An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
Chapter 9
Problem Solutions
9.3
We have
9.1
In the forward bias
eV
I f ≈ I S exp
kT
Then
F I
H K
eV I
F
exp
I
H kT K = expL e aV − V fO
I
=
⋅
PQ
I
I
F eV I MN kT
exp
H kT K
or
F kT I lnFG I IJ
V −V =
H e K HI K
Ip
=
I
1
f1
f2
Ip
I
For
If2
τ nO
1
Dp
Nd
τ pO
+
2
12.4
16
10
1
35
10
10
1
+
−6
−7
10
12.4
16
10
−7
or
= 10 ⇒ V1 − V2 = 59.6 mV ≈ 60 mV
Ip
=
I
(b)
If1
Na
Na
(a)
If2
Dn
=
f2
For
1
1
2
If1
τ pO
So
f1
1
Nd
2
2
2
S
Dp
Let Dn = 35 cm / s and D p = 12.4 cm / s
S
1
1
111
. x10
5.916
−12
. x10
+ 111
−12
Na
= 100 ⇒ V1 − V2 = 119.3mV ≈ 120 mV
Then
_______________________________________
b g
N a cm
9.2
LM F I − 1OP
N H kT K Q
or we can write this as
I
F eV I
+ 1 = exp
H kT K
I
so that
F kT I lnFG I + 1IJ
V =
H e K HI K
I = I S exp
eV
−3
Ip I
10
15
0.995
10
16
0.9995
10
17
≈ 1.0
10
18
≈ 1.0
_______________________________________
S
9.4
The cross-sectional area is
A=
S
I
=
10 x10
J ≈ J S exp
V = (0.0259) ln(1 − 0.90) ⇒
−4
= 5 x10 cm
2
20
J
We have
In reverse bias, I is negative, so at
I
= −0.90 , we have
IS
−3
FG V IJ ⇒ 20 = J expF 0.65 I
H 0.0259 K
HV K
D
S
t
so that
or
J S = 2.52 x10
We can write
V = −59.6 mV
_______________________________________
J S = eni
2
LM 1 ⋅
NN
a
144
−10
A / cm
Dn
τ nO
2
+
1
Nd
⋅
Dp
τ pO
OP \
Q
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
We have
Dp
μp
1
=
=
Dn
μ n 2.4
We want
1
Dn
⋅
τ nO
Na
1
Dn
⋅
1
+
τ nO
Na
= 0.10
Jn
τ pO
Nd
1
Na
5 x10
25
1
⋅
Na
5 x10
+
−7
1
1+
0.1
FG N IJ
0.1 H N K
1
⋅
2.4
25
⋅
τ pO
1
=
Jn + J p
or
1
1
=
so
Dp
⋅
τ nO
and
a
d
or
−7
Jn
10
⋅
Nd
5x10
Jn + J p
−7
1
=
FG N IJ
HN K
1 + (2.04)
a
d
7.07 x10
=
b4.47 x10 g
Na
7.07 x10 +
3
(b)
Using Einstein’s relation, we can write
= 0.10
3
Nd
which yields
Na
3
eμ n
Jn
=
Jn + J p
= 14.24
eμ n
b
−10
L 1
×M
N (14.24) N
= 1.6 x10
−19
25
⋅
5x10
d
gb15. x10 g
+
−7
10
1
Nd
⋅
10
5x10
−7
eμ n N d +
OP
Q
N d = 7.1x10 cm
14
−3
Ln
−3
N a = 1.01x10 cm
_______________________________________
Jn + J p
⋅
=
Dn
τ nO
ni
+
Na
1
=
1+
D pτ nO
Dnτ pO
I S = Aeni ⋅
2
ni
⋅
Dp
τ pO
bσ σ g
bσ σ g + 4.90
n
n
p
p
1
Dp
Nd
τ pO
b gb1.6x10 gb15. x10 g ⋅ 101
2
⋅
= 4.90
0.1
+
Na
2
⋅
=
2.4
=
For a silicon p n junction,
Lp
2
τ nO
Nd
9.6
eD p pnO
Ln
Dn
ni
_______________________________________
Ln
+
Lp
2
⋅
⋅ eμ p N a
Lp
Then
eDn n pO
eDn n pO
Ln
D pτ pO
Jn
=
Na
Dnτ nO
=
Lp
9.5
(a)
Jn + J p
eμ p
+
Also
and
16
ni
We have
σ n = eμ n N d and σ p = eμ p N a
We find
Jn
⋅
eμ n N d
=
2
Na
2
Ln
J S = 2.52 x10
⋅
Ln
Nd
Now
2
ni
= 10
ni
Nd
−4
−19
10
16
or
I S = 3.94 x10
FG N IJ
HN K
Then
a
d
145
2
−15
A
12
10
−7
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
I D = I S exp
FG V IJ = b3.94 x10 g expF 0.50 I
H 0.0259 K
HV K
Also
−15
a
FG N IJ
Hn K
F 10 IJ ⇒
= (0.0259) lnG
H 15. x10 K
n-side: E F − E Fi = kT ln
t
or
d
i
−7
I D = 9.54 x10 A
_______________________________________
17
10
E F − E Fi = 0.407 eV
9.7
We want
Jn
(b)
We can find, for the minority carriers:
p-region:
= 0.95
Jn + J p
2
eDn n pO
=
Dn = (1250)(0.0259) = 32.4 cm / s
n-region:
Dn
Ln
eDn n pO
eD p pnO
+
Ln
Lp
Dp = (320)(0.0259) = 8.92 cm / s
Ln N a
=
Dn
+
Ln N a
2
Dp
Now
Lp N d
J S = eni
2
Dn
=
We obtain
Ln =
Dnτ nO =
×
b
g
b
10
2
32.4
15
−11
10
A / cm
−6
1
+
10
8.29
17
10
−7
OP
Q
2
−4
−11
I S = 4.42 x10
We find
15.8
25 10 N a
+ ⋅
15.8 10 N d
which yields
Na
= 0.083
Nd
_______________________________________
FG IJ
H K
I = I S exp
FG V IJ
HV K
b
= 4.42 x10
or
−15
A
a
t
−15
0.5 I
g expFH 0.0259
K
I = 1.07 μA
(c)
The hole current at the space charge edge is
F N IJ
E − E = kT lnG
Hn K
F 5x10 IJ ⇒
= (0.0259) lnG
H 15. x10 K
I P ∝ eni ⋅ A ⋅
2
a
Fi
τ pO
OP
Q
or
25
(a) p-side:
Nd
gb15. x10 g
LM 1
N 5x10
I S = AJ S = 10
Then
9.8
Dp
b gb4.42 x10 g
g
L p = 10 μm
0.95 =
−19
J S = 4.42 x10
Then
(10) 0.1x10 −6 ⇒
Dpτ pO =
τ nO
1
+
or
(25) 0.1x10 −6 ⇒
Ln = 15.8 μm
Lp =
= 1.6 x10
Dp N a
+
⋅
Ln
Lp N d
Dn
Dn
a
b
Ln
LM 1
NN
F
b
i
= 1.6 x10
15
10
−19
1
Dp
Nd
τ pO
gb15. x10 g b10 gFH 101 IK
10
I P ∝ 3.28 x10
146
−4
17
or
E Fi − E F = 0.329 eV
2
−16
A
8.29
10
−7
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
Then
Ip
3.28 x10
−16
I D = I S exp
Ip
=
⇒
= 0.0742
−15
4.42 x10
I
I
_______________________________________
9.9
t
I D = 4.36 x10
a
−15
+
FG eD p IJ = AF e D ⋅ n I
H L K GH τ N JK
L
b2.4 x10 g OP
49
= b10 gMb1.6 x10 g
⋅
5 x10
10
MN
PQ
9.11
(a) We find
2
nO
p
p
i
pO
d
13
−4
−6
and
16
(a)
For Va = +0.2 V ,
b
I = 2.89 x10
or
−9
b
I = 2.89 x10
−9
≈ −2.89 x10
Also
0.2 I O
gLMNexpFH 0.0259
K − 1PQ
pnO =
J pO =
0.2 I O
gLMNexpFH 0.−0259
K − 1PQ
eDp pnO
= 2.25 x10 cm
5
15
−3
b1.6x10 g(12.4)b2.25x10 g
=
b111. x10 g
−19
5
−4
Lp
−9
J pO = 4.02 x10
−4
A / cm
1
Dn
Na
τ nO
1.6 x10
−19
10
(b)
We have
Dn = μ n
and
Ln =
−14
A
F kT I = (1350)(0.0259) = 35 cm / s
H eK
D τ = (35)b0.4 x10 g ⇒
2
−6
n
nO
Ln = 37.4 μm
gb
15
. x10
16
10
g
Also
b15. x10 g
=
2
25
10
10
2
n pO =
−6
or
−15
2
2
I pO = 4.02 x10
A
I S = 18
. x10
−10
For A = 10 cm , then
+
b gb
=
10
Nd
2
or
For an n p silicon diode
−4
10
ni
Then
9.10
10
b15. x10 g
=
2
I = − I S = −2.89 nA
_______________________________________
2
pO
L p = 111
. μm
or
I S = Aeni ⋅
p
A
I = 6.52 μA
(b)
For Va = −0.2 V ,
2
−6
Lp =
or
−9
F kT I = (480)(0.0259) = 12.4 cm / s
H eK
D τ = (12.4)b0.1x10 g ⇒
Dp = μ p
2
−19
I S = 2.89 x10
A
I D = − I S = −18
. x10 A
_______________________________________
For a p n diode,
p
−7
(b)
For Va = −0.5 V
t
IS = A
−15
a
or
LM FG V IJ − 1OP
N HV K Q
I = I S exp
FG V IJ = b18. x10 g expF 0.5 I
H 0.0259 K
HV K
ni
J nO =
(a)
For Va = 0.5 V
147
= 4.5 x10 cm
4
−3
Na
5x10
eDn n pO
b1.6x10 g(35)b4.5x10 g
=
b37.4 x10 g
Then
A
15
2
Ln
−19
4
−4
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
or
J nO = 6.74 x10
−4
A / cm
I n = 7.13 x10 − 6.10 x10
2
For A = 10 cm , then
2
I nO = 6.74 x10
(c)
a
−9
9.12
(a) The excess hole concentration is given by
δpn = pn − pnO
d
i
15
15
10
2
ni
pnO =
Then for
1
Va = Vbi = 0.309 V
2
We find
eVa
pn = pnO exp
kT
pn = 3.42 x10 cm
10
or
g
LM F 0.610 I − 1OP expF − x I
N H 0.0259 K Q H 2.83x10 K
F − x I cm
δp = 3.81x10 exp
H 2.83x10 K
a
−3
−4
(b)
We have
a f
−15
d δpn
J p = − eDp
A
=
F eV I expLM −a x − x f OP
H kT K N L Q
dx
b
g expF − x I
g H 2.83x10 K
eDp 3.81x10
The hole current is
a
−4
n
or
b2.83x10
−4
14
−4
−4
At x = 3 x10 cm ,
n
b1.6x10 g(8)b3.81x10 g expF −3 I
=
H 2.83K
2.83 x10
−19
p
Jp
The electron current is given by
In = I − I p
14
−4
or
b
g
F 0.309 I expLM −a x − x f OP
× exp
H 0.0259 K N L Q
J p = 0.597 A / cm
−14
(c)
We have
n
p
2
4
14
b g FH eVkT IK
F 0.309 I
= b4.02 x10 + 6.74 x10 g exp
H 0.0259 K
= 7.13 x10 − 4.02 x10
g
× exp
−3
−14
b
δpn = 2.25 x10
(d)
The total current is
I = I pO + I nO exp
b
(8) 0.01x10 −6 ⇒
Dpτ pO =
Then
or
At x = xn +
10
−3
16
L p = 2.83 μm
5
1
4
Nd
Lp =
2
10
and
F I
H K
F 0.309 I
= b2.25 x10 g exp
H 0.0259 K
−9
a
p
We find
Vbi = 0.617 V
I p = I pO exp
LM F eV I − 1OP expFG − x IJ
N H kT K Q H L K
b15. x10 g = 2.25x10 cm
=
= pnO exp
2
or
I = 7.13 x10
g expFH −21IK
I n = 3.43 x10 A
_______________________________________
A
2
−9
−9
or
−15
FG N N IJ
H n K
L b5x10 gb10 g OP
= (0.0259) ln M
MN b15. x10 g PQ
Vbi = Vt ln
b
−9
−11
J nO =
Lp
148
eDn n pO
Ln
exp
F eV I
H kT K
a
2
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
LM n (0) OP
Nn Q
F 10 IJ
= (0.0259) lnG
H 2.25x10 K
We can determine that
Va = Vt ln
n pO = 4.5x10 cm and Ln = 10.7 μm
−3
3
p
pO
Then
b1.6x10 g(23)b4.5x10 g expF 0.610 I
=
H 0.0259 K
10.7 x10
−19
J nO
15
3
4
−4
or
or
J nO = 0.262 A / cm
We can also find
J pO = 1.72 A / cm
Va = 0.635 V
_______________________________________
2
2
Then, at x = 3 μm ,
a f
9.14
The excess electron concentration is given by
δn p = n p − n pO
a f
J n 3 μm = J nO + J pO − J p 3 μm
LM F eV I − 1OP expFG − x IJ
N H kT K Q H L K
= 0.262 + 1.72 − 0.597
or
a f
= n pO exp
n
J n 3 μm = 1.39 A / cm
_______________________________________
2
The total number of excess electrons is
z
∞
N p = A δn p dx
9.13
(a) From Problem 9.9 (Ge diode)
Low injection means
pn (0) = (0.1) N d = 10 cm
Now
15
2
ni
pnO =
b2.4 x10 g
=
13
10
Nd
0
We may note that
z expFGH −Lx IJK dx = − L expFGH −Lx IJK
Then
LM F eV I − 1OP
N = AL n exp
N H kT K Q
∞
−3
= 5.76 x10 cm
10
−3
n
n
2
t
2
n pO =
n
t
Then
nO
10
Va = 0.253 V
or
(b)
For Problem 9.10 (Si diode)
2
n pO =
Na
b15. x10 g
=
10
10
16
b15. x10 g
=
10
8 x10
Na
15
2
= 2.81x10 cm
4
Np
−3
−3
−4
4
a
Then we find the total number of excess
electrons in the p-region to be:
2
= 2.25x10 cm
4
(a) Va = 0.3 V , N p = 1.78 x10
−3
Then
149
−3
b gb59.2 x10 gb2.81x10 g
LM F eV I − 1OP
× exp
N H kT K Q
LM F eV I − 1OP
= 0.166 exp
N H kT K Q
a
or
ni
ni
N p = 10
15
15
pO
Dn = 35 cm / s and Ln = 59.2 μm
Also
a
n p (0) = (0.1) N a = 10 cm
= Ln
0
We can find
FG V IJ
HV K
or
L p (0) OP
V = V ln M
Np Q
F 10 IJ
= (0.0259) lnG
H 5.76x10 K
a
n
a
p
We have
pn (0) = pnO exp
∞
n
0
2
16
a
4
(b) Va = 0.4 V , N p = 8.46 x10
5
(c) Va = 0.5 V , N p = 4.02 x10
7
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
Similarly, the total number of excess holes in the
n-region is found to be:
eVa
N n = AL p pnO exp
−1
kT
We find that
which yields
E g 2 = 0.769 eV
LM F I OP
N H K Q
_______________________________________
9.16
(a) We have
Dp = 12.4 cm / s and L p = 111
. μm
2
Also
b15. x10 g
=
10
pnO
10
Then
I S = Aeni
2
2
= 2.25x10 cm
4
16
b
N n = 2.50 x10
−2
τ nO
+
1
Dp
Nd
τ pO
OP
Q
which can be written in the form
2
I S = C ′ni
F T I expFG − E IJ
= C ′N N
H 300K H kT K
F − E IJ
= CT expG
H kT K
3
a
g
CO
So
or
(b) Va = 0.4 V , N n = 1.27 x10
Dn
a
−3
gLMNexpFH eVkT IK − 1OPQ
(a) Va = 0.3 V , N n = 2.68 x10
LM 1
NN
3
IS
5
(c) Va = 0.5 V , N n = 6.05x10
_______________________________________
VO
g
3
6
(b)
Taking the ratio
9.15
F eV I ∝ expFG − E IJ expF eV I
H kT K H kT K H kT K
Then
F eV − E IJ
I ∝ expG
H kT K
so
F eV − E IJ
expG
H kT K
I
=
I
F eV − E IJ
expG
H kT K
or
I
F eV − eV − E + E IJ
= expG
H
K
I
kT
We have
10 x10
F 0.255 − 0.32 − 0.525 + E IJ
= expG
H
K
10 x10
0.0259
or
F E − 0.59 IJ
10 = expG
H 0.0259 K
Then
E = 0.59 + (0.0259) lnb10 g
I ∝ ni exp
2
g
a
a
IS2
a
I S1
g
2
2
g
1
1
3
g
a1
FG − E IJ
F T I H kT K
=G J ⋅
H T K expFG − E IJ
H kT K
F T I L F 1 − 1 IJ OP
= G J ⋅ exp M+ E G
H T K N H kT kT K Q
exp
3
2
g
1
1
For T1 = 300 K , kT1 = 0.0259 ,
g1
1
a2
2
a1
1
For T2 = 400 K , kT2 = 0.03453 ,
g2
a2
g1
1
= 38.61
kT1
1
= 28.96
kT2
(i) Germanium, E g = 0.66 eV
g2
IS2
2
F 400I ⋅ exp (0.66)(38.61 − 28.96)
H 300K
3
=
I S1
−3
or
g2
−6
3
2
IS2
= 1.38 x10
3
I S1
g2
(ii)Silicon, E g = 112
. eV
IS2
3
g2
I S1
150
F 400I ⋅ exp (112. )(38.61 − 28.96)
H 300K
3
=
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
(c) GaAs
or
IS2
Let Dn = 220 cm / s and Dp = 10.4 cm / s
2
= 117
. x10
5
I S1
_______________________________________
9.17
J S = eni
2
LM 1
NN
Dn
+
τ nO
a
and
ni = N c N v exp
2
1
Dp
Nd
τ pO
Then
b
×
or
FG − E IJ
H kT K
g
35
10
−6
+
1
12.4
15
5 x10
10
T(K)
gn
n bcm g
200
7.68 x10
4
2.21x10
300
6.95x10
9
181
. x10
J S = 3.752 x10
Then
−31
−7
OP
Q
200
1.38
300
18
. x10
b
J S A / cm
2.73 x10
8.53x10
g
b
or
−19
kT
−3
b
T(K)
10
gn
n bcm g
J S = 7.419 x10
Then
16
−31
−6
+
49.2
15
5x10
10
i
b
300
2.31x10
3.97 x10
400
8.60 x10
14
0.549
15
49.0
4
FTI
H 300K
FG eD n
H L
n
pO
+
eDp pnO
Lp
n
FG D + D IJ
HLN L N K
L1 D + 1
= AeN N M
NN τ N
g
a
p
d
V
a
which becomes
−4
151
IJ
K
p
n
n
C
−10
13
3.46 x10
2
0.5
=
4
= 0.05429 = (0.0259)
n
J S A / cm
4
b g lnb10 g
ln 10
IS = A
i
8.13 x10
Va
=
2
2.16 x10
500
a
= Aeni
10
200
−7
OP
Q
2
−3
−8
a
e
which yields
T = 629 K
Second conditon:
2
i
1
2.95 x10
or
gbn g
101
−12
S
e
2
L 1
×M
N 5x10
g
or
Let Dn = 101 cm / s and Dp = 49.2 cm / s
J S = 1.6 x10
=
Ir
kT
Then
2
−18
11
F eV I
H kT K = expF eV I = 10
H kT K
J
J S exp
If
(b) Germanium
2
1.21x10
4.02 x10
2.81x10
−6
13
500
OP
Q
−31
9
500
−11
2.13x10
2.38 x10
J S A / cm
6
3.28 x10
9.18
One condition:
−21
12
400
2
−7
_______________________________________
2
−3
10
7.12 x10
400
i
i
5x10
15
b
−3
gbn g
10.4
1
+
2
i
2
16
10
−6
i
2
LM 1
N 5x10
220
16
−31
i
b
2
i
S
Let Dn = 35 cm / s and Dp = 12.4 cm / s
J S = 1.6 x10
gbn g
LM 1
N 5x10
or
J = b3.738 x10 gn
Then
n bcm g
T(K)
OP
Q
2
−19
−19
×
(a) Silicon
Then
b
J S = 1.6 x10
2
nO
d
Dp
τ pO
OP expFG − E IJ
Q H kT K
g
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
b gb1.6x10 gb2.8x10 gb1.04 x10 g
F 1 25 + 1 10 IJ expFG − E IJ
×G
H 5x10 10 10 10 K H kT K
−6
10 = 10
−4
−19
19
We then obtain
19
LM FG V IJ − 1OP
N HV K Q
B=
expb − x L g 1 − expb −2W L g
which can be written in the form
L FV I O
p MexpG J − 1P ⋅ exp a x + W f L
N HV K Q
B=
expbW L g − expb −W L g
Also
L FV I O
− p MexpG J − 1P ⋅ exp −a x + W f L
N HV K Q
A=
expbW L g − expb −W L g
The solution can now be written as
L FV I O
p MexpG J − 1P
N HV K Q
δp =
FW I
2 sinhG J
HL K
R L a x + W − xf OP − expLM −a x + W − xf OPUV
× Sexp M
T N L Q N L QW
−7
18
or
exp
FG + E IJ = 4.66x10
H kT K
g
t
−7
15
n
10
b
Eg
10
10
dx
2
p
p
t
n
p
n
n
n
n
FG x + W − x IJ
L FV I O H L K
δp = p MexpG J − 1P ⋅
N H V K Q sinhFG W IJ
HL K
(b)
d aδp f
J = − eD
dx
L FV I O
− eD p MexpG J − 1P
N HV K Q
=
FW I
sinhG J
HL K
F −1I F x + W − x IJ
×G J coshG
HL K H L K
n
t
p
n
n
n
n
n
n
n
a
p
t
n
nO
n
p
p
p
x = xn
a
p
n
p
nO
t
p
n
p
p
n
Then, from the first boundary condition, we
obtain
LM FG V IJ − 1OP
N HV K Q
= − B exp −a x + 2W f L + B expb − x L g
= B expb − x L g 1 − expb −2W L g
p
Then
a
Jp =
t
n
n
p
p
n
n
n
p
nO
n
n
sinh
a
n
p
or finally,
n
n
n
n
g b g
The boundary condition at x = x gives
L FV I O
δp a x f = p MexpG J − 1P
N HV K Q
= A expb + x L g + B expb − x L g
and the boundary condition at x = x + W gives
δp a x + W f = 0
= A exp a x + W f L + B exp −a x + W f L
From this equation, we have
A = − B exp −2a x + W f L
n
p
a
b
pnO exp
n
p
2
n
p
nO
δpn = A exp + x L p + B exp − x L p
n
n
n
n
Lp
n
p
t
=0
n
p
a
The general solution is
n
n
nO
9.19
(a) We can write for the n-region
a f − δp
n
n
110
.
=
T = 519 K
This second condition yields a smaller
temperature, so the maximum temperature is
T = 519 K
_______________________________________
2
p
t
Then
d δpn
n
a
g lnb4.66x10 g
or
FTI
kT = 0.04478 eV = (0.0259)
H 300K
ln 4.66 x10
p
nO
For E g = 110
. eV ,
kT =
a
pnO exp
g
p
eD p pnO
Lp
coth
n
p
x = xn
FG W IJ ⋅ LexpFG V IJ − 1O
H L K MN H V K PQ
n
a
p
t
_______________________________________
p
152
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
9.20
I D ∝ ni exp
2
9.21
(a) From Problem 9.20
E g − eVD1 E g − eVD 2
=
kT1
kT2
FG V IJ
HV K
D
t
For the temperature range 300 ≤ T ≤ 320 K ,
neglect the change in N C and N V
So
− Eg
eVD
I D ∝ exp
⋅ exp
kT
kT
Assuming VD1 and T1 are known, then
F kT I LM E FG 1 − 1 IJ + e V OP
H e K N H kT kT K kT Q
We have
FV I
I = I expG J
HV K
FG IJ F I
H K H K
L −b E − eV g OP
∝ exp M
N kT Q
g
VD 2 =
E g − eVD1
D
kT2 = 0.02676 eV ,
T = 320 K
kT3 = 0.02763 eV ,
So, for T = 310 K ,
112
. − 0.60
=
kT3
VD1 = 0.6560 V
So we can write
F kT I LM E − 43.24 + 25.33OP
H e K N kT
Q
F kT I LM E − 17.91OP
=
H e K N kT Q
VD 2 =
g
2
2
or
VD 2
g
2
2
Now
T ( K ) = TC + 273
= 0.02676 V
e
We find
T (° C )
20
65
110
155
200
= 0.02763 V
e
112
. − VD 2
T(K)
293
338
383
428
473
kT (eV )
0.02530
0.02918
0.03307
0.03695
0.04084
(c)
For I D = 1 mA , we find VD1 = 0.7156 V
Then
Eg
kT2
VD 2 =
− 15.61
e
kT2
For T = 320 K ,
=
−15
or
0.0259
0.02676
which yields
VD 2 = 0.5827 V
112
. − 0.60
D1
S
= 0.0259 V
kT2
FG I IJ = (0.0259) lnFG 0.1x10 IJ
H 10 K
HI K
−3
VD1 = Vt ln
e
T = 310 K
S
t
kT1
kT2
We have
T = 300 K , VD1 = 0.60 V and
kT1
1
(b)
For I D = 0.1 mA , then
E g − eVD 2
kT1 = 0.0259 eV ,
1
D
D
g OP
Q
g OP
Q
=
D1
g
2
Taking the ratio of currents, but maintaining I D
a constant, we have
− E g − eVD1
exp
kT1
1=
⇒
− E g − eVD 2
exp
kT2
LM b
N
LM b
N
2
F I LM
H KN
112
. − VD 3
0.0259
0.02763
which yields
VD 3 = 0.5653 V
_______________________________________
153
OP
Q
VD 2 (V )
0.6669
0.5974
0.5277
0.4582
0.3886
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
We find
T (° C )
20
65
110
155
200
9.23
(a)
We have
VD 2 (V )
0.7251
0.6645
0.6038
0.5432
0.4825
FG −φ IJ
HV K
F −0.68 I
= (120)(300) exp
H 0.0259 K
J ST = A T exp
*
2
or
J ST = 4.28 x10
account, then VD 2 is no longer a linear function
of temperature.
_______________________________________
For I = 10
9.22
We have
FG N IJ
HN K
= (0.0259) ln
FG 2.8x10 IJ
H 10 K
2
5 x10
−4
= 2 A / cm
2
n
F 2 I
H 4.28x10 K
−5
Bn
F
H
−8
A / cm
2
I
K
F
H
2
For I = 100 mA ⇒ J n = 200 A / cm
And
200
Va = (0.0259) ln
−5
4.28 x10
or
Va = 0.398 V
2
F
H
I
K
2
2
I
K
(b)
For T = 400 K , φ Bn = 0.68 V
Now
(d)
F eV I
H kT K
or
F J IJ = (0.0259) lnF 2 I
V = V lnG
H 1.29 x10 K
HJ K
a
J ST = (120)(400) exp
2
LM −0.68 OP
N (0.0259)a400 300f Q
or
n
J ST = 5.39 x10
For I = 1 mA ,
−8
t
−3
For I = 10 mA ⇒ J n = 20 A / cm
And
20
Va = (0.0259) ln
−5
4.28 x10
or
Va = 0.338 V
For silicon, A = 120 A / cm / ° K
Then
−0.89
2
J ST = (120)(300) exp
0.0259
or
a
10
Va = 0.278 V
*
J n = J ST exp
FG J IJ
HJ K
2
or
F −eφ I
H kT K
J ST = 1.29 x10
A ⇒ Jn =
16
(b)
Vbi = φ Bn − φ n = 0.89 − 0.206
or
Vbi = 0.684 V
*
A / cm
= (0.0259) ln
φ n = 0.206 V
J ST = A T exp
−5
ST
19
or
(c)
−3
Va = Vt ln
C
d
Bn
t
(d) If E g versus temperature is taken into
(a) φ n = Vt ln
2
ST
or
−2
A / cm
2
F 400I lnF 2 I
H 300K H 5.39 x10 K
Va = 0.488 V
_______________________________________
Va = (0.0259)
or
154
−2
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
Va = 0.125 V
For I = 10 mA ,
9.25
(a) For GaAs, A = 112
. A / cm / K
*
F 400I lnLM 20 OP
H 300K N 5.39 x10 Q
Va = (0.0259)
I ST = A A T exp
*
−2
For I = 100 mA ,
F I lnF
I
H 300K H 5.39 x10 K
FG −φ IJ
HV K
BO
−4
2
I ST = 1.36 x10
Then
or
J n = J ST exp
2
2
or
2
A / cm
2
0.65 I
b g(120)(300) expFH 0−.0259
K
= 10
t
−10
a
t
Bn
J ST = 3.83 x10
Now
V I
g expFH 0.0259
K
I ST = A A T exp
2
F −φ IJ
= A T expG
HV K
F −0.86 I
= (112
. )(300) exp
H 0.0259 K
2
−14
*
*
*
A
(b) For Silicon, A = 120 A / cm / K
9.24
2
−14
I D = 3.83x10
Va = 0.284 V
_______________________________________
b
2
−8
I D = 1.36 x10
FG V IJ
HV K
a
−8
A
V I
g expFH 0.0259
K
a
We find
t
and we can write, for J n = 5 A / cm
2
F J IJ
V = V lnG
HJ K
F 5 I
= (0.0259) ln
H 3.83x10 K
n
a
2
b
or
J ST
−4
I ST = 3.83 x10
Then
−2
*
BO
or
200
(a) For GaAs, A = 112
. A / cm / K
FG −φ IJ
HV K
0.86 I
b g(112. )(300) expFH 0−.0259
K
= 10
Va = 0.204 V
Va = (0.0259)
2
2
t
or
400
2
t
ST
−10
or
Va = 0.6033 V
Va (V )
GaAs
I D ( A)
0.1
0.2
0.3
0.4
0.5
182
. x10
−11
8.65 x10
−9
4.11x10
−7
1.95x10
−6
9.27 x10
Silicon
I D ( A)
−12
−7
6.46 x10
−5
3.07 x10
−3
1.46 x10
−2
6.93x10
3.29
_______________________________________
(b)
For J n = 10 A / cm
Va = (0.0259) ln
9.26
For the Schottky diode,
2
F 10 I = 0.6212 V
H 3.83x10 K
−8
−4
J ST = 3x10 A / cm , A = 5x10 cm
For I = 1 mA ,
−10
so
ΔVa = 0.6212 − 0.6033 ⇒
J=
10
−3
5 x10
We have
ΔVa = 17.9 mV
_______________________________________
2
−4
Va = Vt ln
= 2 A / cm
FG J IJ
HJ K
ST
155
2
2
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
= (0.0259) ln
F 2 I
H 3x10 K
I PN = I S exp
−8
−12
Then
For the pn junction, J S = 3 x10 A / cm
Then
2
Va = (0.0259) ln
−12
3x10
or
Va = 0.705 V (pn junction diode)
_______________________________________
2
I
K
8 x10
−4
Then
Va = Vt ln
2
. A / cm
= 15
−3
S
= (0.0259) ln
2
−8
or
2
I S ≈ 0.5x10
F 15. I = 0.684 V
H 5x10 K
−3
−3
exp
FG 0.239 IJ
H 0.0259 K
I PN = 1.02 x10
−8
A (pn junction diode)
(b) Diodes in Series:
We obtain,
Va S = (0.0259) ln
FG V IJ
HV K
FG 0.5x10 IJ
H 5x10 K
−3
−8
or
a
b
−12
A (Schottky diode)
or
−12
Va S = 0.239 V (Schottky diode)
t
so
−12
S
For the Schottky diode, the applied voltage will
be less, so
Va = 0.684 − 0.265 = 0.419 V
We have
I = AJ ST exp
ST
−3
I PN = 10
S
a
FG 0.5x10
H I +I
F 0.5x10 IJ = 0.239 V
= (0.0259) lnG
H 5x10 + 10 K
Now
F 0.239 I
I = 5x10 exp
H 0.0259 K
Va = Vt ln
and
FG J IJ
HJ K
f FGH VV IJK
IJ
K
t
or
−3
−4
A , Va S = Va PN
−8
J S = 5x10 A / cm , A = 8 x10 cm
For I = 1.2 mA ,
1.2 x10
a
−3
−3
0.5x10 = I ST + I S exp
9.27
For the pn junction diode,
J=
(pn junction diode)
We have I S + I PN = 0.5x10
Va = 0.467 V (Schottky diode)
−12
a PN
t
or
F
H
FG V IJ
HV K
1.2 x10 = A 7 x10
−8
and
g expFH 0.0259 IK
0.419
Va PN = (0.0259) ln
which yields
−3
FG 0.5x10 IJ
H 10 K
−3
−12
or
A = 1.62 x10 cm
_______________________________________
2
Va PN = 0.519 V
(pn junction diode)
_______________________________________
9.28
(a) Diodes in parallel:
We can write
Va S
I S = I ST exp
(Schottky diode)
Vt
and
9.29
(a) For I = 0.8 mA , we find
FG IJ
H K
J=
0.8 x10
7 x10
We have
Va = Vt ln
−3
. A / cm
= 114
−4
FG J IJ
HJ K
S
156
2
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
For the pn junction diode,
114
.
Va = (0.0259) ln
−12
3x10
or
Va = 0.691 V
F
H
For the Schottky diode,
114
.
Va = (0.0259) ln
−8
4 x10
or
Va = 0.445 V
F
H
J ST (400)
I
K
b
I = 7 x10
I
K
g
kT
Also
E O
F 400I expLM − E
=
+
H 300K N (0.0259)a400 300f 0.0259 PQ
or
LM 112. − 112. OP
= 2.37 exp
N 0.0259 0.03453 Q
b
I = 7 x10
−4
F e IbI
2 H kT K
1
τ pO + I nOτ nO
I pO + I nO = 2 x10
−3
A
Then
b2 x10 gb10 g ⇒
C =
−3
−6
2(0.0259)
d
Cd = 3.86 x10
Then
Y = g d + jωCd
or
−12
5
g
−6
.691 I
gb117. x10 gb3x10 g expFH 0.003453
K
−8
F
b
Y = 0.0772 + jω 3.86 x10
−8
g
_______________________________________
I = 121 mA
F
H
pO
τ pO = τ nO = 10 s
g
5
For the Schottky diode
− eφ BO
2
J ST ∝ T exp
kT
Now
J ST (400)
0.0259
We have
3
J S (300)
Now
−3
g d = 0.0772 S
Cd =
g
2 x10
⋅ ID =
or
J S (300)
or
e
gd =
F T I expFG − E IJ
H 300K H kT K
= 117
. x10
−8
3
9.30
Then
J S (400)
We find
J S (400)
.445 I
gb4.86x10 gb4 x10 g expFH 0.003453
K
I = 53.8 mA
_______________________________________
3
2
−4
3
or
(b)
For the pn junction diode,
J S ∝ ni ∝
= 4.86 x10
J ST (300)
and so
9.31
I
K
+
For a p n diode
gd =
I DQ
,
Vt
Cd =
I DQτ pO
2Vt
Now
J ST (300)
φ
OP
F 400I expLM −φ
=
+
H 300K N (0.0259)a400 300f 0.0259 Q
or
LM 0.82 − 0.82 OP
= 1.78 exp
N 0.0259 0.03453 Q
gd =
2
BO
BO
and
10
−3
Now
We obtain
157
−2
S
b10 gb10 g = 1.93x10
−3
Cd =
= 3.86 x10
0.0259
−7
2(0.0259)
−9
F
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
Z=
1
1
=
g d − jωCd
=
I no = Aeni
2
Y g d + jωCd
g d + ω Cd
We have ω = 2π f ,
We find:
f = 10 kHz : Z = 25.9 − j 0.0814
f = 100 kHz : Z = 25.9 − j 0.814
f = 1 MHz : Z = 23.6 − j 7.41
f = 10 MHz : Z = 2.38 − j 7.49
_______________________________________
2
2
2
Dn
τ no
1
⋅
Na
b10 gb1.6x10 gb15. x10 g
=
−4
I no
−19
5 x10
10
or
a
I no = 4.098 x10
d
R
a
bi
Cd =
R
15
17
exp
FG V IJ
HV K
a
15
17
FG V IJ
HV K
a
FV I
4.098 x10 gb10 g expG J
b
HV K
2(0.0259)
1
−15
or
Cd = 7.911x10
−6
a
−20
exp
FG V IJ
HV K
a
t
1/ 2
We find
−4
Va (V )
Cd ( F )
R
15
bi
10
−18
0.1
3.76 x10
17
0.2
1.79 x10
2
0.3
8.49 x10
0.4
4.03x10
0.5
1.92 x10
0.6
9.10 x10
0.7
4.32 x10
VR (V )
C j ( pF )
10
8
6
4
2
0
-0.1
-0.2
-0.3
-0.4
0.60
0.672
0.765
0.913
1.20
2.31
2.49
2.71
3.01
3.43
0.75
−16
−15
−13
−11
−10
2.98 x10
−8
−7
(b) We want to set
x10 I
b10 gFGH 3.945
J
V −V K
−16
−4
bi
1/ 2
= 7.911x10
a
−20
exp
FG V IJ
HV K
a
t
where
Vbi = 0.7363 V
By trial and error, we find Va = 0.458 V and
Cd = C j ≅ 3.77 pF
+
Diffusion Capacitance; n p junction
_______________________________________
aI τ f
no
−6
t
1/ 2
j
2Vt
10
t
Then
−14
−16
−15
d
−4
Cd =
32.4
15
t
−19
1
2
× exp
j
We find
t
2
1/ 2
bi
a
Dn = (1250)(0.0259) = 32.4 cm / s
Then
L e ∈ N N OP
C = AM
N 2aV + V fa N + N f Q
L b1.6x10 g(11.7)b8.85x10 g
= b10 gM
2aV + V f
N
b5x10 gb10 g OP
×
b5x10 + 10 g Q
or
F 3.945x10 IJ
C = b10 gG
H V +V K
where
L b5x10 gb10 g OP = 0.7363 V
V = (0.0259) ln M
MN b15. x10 g PQ
bi
FG V IJ
HV K
We find
9.32
Depletion capacitance
s
exp
no
where
158
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
9.33
−3
0.518 x10 = 5.70 x10
+
For a p n diode, I pO >> I nO , then
Cd =
FG 1 IJ b I τ g
H 2V K
pO
= 2.5x10
(c)
d
τ pO = 2(0.0259) 2.5 x10
−6
rd =
i
−7
b
−6
1
=
D
⇒
rd
0.0259
0.518 x10
−3
rd = 50 Ω
_______________________________________
τ pO = 1.3x10 s
Cd = 2.5 x10
a
or
or
At 1 mA ,
F e II
H kT K
gd =
F/A
2Vt
Then
F V I
H 0.0259 K
Va = 0.594 V
Now
−6
exp
We find
pO
t
τ pO
−14
9.35
(a) p-region
ρpL
L
L
Rp =
=
=
A
σ p A A eμ p N a
gb10 g ⇒
−3
−9
Cd = 2.5 x10 F
_______________________________________
b
g
so
9.34
(a)
F e I Ab I
C =
2 H kT K
1
d
pO
τ pO + I nOτ nO
g
so
10
−12
F e I Aa I τ f
2 H kT K
1F 1 I
=
b10 ga I fb10 g
2 H 0.0259 K
ρn L
Rn =
=
A
L
σnA
Rn =
or
I nO = I D = 0.518 mA
eDn n pO
I nO = A
Ln
a
t
= 2.25x10 cm
4
−3
Then
0.518 x10
−19
4
−3
1.6 x10 (25) 2.25x10 10
V
exp a
=
−4
15.8 x10
Vt
or
b
g b
0.10
b10 gb1.6x10 g(1350)b10 g
−2
−19
15
(b)
V = IR ⇒ 0.1 = I (72.3)
or
I = 1.38 mA
_______________________________________
Na
−3
f
R = 72.3 Ω
Dnτ nO = 15.8 μm and
ni
A eμ n N d
Rn = 46.3 Ω
The total series resistance is
R = R p + Rn = 26 + 46.3 ⇒
FV I
expG J
HV K
2
n pO =
a
L
or
We find
Ln =
=
so
−7
nO
(b)
16
n-region
nO
−3
−19
R p = 26 Ω
1
nO
b10 gb1.6x10 g(480)b10 g
−2
or
+
For a one-sided n p diode, I nO >> I pO , then
Cd =
0.2
Rp =
gb g FG IJ
H K
9.36
R=
159
ρ n L( n )
A(n)
+
ρ p L( p )
A( p)
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
b g + (0.1)b10 g
(0.2) 10 −2
=
2 x10
or
F 0.539 x10 IJ
V = (0.0259) lnG
H 2 x10 K
−2
−5
2 x10
−3
V = I D R + Vt ln
FG I IJ
HI K
9.38
FG 1 IJ expFG V IJ
HV K H V K
r
dV
or
1
10
F 0.020 I
=
⋅ exp
H 0.0259 K
0.0259
r
D
(a) (i) For I D = 1 mA
=
dI D
d
b g(150) + (0.0259) lnFGH 1010 IJK
−3
or
1
(a)
S
V = 10
⇒
Va = 0.443 V
_______________________________________
R = 150 Ω
We can write
−11
a
−5
−3
= IS
a
a
t
t
−13
−10
d
V = 0.567 V
which yields
rd = 1.2 x10 Ω
11
(ii) For I D = 10 mA
b
(b)
For Va = −0.020 V ,
g(150) + (0.0259) lnFGH 1010x10 IJK
−3
V = 10 x10
−3
1
−10
or V = 1.98 V
10
=
−13
⋅ exp
0.0259
rd
F −0.020I
H 0.0259 K
or
(b)
For R = 0
(i) For I D = 1 mA
rd = 5.6 x10 Ω
_______________________________________
11
F 10 IJ ⇒
V = (0.0259) lnG
H 10 K
−3
9.39
Ideal reverse-saturation current density
eDn n pO eDp pnO
JS =
+
Ln
Lp
−10
V = 0.417 V
(ii) For I D = 10 mA
F 10x10 IJ ⇒
V = (0.0259) lnG
H 10 K
−3
We find
−10
2
n pO =
V = 0.477 V
_______________________________________
and
ni
b18. x10 g
=
6
Na
10
b18. x10 g
=
6
9.37
rd = 48 Ω =
1
gd
pnO
⇒ g d = 0.0208
e
kT
⋅ I D ⇒ I D = (0.0208)(0.0259)
I D = 0.539 mA
Dnτ nO =
Lp =
D pτ pO
I D = I S exp
JS =
FG V IJ ⇒ V = V lnFG I IJ
HV K
HI K
a
t
−4
= 3.24 x10 cm
−3
(200) 10 −8 = 14.1 μm
−8
−19
−4
−4
2.45x10
S
so
so
160
−3
2
−19
D
a
t
b1.6x10
−4
= 3.24 x10 cm
16
b g
= (6)b10 g = 2.45 μm
g(200)b3.24 x10 g
14.1x10
b1.6x10 g(6)b3.24 x10 g
+
Ln =
Then
or
Also
16
Also
We have
gd =
10
2
−4
−4
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
J S = 8.62 x10
−18
A / cm
2
−4
d
i
J gen
16
or
2
6
b1.6x10 gb114. x10 gn
=
2b5 x10 g
J = n b182
. x10 g
−19
2
16
i
−17
LM 2 ∈aV + V f FG N + N IJ OP
N e H N N KQ
L 2(131. )b8.85x10 g(116. + 5)
=M
1.6 x10
N
L 10 + 10 OPOP
×M
N b10 gb10 g QQ
−31
185
. x10 ni = 182
. x10
which yields
1/ 2
bi
i
When J S = J gen ,
Vbi = 116
. V
W=
−4
−7
gen
or
And
R
a
d
a
13
We have
ni = N C N V exp
2
−19
16
16
16
16
1/ 2
FG − E IJ
H kT K
g
b
9.88 x10
13
g b
2
−4
b1.6x10 gb18. x10 gb1.34 x10 g
=
2b10 g
J gen
−4
b
or
J gen = 1.93x10
−9
A / cm
J S = J gen = 182
. x10
2
(b)
J S exp
LM 1 D + 1
NN τ N
L 1 25
= n b1.6 x10 gM
N10 5x10
2
Dp
n
a
2
FG V IJ = J
HV K
a
gen
exp
t
J S = eni
i
−17
nO
τ pO
d
OP
Q
At T = 300 K
b
. x10
J S = 15
+
10
and
−7
10
16
−3
A / cm
FG V IJ
H 2V K
a
t
g b185. x10 g
2
J S = 4.16 x10
1
13
−31
or
19
16
1.04 x10
10
5 x10
−7
b
J gen = 15
. x10
OP
Q
10
−11
A / cm
2
gb182. x10 g ⇒
−17
or
J gen = 2.73 x10
or
Then we can write
161
−7
19
gb9.88x10 g ⇒
J S = J gen = 18
. x10
Generation current dominates in GaAs reversebiased junctions.
_______________________________________
9.40
(a) We can write
19
By trial and error, we find
T = 505 K
At this temperature
−8
gFH 300T IK
gb
L −112. OP
× exp M
N (0.0259)aT 300f Q
= 2.8 x10
W = 1.34 x10 cm
6
−3
Then
or
−19
−17
ni = 9.88 x10 cm
d
−14
Then
g
For Vbi + VR = 5 V , we find W = 114
. x10 cm
So
FG N N IJ
H n K
L b10 gb10 g OP
= (0.0259) ln M
NM b18. x10 g PQ
a
−31
We also have
en W
J gen = i
2τ O
Reverse-biased generation current density
en W
J gen = i
2τ O
We have
Vbi = Vt ln
b
J S = ni 185
. x10
2
A / cm
2
2
3
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
F V IJ = J
expG
H 2V K J
=
gen
a
t
2.73x10
so that
b
Va = 2(0.0259) ln 6.56 x10
3
g
−14
= 6.56 x10
−11
4.16 x10
S
L 2(131. )b8.85x10 g(1.28 + 5)
=M
1.6 x10
N
F 10 + 10 I OP
×G
H b10 gb10 gJK Q
−7
3
−19
⇒
Va = 0.455 V
(Neglects change in space charge width)
_______________________________________
JS
Dn
2
τ nO
i
a
1
Dp
Nd
τ pO
+
OP
Q
I gen
b1.6x10 gb18. x10 gb0.427 x10 gb10 g
=
2b10 g
b
J S = 1.6 x10
gb18. x10 g LMN10
+
10
1
10
I R ≈ 6.15 x10
so
17
518
.
10
−8
OP
Q
I D = I S exp
A / cm
2
−3
−22
a
For Va = 0.5 V
−19
I S = 5.75x10
b
I D = 5.75x10
−22
A
−22
−17
A
0.5 I
g expFH 0.0259
K
or
We also have
en WA
I gen = i
2τ O
Now
I D = 1.39 x10
−13
A
Recombination current
For Va = 0.3 V :
L 2(131. )b8.85x10 g(1.28 − 0.3) FG 2 x10
W=M
H 10
1.6 x10
N
FG N N IJ
H n K
−14
d
−19
2
L b10 gb10 g OP
= (0.0259) ln M
MN b18. x10 g PQ
i
17
or
17
6
−4
W = 0.169 x10 cm
2
Then
or
I rec =
Vbi = 1.28 V
Also
L 2 ∈aV + V f FG N + N IJ OP
W=M
N e H N N KQ
bi
A
FG V IJ = b5.75x10 g expF 0.3 I
H 0.0259 K
HV K
I D = 6.17 x10
or
a
−13
t
b gb5.75x10 g
Vbi = Vt ln
−13
or
−19
I S = AJ S = 10
+ 6.15 x10
Forward Bias: Ideal diffusion current
For Va = 0.3 V
−8
or
J S = 5.75 x10
−22
or
77.7
17
A
I R = I S + I gen = 5.75x10
2
6
−13
The total reverse-bias current
Dp = (200)(0.0259) = 518
. cm / s
−19
−3
−8
I gen = 6.15x10
2
1
−4
6
or
Dn = (3000)(0.0259) = 77.7 cm / s
2
17
−4
We find
Then
17
1/ 2
W = 0.427 x10 cm
−19
L1
= en M
NN
17
or
so
9.41
(a) We can write
17
R
a
1/ 2
d
a
d
162
eniWA
2τ O
exp
FG V IJ
H 2V K
a
t
34
17
IJ OP
KQ
1/ 2
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
b1.6x10 gb18. x10 gb0.169 x10 gb10 g
=
2b10 g
L 0.3 OP
× exp M
N 2(0.0259) Q
−19
−4
6
For Va = 0.5 V
−3
−8
I rec
−11
−14
−19
34
4
17
IJ OP
KQ
From Problem 9.41, I S = 5.75x10
So diffusion current is
1/ 2
b
I D = 5.75x10
−22
−22
A
g expFGH VV IJK
a
t
or
Now, the recombination current is
FG V IJ
H 2V K
2τ
b1.6x10 gb18. x10 gb10 gW expFG V IJ
=
H 2V K
2b10 g
or
F V IJ
I = b1.44 x10 gW expG
H 2V K
−4
W = 0.150 x10 cm
I rec =
b1.6x10 gb18. x10 gb0.15x10 gb10 g
2b10 g
L 0.5 OP
× exp M
N 2(0.0259) Q
−19
−13
9.42
L 2(131. )b8.85x10 g(1.28 − 0.5) FG 2 x10
W=M
H 10
1.6 x10
N
I rec =
1.39 x10
−9
Ratio = 2.42 x10
_______________________________________
A
For Va = 0.5 V
Then
=
ID
or
I rec = 7.97 x10
3.36 x10
−4
6
−3
eniWA
a
exp
o
t
−19
−8
−3
6
a
−8
t
−8
or
a
rec
I rec = 3.36 x10
−9
A
t
The space charge width is
Total forward-bias current:
For Va = 0.3 V ;
I D = 6.17 x10
−17
L 2(131. )b8.85x10 gaV − V f
W=M
1.6 x10
N
F 10 + 10 I OP
×G
H b10 gb10 gJK Q
or W = b1.702 x10 g V − V
−14
+ 7.97 x10
bi
−11
or
I D ≈ 7.97 x10
−11
A
For Va = 0.5 V
I D = 1.39 x10
−13
+ 3.36 x10
−9
bi
−9
9
ID
=
6.17 x10
17
17
1/ 2
a
Vbi = 1.282 V
A
Ratio = 1.07 x10
Forward bias: Ratio of recombination to ideal
diffusion current:
For Va = 0.3 V
We find
Va (V )
W ( μm)
0.1
0.185
184
. x10
0.3
0.169
7.97 x10
I rec ( A)
−12
0.151
3.38 x10
0.7
0.130
1.38 x10
0.8
0.118
8.66 x10
0.9
0.105
5.31x10
1.0
0.0904
315
. x10
−17
11
.
0.0726
1.74 x10
Ratio = 1.29 x10
−11
0.5
−11
7.97 x10
17
where
(b)
Reverse-bias; ratio of generation to ideal
diffusion current:
I gen 6.15x10 −13
=
−22
IS
5.75 x10
I rec
17
−5
or
I D ≈ 3.36 x10
a
−19
−9
−7
−7
−6
−5
−4
I D ( A)
2.73x10
6.17 x10
1.39 x10
314
. x10
−20
−17
−13
−10
−8
1.49 x10
−7
7.10 x10
3.37 x10
1.60 x10
−5
−3
_______________________________________
6
163
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
We find
Va (V )
W (cm)
p
0.1
1.24 x10
pO
0.2
113
. x10
0.3
1.01x10
9.43
Diffusion current
LM 1 D + 1 D OP
NN τ N τ Q
= b10 gb1.6 x10 gb15
. x10 g
L 1 25 + 1 10 OP
×M
N10 5x10 10 5x10 Q
I S = Aeni
2
n
a
nO
−4
d
−19
2
10
−7
16
−7
16
I rec ( A)
−7
1.29 x10
−7
8.73 x10
0.5
7.10 x10
0.6
4.95x10
4.73x10
−8
2.65x10
−8
1.98 x10
−12
4.46 x10
−11
−12
−10
2.12 x10
−10
1.27 x10
−13
9.39 x10
−12
7.94 x10
−8
0.4
−13
2.05x10
−7
I D ( A)
1.01x10
−9
−8
−6
4.79 x10
−5
or
_______________________________________
−15
I S = 4.16 x10 A
Then the diffusion current is
FV I
I = 4.16 x10 expG J
HV K
Recombination current
AeWn
F V IJ
I =
expG
H 2V K
2τ
b10 gb1.6x10 gb15. x10 gW expFG V IJ
=
H 2V K
2b5x10 g
or
F V IJ
I = b2.4 x10 gW expG
H 2V K
The space charge width is
L 2 ∈ aV − V f FG N + N IJ OP
W=M
N e H N N KQ
L 2(11.7)b8.85x10 gaV − V f
=M
1.6 x10
N
F 10 + 10 I OP
×G
H b10 gb10 gJK Q
−15
9.44
Diffusion current
a
LM 1
NN
D
I S = Aeni
t
i
2
t
−19
a
Then
−7
t
b
I S = 5 x10
−3
a
rec
a
a
I S = 3.087 x10
d
a
−19
or
1/ 2
−14
bi
10
−8
OP
Q
A
−21
a
exp
i
a
o
t
−3
−19
6
a
−8
a
t
16
10
5.70
rec
1/ 2
16
Vbi
16
17
t
b2.589 x10 gaV − V f
L b10 gb10 g OP = 0.6946 V
= (0.0259) ln M
MN b15. x10 g PQ
W=
Also
16
−21
I D = 3.087 x10
a
16
−8
2
FG V IJ
HV K
Recombination current
Aen W
F V IJ
I =
expG
H 2V K
2τ
b5x10 gb1.6x10 gb18. x10 gW expFG V IJ
=
H 2V K
2b10 g
or
F V IJ
I = b7.2 x10 gW expG
H 2V K
From Problem 9.42, we have
W = b1.702 x10 g V − V
−14
16
6
So
d
bi
−19
or
1/ 2
bi
τ pO
gb1.6x10 gb18. x10 g
L 1 90.65 + 1
×M
N10 10 10
17
t
s
Nd
OP
Q
Dp = (220)(0.0259) = 5.70
10
−7
Dp
We find
Dn = (3500)(0.0259) = 90.65
a
−4
1
+
τ nO
a
rec
o
Dn
−8
2
a
rec
t
−5
bi
where
Vbi = 1.282 V
164
a
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
We find
Va (V )
c h
I rec ( A)
W μm
0.2
0.177
0.4
6.06 x10
0.160
−11
−9
2.60 x10
−7
0.6
0.141
1.09 x10
0.8
0.118
4.33x10
0.9
0.105
2.66 x10
1.0
0.0904
11
.
158
. x10
0.0726
−6
−5
−4
8.72 x10
9.46
I D ( A)
−4
6.97 x10
157
. x10
3.55x10
J S = eni
2
−18
b
= 1.6 x10
−11
8.02 x10
Dp
Nd
τ pO
16
10
1
+
−4
8.60 x10
18
2
10
−6
3.81x10
OP
Q
gb15. x10 g LMN 3x101
−19
10
−3
18
−7
6
10
−7
OP
Q
or
J S = 1.64 x10
Now
J D = J S exp
z
−11
A / cm
2
FG V IJ
HV K
D
t
W
Also
J = 0 = JG − J D
or
J gen = eGdx
0
19
−3 −1
In this case, G = g ′ = 4 x10 cm s , that is a
constant through the space charge region. Then
J gen = eg ′W
−3
0 = 25 x10 − 1.64 x10
a
or
15
VD
2
R
a
9.47
d
−14
15
15
15
30 =
gb g
gN
b
−19
2
B
which yields
W = 2.35 x10 cm
Then
16
−4
19
or
−3
A / cm
−3
N B = N d = 1.73 x10 cm
_______________________________________
gb4 x10 gb2.35x10 g
J gen = 15
. x10
b
(11.7) 8.85x10 −14 4 x105
2 1.6 x10
−4
−19
2eN B
or
1/ 2
or
J gen = 1.6 x10
∈ ε crit
2
VB =
−19
15
9
t
VD = 0.548 V
_______________________________________
d
a
9
D
so
1/ 2
bi
D
t
i
10
FG V IJ
HV K
FG V IJ = 152. x10
HV K
= V lnb152
. x10 g
exp
d
2
15
exp
t
FG N N IJ
H n K
L b5x10 gb5x10 g OP = 0.659 V
= (0.0259) ln M
MN b15. x10 g PQ
and
L 2 ∈aV + V f FG N + N IJ OP
W=M
N e H N N KQ
L 2(11.7)b8.85x10 g(0.659 + 10)
=M
1.6 x10
N
F 5x10 + 5x10 I OP
×G
H b5x10 gb5x10 gJK Q
Vbi = Vt ln
−11
which yields
We find
b
1
+
τ nO
−8
_______________________________________
9.45
We have
Dn
a
−14
181
. x10
LM 1
NN
9.48
For the breakdown voltage, we need
2
−3
N d ≈ 3x10 cm and for this doping, we find
15
_______________________________________
2
μ p ≈ 430 cm / V − s . Then
165
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
b518. x10 g = LMN 2(11.71.)6bx810.85x10 g
F 5x10 IJ F 1 I OP
× aV + V fG
H 5x10 K H 5x10 + 5x10 K Q
which yields
2.68 x10 = 1.29 x10 aV + V f
Dp = (430)(0.0259) = 1114
. cm / s
−14
2
−19
+
For the p n junction,
J S = eni ⋅
1
Dp
Nd
τ pO
2
16
bi
b1.6x10 gb15. x10 g
=
−19
10
3 x10
2
−5
2
1114
.
15
10
−10
A / cm
16
−9
16
16
−10
bi
−7
R
so
Vbi + VR = 20.7 ⇒ VR = 19.9 V
_______________________________________
or
J S = 1.27 x10
R
2
Then
FV I
I = J A expG J
HV K
F 0.65 I
2 x10 = b1.27 x10 g A exp
H 0.0259 K
9.51
+
a
For a silicon p n junction with
S
t
−3
N d = 5x10 cm
15
−10
and VB ≈ 95 V
Neglecting Vbi compared to VB
L 2 ∈V OP
x ≈M
N eN Q
L 2(11.7)b8.85x10 g(95) OP
=M
N b1.6x10 gb5x10 g Q
1/ 2
Finally
B
−4
A = 1.99 x10 cm
_______________________________________
2
n
d
−14
9.49
−19
+
−3
GaAs, n p , and N a = 10 cm
From Figure 9.30
VB ≈ 75 V
_______________________________________
16
1/ 2
15
or
xn ( min) = 4.96 μm
_______________________________________
9.52
We find
9.50
ε
−3
max
=
L b10 gb10 g OP = 0.933 V
= (0.0259) ln M
MN b15. x10 g PQ
eN d xn
18
∈
Vbi
∈
Now
10
We can write
xn =
ε
max
ε
eN d
b4 x10 g(11.7)b8.85x10 g
=
b1.6x10 gb5x10 g
max
−19
so
16
6
−5
xn = 518
. x10 cm
LM b5x10 gb5x10 g OP = 0.778 V
MN b15. x10 g PQ
Now
L 2 ∈aV + V f FG N IJ FG 1 IJ OP
x =M
N e H N KH N + N KQ
16
18
n
−14
−6
2
xn =
Then
a
n
a
LM 2 ∈aV + V f FG N IJ FG 1 IJ OP
N e H N KH N + N KQ
bi
R
a
d
1/ 2
d
b1.6x10 gb10 gx
(11.7)b8.85x10 g
xn = 6.47 x10 cm
Now
16
10
R
∈
which yields
We find
Vbi = (0.0259) ln
2
eN d xn
−19
10 =
or
bi
=
−14
5
18
d
or
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a
d
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An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
b6.47 x10 g = LMN 2(11.71.)6bx810.85x10 g
F 10 IJ F 1 I OP
×aV + V fG
H 10 K H 10 + 10 K Q
−14
−6
2
ts
erf
−19
=
τ pO
IF
IF + IR
1
=
1+
R
18
18
18
tS
We find
9.53
(b)
erf
bi
If
1/ 2
R
erf
Assume Vbi << VR
(a)
For x p = 75 μm
b
tS
=
τ pO
τ pO
b75x10 g = b1.6x10 gb10 ggV
2(11.7) 8.85 x10
2
= 0.956
= 1.0 , then
which yields
ts
Then
τ pO
IF
a
−4
IR
tS
= 0.978 ⇒
τ pO
Assume silicon: For an n p junction
= 0.833
τ pO
tS
+
LM 2 ∈aV + V f OP
N eN Q
1 + 0.2
or
which yields
Vbi + VR = 6.468 V
or
VR = 5.54 V
_______________________________________
xp =
1
IF
18
bi
IR
=
−19
−14
1
1+1
= 0.5
= 0.228
_______________________________________
R
15
9.56
We want
tS
= 0.2
which yields VR = 4.35 x10 V
(b)
For x p = 150 μm , we find
3
τ pO
VR = 1.74 x10 V
From Figure 9.30, the breakdown voltage is
approximately 300V. So, in each case,
breakdown is reached first.
_______________________________________
4
Then
erf
tS
1
=
τ pO
1+
= erf
IR
0.2
IF
where
9.54
Impurity gradient
a=
2 x10
erf 0.2 = erf (0.447) = 0.473
We obtain
IR
I
1
=
− 1 ⇒ R = 111
.
I F 0.473
IF
18
= 10 cm
22
−4
−4
2 x10
From Figure 9.30
VR = 15 V
_______________________________________
We have
FG −t IJ
H τ K = 1 + (0.1)FG I IJ = 111.
HI K
F t IJ
πG
Hτ K
2
exp
9.55
erf
IR
= 0.2
IF
Then we have
(a) If
t2
τ pO
+
pO
2
pO
167
R
F
An Introduction to Semiconductor Devices
Chapter 9
Solutions Manual
Problem Solutions
______________________________________________________________________________________
By trial and error
t2
= 0.65
τ pO
_______________________________________
9.57
C j = 18 pF at VR = 0
C j = 4.2 pF at VR = 10 V
We have τ nO = τ pO = 10
−7
s , I F = 2 mA
And
IR ≈
So
VR
R
=
10
= 1 mA
10
FG
H
t S ≈ τ pO ln 1 +
IJ = b10 g lnF1 + 2 I
H 1K
I K
IF
−7
R
or
−7
t S = 11
. x10 s
Also
18 + 4.2
Cavg =
= 111
. pF
2
The time constant is
b gb111. x10 g = 111. x10
τ S = RCavg = 10
4
−12
−7
s
Now
Turn-off time =
t S + τ S = (11
. + 111
. ) × 10
−7
s
Or
−7
2.21x10 s
_______________________________________
168