 =10kΩ R2 = 200kΩ C = 270 pF

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ii( ) Rin = R1 =10 kΩ ΔV = −ICΔT C =

5V10kΩ

110V

1ms( ) = 0.05µF

t (msec)vO 2 4 6 8

-10V

Page 567

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vO = −RCdvSdt

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

Page 569

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i( ) AvA = AvB = AvC = −R2R1

= −68kΩ2.7kΩ

= −25.2 RinA = RinB = RinC = R1 = 2.7 kΩ

The op - amps are ideal : RoutA = RoutB = RoutC = 0

ii( ) Av = AvA AvB AvC = −25.2( )3

= −16,000 Rin = RinA = 2.7 kΩ Rout = RoutC = 0

Page 570

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Av = −25.2( )3 2.7kΩ

Rout + 2.7kΩ

2

≥ 0.99 25.2( )3

2.7kΩRout + 2.7kΩ

2

≥ 0.99

2.7kΩRout + 2.7kΩ

≥ 0.9950 Rout ≥13.6 Ω

Page 574

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i( ) Av 0( ) = +1 Av s( ) =ωo

2

s2 + s 2ωo +ωo2

Av jω( ) =ωo

2

jω 2ωo +ωo2 −ω2

Av jωH( ) =1

2 →

ωo2

ωo2 −ωH

2( )2

+ 2ωH2ωo

2

=1

2 → 2ωo

4 =ωo4 +ωH

4 →ωo =ωH

ii( ) 12

=C1C2

R2

2R→C1 = 2C2 →C2 =

1

2 2.26kΩ( ) 20000π( )= 4.98nF

C2 = 0.005 µF C1 = 0.01 µF

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iii( ) To decrease the cutoff frequency from 5kHz to 2 kHz, we must increase the

resistances by a factor of 5kHz2kHz

= 2.50→ R1 = R2 = 2.50 2.26kΩ( ) = 5.65 kΩ

iv( ) 12

=CC

R1R2R1 + R2

→ R12 + 2R1R2 + R2

2 = 2R1R2 → R12 = −R2

2 - - can' t be done!

Q =R1R2

R1 + R2

dQdR2

=1

R1 + R2( )2

R1 R1 + R2( )2 R1R2

− R1R2

= 0→ R2 = R1 →Qmax =12

Page 575

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SC1Q =

C1Q

dQdC1

=C1Q

1

2 C1C2

R1R2R1 + R2

=C1Q

Q2C1

= 0.5

SR2Q =

R2Q

dQdR2

R1 = R2 →Q =12

C1C2

→ SR2Q = 0

Page 577

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i( ) Av jωo( ) = K−ωo

2

−ωo2 + j 3− K( )ωo2 +ωo2

=K

3− K Av jωo( ) =

K3− K

∠90o

ii( ) fo =1

2π 10kΩ 20kΩ( ) 0.0047µF( ) 0.001µF( )= 5.19 kHz

Q =10kΩ20kΩ

4.7nF +1.0nF

4.7nF 1.0nF( )+ 1− 2( )

20kΩ 1.0nF( )10kΩ 4.7nF( )

−1

= 0.829

iii( ) SKQ = KQdQdK

Q =1

3− K

dQdK

=−1

3− K( )2−1( ) = Q2 SKQ = KQ

dQdK

= KQ

Q =1

3− K→ KQ = 3Q −1 SK

Q = 3Q −1 =1.12

Page 578

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Rth = 2kΩ 2kΩ =1kΩ fo =1

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

12

82kΩ1kΩ

= 4.53

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ii( ) ABP jωo( ) = KQ = R2R1 10 =

294kΩR1

→ R1 = 29.4 kΩ

iii( ) fo =1

2πRC=

1

2π 40.2kΩ( ) 2nF( )=1.98 kHz BW =

12πR2C

=1

2π 402kΩ( ) 2nF( )=198 Hz

ABP jωo( ) = −R2R1

= −402kΩ20.0kΩ

= −20.1

iv( ) Blindly using the equations at the top of page 580 yields

fomin =

12πRC

=1

2π 1.01( ) 29.4kΩ( ) 1.02( ) 2.7nF( )=1946 Hz

fomax =

12πRC

=1

2π 0.99( ) 29.4kΩ( ) 0.98( ) 2.7nF( )= 2067 Hz

BW min =1

2πR2C=

1

2π 1.01( ) 294kΩ( ) 1.02( ) 2.7nF( )=195 Hz

BW max =1

2πR2C=

1

2π 0.99( ) 294kΩ( ) 0.98( ) 2.7nF( )= 207 Hz

ABPmin = −

R2R1

= −294kΩ 1.01( )

14.7kΩ 0.99( )= −20.4 ABP

max = −R2R1

= −294kΩ 0.99( )14.7kΩ 1.01( )

= −19.6

The W/C results are similar if R and C are not the same for example where ωo =1

RA RBCACB.

Page 583

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i( ) - a( ) R1 = R2 = 5 2.26kΩ( ) =11.3 kΩ C1 =0.02µF

5= 0.004 µF C2 =

0.01µF5

= 0.002 µF

fo =1

2π 11.3kΩ( ) 11.3kΩ( ) 0.004µF( ) 0.002µF( )= 4980 Hz

Q =11.3kΩ11.3kΩ

0.004µF( ) 0.002µF( )0.004µF + 0.002µF

= 0.471

b( ) R1 = R2 = 0.885 2.26kΩ( ) = 2.00 kΩ C1 = 0.02µF0.885 = 0.0226 µF C2 =0.01µF0.885

= 0.0113 µF

fo =1

2π 2.00kΩ( ) 2.00kΩ( ) 0.0226µF( ) 0.0113µF( )= 4980 Hz

Q =2.00kΩ2.00kΩ

0.0226µF( ) 0.0113µF( )0.0226µF + 0.0113µF

= 0.471

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Page 583

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ii( ) fo =1

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

82kΩ1kΩ

0.02µF( ) 0.02µF( )0.02µF + 0.02µF

= 4.53

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

4= 0.005 µF

fo =1

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

82kΩ1kΩ

0.005µF( ) 0.005µF( )0.005µF + 0.005µF

= 4.53

Page 585

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The diode will conduct and pull the output up to vO = vS =1.0 V . v1 = vO + vD =1.0 + 0.6 =1.6 V

For a negative input, there is no path for current through R, so vO = 0 V . The op - amp

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

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

Page 587

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i( ) vS = 2 V : Diode D1 conducts, and D2 is off. The negative input is a virtual ground.v1 = −vD2 = −0.6 V . The current in R is 0, so vO = 0 V .

vS = −2 V : Diode D2 conducts, and D1 is off. The negative input is a virtual ground.

vO = −R2R1

vS = −68kΩ22kΩ

−2V( ) = +6.18 V v1 = vO + vD1 = 6.78 V .

vS =15V−3.09

= −4.85 V v1 = vO + vD1 = 6.78 V . When vO = 15 V , vD2 = -15.6 V , so VZ =15.6 V .

ii( ) vO = 20kΩ20kΩ10.2kΩ3.24kΩ

2Vπ

= 2.00 V

Page 589

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V− = −R1

R1 + R2VEE = −

1kΩ1kΩ+ 9.1kΩ

10V = −0.990 V V+ =1kΩ

1kΩ+ 9.1kΩ10V = +0.990 V

Vn = 0.990V − −0.990V( ) =1.98 V

Page 591

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T = 2 10kΩ( ) 0.001µF( ) ln 1+ 0.51− 0.5

= 21.97µs f =

1T

= 45.5 kHz

48

Page 594

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β = −22kΩ

22kΩ+18kΩ= 0.550 T = 11kΩ( ) 0.002µF( ) ln

1+0.75

1− 0.550

= 20.4 µs

Tr = 11kΩ( ) 0.002µF( ) ln1+ 0.55

5V5V

1−0.75

=13.0 µs