Microelectronic Circuit Design Fourth Edition

Microelectronic Circuit Design Fourth Edition - Part I Solutions to Exercises CHAPTER 1 Page 11 5.12V 5.12V VLSB = 10 = = 5.00 mV 2 bits 1024bits 11000100012 = 29 + 28 + 2 4 + 20 = 78510 or

€

€

(

5.12V = 2.560V 2 VO = 785(5.00mV ) = 3.925 V VMSB =

)

VO = 2−1 + 2−2 + 2−6 + 2−10 5.12V = 3.925 V

Page 12 5.0V 5.00V 1.2V VLSB = 8 = = 19.53 mV N= 256bits = 61.44bits 5.00V 2 bits 256bits 61 = 32 + 16 + 8 + 4 + 1 = 25 + 2 4 + 23 + 22 + 20 = 001111012 Page 12 The dc component is VA = 4V. The signal consists of the remaining portion of vA: va = (5 sin 2000πt + 3 cos 1000 πt) Volts. Page 23 vo = −5cos 2000πt + 25o = − −5sin 2000πt + 25o − 90 o = 5sin 2000πt − 65o

) [

(

Vo = 5∠ − 65

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Page 25 R Av = − 2 R1

o

|

Vi = 0.001∠0

−5=−

o

)]

(

(

)

5∠ − 65o Av = = 5000∠ − 65o o 0.001∠0

R2 → R2 = 50 kΩ 10kΩ

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1

©R. C. Jaeger and T. N. Blalock 08/09/10

Page 26 vs = 0.5sin (2000πt ) + sin (4000πt ) + 1.5sin (6000πt )

[

]

The three spectral component frequencies are f1 = 1000 Hz

f2 = 2000 Hz

f3 = 3000 Hz

(a ) The gain of the band - pass filter is zero at both f and f . At f , V = 10(1V ) = 10V , and v = 10.0sin (4000πt ) volts. (b) The gain of the low - pass filter is zero at both f and f . At f , V = 6(0.5V ) = 3V , and v = 3.00sin (2000πt ) volts. 1

3

2

o

o

2

3

2

o

o

Page 27 €

39kΩ(1− 0.1) ≤ R ≤ 39kΩ(1+ 0.1)

or

3.6kΩ(1− 0.01) ≤ R ≤ 3.6kΩ(1+ 0.01)

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Page 29 VI2 P= R1 + R2 P

max

=

P nom =

(1.1x15)

35.1 kΩ ≤ R ≤ 42.9 kΩ or

152 = 4.17 mW 54kΩ

2

0.95x54kΩ

3.56 kΩ ≤ R ≤ 3.64 kΩ

= 5.31 mW

P

min

=

(0.9x15)

2

1.05x54kΩ

= 3.21 mW

Page 33 €

 10−3  R = 10kΩ 1+ o (−55 − 25) oC = 9.20 kΩ C  

 10−3  R = 10kΩ 1+ o (85 − 25) oC = 10.6 kΩ C  

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2


CHAPTER 2 Page 47 ni =

(2.31x10

30

K cm −3

−6

)

  −0.66eV  (300K ) exp 8.62x10−5 eV / K 300K  = 2.27x1013 cm3 ( )  3

(

)

Page 47 €

ni =

(1.08x10

31

ni =

(1.08x10

31

K cm

−6

K cm

−6

−3

−3

)

  −1.12eV  (50K ) exp 8.62x10−5 eV / K 50K  = 4.34x10−39 cm3 ( ) 

)

  −1.12eV  (325K ) exp 8.62x10−5 eV / K 325K  = 4.01x1010 cm3 ( ) 

3

(

)

3

(

)

3

 −2 m  cm 3 10 L = 10  → L = 6.13x10 m −39  cm  4.34x10  3

Page 49 €

v p = µ p E = 500 E=

€

cm2  V  3 cm 10  = 5.00x10 V − s  cm  s

vn = −µn E = −1350

cm2  V  6 cm 1000  = −1.35x10 V − s cm  s

V 1 V V = = 5.00x103 −4 L 2x10 cm cm

Page 49 (a) From Fig. 2.5 : The drift velocity for Ge saturates at 6x106 cm / sec. At low fields the slope is constant. Choose E = 100 V/cm v v 4.3x105 cm / s cm2 2.1x105 cm / s cm2 µn = n = = 4300 µp = p = = 2100 E 100V / cm s E 100V / cm s 7 (b) The velocity peaks at 2x10 cm / sec

µn =

vn 8.5x105 cm / s cm2 = = 8500 E 100V / cm s

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3


Page 51   3 −1.12  = 5.40x1024 / cm 6 | ni = 2.32x1012 / cm3 ni2 = 1.08x1031 (400) exp −5 8.62x10 400  ( )

ρ=

1 1 = = 1450 Ω − cm −19 12 σ 1.60x10 2.32x10 (1350) + 2.32x1012 (500)

[(

)

(

]

)

  3 −1.12  = 1.88x10−77 / cm6 | ni = 4.34x10−39 / cm3 ni2 = 1.08x1031 (50) exp −5  8.62x10 (50) 

ρ=

1 1 = = 1.69x1053 Ω − cm σ 1.60x10−19 4.34x10−39 (6500) + 4.34x10−39 (2000)

[(

)

(

)

]

Page 55 €

  3 −1.12  = 5.40x1024 / cm 6 ni2 = 1.08x1031 (400) exp −5  8.62x10 (400)  p = N A − N D = 1016 − 2x1015 = 8x1015

holes cm3

n=

ni2 5.40x1024 electrons = = 6.75x108 15 p 8x10 cm3

−−− Antimony (Sb) is a Column - V element, so it is a donor impurity. p=

ni2 1020 holes = = 5.00x103 16 n 2x10 cm3

n = N D = 2x1016

electrons cm 3

n > p → n - type silicon

Page 56 €

Reading from the graph for NT = 1016/cm3, 1230 cm2/V-s and 405 cm2/V-s. Reading from the graph for NT = 1017/cm3, 800 cm2/V-s and 230 cm2/V-s.

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Page 58

(

1 = 0.001 Ω − cm σ

)

σ = 1000 = 1.60x10−19 µn n → un n = 6.25x1021 / cm3 = 6.25x1019 (100) | ρ = −−− (a) N D = 2x1016 / cm 3

µn = 92 +

| N A = 0 / cm3

1270 0.91

 2x10  1+   17 1.3x10  16

| N T = 2x1016 / cm 3

= 1170 cm 2 /V − s | µ p = 48 +

| 447 0.76

 2x10  1+   16  6.3x10  16

= 363 cm2 /V − s

(b) N T = N D + N A = 2x1016 / cm 3 + 3x1016 / cm3 = 5x1016 / cm3

µn = 92 +

€

1270 0.91

 5x1016  1+   17 1.3x10 

= 990 cm 2 /V − s | µ p = 48 +

447 0.76

 5x1016  1+   16  6.3x10 

= 291 cm2 /V − s

Page 59 Boron (B) is a Column - III element, so it is an acceptor impurity. p = N A − N D = 4x1018

holes cm3

| n=

ni2 1020 electrons = = 25 18 p 4x10 cm3

| p - type material

4x1018 and mobilities from Fig. 2.8 : µn = 150 cm 2 /V − s | µp = 70 cm2 /V − s 3 cm 1 1 ρ≅ = = 0.022 Ω − cm qµ p p 1.60x10−19 4x1018 (70) NT =

(

)

--Indium (In) is a Column - III element, so it is an acceptor impurity. holes p = N A − N D = 7x10 cm3 19

ni2 1020 electrons | n= = = 1.43 | p - type material 19 p 7x10 cm3

7x1019 and mobilities from Fig. 2.8 : µn = 100 cm 2 /V − s | µp = 50 cm2 /V − s 3 cm 1 1 ρ= = = 1.79 mΩ − cm −19 σ 1.60x10 7x1019 (50) NT =

(

)

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5


Page 60 −23 1.38x10−23 (300) kT 1.38x10 (50) VT = = = 4.31 mV | VT = = 25.8 mV q 1.602x10−19 1.602x10−19 VT =

1.38x10−23 (400) 1.602x10−19

= 34.5mV

−−− Dn =

 kT cm2  cm 2 µn = 25.8mV 1362  = 35.1 q V − s s 

| Dp =

 kT cm2  cm2 µ p = 25.8mV  495  = 12.8 q V − s s 

−−− jn = qDn

 cm2  1016  10 4 µm  dn A = 1.60x10−19 C 20  3   = 320 2 dx s  cm − µm  cm  cm 

j p = −qD p

 cm2  1016  10 4 µm  dp A = −1.60x10−19 C 4  3   = −64 2 dx cm  s  cm − µm  cm 

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CHAPTER 3 Page 79  2x1018 1020 N N  A D φ j = VT ln 2  = 0.025V ln  1020  ni  

( )  = 1.05 V  

(

)

2(11.7) 8.85x10−14  1 2εs  1 1  1  wdo = + + 1.05) = 2.63x10−6 cm = 0.0263 µm  φ j =  −19 18 20 ( q  N A ND  1.60x10 10   2x10

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Page 80 qN A x p 1 0 Emax = ∫ −qN A dx = εs −x p εs Emax =

Emax

1 =− εs

( )( 11.7(8.854x10 F / cm)

1.6x10−19 C 1017 / cm3 1.13x10−5 cm −14

xn

∫ qN

D

dx =

0

qN D x n εs

For the values in Ex.3.2 :

) = 175 kV

cm

−−− 2(1.05V )

kV cm 2.63x10 cm −1  2x1018  x p = 0.0263µm1+  = 0.0258 µm 1020  

Emax =

−6

= 798

−1

 1020  −4 xn = 0.0263µm1+  = 5.16x10 µm 18  2x10 

Page 83 1.381x10 (300) kT 300 n =1 = 0.0259 | T = = 291 K −19 q 1.03 1.602x10 −23

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Page 85 €

  0.55     0.70   −15 iD = 40x10−15 Aexp  −1 = 143 µA iD = 40x10 Aexp  −1 = 57.9 mA   0.025    0.025    6x10−3 A  VD = (0.025V ) ln1+  = 0.643 V −15  40x10 A  −−−   −0.04     −2.0   −15 iD = 5x10−15 Aexp  −1 = −3.99 fA | iD = 5x10 Aexp  −1 = −5.00 fA   0.025     0.025  

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Page 87  40x10−6 A   I  D a V = V ln 1+ = 0.025V ln 1+  = 0.593 V ( ) BE T  I  −15  2x10 A  S  400x10−6 A   I  VBE = VT ln1+ D  = 0.025V ln1+  = 0.651 V ΔVBE = 57.6 mV 2x10−15 A   IS    40x10−6 A   I  (b) VBE = VT ln1+ ID  = 0.0258V ln1+ 2x10−15 A  = 0.612 V   S −6  400x10 A   I  VBE = VT ln1+ D  = 0.0258V ln1+  = 0.671 V ΔVBE = 59.4 mV 2x10−15 A   IS  

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Page 89 i  ln D  = ln (iD ) − ln ( I S ) | Assume iD is constant, and EG = EGO  IS   E  dvD k  iD  kT 1 dI S v d 1 dI S = ln  − = − VT | I S = Kni2 = KBT 3 exp − GO  dT q  I S  q I S dT T I S dT  kT   E   E  E  dI S = 3KBT 2 exp− GO  + KBT 3 exp − GO  + GO2  dT  kT   kT  kT  1 dI S 3 EGO 3 qVGO 3 VGO dvD v d 1 dI S v d − VGO − 3VT = + 2= + = + | = − VT = 2 I S dT T kT T kT T VT T dT T I S dT T Page 90

€

wd = 0.113µm 1+

2(V + φ j ) 2(10.979V ) 10V kV = 0.378 µm | Emax = = = 581 −4 0.979V wd cm 0.378x10 cm

−−− I S = 10 fA 1+

€

10V = 36.7 fA 0.8V

Page 93 11.7 8.854x10−14 F / cm nF nF C jo = = 91.7 2 | C j(0V ) = 91.7 2 10−2 cm 1.25x10−2 cm = 11.5 pF −4 0.113x10 cm cm cm 11.5 pF C j (5V ) = = 4.64 pF 5V 1+ 0.979V −−−

(

)

(

)(

)

10−5 A −8 8x10−4 A −8 5x10−2 A −8 CD = 10 s = 4.00 pF | CD = 10 s = 320 pF | CD = 10 s = 0.02 µF 0.025V 0.025V 0.025V

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Page 98 5V = 1 mA; I D = 0, VD = 5V 5kΩ The intersection of the two curves occurs at the Q - pt : (0.88 mA, 0.6 V) Two points on the load line : VD = 0, I D =

Page 100 €

10 = 10 4 I D + VD

 I   I  | VD = VT ln1+ D  | 10 = 10 4 I D + 0.025ln1+ D   IS   IS 

Page 101 fzero(@(id) (10-10000*id-0.025*log(1+id/1e-13), 5e-4)

€

ans = 9.4258e-004 Page 102 fzero(@(id) (10-10000*(1e-14)*(exp(40*vd)-1)-vd), 0.5) ans = 0.6316 Page 109 From the answer, the diodes are on,on,off. 10V - VB VB − 0.6V − (−20V ) VB − 0.6V − (−10V ) = + = 0 → VB = 1.87 V 2.5kΩ 10kΩ 10kΩ 1.87 − 0.6 − (−20V ) 1.87 − 0.6 − (−10V ) I D1 = = 2.13 mA | I D2 = = 1.13 mA | VD3 = −(1.87 − 0.6) = −1.27 V 10kΩ 10kΩ I D1 > 0, I D2 > 0, VD3 < 0. These results are consistent with the assumptions. I1 = I D1 + I D2

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Page 111 5kΩ 1kΩ Rmin = = 1.67 kΩ | VO = 20V = 3.33 V (VZ is off) | VO = 5 V (VZ is conducting) 20 5kΩ + 1kΩ −1 5 Page 112 VL − 20V VL − 5V V 6.25V − 5V + + L = 0 → VL = 6.25 V | I Z = = 12.5mA 1kΩ 0.1kΩ 5kΩ 0.1kΩ 2

PZ = 5V (12.5mA) + 100Ω(12.5mA) = 78.1 mW −−− Line Regulation =

€

0.1kΩ mV = 19.6 0.1kΩ + 5kΩ V

Load Regulation = 0.1kΩ 5kΩ = 98.0 Ω

9


Page 118 Vdc = VP − Von = 6.3 2 −1 = 7.91 V ΔT =

I dc =

7.91V = 15.8 A 0.5Ω

 0.527V  1 V 1 2 r = 2  = 0.912 ms ω VP 2π (60)  8.91V 

Vr =

I dc 15.8 A 1 T= s = 0.527 V C 0.5F 60

θ c = 120π (0.912ms)

180 o = 19.7 o π

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

I dc =

13.1V = 6.57 A 2Ω

 0.1V  1 V 1 2 r = 2  = 0.316 ms ω VP 2π (60) 14.1V 

C=

I dc 6.57 A 1 T= s = 1.10 F Vr 0.1F 60

θ c = 120π (0.316ms)

180 o = 6.82 o π

Page 120 €

−23 kT  I D  1.38x10 J / K (300K )  48.6 A  At 300K : VD = ln1+  = ln1+ −15  = 0.994 V q  IS  1.60x10−19 C  10 A  −23 kT  I D  1.38x10 J / K (273K + 50K )  48.6 A  At 50 C : VD = ln1+  = ln1+ −15  = 1.07 V q  IS  1.60x10−19 C  10 A   48.6 A  kT  I D  Note : For VT = 0.025V , VD = ln1+  = 0.025V ln1+ −15  = 0.961 V q  IS   10 A  o

Page 127 €

Vrms =

Vdc + Von 2

=

15 + 1 2

= 11.3 V | C =

I dc 2A 1 T= s = 0.222F = 222,000 µF Vr 60 0.01(15V )

I SC = ωCVP = 2π (60Hz )(0.222F )(16V ) = 1340 A ΔT =

(0.01)(15) = 0.363 ms | I = I 2T = 2 A 2s 1 V 1 2 r = 2 = 184 A P dc ω VP 2π (60) 16 ΔT 60(0.363ms)

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CHAPTER 4 Page 153

(

)

14 εox cm2 3.9 8.854x10 F / cm C µA K = µn = 500 = 69.1x10−6 2 = 69.1 2 −7 Tox V −s 25x10 cm V −s V --W µA  20µm  µA µA  60µm  µA Kn = Kn' = 50 2   = 1000 2 Kn = 50 2   = 1000 2 L V  1µm  V V  3µm  V µA  10µm  µA Kn = 50 2   = 2000 2 V  0.25µm  V --For VGS = 0V and 1V , VGS < VTN and the transistor is off. Therefore I D = 0. ' n

VGS - VTN = 2 -1.5 = 0.5V and VDS = 0.1 V → Triode region operation W V  µA  10µm  0.1 2 I D = Kn' VGS − VTN − DS VDS = 25 2   2 −1.5 − 0.1V = 11.3 µA L 2  2 V  1µm  VGS - VTN = 3 -1.5 = 1.5V and VDS = 0.1 V → Triode region operation W V  µA  10µm  0.1 2 I D = Kn' VGS − VTN − DS VDS = 25 2   3 −1.5 − 0.1V = 36.3 µA L 2  2 V  1µm  Kn' = 25

µA  100µm  µA  = 250 2 2  V  1µm  V

Page 154 €

1 1 = 4.00 kΩ | Ron = = 1.00 kΩ µA µA ' W Kn (VGS − VTN ) 250 2 (2 −1) 250 2 (4 −1) L V V 1 1 VGS = VTN + = 1V + = 3.00 V µA Kn Ron 250 2 (2000Ω) V Ron =

1

=

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11


Page 157 VGS - VTN = 5 -1 = 4 V and VDS = 10 V → Saturation region operation 2 2 Kn 1 mA VGS − VTN ) = 5 −1) V 2 = 8.00 mA ( 2 ( 2 2V W W 1mA 25 Kn = Kn' → = = W = 25L = 8.75 µm L L 40µA 1

ID =

−−− Assuming saturation region operation, 2 2 Kn 1 mA VGS − VTN ) = 2.5 −1) V 2 = 1.13 mA ( 2 ( 2 2V mA g m = Kn (VGS − VTN ) = 1 2 (2.5 −1)V = 1.50 mS V 2I D 2(1.125mA) Checking : g m = = 1.50 mS VGS − VTN (2.5 −1)V

ID =

€

Page 158 VGS - VTN = 5 -1 = 4 V and VDS = 10 V → Saturation region operation 2 2 Kn 1 mA VGS − VTN ) (1+ λVDS ) = 5 −1) V 2 1+ 0.02(10) = 9.60 mA ( 2 ( 2 2V 2 2 K 1 mA I D = n (VGS − VTN ) (1+ λVDS ) = 5 −1) V 2 1+ 0(10) = 8.00 mA 2 ( 2 2V --VGS - VTN = 4 -1 = 3 V and VDS = 5 V → Saturation region operation

[

ID =

[

]

]

2 2 2 Kn 25 µA VGS − VTN ) (1+ λVDS ) = 4 −1 V 1+ 0.01(5) = 118 µA ( ( ) 2 2 V2 2 2 K 25 µA I D = n (VGS − VTN ) (1+ λVDS ) = 5 −1) V 2 1+ 0.01(10) = 220 µA 2 ( 2 2 V

[

ID =

[

]

]

Page 159 €

Assuming pinchoff region operation, I D = VGS = VTN +

2 2 Kn 50 µA 0 − VTN ) = +2V ) = 100 µA ( 2 ( 2 2 V

2(100 µA) 2I D = 2V + = 4.00 V Kn 50µA /V 2

−−− Assuming pinchoff region operation, I D =

€

2 2 Kn 50 µA VGS − VTN ) = 1− −2V = 225 µA ( ( ) 2 2 V2

[

12

]


Page 161

( V + 0.6V − 0.6V ) = 1+ 0.75( = 1+ 0.75( 1.5 + 0.6 − 0.6 ) = 1.51 V | V

VTN = VTO + γ VTN

)

0 + 0.6 − 0.6 = 1 V

SB

TN

(

)

= 1+ 0.75 3 + 0.6 − 0.6 = 1.84 V

---

(a ) V

is less than the threshold voltage, so the transistor is cut off and ID = 0.

( b) V

- VTN = 2 -1 = 1 V and VDS = 0.5 V → Triode region operation

GS

GS

 V  mA  0.5  2 I D = KnVGS − VTN − DS VDS = 1 2  2 −1− 0.5V = 375 µA 2  2  V   (c) VGS - VTN = 2 -1 = 1 V and VDS = 2 V → Saturation region operation 2 2 Kn mA VGS − VTN ) (1+ λVDS ) = 0.5 2 (2 −1) V 2 1+ 0.02(2) = 520 µA ( 2 V

[

ID =

€

]

Page 163 (a ) VGS is greater than the threshold voltage, so the transistor is cut off and I D = 0.

( b) V

GS

- VTN = -2 + 1 = 1 V and VDS = 0.5 V → Triode region operation

 V  mA  −0.5  2 I D = KnVGS − VTN − DS VDS = 0.4 2  −2 + 1− (−0.5)V = 150 µA 2 2 V    

(c) V

GS

ID =

- VTN = −2 + 1 = 1 V and VDS = 2 V → Saturation region operation

2 2 Kn 0.4 mA VGS − VTN ) (1+ λVDS ) = −2 + 1) V 2 1+ 0.02(2) = 208 µA ( 2 ( 2 2 V

[

]

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13


Page 167  µF  CGC = 200 2  5x10−6 m 0.5x10−6 m = 0.500 fF m  

(

)(

)

 CGC pF  −6 + CGSOW = 0.25 fF +  300  5x10 m = 1.75 fF 2 m    2 pF  −6 Saturation region : CGS = CGC + CGSOW = 0.333 fF +  300  5x10 m = 1.83 fF 3 m    pF  −6 CGD = CGSOW =  300  5x10 m = 1.50 fF m  Triode region : CGD = CGS =

(

€

(

)

(

)

)

Page 169 KP = Kn = 150U | LAMBDA = λ = 0.0133 | VTO = VTN = 1 | PHI = 2φ F = 0.6 W = W = 1.5U | L = L = 0.25U Page 170

€

 1µm 2 Circuits/cm ∝ α =   = 16  0.25µm  2

2

Power - Delay Product ∝

€

1 1 1 = 3= → 64 times improvement 3 64 α 4

Page 171 ε W /α  v DS  * * iD* = µn ox  vGS − VTN − v DS = α iD | P = VDD iD = VDD (α iD ) = α P Tox / α L / α  2  P* αP P = = α3 * A A (W / α )( L / α ) −−−

(a ) fT =

1 500cm2 /V − s (1V ) = 7.96 GHz | 2 −4 2π 10 cm

(

)

(b) fT =

1 500cm2 /V − s (1V ) = 127 GHz 2π 0.25x10−4 cm 2

(

)

−−−   5 V  −5 V V V  −4 ≥ 105 → L = 105  10 cm = 10 V | L = 10  10 cm = 1 V L cm  cm   cm 

(

)

(

)

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14


Page 172 (a ) From the graph for VGS = 0.25 V , I D ≅ 10−18 A |

(b) From the graph for V = V − 0.5 V , I ≅ 3x10 (c) I = (10 transistors)(3x10 A/transistors) = 3 µA GS

9

TN

D

−15

A

-15

Page 176 2

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Active area = 10Λ(12Λ) = 120Λ2 = 120(0.125µm ) = 1.88 µm 2 L = 2Λ = 0.250 µm | W = 10Λ = 1.25 µm 2

Gate area = 2Λ(10Λ) = 20Λ2 = 20(0.125µm) = 0.313 µm 2 2

Transistor area = (10Λ + 2Λ + 2Λ)(12Λ + 2Λ + 2Λ) = 224Λ2 = 224(0.125µm ) = 3.50 µm 2

(10 µm) N= 4

3.50µm

€

2

2

= 28.6 x 106 transistors

Page 180 Assume saturation region operation and λ = 0. Then I D is indpendent of VDS , and I D = 50 µA. VDS = VDD − I D RD = 10 − 50kΩ(50µA) = 7.50 V. VDS ≥ VGS − VTN , so our assumption was correct. Q - Point = (50.0 µA, 7.50 V ) --270kΩ 10V = 2.647 V | REQ = 270kΩ 750kΩ | Assume saturation region. 270kΩ + 750kΩ 2 2 25x10−6 A ID = 2.647 −1 V = 33.9 µA | VDS = VDD − I D RD = 10 −100kΩ(33.9µA) = 6.61 V. ( ) 2 V2 VDS ≥ VGS − VTN , so our assumption was correct. Q - Point = (33.9 µA, 6.61 V ) VEQ =

--2 30x10−6 A 3 −1) V 2 = 60.0 µA 2 ( 2 V = VDD − I D RD = 10 −100kΩ(60.0µA) = 4.00 V. VDS ≥ VGS − VTN , so our assumption was correct.

VGS does not change : VGS = 3.00 V | I D = VDS

Q - Point = (60.0 µA, 4.00 V )

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Page 181 2 25x10−6 A 3 −1.5) V 2 = 28.1 µA 2 ( 2 V = VDD − I D RD = 10 −100kΩ(28.1µA) = 7.19 V. VDS ≥ VGS − VTN , so our assumption was correct.

VGS does not change : VGS = 3.00 V | I D = VDS

Q - Point = (28.1 µA, 7.19 V ) --VDS = 10 − VDS =

( ) (3 −1) (1+ 0.01V ) → V 2

25x10−6 105

10 − 5 V = 4.76 V 1.05

2

DS

ID =

DS

= 10 − 5(1+ 0.01VDS )

2 25x10−6 3 −1) 1+ 0.01(4.76) = 52.4 µA ( 2

[

]

Page 182 €

10V = 150µA. 66.7kΩ = 3 - V curve at I D = 50 µA, VDS = 6.7 V.

For I D = 0, VDS = 10 V. For VDS = 0, I D = The load line intersects the VGS

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Page 185 Upper group  2  2V V + VGS  − 2VTN  + VTN2 − EQ = 0 Kn RS  Kn RS  2 GS

2

 1   1  2V | VGS = − − VTN  ±  − VTN  − VTN2 + EQ Kn RS  Kn RS   Kn RS 

2

 1  2VTN 2V 1 1 VGS = VTN − ±  + VTN2 − VTN2 + EQ = VTN +  − Kn RS Kn RS Kn RS  Kn RS  Kn RS

( 1+ 2K R (V n

S

EQ

)

− VTN ) −1

−−− Assume saturation region operation. ID =

1 2Kn RS2

(

)

2

1+ 2Kn RS (VEQ − VTN ) −1 =

1 2(30µA)(39kΩ)

2

(

)

2

1+ 2(30µA)(39kΩ)(4 −1) −1 = 36.8 µA

VDS = 10 −114kΩ(36.8µA) = 5.81 V | Saturation region is correct. | Q - Point : (36.8 µA, 5.81 V ) −−− Assume saturation region operation. I D =

1 2(25µA)(39kΩ)

2

(

)

2

1+ 2(25µA)(39kΩ)(4 −1.5) −1 = 26.7 µA

VDS = 10 −114kΩ(26.7µA) = 6.96 V | Saturation region is correct. | Q - Point : (26.7 µA, 6.96 V ) Assume saturation region operation. I D =

1 2(25µA)(62kΩ)

2

(

)

2

1+ 2(25µA)(62kΩ)(4 −1) −1 = 36.8 µA

VDS = 10 −137kΩ(25.4µA) = 6.52 V | Saturation region is correct. | Q - Point : (25.4 µA, 6.52 V )

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17


Page 185 Lower group R1 1MΩ VEQ = VDD = 10V = 4 V | Assume saturation region operation. R1 + R2 1MΩ + 1.5MΩ ID =

1 2Kn RS2

(

)

2

1+ 2Kn RS (VEQ − VTN ) −1 =

1 2(25µA)(1.8kΩ)

2

(

)

2

1+ 2(25µA)(1.8kΩ)(4 −1) −1 = 99.5 µA

VDS = 10 − 40.8kΩ(99.5µA) = 5.94 V | Saturation region is correct. | Q - Point : (99.5 µA, 5.94 V ) VEQ = ID =

R1 1.5MΩ VDD = 10V = 6 V | Assume saturation region operation. R1 + R2 1.5MΩ + 1MΩ 1

2(25µA)(22kΩ)

2

(

)

2

1+ 2(25µA)(22kΩ)(6 −1) −1 = 99.2 µA

VDS = 10 − 40kΩ(99.2µA) = 6.03 V | Saturation region is correct. | Q - Point : (99.2 µA, 6.03 V ) −−− I Bias =

VDD V 10V → R1 + R2 = DD = = 5 MΩ R1 + R2 I Bias 2µA

VEQ =

V R1 4V VDD → R1 = ( R1 + R2 ) EQ = 5MΩ = 2 MΩ R1 + R2 VDD 10V

R2 = 5MΩ − R1 = 3 MΩ | REQ = R1 R2 = 1.2 MΩ

Page 187 €

(

VGS = 6 − 22000I D VSB = 22000I D VTN = 1+ 0.75 VSB + 0.6 − 0.6

)

ID =

2 25µA VGS − VTN ) ( 2

Spreadsheet iteration yields ID = 83.2 µA. Page 183 €

[

(

)

]

Equation 4.55 becomes 6 - 1 + 0.75 VSB + 0.6 − 0.6 + 2.83 − VSB = 0. VSB2 − 6.065VSB + 7.231 = 0 → VSB = 1.63V. RS =

1.63V = 16.3 kΩ →16 kΩ 10−4 A

RD =

(10 − 6 −1.63)V = 23.7kΩ → 24 kΩ 10−4 A

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Page 189 Assume saturation region operation. 10 − 6 =

(

25x10−6 62x10 4 2

) (V

GS

2

+ 1) − VGS → VGS2 + 0.710VGS − 4.161 = 0 → VGS = −2.426V , + 1.716V

2 25x10 −2.426 + 1) = 25.4 µA | VDS = − 10 −137kΩ(25.4µA) = −6.52 V ( 2 Saturation region is correct. | Q - Point : (25.4 µA, - 6.52 V ) −6

ID =

€

[

]

Page 195 (a) VDS = 3 V , VGS − VP = −2 − (−3.5) = +1.5 V. The transistor is pinched off. 2

  −2V   V 2 I D = I DSS 1− GS  = 1mA 1−   = 184 µA | Pinchoff requires VDS ≥ VGS − VP = +1.5 V  VP    −3.5V 

(b)

VDS = 6 V , VGS − VP = −1− (−3.5) = +2.5 V. The transistor is pinched off. 2

2   −1V   VGS  I D = I DSS 1−  = 1mA 1−   = 510 µA | Pinchoff requires VDS ≥ VGS − VP = +2.5 V  VP    −3.5V 

(c)

VDS = 0.5 V , VGS − VP = −2 − (−3.5) = +1.5 V. The transistor is in the triode region.

ID =

2(1mA)  2I DSS  VDS  0.5 V − V − V = −2 + 3.5 −    0.5 = 51.0 µA GS P DS 2 2  2 VP2    −3.5 ( )

Pinchoff requires VDS ≥ VGS − VP = +1.5 V −−−

(a )

VDS = 0.5 V , VGS − VP = −2 − (−4) = +2 V. The transistor is in the triode region.

ID =

(b)

2(0.2mA)  2I DSS  VDS  0.5 V − V − V = −2 + 4 −    0.5 = 21.9 µA GS P DS 2 2  2  VP2  (−4) 

VDS = 6 V , VGS − VP = −1− (−4) = +3 V. The transistor is pinched off. 2

  −1V   V 2 I D = I DSS 1− GS  = 0.2mA 1−   = 113 µA  VP    −4V 

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19


Page 197 (a) VDS = −3 V , VGS − VP = 3 − 4 = −1 V. The transistor is pinched off.  VGS 2  3V 2 I D = I DSS 1−  = 2.5mA 1−  = 156 µA | Pinchoff requires VDS ≤ VGS − VP = −1 V  4V   VP 

( b)

VDS = −6 V , VGS − VP = 1− (4) = −3 V. The transistor is pinched off.

 VGS 2  1V 2 I D = I DSS 1−  = 2.5mA 1−  = 1.41 mA | Pinchoff requires VDS ≤ VGS − VP = −3 V  4V   VP 

(c)

VDS = −0.5 V , VGS − VP = 2 − (4) = −2 V. The transistor is in the triode region.

ID =

2(2.5mA)  2I DSS  VDS  −0.5  V − V − V = 2 − 4 −    (−0.5) = 273 µA GS P DS 2  2  VP2  42 

Pinchoff requires VDS ≤ VGS − VP = −2 V Page 198 €

BETA =

I DSS 2.5mA = = 0.625 mA | VTO = VP = −2 V | LAMBDA = λ = 0.025 V -1 2 2 VP (−2)

−−− BETA =

I DSS 5mA = 2 = 1.25 mA | VTO = VP = 2 V | LAMBDA = λ = 0.02 V -1 2 VP 2

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Page 200 VTO = VP = −5 V | BETA =

I DSS 5mA = = 0.2 mA | LAMBDA = λ = 0.02 V -1 VP2 (−5)2

−−− VGS = VG − VS = −IG RG − I S RS = 0 − I D RS = −I D RS 2

2

V   V  2 K K K VGS = − n (VGS − VTN ) RS = − n VTN2 RS  GS −1 = −I DSS RS 1− GS  for I DSS = n VTN2 and VP = VTN 2 2 2  VTN   VP  2

2

 V   V  2 0.4mA VGS = − −5) (1kΩ)1− GS  = 51+ GS  → VGS2 −15VGS + 25 = 0 and the rest is identical. ( 2 5   −5   --Assuming pinchoff, I D = 1.91 mA and VDS = 9 −1.91mA(2kΩ + 1kΩ) = 3.27 V. VGS − VP = 3.09 V , VDS > VGS − VP , and pinchoff is correct. --2

 V   VGS 2 −3 3 2 GS Assuming pinchoff, VGS = −I DSS RS 1−  = − 5x10 2x10 1+  → VGS + 12.5VGS + 25 = 0 5    VP 

(

)(

)

2

2  V   −2.5  GS VGS = −2.5 V | I D = I DSS 1−  = 5mA1−  = 1.25 mA −5    VP 

VDS = 12 −1.25mA(2kΩ + 2kΩ) = 7.00 V. VGS − VP = −2.5 − (−5) = +2.5 V , VDS > VGS − VP , and pinchoff is correct. Q - Point : (1.25 mA, 7.00 V ) ---

(a ) V

G

= −IG RG = −10nA(680kΩ) = −6.80 mV. 2

2  VGS   VGS  −3 3 2 VGS = VG − VS = −IG RG − I DSS RS 1−  = −.00680 − 5x10 10 1+  → VGS −15VGS + 25 = 0 V 5    P 

(

)( )

The value of VG is insignificant with respect to the constant term of 25. So the answers are the same to 3 significant digits.

(b) V

G

= −IG RG = −1µA(680kΩ) = 0.680 V and now cannot be neglected.

2  V 2  VGS  −3 3 2 GS VGS = VG − VS = −IG RG − I DSS RS 1−  = −0.680 − 5x10 10 1+  → VGS −15VGS + 28.4 = 0 V 5    P 

(

)( )

2

 V   −2.226 2 GS VGS = −2.226 V | I D = I DSS 1−  = 5mA1−  = 1.54 mA −5    VP  VDS = 12 −1.54mA(2kΩ + 1kΩ) = 7.38 V | Q - Point : (1.54 mA, 7.38 V )

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CHAPTER 5 Page 223

(a )

βF =

αF 0.970 0.993 0.250 = = 32.3 | β F = = 142 | β F = = 0.333 1− α F 1− 0.970 1− 0.993 1− .250

(b)

αF =

βF 40 200 3 = = 0.976 | α F = = 0.995 | α F = = 0.750 β F + 1 41 201 4

Page 225 €

  0.700   −9.30  10−15 A   −9.30   iC = 10−15 Aexp − exp exp    −  −1 = 1.45 mA 0.5   0.025    0.025    0.025    0.700   −9.30  10−15 A   0.700   iE = 10−15 Aexp − exp exp    +  −1 = 1.46 mA  0.025  100   0.025     0.025  10−15 A   0.700   10−15 A   −9.30   iB = exp exp  −1 +  −1 = 14.5 µA 100   0.025   0.5   0.025  

Page 227 €

  0.750   0.700  10−16 A   0.700   iC = 10−16 Aexp exp  − exp  −  −1 = 563 µA 0.4   0.025    0.025    0.025    0.750   0.700  10−16 A   0.750   iE = 10−16 Aexp − exp exp    +  −1 = 938 µA 75   0.025    0.025    0.025 

iB =

10−16 A   0.750   10−16 A   0.700   exp exp  −1 +  −1 = 376 µA 75   0.025   0.4   0.025  

−−−   0.750   −2  iT = 10−15 Aexp  − exp  = 10.7 mA  0.025   0.025    0.750   −4.25  iT = 10−16 Aexp  − exp  = 1.07 mA  0.025    0.025 

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Page 230  10−4 A  I  VBE = VT ln C + 1 = 0.025V ln −16 + 1 = 0.691 V  IS   10 A   10−3 A  I  VBE = VT ln C + 1 = 0.025V ln −16 + 1 = 0.748 V  IS   10 A  Page 231 €

€

npn :VBE > 0, VBC < 0 → Forward − Active Region | pnp :VEB > 0, VCB > 0 → Saturation Region Page 233 αF 0.95 βF = = = 19 1− α F 0.05 VBE = 0, VBC

I E = I S = 0.100 fA

IB = −

IS 10−16 A =− = −0.300 fA βR 0.333

VBE

Microelectronic Circuit Design Fourth Edition - Part I Solutions to ...

Microelectronic Circuit Design Fourth Edition - Part I Solutions to ...

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