Microelectronics Circuit Analysis and Design Donald Neamen 4th Solutions PDF

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Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 1 1.1 − Eg / 2 kT ni = BT 3 / 2 e (a) Silicon ⎡ −1.1 ⎤ ni = ( 5.23 × 1015 ) ( 250 ) 3/ 2 (i) exp ⎢ ⎥ ...

Description

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 1 1.1

ni = BT 3 / 2 e (a) Silicon

− Eg / 2 kT

⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦ = 2.067 × 1019 exp [ −25.58] ni = 1.61× 108 cm −3

(i)

ni = ( 5.23 × 1015 ) ( 250 )

(ii)

ni = ( 5.23 × 1015 ) ( 350 )

(b)

GaAs

(i)

ni = ( 2.10 × 1014 ) ( 250 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦ = 3.425 × 1019 exp [ −18.27 ] ni = 3.97 ×1011 cm −3 3/ 2

3/ 2

⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦

= ( 8.301× 1017 ) exp [ −32.56] ni = 6.02 × 103 cm −3

(ii)

ni = ( 2.10 × 1014 ) ( 350 )

3/ 2

⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦

= (1.375 × 1018 ) exp [ −23.26] ni = 1.09 × 108 cm −3

______________________________________________________________________________________ 1.2

a.

⎛ − Eg ⎞ ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠ ⎛ ⎞ −1.1 1012 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎟ −6 ⎝ 2(86 × 10 )(T ) ⎠

⎛ 6.40 × 103 ⎞ 1.91× 10−4 = T 3 / 2 exp ⎜ − ⎟ T ⎝ ⎠ By trial and error, T ≈ 368 K b. ni = 109 cm −3 ⎛ ⎞ −1.1 ⎟ 109 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎜ 2 ( 86 × 10−6 ) (T ) ⎟ ⎝ ⎠ ⎛ 6.40 × 103 ⎞ 1.91× 10−7 = T 3 / 2 exp ⎜ − ⎟ T ⎝ ⎠ By trial and error, T ≈ 268° K ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.3

Silicon (a)

ni = ( 5.23 × 1015 ) (100 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) (100 ) ⎥⎦

= ( 5.23 × 1018 ) exp [ −63.95] ni = 8.79 ×10−10 cm −3

(b)

ni = ( 5.23 × 1015 ) ( 300 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦

= ( 2.718 × 1019 ) exp [ −21.32] ni = 1.5 × 1010 cm −3

(c)

ni = ( 5.23 × 1015 ) ( 500 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 500 ) ⎥⎦

= ( 5.847 × 1019 ) exp [ −12.79] ni = 1.63 × 1014 cm −3

Germanium. (a)

ni = (1.66 × 1015 ) (100 )

3/ 2

⎡ ⎤ −0.66 ⎥ = (1.66 × 1018 ) exp [ −38.37 ] exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) (100 ) ⎥⎦

ni = 35.9 cm −3 (b)

ni = (1.66 × 1015 ) ( 300 )

3/ 2

⎡ ⎤ −0.66 ⎥ = ( 8.626 × 1018 ) exp [ −12.79] exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦

ni = 2.40 × 1013 cm −3 (c)

ni = (1.66 × 1015 ) ( 500 )

3/ 2

⎡ ⎤ −0.66 ⎥ = (1.856 × 1019 ) exp [ −7.674] exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 500 ) ⎥⎦

ni = 8.62 ×1015 cm −3 ______________________________________________________________________________________ 1.4

(a) n-type; no = 10

15

(

)

(

)

n2 2.4 × 1013 cm ; po = i = no 1015 −3

2

2

= 5.76 × 1011 cm −3

ni2 1.5 × 1010 = = 2.25 × 10 5 cm −3 no 1015 ______________________________________________________________________________________

(b) n-type; no = 1015 cm −3 ; po =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.5

(a) p-type; p o = 1016 cm −3 ; no =

(

ni2 1.8 × 10 6 = po 1016

)

(

2

= 3.24 × 10 − 4 cm −3

)

2

ni2 2.4 × 1013 = = 5.76 × 1010 cm −3 po 1016 ______________________________________________________________________________________

(b) p-type; p o = 1016 cm −3 ; no =

1.6

(a) (b)

n-type no = N d = 5 × 1016 cm −3 10 ni2 (1.5 × 10 ) po = = = 4.5 × 103 cm −3 no 5 × 1016 2

(c)

no = N d = 5 × 1016 cm −3

From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3

( 3.97 × 10 ) =

11 2

= 3.15 × 106 cm −3 5 × 1016 ______________________________________________________________________________________ po

1.7

(a) p-type; p o = 5 × 1016 cm −3 ; no =

(

)

(

)

ni2 1.5 × 1010 = po 5 × 1016

2

= 4.5 × 10 3 cm −3

2

ni2 1.8 × 10 6 = = 6.48 × 10 −5 cm −3 po 5 × 1016 ______________________________________________________________________________________

(b) p-type; p o = 5 × 1016 cm −3 ; no =

1.8 (a) Add boron atoms (b) N a = po = 2 × 1017 cm −3

(

)

2

ni2 1.5 × 1010 = = 1.125 × 10 3 cm −3 po 2 × 1017 ______________________________________________________________________________________

(c) no =

1.9

(a)

no = 5 × 1015 cm −3 10 n 2 (1.5 × 10 ) po = i = ⇒ po = 4.5 × 104 cm −3 no 5 × 1015 2

(b)

n o > p o ⇒ n-type

(c) no ≅ N d = 5 × 1015 cm −3 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.10 a.

b.

Add Donors N d = 7 × 1015 cm −3 Want po = 106 cm −3 = ni2 / N d

So ni2 = (106 )( 7 × 1015 ) = 7 × 10 21 ⎛ − Eg ⎞ = B 2T 3 exp ⎜ ⎟ ⎝ kT ⎠

⎛ ⎞ 2 −1.1 ⎟ 7 × 1021 = ( 5.23 × 1015 ) T 3 exp ⎜ 6 − ⎜ ( 86 × 10 ) (T ) ⎟ ⎝ ⎠ By trial and error, T ≈ 324° K ______________________________________________________________________________________

1.11 (a) I = Aσ Ε = 10 −5 (1.5)(10) ⇒ I = 0.15 mA

( )

(

)

Iρ 1.2 × 10 −3 (0.4) = 2.4 V/cm = A ρ 2 × 10 − 4 ______________________________________________________________________________________

(b) I =

AΕ

⇒Ε=

(

)

1.12

J 120 −1 = = 6.67 (Ω − cm) Ε 18 σ (6.67) σ ≅ eμ n N d ⇒ N d = = = 3.33 × 1016 cm −3 eμ n 1.6 × 10 −19 (1250) ______________________________________________________________________________________ J =σΕ ⇒σ =

(

)

1.13

1 1 1 ⇒ Nd = = = 7.69 × 1015 cm −3 eμ n N d eμ n ρ 1.6 × 10 −19 (1250)(0.65) Ε (b) J = ⇒ Ε = ρ J = (0.65)(160 ) = 104 V/cm ρ ______________________________________________________________________________________ (a) ρ ≅

(

)

1.14

σ 1.5 = = 9.375 × 1015 cm −3 eμ n 1.6 × 10 −19 (1000) σ 0.8 = = 1.25 × 1016 cm −3 (b) N a = −19 eμ p 1.6 × 10 (400) ______________________________________________________________________________________ (a) σ ≅ eμ n N d ⇒ N d =

(

(

)

)

1.15 (a) For n-type, σ ≅ eμ n N d = (1.6 × 10 −19 ) ( 8500 ) N d

For 1015 ≤ N d ≤ 1019 cm −3 ⇒ 1.36 ≤ σ ≤ 1.36 × 104 ( Ω − cm )

−1

(b) J = σ E = σ ( 0.1) ⇒ 0.136 ≤ J ≤ 1.36 ×103 A / cm2 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.16 Dn = (0.026)(1250) = 32.5 cm 2 /s; D p = (0.026 )(450 ) = 11.7 cm 2 /s J n = eDn

⎛ 1016 − 1012 dn = 1.6 × 10 −19 (32.5)⎜⎜ dx ⎝ 0 − 0.001

(

J p = −eD p

)

⎞ ⎟⎟ = −52 A/cm 2 ⎠

⎛ 1012 − 1016 dp = − 1.6 × 10 −19 (11.7 )⎜⎜ dx ⎝ 0 − 0.001

(

)

⎞ ⎟⎟ = −18.72 A/cm 2 ⎠

Total diffusion current density J = −52 − 18.72 = −70.7 A/cm 2 ______________________________________________________________________________________ 1.17 J p = −eD p

dp dx

⎛ −1 ⎞ ⎛ −x ⎞ = −eD p (10 15 ) ⎜ ⎟ exp ⎜ ⎟ ⎜ Lp ⎟ ⎜ Lp ⎟ ⎝ ⎠ ⎝ ⎠

(1.6 ×10 ) (15) (10 ) exp ⎛ − x ⎞ −19

Jp =

15

⎜⎜ ⎟⎟ ⎝ Lp ⎠

10 × 10 −4

J p = 2.4 e

− x / Lp

J p = 2.4 A/cm2

(a)

x=0

(b)

x = 10 μ m

J p = 2.4 e−1 = 0.883 A/cm 2

x = 30 μ m J p = 2.4 e−3 = 0.119 A/cm 2 ______________________________________________________________________________________ (c)

1.18 a.

N a = 1017 cm −3 ⇒ po = 1017 cm −3 n 2 (1.8 × 10 no = i = 1017 po

b.

)

6 2

⇒ no = 3.24 × 10−5 cm −3

n = no + δ n = 3.24 × 10−5 + 1015 ⇒ n = 1015 cm −3 p = po + δ p = 1017 + 1015 ⇒ p = 1.01× 1017 cm −3

______________________________________________________________________________________ ⎛N N 1.19 Vbi = VT ln⎜⎜ a 2 d ⎝ ni

(a) (i)

⎞ ⎟ ⎟ ⎠

(

)(

⎡ 5 × 1015 5 × 1015 Vbi = (0.026 ) ln ⎢ 2 ⎢⎣ 1.5 × 1010

(

(

)

)⎤⎥ = 0.661 V ⎥⎦

)( ) ( ) ⎡ (10 )(10 ) ⎤ = (0.026 ) ln ⎢ ⎥ = 0.937 V ⎣⎢ (1.5 × 10 ) ⎦⎥

(ii)

⎡ 5 × 1017 1015 ⎤ Vbi = (0.026 ) ln ⎢ ⎥ = 0.739 V 10 2 ⎥⎦ ⎣⎢ 1.5 × 10

(iii)

Vbi

18

18

10 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(

)(

⎡ 5 × 1015 5 × 1015 Vbi = (0.026 ) ln ⎢ 2 ⎢⎣ 1.8 × 10 6

(b) (i)

(

(

)

)⎤⎥ = 1.13 V ⎥⎦

)( )⎤⎥ = 1.21 V ( ) ⎥⎦ ⎡ (10 )(10 ) ⎤ = (0.026 ) ln ⎢ ⎥ = 1.41 V ⎢⎣ (1.8 × 10 ) ⎥⎦

(ii)

⎡ 5 × 1017 1015 Vbi = (0.026 ) ln ⎢ 2 ⎢⎣ 1.8 × 10 6

(iii)

Vbi

18

18

6 2

______________________________________________________________________________________ 1.20 ⎛N N Vbi = VT ln⎜⎜ a 2 d ⎝ ni

or

⎞ ⎟ ⎟ ⎠

(n ) exp⎛⎜ V

(1.5 × 10) exp⎛ 0.712 ⎞ = 1.76 × 1016 cm −3 bi ⎞ ⎜ ⎟ ⎜ V ⎟⎟ = 1016 Nd ⎝ 0.026 ⎠ ⎝ T ⎠ ______________________________________________________________________________________ Na =

1.21

2 i

2

⎡ N a (1016 ) ⎤ ⎛N N ⎞ ⎥ Vbi = VT ln ⎜ a 2 d ⎟ = ( 0.026 ) ln ⎢ 10 2 ⎢⎣ (1.5 × 10 ) ⎥⎦ ⎝ ni ⎠

For N a = 1015 cm −3 , Vbi = 0.637 V For N a = 1018 cm −3 , Vbi = 0.817 V

______________________________________________________________________________________ 1.22

⎛ T ⎞ kT = (0.026) ⎜ ⎟ ⎝ 300 ⎠ 200 250 300 350 400 450 500

kT 0.01733 0.02167 0.026 0.03033 0.03467 0.0390 0.04333

(T)3/2 2828.4 3952.8 5196.2 6547.9 8000.0 9545.9 11,180.3

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ ⎞ −1.4 ⎟ ni = ( 2.1× 1014 )(T 3 / 2 ) exp ⎜ ⎜ 2 ( 86 × 10−6 ) (T ) ⎟ ⎝ ⎠ ⎛N N ⎞ Vbi = VT ln ⎜ a 2 d ⎟ ⎝ ni ⎠ T ni 200 1.256 250 6.02 × 103 300 1.80 × 106 350 1.09 × 108 400 2.44 × 109 450 2.80 × 1010 500 2.00 × 1011

Vbi 1.405 1.389 1.370 1.349 1.327 1.302 1.277

______________________________________________________________________________________ 1.23

⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠

−1/ 2

⎡ (1.5 × 10 16 )( 4 × 10 15 ) ⎤ ⎥ = 0.684 V Vbi = ( 0.026 ) ln ⎢ ⎢⎣ (1.5 ×10 10 ) 2 ⎥⎦ 1 ⎞ ⎛ C j = ( 0.4 ) ⎜ 1 + ⎟ 0.684 ⎝ ⎠

−1/ 2

(a)

3 ⎞ ⎛ C j = ( 0.4 ) ⎜ 1 + ⎟ ⎝ 0.684 ⎠

−1/ 2

(b)

= 0.255 pF

= 0.172 pF −1/ 2

5 ⎞ ⎛ = 0.139 pF C j = ( 0.4 ) ⎜ 1 + ⎟ ⎝ 0.684 ⎠ ______________________________________________________________________________________

(c)

1.24

(a)

⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠

−1 / 2

5 ⎞ ⎛ For VR = 5 V, C j = (0.02) ⎜ 1 + ⎟ ⎝ 0. 8 ⎠

−1 / 2

⎛ 1. 5 ⎞ For VR = 1.5 V, C j = (0.02) ⎜1 + ⎟ ⎝ 0. 8 ⎠

= 0.00743 pF −1 / 2

= 0.0118 pF

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 0.00743 + 0.0118 = 0.00962 pF 2 vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ

C j (avg ) =

where τ = RC = RC j (avg ) = (47 × 103 )(0.00962 × 10−12 ) or

τ = 4.52 ×10−10 s Then vC ( t ) = 1.5 = 0 + ( 5 − 0 ) e−ti / τ

5 + r /τ ⎛ 5 ⎞ = e 1 ⇒ t1 = τ ln ⎜ ⎟ 1.5 ⎝ 1.5 ⎠ −10 t1 = 5.44 × 10 s (b)

For VR = 0 V, Cj = Cjo = 0.02 pF −1/ 2 ⎛ 3.5 ⎞ For VR = 3.5 V, C j = ( 0.02 ) ⎜ 1 + = 0.00863 pF ⎟ ⎝ 0.8 ⎠ 0.02 + 0.00863 C j (avg ) = = 0.0143 pF 2 τ = RC j ( avg ) = 6.72 ×10−10 s vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ

(

3.5 = 5 + (0 − 5)e − t2 /τ = 5 1 − e − t2 /τ

so that t2 = 8.09 × 10

−10

)

s

______________________________________________________________________________________ 1.25

⎛ V C j = C jo ⎜⎜1 + R ⎝ Vbi

⎞ ⎟⎟ ⎠

−1 / 2

(

)( ) )

⎡ 5 × 1015 1017 ⎤ ; Vbi = (0.026 ) ln ⎢ ⎥ = 0.739 V 10 2 ⎥⎦ ⎣⎢ 1.5 × 10

(

For V R = 1 V, Cj =

0.60 1 1+ 0.739

= 0.391 pF

For VR = 3 V, Cj =

0.60 3 1+ 0.739

= 0.267 pF

For V R = 5 V, 0.60

Cj =

(a)

fo =

(b) f o =

1 2π LC

(

5 1+ 0.739

=

(

= 0.215 pF

2π 1.5 × 10

2π 1.5 × 10

1 −3

)(0.391× 10 ) −12

1 −3

)(0.267 × 10 ) −12

⇒ f o = 6.57 MHz

⇒ f o = 7.95 MHz

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

fo =

1

⇒ f o = 8.86 MHz 2π 1.5 × 10 0.215 × 10 −12 ______________________________________________________________________________________ (c)

(

−3

)(

)

1.26

a.

⎡ ⎛V ⎞ ⎤ ⎛V ⎞ I = I S ⎢ exp ⎜ D ⎟ − 1⎥ − 0.90 = exp ⎜ D ⎟ − 1 ⎝ VT ⎠ ⎦ ⎝ VT ⎠ ⎣ ⎛V ⎞ exp ⎜ D ⎟ = 1 − 0.90 = 0.10 ⎝ VT ⎠

VD = VT ln ( 0.10 ) ⇒ VD = −0.0599 V

b. IF IR

⎡ ⎛ VF ⎢ exp ⎜ I ⎝ VT = S ⋅⎣ IS ⎡ ⎛ VR ⎢exp ⎜ ⎝ VT ⎣ =

⎞ ⎤ ⎛ 0.2 ⎞ ⎟ − 1⎥ exp ⎜ ⎟ −1 0.026 ⎠ ⎠ ⎦ ⎝ = ⎞ ⎤ exp ⎛ −0.2 ⎞ − 1 ⎜ ⎟ ⎟ − 1⎥ ⎝ 0.026 ⎠ ⎠ ⎦

2190 −1

IF = 2190 IR

______________________________________________________________________________________

⎡ ⎛V 1.27 I D = I S ⎢exp⎜⎜ D ⎣⎢ ⎝ VT (a) (i) (ii) (iii) (iv)

⎞ ⎤ ⎟⎟ − 1⎥ ⎠ ⎦⎥

(

)

(

)

(

)

(

)

(

)

⎛ 0.3 ⎞ I D = 10 −11 exp⎜ ⎟ ⇒ 1.03 μ A ⎝ 0.026 ⎠ ⎛ 0.5 ⎞ I D = 10 −11 exp⎜ ⎟ ⇒ 2.25 mA ⎝ 0.026 ⎠ ⎛ 0.7 ⎞ I D = 10 −11 exp⎜ ⎟ ⇒ 4.93 A ⎝ 0.026 ⎠ ⎡ ⎛ − 0.02 ⎞ ⎤ −12 A I D = 10 −11 ⎢exp⎜ ⎟ − 1⎥ = −5.37 × 10 ⎣ ⎝ 0.026 ⎠ ⎦

(v)

⎡ ⎛ − 0.20 ⎞ ⎤ −11 I D = 10 −11 ⎢exp⎜ ⎟ − 1⎥ ≅ −10 A 0 . 026 ⎠ ⎦ ⎣ ⎝

(vi)

I D = − 10 −11 A

(b) (i) (ii) (iii)

(

)

(

)

(

)

(

)

⎛ 0.3 ⎞ I D = 10 −13 exp⎜ ⎟ ⇒ 0.0103 μ A ⎝ 0.026 ⎠ ⎛ 0.5 ⎞ I D = 10 −13 exp⎜ ⎟ ⇒ 22.5 μ A ⎝ 0.026 ⎠ ⎛ 0.7 ⎞ I D = 10 −13 exp⎜ ⎟ ⇒ 49.3 mA ⎝ 0.026 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(

)

(iv)

⎡ ⎛ − 0.02 ⎞ ⎤ −14 I D = 10 −13 ⎢exp⎜ A ⎟ − 1⎥ = −5.37 × 10 0 . 026 ⎝ ⎠ ⎣ ⎦

(v)

I D ≅ −10 −13 A

I D ≅ −10 −13 A (vi) ______________________________________________________________________________________ ⎛I 1.28 V D = VT ln⎜⎜ D ⎝ IS

⎞ ⎟ ⎟ ⎠

⎛ 10 × 10 −6 (a) (i) V D = (0.026 ) ln⎜⎜ −11 ⎝ 10

⎞ ⎟⎟ = 0.359 V ⎠

⎛ 100 × 10 −6 V D = (0.026 ) ln⎜⎜ −11 ⎝ 10

⎞ ⎟⎟ = 0.419 V ⎠

⎛ 10 −3 ⎞ V D = (0.026) ln⎜⎜ −11 ⎟⎟ = 0.479 V ⎝ 10 ⎠ ⎡ ⎛ V ⎞ ⎤ (ii) − 5 × 10 −12 = 10 −11 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = −0.018 V ⎣ ⎝ 0.026 ⎠ ⎦ ⎛ 10 × 10 −6 (b) (i) V D = (0.026 ) ln⎜⎜ −13 ⎝ 10

⎞ ⎟⎟ = 0.479 V ⎠

⎛ 100 × 10 −6 V D = (0.026 ) ln⎜⎜ −13 ⎝ 10 ⎛ 10 −3 V D = (0.026) ln⎜⎜ −13 ⎝ 10

⎞ ⎟⎟ = 0.539 V ⎠

⎞ ⎟⎟ = 0.599 V ⎠

⎡ ⎛ V ⎞ ⎤ (ii) − 10 −14 = 10 −13 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = −0.00274 V ⎣ ⎝ 0.026 ⎠ ⎦ ______________________________________________________________________________________

1.29

(a)

(b)

VD

⎛ 0.7 ⎞ 10−3 = I S exp ⎜ ⎟ ⎝ 0.026 ⎠ I S = 2.03 × 10 −15 A I D ( A ) ( n = 1)

I D ( A )( n = 2 )

0.1 9.50 ×10 1.39 ×10 −14 0.2 4.45 ×10 −12 9.50 ×10 −14 0.3 2.08 ×10 −10 6.50 ×10 −13 − 9 0.4 9.75 ×10 4.45 ×10 −12 0.5 4.56 ×10 −7 3.04 ×10 −11 − 5 0.6 2.14 ×10 2.08 ×10 −10 0.7 10 −3 1.42 ×10 −9 ______________________________________________________________________________________ −14

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.30 (a) I S = 10 −12 A

VD(v) 0.10 0.20 0.30 0.40 0.50 0.60 0.70 (b)

ID(A) 4.68 ×10−11 2.19 ×10−9 1.03 ×10−7 4.80 ×10−6 2.25 ×10−4 1.05 ×10−2 4.93 ×10−1

log10ID −10.3 −8.66 −6.99 −5.32 −3.65 −1.98 −0.307

I S = 10 −14 A

VD(v) ID(A) log10ID −13 −12.3 0.10 4.68 ×10 − 11 −10.66 0.20 2.19 ×10 − 9 −8.99 0.30 1.03 ×10 −8 −7.32 0.40 4.80 ×10 − 6 −5.65 0.50 2.25 ×10 −4 −3.98 0.60 1.05 ×10 − 3 −2.31 0.70 4.93 ×10 ______________________________________________________________________________________ 1.31 a. ⎛ V − VD1 ⎞ ID2 = 10 = exp ⎜ D 2 ⎟ I D1 ⎝ VT ⎠ ΔVD = VT ln (10) ⇒ ΔVD = 59.9 mV ≈ 60 mV

b.

ΔVD = VT ln (100 ) ⇒ ΔVD = 119.7 mV ≈ 120 mV

______________________________________________________________________________________ 1.32

⎛ 2 ⎞ (a) (i) V D = (0.026) ln⎜ ⎟ = 0.539 V −9 ⎝ 2 × 10 ⎠ ⎛ 20 ⎞ (ii) V D = (0.026) ln⎜ ⎟ = 0.599 V −9 ⎝ 2 × 10 ⎠ ⎛ 0.4 ⎞ (b) (i) I D = 2 × 10 −9 exp⎜ ⎟ ⇒ 9.60 mA ⎝ 0.026 ⎠ ⎛ 0.65 ⎞ (ii) I D = 2 × 10 −9 exp⎜ ⎟ ⇒ 144 A ⎝ 0.026 ⎠ ______________________________________________________________________________________

(

)

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.33 ⎛I ⎞ ⎛ 2 × 10−3 ⎞ = 0.6347 V VD = Vt ...