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

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 10

15

5

-cm > 10 -cm, and silicon dioxide is an insulator.

2.3

2.4
1.8 2 cm
 2.82 x10 6  cm

 144 
4
4
A
5x10 cm 1x10 cm
L

a R  

L

b R  

 2.82x10  cm
6

A

1.8 2 cm

5x10

4

 287 
4
cm0.5x10 cm

2.5

a R  L  1.66x106  cm

1.8 2 cm

A 
b R  L  1.66 x106  cm

5x10 



5x10

A

4

4

 94.5 

cm1x10 4 cm
1.8 2 cm
169 
cm0.5x10 4 cm

2.6

2-1
©R. C. Jaeger & T. N. Blalock 3/23/15


2

= BT 3

ni

1010 

æ -E ö
÷
è kT ø

expç

31
B=1.08x10

G

1.12 ö÷
è 8.62x10 T ø
Using a spreadsheet, solver, or MATLAB yields T=305.23K
2

31

3

æ

=1.08x10 T expç-

5

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

1.69 x 1013/cm3.

6.73 x109/cm3

For germanium, B = 2.31 x 1030 and EG = 0.66 eV:
ni = 2.63 x10-4/cm3

2.93 x 1015/cm3.

2.27 x1013/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));
15

3

ni = 10 /cm for T = 602 K
15

b N

15
=10

3
cm

, n =10

D

p=

n

n

15

cm

3

10

15 2

+



:n =

10

i

10

2

i

2

30

=

15

=6.18x10



15
+ 4 10




2
15
=1.62 x10

/ cm 3

2
14

/ cm

3

1.62 x10

c At room temperature, ND >> ni2.
\n = ND =1015 electrons / cm3 and p = ni2 = 1020 =105 holes / cm3
n 1015
2.9

6

3

T = 300 K and EG = 1.42 eV: ni = 2.21 x10 /cm

2-2
©R. C. Jaeger & T. N. Blalock 3/23/15

T = 100 K: ni = 6.03 x 10

-19

3

/cm

10

3

10

3

T = 450 K: ni = 3.82 x10 /cm
6

3

T = 300 K and EG = 1.42 eV: ni = 2.21 x10 /cm
T = 100 K: ni = 6.03 x 10

-19

3

/cm

T = 450 K: ni = 3.82 x10 /cm

2-3
©R. C. Jaeger & T. N. Blalock 3/23/15


2.10

cm 2 öæ
Vö
æ
÷ç-2000 ÷ = +1.40x10
v = - E = ç-700 V
n

è

n

- s øè

v = + E = ç+250 V
è

p

Vö
cm
÷ = -5.00 x10 5 s

öæ

֍-2000
- s øè

cm ø

p

n

j

s

cm ø

cm 2

æ

cm
6

n

=

= -qnv



æ

19
-1.60 x10

1

17

ç

C

3 öæ
֍

10

6

cm ö

A2

4

÷

= 2.24x10

1.40 x10

ø
cm è

è

s ø
cm
1 öæ
cm ö
A
j p = qnv p = 1.60 x1019 C æ10 3 3 ÷ç -5.00 x105
÷ = -8.00 x1011
è cm øè
s ø
cm


2.11
j

n

n
= qnv

=

2

ç



æ

19

ç

C

1.60x10

è
j p = qnv p = 1.60x1019 C


n

6

I = j · Area =1.60x10

æ
ç

è
A
cm

1 öæ

cm ö =1.60x10

A 2 =1.60 MA
cm øè
s ø
cm
cm
1 öæ cm ö
A
pA
= 160
102
107
÷ =1.60x1010
2
2
cm øè
s ø
cm
cm


2

4
4
10 cm 25x10 cm = 400 mA
18

3

7

֍

6

÷

10

10

2

3 ֍

2.12

2.13
2
v = j = 2500A / cm = 2.5x10 5 cm
2
Q 0.01C / cm

s

2.14
æ

vn = - E = - 1000

cm 2 öæ

ç
è

n

æ

v p = + E = +ç 400
è

V ö

֍ -1500
V - s øè

cm

÷ = +1.50 x106

s

cm ø

cm 2 öæ
V ö
cm
5
÷ = -6.00x10 s
V - s ֍-1500
øè

cm ø

p

jn = -qnv n = -1.60 x1019 C
jp = qnv

p =



1.60 x10

æ
ç

1
103

è

æ

19
C

ç

17
10

è

öæ
3

֍ +1.50 x106

cm øè
1 3 öæ

s
5 cm ö

-6.00x10

֍

cm øè

cm ö

÷

s

ø

÷

A

= -2.40 x10 10

cm

ø
3

A

= -9.60x10

2

2

cm

2-4

©R. C. Jaeger & T. N. Blalock 3/23/15


2.15
a  E =

5V

=10, 000

4

5x10 cm

b

V

 V =

ç
10

è

cm

V ö 
5x10

æ

÷

5

4 cm

= 50 V

cm ø

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

1

for an insulator
10

5

W- cm

1

10

£
= 2.497x10
2
= q n+  p 
cm
cm 3
ö
19 
æ

1.602x10 C 1800+ 700 è v - sec ø
20
æ -E ö
G
5.152
x10
with
2
3
ni =
=
BT
kT
ø
è
6
cm

-5
B =1.08x1031 K 3cm6, k =8.62x10 eV/K and E G =1.12eV
Using MATLAB as in Problem 2.6 yields T ≤ 316.6 K.
ç

÷

÷

expç

2.17

p i 
i  n
 n i
+ p 
For intrinsic silicon,  = q  n +  n = qn 

 ³1000 W- cm1 for a conductor
n=
i

n2
i

=



q n + p 
1.203x10 39

³

1000 W- cm



1

 cm
1.602x10 C 120 + 60 v - sec
æ E ö
= BT 3 expç - ÷ with
19

2

= 3.468x1019
cm3

G

cm

6

è

kT ø

G =1.12eV
B =1.08x1031 K 3cm6, k = 8.62x10-5eV/K and E
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.

2-5
©R. C. Jaeger & T. N. Blalock 3/23/15


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

j
E = 

A ö

= j = ç 5000 cm
è

÷
2

ø

V
0.02W- cm=100 cm
, a small electric field

2.23
16

æ10

N= ç

è

atoms ö
3

cm

÷0.180m2m0.5mç

ø

æ104 cm ö3

÷ =1800 atoms

è m ø

2-6
©R. C. Jaeger & T. N. Blalock 3/23/15


2.24
18
3
(a) Since boron is an acceptor, NA = 7 x 10 /cm . Assume ND = 0, since it is not specified.
The material is p-type.
10
3
18
3
(b) At room temperature, ni =10 / cm and NA - ND = 7x10 / cm >> 2ni
2

20

6

n
10 / cm
3
So p = 7x10 / cm and n =
=
18
3 =14.3 / cm
p 7x10 / cm
æ
ö
1.12
(c) At 200K, n i2 =1.08x10 31 2003 expç-ç
÷÷ = 5.28x10 9 / cm 6
5
18

3

i

200ø

è 8.62x10

9

n = 7.27x10 / cm
4

3

i

N A - N D >> 2n , so p = 7x10

18

i

/ cm 3 and n = 5.28x10
= 7.54x10 10 / cm 3
18
7x10

2.25
17
3
(a) Since arsenic is a donor, ND = 3 x 10 /cm . Assume NA = 0, since it is not specified. The
material is n-type.

2.26
18
3
18
3
(a) Arsenic is a donor, and boron is an acceptor. ND = 3 x 10 /cm , and NA = 8 x 10 /cm .
Since NA > ND, the material is p-type.
10

18
3
3
(b) At room temperature, n i =10 / cm and N A - N D = 5x10 / cm >> 2n i
n 2 10 20 / cm 6
18
3
3
=
So p = 5x10 / cm and n =
18
3 = 20.0 / cm
p 5x10 / cm
i

2.27
17
3
(a) Phosphorus is a donor, and boron is an acceptor. ND = 2 x 10 /cm , and NA = 6 x
17
3
10 /cm . Since NA > ND, the material is p-type.
10
17
3
/ cm 3 and N - N
>> 2n
(b) At room temperature, n =10
= 4x10 / cm
A

i
17

So p = 4x10

3

/ cm and n =

n2
i

p

=

10 20 / cm

D

i

6

17

3

= 250 / cm

3

4x10 / cm

2-7
©R. C. Jaeger & T. N. Blalock 3/23/15


2.28

2.29

2.30
16
3
ND = 5 x 10 /cm . Assume NA = 0, since it is not specified.
16
N D  NA : The material is
D
A
n 2 10 20
16
3
3
3
n=

16 2x10 / cm
5x10 / cm | p  n

5x10
n-ty pe.

|

N

N

 5x10

/ cm

 2n  2x10

/ cm

10

3

3

i

i

N D  N A 5x10

16

/ cm

| Using the equations in Fig. 2.8, n 885

3

1
1

 
öæ
cm2
q n
 æ

ç885
19
n
1.602x10 C è V  s ÷ç

16

5x10

øè cm

3

ö  0.141  cm

÷
ø

2.31
18
3
NA = 2.5x10 /cm . Assume ND = 0, since it is not specified.
18
N  N : The material is p-type. | N  N
/ cm
 2.5x10
A

D

A

p=2.5x10 18 / cm
N D  NA

 2.5x10

18

/ cm

1020

18  40 / cm
p 2.5x10

1

p

1.602x10

æ
19

cm 2 öæ 2.5x10

C ç58.7
֍
3
V  s øè cm
è

18

10

/ cm3

3

| Using the equations from Fig. 2.8, n 187

1

 q p 

 2n  2x10

i

i

3

3

D

n2

| n=

3

cm 2
cm 2
and p 198
V s
Vs

cm 2
cm 2
and p  58.7
Vs

Vs

ö  42.5 m cm
÷
ø

2-8

©R. C. Jaeger & T. N. Blalock 3/23/15


2.32
19
3
Indium is from column 3 and is an acceptor. NA = 8 x 10 /cm . Assume ND = 0, since it is not
specified.
N A  ND : material is p-type | N A  ND  8x1019 / cm3  2ni  2x1010 / cm3
n 2 10 20
3
p  8x10 19 / cm 3 | n = i 
19 1.25 / cm
p 8x10
cm 2
cm 2
19
3
N D  N A  7x10 / cm | Using Fig. 2.8, n  66.2
and p  46.1
Vs
Vs

1
1
1.69 m cm

   q p
æ
cm 2 öæ8x10 19 ö
p
1.602x10 19 C ç46.1
֍
÷
3
V  s øè cm ø
è
2.33
16
3
16
3
Phosphorus is a donor: N D  4.5x10 / cm | Boron is an acceptor: N A  5.5x10
/ cm
N  N : The material is p-type. | N  N 10 16 / cm 3  2n  2 x10 10 / cm 3
A

D

A

16

n

3

p 10 / cm | n 
ND  NA



1

q n
n

2

D

i

20

10
4
3
 16 10 / cm
p 10
i

cm 2

cm 2
and p 153
10 / cm | Using Fig. 2.8,  n  727
V s
Vs
1
ö  4.08  cm

cm öæ 10
æ
2
1.602x10 19 C 153
֍
÷
3
V  s øè cm ø
è
17

3

16

ç

2.34

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

NA



p

p p

1018

70.8

7.08 x 1019

2 x1018

61.0

1.22 x 1020

1.90 x 1018

61.6

1.17 x 1020

1.89 x 1018

61.6

1.16 x 1020

2-9
©R. C. Jaeger & T. N. Blalock 3/23/15


2.35

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

NA

p p

1016

318

3.18 x 1018

4 x 1016

214

8.55 x 1018

7.5 x 1016

170

1.28 x 1019

7.2 x 1016

173

1.25 x 1019

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:
ND

n

n n

1015

1360

1.36 x 1018

1.5 x 1015

1340

2.01 x 1018

1.6 x 1015

1340

2.14 x 1018

1.55 x 1015

1340

2.08 x 1018

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 ND  N A =0, n=p=ni, and   q n ni  pni   qni n  p 

NA  ND 1020 / cm3 which yields p  45.9 and n  64.3 using the
equations from Fig. 2.8.
1
1
6
2
æ cm ö  5.66x10 -cm
 qni n  p £
19
10

3
cm
÷
è v  sec ø
1.602x10

64.3  45.9ç

C10

2-10
©R. C. Jaeger & T. N. Blalock 3/23/15


2.39 (a)

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

n

n n

1019

108

1.08 x 1021

7 x 1019

67.5

4.73 x 1021

1 x 1021

64.3

6.43 x 1021

9.67 x 1019

64.5

6.24 x 1021

(b)

An iterative solution is required using the equations in Fig. 2.8 and trial and error:
NA
1 x1020

p
45.9

p p
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 1020

45.7

6.26 x 1021

2 - 11
©R. C. Jaeger & T. N. Blalock 3/23/15


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:
NA

p

p p

1 Ohm-cm

2.51x1016

249

6.25 x 1018

0.25 Ohm-cm

2.2 x 1017

147

2.5 x 1018

17

Additional acceptor concentration = 2.2 x 10 - 2.5 x 10
16
3
10 /cm (b) If donors are added:

16

= 1.95 x

ND

ND + NA

n

ND - NA

n n

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

765

3.5 x 1016

2.68 x 1019

772

3.24 x 1016

2.50 x 1019

8.5 x 1016

6 x 1016
5.74 x 1016

8.24 x 1016

16

3

So ND = 5.7 x 10 /cm must be added to change achieve a resistivity of 0.25 ohm-cm. The
silicon is converted to n-type material.

2-12
©R. C. Jaeger & T. N. Blalock 3/23/15

2.41

16

3

2

/cm and p = 318 cm /V-s from equations with Fig. 2.8.
0.509


  qp p  qp NA  1.602x10 19 C3181016 
Boron is an acceptor: NA = 10

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

  q n

n

n

D

:

4.5 cm

|  n   N  N

n

 cm

-1

1


A

2.81x10


1.602x10 19 C V  cm  s
19

Using trial and error:
ND

ND + NA

n

ND - NA
7x

n n

8 x 1016

9 x 1016

752

1016
4x

5.26 x 1019

5 x 1016

6 x 1016

845

1016
3x

3.38 x 1019

4 x 1016

5 x 1016

885

1016
3.2 x

2.66 x 1019

4.2 x 1016

5.2 x 1016

877

1016

2.81 x 1019

2.42
Phosphorus is a donor: N

D

= 10

  q n  q N  1.602x10
n

n

19



D

16

3
2
/cm and  = 1180 cm /V-s from Fig. 2.8.
n

16

C 1180 10



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

-1

 1.89
  cm

:
1

3.12x1019
5 cm
  qp p |  p   N  N 

19
p

A
D
p
1.602x10 C V  cm  s
Using trial and error:
NA

ND + NA
1.10E+17
2.10E+17
3.60E+17
3.40E+17

1.00E+17
2.00E+17
3.50E+17
3.30E+17



NA - ND
9.00E+16
1.90E+17
3.40E+17
3.20E+17

p

147
116

95.6
97.4

p p
1.33E+19
2.20E+19
3.25E+19
3.12E+19

2.43

T (K)

50

75

100

150

200

250

300

350

400

VT (mV)

4.31

6.46

8.61

12.9

17.2

21.5

25.8

30.1

34.5


2 - 13
©R. C. Jaeger & T. N. Blalock 3/23/15


2.44
æ

j = -qD

ç

-

dn ö

÷=

è

n

dx ø



j = 1.602x10

19

dn
qV  dx
T

n

æ



cm öæ

ç

C 0.025V

è

0 -10

2 ֍

350

ö 1

V - s øè 0.25x10

4

kA

÷

18

- 0 ø cm

4

= -56.1
cm

2

2.45

2.46

2.47
At x = 0:
drift

jn

= qnnE = 1.60 x10

19

16 öæ
V ö
A
֍10 3 ֍+25
÷ =14.0
2
cm
V - s øè cm øè
cm ø

æ

Cç350

cm

è

j

drift

p

= q pE = 1.60 x10
dn
= qDn

diff

jp



p

diff

jn

19



= 1.60 x10

dx
dp

= -qDp

dx

= -1.60 x10

æ

C

2

öæ

ç

19

öæ

ö

֍

÷

2
18
150 cm öæ1.01x10
+25 V = +606 A
è V - s øè cm3 øè cm ø
cm2
cm2 öæ 104 -1016 ö
A
æ
֍

Cç350 × 0.025
֍
4
4 ÷ = -70.0
2
s øè 2 x10 cm ø
cm
è
ö
cm2 öæ1018 -1.01x1018
æ
19
Cç150 × 0.025

è

֍
s øè

4

2x10 cm

4

A

÷ = 30.0
ø

2

cm

A

jT =14.0 + 607 - 70.0 + 30.0 = +580 cm

2

2-14
©R. C. Jaeger & T. N. Blalock 3/23/15

At x=1 m assuming linear distributions:




18

p 1m =1.005x10 / cm
drift

jn

= qnnE = 1.60 x10



3



æ

19

diff
n



= qD dx = 1.60 x10

j

dp

A

Cç350
֍
÷ = +7.00
3 ֍+25
2
cm A
è
V - s øè cm øè öæ cm ø V ö



n

ö

V

öæ1.005x10

cm

ç

è

ç

æ

19

15

2

j drift = qp pE = 1.60 x10 19 C  150
dn

3

öæ 5x10

cm

æ

p

15

, n 1m = 5x10 / cm öæ

2

18

֍
V - s øè

3

֍ +25

cm 10 øè
öæ

cm

2

4

֍

4

-10

÷ = +603 2
cm
cm ø
ö

16

p

2.48

2

C è 350 × 0.025 s øè 2 x10 cm ø = -70.0 cm
æ

öæ10

cm
2

ç

ö

-1.01x10
18

= -1.60 x10 19 C 150 × 0.025
֍
4
4
dx
è
s øè 2x10 cm

j + 7.00 + 603 - 70.0 + 30.0 = -570 A
T
cm 2
j diff = -qD p

A

4÷

18

A
÷ = 30.0 2
cm
ø

NA = 2ND

2.49

2.50

2 - 15
©R. C. Jaeger & T. N. Blalock 3/23/15


2.51
+
An n-type ion implantation step could be used to form the n region following step (f) in Fig. 2.17.
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
©R. C. Jaeger & T. N. Blalock 3/23/15