Download PDF solution manual for microelectronic circuit design 5th edition by jaege

2.19 Since Ge is also from column IV, acceptors come from column III and donors come from column V.. Thus silicon has an extra electron and will act as a donor impurity.. Thus silicon i

CHAPTER 2 Solution Manual for Microelectronic Circuit

Design 5th Edition by Jaege

2.1

Based upon Table 2.1, a resistivity of 2.82 -cm < 1 m-cm, and aluminum is a conductor

2.2

Based upon Table 2.1, a resistivity of 1015-cm > 105-cm, and silicon dioxide is an insulator

2.3

2.4

a R L 2.82 x10 6  cm 1.8 2 cm



5x10 cm 1x10 cm

b R L 2.82x10 6  cm 1.8 2 cm

 287 

5x10 0.5x10 cm

2.5

a R L  1.66x106 cm



1.8 2 cm  94.5 

5x10 4 cm1x10 4 cm

b R  L  1.66 x106 cm



1.8 2 cm 169 

5x10 4 cm0.5x10 4 cm

2.6

2 - 1

©R C Jaeger & T N Blalock 3/23/15

2 = BT 3 æ - E ö 31

n i expç G ÷ B=1.08x10

8.62x10

 5

Using a spreadsheet, solver, or MATLAB yields T=305.23K

Define an M-File:

function f=temp(T)

f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T));

Then: fzero('temp',300) | ans = 305.226 K

2.7

For silicon, B = 1.08 x 1031 and EG = 1.12 eV:

ni = 5.07 x10-19/cm3 6.73 x109/cm3 1.69 x 1013/cm3

For germanium, B = 2.31 x 1030 and EG = 0.66 eV:

ni = 2.63 x10-4/cm3 2.27 x1013/cm3 2.93 x 1015/cm3

2.8

(a) Define an M-File:

function f=temp(T)

ni=1E15;

f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T));

ni = 1015/cm3 for T = 602 K

15 + 15

D

2

p =

n 2

=

10 30

=

/ cm

i

c At room temperature, N D >> n i2.

\n = ND =1015 electrons / cm3 and p = ni2 = 1020 =105 holes / cm3

n 1015

2.9

T = 300 K and EG = 1.42 eV: ni = 2.21 x106/cm3

2-2

T = 100 K: ni = 6.03 x 10-19/cm3 T = 450 K: ni = 3.82 x1010/cm3

T = 300 K and EG = 1.42 eV: ni = 2.21 x106/cm3

T = 100 K: ni = 6.03 x 10-19/cm3 T = 450 K: ni = 3.82 x1010/cm3

2 - 3

2.10

æ cm öæ

V ö cm

= - E = ç-700

÷ç-2000 ÷ = +1.40x10 6

n n è - s øè cm ø

æ cm 2 öæ V ö cm

= + E = ç+250

÷ç-2000 ÷ = -5.00 x10 5

p p

è - s øè cm ø

n n   19 æ 17 1 3 öæ 6 cm ö 4 A

2

j = -1.60 x10 C  ç ÷ç ÷ = 2.24x10

= -qnv 10

cm

1.40 x10

è

ø

j = qnv = 1.60 x10 19 C æ 3

1 öæ

-5.00 x105 cm ö

= -8.00 x10 11 A

cm

s

cm

è øè ø

æ

öæ

7 cmö

j n n   19 C ç 18 1 3 ÷ç ÷ =1.60x10 6 A 2 =1.60 MA 2

= qnv = 1.60x10 10 10

è cm øè

j = qnv = 1.60x10 19 C æ 1 öæ cm ö

=1.60x10 10 A

= 160 pA

cm

3 ֍

s

è øè ø

n

6 A 2  4  4 

I = j · Area =1.60x10

10 cm 25x10 cm = 400 mA

2.12

2.13

v = j = 2500A / cm 2

= 2.5x10 5 cm

2

2.14

v n

v p

j n

j p

=

=

=

=

- E = - 1000 ÷ç -1500 ÷ = +1.50 x106

n ç

s

æ cm 2 öæ V ö cm

+ E = +ç ÷ç-1500 ÷ = -6.00x10 5

400

p

-qnv = -1.60 x10 19 C æ 1 öæ +1.50 x106 cm ö

= -2.40 x10  10

A

p   19 æ 17 1 3 öæ 5 cmö 3 A

2

qnv = 1.60 x10 C  ç 10 ÷ç -6.00x10 ÷ = -9.60x10

2.15

æ ö

 E = 5V =10, 000 V  V = 5 V  4 = 50 V a b ç ÷ 5x10 cm  4 10 5x10 cm cm è cm ø

2.16 For intrinsic  £10  5W-n i = q n+  silicon,  = q  n i p i  i  n +  p 

n +  n = qn

cm 1 for an insulator

10 5 W- cm1 10

p £

æ

ö = 2.497x10 

  cm 2 cm 3  19

C + 700  ç ÷ 1.602x10 1800 è v - sec ø

= 5.152 x10 20 n 2 = BT 3 i cm 6

B =1.08x1031 K  3cm 6, æ - E ö expç G ÷ è kT ø k =8.62x10 with -5 eV/K and E G =1.12eV Using MATLAB as in Problem 2.6 yields T ≤ 316.6 K 2.17  n i p i  i  n For intrinsic silicon,  = q  n +  n = qn   ³1000 W- cm 1 for a conductor 

1000 W- cm  1 n = ³

q n + p

2 i

 19  + 60  cm

1.602x10 C 120 v - sec 2 = 1.203x10 39 = BT 3 æ - E ö

B =1.08x1031 K  3cm 6, k = 8.62x10-5eV/K and E

+ p 

= 3.468x1019

cm3

G =1.12eV

This is a transcendental equation and must be solved numerically by iteration Using the HP solver routine or a spread sheet yields T ≥ 2579.3 K Note that this temperature is far above the melting temperature of silicon

2.18

No free electrons or holes (except those corresponding to ni)

2.19

Since Ge is also from column IV, acceptors come from column III and donors come from

column V (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi

2.20

(a) Gallium is from column 3 and silicon is from column 4 Thus silicon has an extra electron and will act as a donor impurity

(b) Arsenic is from column 5 and silicon is from column 4 Thus silicon is deficient in one electron and will act as an acceptor impurity

2.21

(a) Germanium is from column IV and indium is from column III Thus germanium has one extra electron and will act as a donor impurity

(b) Germanium is from column IV and phosphorus is from column V Thus germanium has one less electron and will act as an acceptor impurity

2.22

æ

ö

2.23

N = ç ÷  0.180 m 2 m 0.5 m ç ÷ =1800 atoms

cm3

2-6

2.24

(a) Since boron is an acceptor, NA = 7 x 1018/cm3 Assume ND = 0, since it is not specified The material is p-type

(b) At room temperature, n i =1010 / cm3 and N A - N D = 7x1018 / cm3 >> 2ni

18 3 n2 1020 / cm6 3

So p = 7x10 / cm and n = i

÷ = 5.28x10 9 / cm 6

8.62x10

n = 7.27x10

- N >>

2.25

(a) Since arsenic is a donor, ND = 3 x 1017/cm3 Assume NA = 0, since it is not specified The material is n-type

2.26

(a) Arsenic is a donor, and boron is an acceptor ND = 3 x 1018/cm3, and NA = 8 x 1018/cm3 Since NA > ND, the material is p-type

(b) At room temperature, n i =10 10 / cm 3 and N A - N D = 5x10 18 / cm 3 >> 2n i

= n

2

= 10

20 / cm 6

So p = 5x10

p

18

/ cm

2.27

(a) Phosphorus is a donor, and boron is an acceptor ND = 2 x 1017/cm3, and NA = 6 x

1017/cm3 Since NA > ND, the material is p-type

10 / cm 3 and N - N 17 / cm 3 >> 2n

(b) At room temperature, n =10

A D

= 4x10

i

and n = n

2

= 10

20 / cm 6

= 250 / cm 3

/ cm

i

So p = 4x10

p

17

/ cm

2 - 7

2.28

2.29

2.30

N D  N A : The material is

3

 2x10 / cm

3

n =

N D

| p 

n 2

/ cm

i

/ cm 3 |

A 5x10

1 

n

1.602x10 C

5x10

Using the

1

æ cm2



ç885

è V  s

2x10 3

/ cm 3

equations in

16 ö 

øè cm ø

Fig 2.8, n 

0.141  cm

885 cm

V 

2

s and  198 cm

2

p

V  s

2.31

N  N : The material is p-type | N  N 18 / cm 3 10

 2.5x10  2n  2x10

 10 20  40 / cm 3

p

2.5x10

Using the equations from Fig 2.8,  187 cm

2

V  s

q1 p  æ 1 cm 2 öæ 2.5x10 18 ö  42.5 m cm

p 1.602x10 C ç58.7 ÷ç ÷

è V  s øè cm

ø

/ cm3

and   58.7 cm

2

p

V  s

2-8

2.32

Indium is from column 3 and is an acceptor NA = 8 x 1019/cm3 Assume ND = 0, since it is not specified

N A

p 

N D

 N D : material is p-type | N A N D  8x1019 / cm3 2n i  2x1010 / cm3

n 2

10 20 1.25 / cm 3

p

19 8x10

Using Fig 2.8,   66.2 cm

2 and   46.1 cm

2

1 

æ

1

öæ

ö 1.69 m cm

cm 8x10

q p

3

è V  s øè cm ø

2.33

Phosphorus is a donor: N  16

/ cm 3 | Boron is an acceptor: N 16 / cm 3

N  N : The material is p-type | N  N 16 / cm 3  2n  10 / cm 3

A D

i

p  16 3

n n

2

20

10

p

10

17 3

Using Fig 2.8,   727 cm

2 and  153 cm

 1  æ

1

öæ

ö  4.08  cm

q n cm

ç

3

2.34

An iterative solution is required Using the equations from Fig 2.8 and trial and error:

2 x1018 61.0 1.22 x 1020

1.90 x 1018 61.6 1.17 x 1020

1.89 x 10 18 61.6 1.16 x 10 20

2 - 9

2.35

An iterative solution is required Using the equations in Fig 2.8 and trial and error:

4 x 1016 214 8.55 x 1018

7.5 x 1016 170 1.28 x 1019

7.2 x 10 16 173 1.25 x 10 19

2.36

Yes, by adding equal amounts of donor and acceptor impurities the mobilities are reduced, but the hole and electron concentrations remain unchanged See Problem 2.39 for example However, it is physically impossible to add exactly equal amounts of the two impurities

2.37

An iterative solution is required Using the equations in Fig 2.8 and trial and error:

1.5 x 1015 1340 2.01 x 1018

1.6 x 1015 1340 2.14 x 1018

1.55 x 10 15 1340 2.08 x 10 18

2.38

Based upon the value of its resistivity, the material is an insulator However, it is not intrinsic because it contains impurities The addition of the impurities has increased the resistivity

Since N D N A =0, n=p=n i, and  q n n i p n i qn i n p 

N A N D 1020 / cm3 which yields p 45.9 and n  64.3 using the

ö

1.602x10 C 10 cm  64.3  45.9  ç ÷

2-10

2.39 (a)

An iterative solution is required Using the equations in Fig 2.8 and trial and error:

7 x 1019 67.5 4.73 x 1021

1 x 1021 64.3 6.43 x 1021

9.67 x 10 19 64.5 6.24 x 10 21

(b)

An iterative solution is required using the equations in Fig 2.8 and trial and error:

1 x1020 45.9 4.96 x 1021

1.2 x1020 45.8 5.93 x 1021

1.4 x1020 45.7 6.17 x 1021

1.37 x 10 20 45.7 6.26 x 10 21

2 - 11

2.40

(a) For the 1 ohm-cm starting material:

To change the resistivity to 0.25 ohm-cm:

Iterative solutions are required using the equations with Fig 2.8 aand trial and error:

1 Ohm-cm 2.51x1016 249 6.25 x 1018

0.25 Ohm-cm 2.2 x 1017 147 2.5 x 1018

3 x 1016 5.5 x 1016 864 0.5 x 1015 4.32 x 1018

5 x 1016 7.5 x 1016 794 2.5 x 1016 1.98 x 1019

6 x 1016 8.5 x 1016 765 3.5 x 1016 2.68 x 1019

5.74 x 10 16 8.24 x 10 16 772 3.24 x 10 16 2.50 x 10 19

So ND = 5.7 x 1016/cm3 must be added to change achieve a resistivity of 0.25 ohm-cm The silicon is converted to n-type material

2-12

2.41

/V-s from equations with Fig 2.8

  qp p  qp NA 1.602x1019 C31810 160.509



Now we add donors until  = 4.5 (  -cm) -1:

 q n  N  N 

n |

n n 1.602x10 C V  cm  s

Using trial and error:

ND ND + NA n ND - NA n n

8 x 1016 9 x 1016 752

7 x

1016 5.26 x 1019

5 x 1016 6 x 1016 845

4 x

1016 3.38 x 1019

4 x 1016 5 x 1016 885

3 x

1016 2.66 x 1019

4.2 x 10 16 5.2 x 10 16 877

3.2 x

10 16 2.81 x 10 19

2.42

Phosphorus is a donor: N

D = 10

16

n

 qn  qN  1.602x1019 C 1180 1016

 1.89

n n D   

Now we add acceptors until  = 5.0 (  -cm) -1:

 qp p | N

 5  cm 1

p   N

3.12x1019

Using trial and error:

3.30E+17 3.40E+17 97.4 3.20E+17 3.12E+19

2.43

2 - 13

©R C Jaeger & T N Blalock 3/23/15

2.44

æ dn ö

j = -qD ç - ÷ = qV  dn dx

n è dx ø

T n

è V - s øè 0.25x10 4 - 0 ø cm 4 cm 2

2.45

2.46

At x = 0:

æ

öæ

16 öæ

cm 2 V ö A

j n drift = qn nE = 1.60 x1019 Cç350 ÷ç 10

÷ç+25 ÷ =14.0

cm

2

è V - s øè cm øè cm ø

jdrift = q pE = 1.60 x1019 C

æ

150 cm2 öæ 1.01x1018 öæ

+25 V ö

= +606 A

cm3

cm2

diff dn 1.60 x10 19 Cæ cm2 öæ 104 -1016 ö A

j n = qD n = ç350 × 0.025

֍

÷ = -70.0

dx

s

 4

cm

4

cm

2

è øè 2 x10 ø

diff dp  19 æ cm2 öæ1018 -1.01x1018ö

A

j p = -qD p = -1.60 x10 Cç150 × 0.025 ÷ç

÷ = 30.0

dx

s

2x10

 4

cm

4

cm

2

è øè ø

j T =14.0 + 607 - 70.0 + 30.0 = +580 cm A2

2-14

At x=1 m assuming linear distributions:

 

18 3  

15 3

, n 1m

/ cm

p 1m =1.005x10 / cm = 5x10 öæ ö

æ cm 2 öæ 5x10 V A drift = qn nE = 1.60 x10 19 Cç350 15

֍+25 = +7.00

j n ÷ç ÷

cm 3 cm 2 è æ V - s øè cm øè öæ cm ø V ö A 2 öæ 1.005x10

drift = qpE = 1.60 x10  19 C  18

+25 = +603

j p p  ç 150 ÷ç 3 ÷ç ÷

cm

cm dn è V - s øè 10 øè cm ø

æ cm 2 öæ 4 -10 ö A diff

= 

 19 

350 × 0.025 16

= -70.0 j n

C ç ÷ç  4 4 ÷ 2 n

= qD dx 1.60 x10 è s øè 2 x10 cm ø cm æ

ö dp cm 2 öæ10 -1.01x10 18 diff = -qD = -1.60 x10  19 150 × 0.025

18

= 30.0 j p C ÷ç  4

4 ÷

p

dx  ç

s 2x10 cm è øè ø

j + 7.00 + 603 - 70.0 + 30.0 = -570 A

2

T

cm

2

A

cm 2

2.48 NA = 2ND

2.49

2.50

2 - 15

2.51

A mask would be used to cover up the opening over the p-type region and leave the opening over the n-type silicon The masking layer for the implantation could just be photoresist

2.52

2-16