课后答案网,用心为你服务!
大学答案 --- 中学答案 --- 考研答案 --- 考试答案
最全最多的课后习题参考答案,尽在课后答案网(www.khdaw.com)!
Khdaw团队一直秉承用心为大家服务的宗旨,以关注学生的学习生活为出发点,
旨在为广大学生朋友的自主学习提供一个分享和交流的平台。
爱校园(www.aixiaoyuan.com) 课后答案网(www.khdaw.com) 淘答案(www.taodaan.com)
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 15
Solutions Manual Problem Solutions
225
Chapter 15
Problem Solutions
15.1
The limit of low injection means that
n N cmB B0 01 01 10 10
16 15 3( ) ( ) ( )= = = −. . b g
Now
I
AeD n
xC
B B
B
=
( )0
=
( ) ( )−
−
0 5 1 6 10 20 10
3 10
19 15
4
. . x
x
b g b g
or
I AC = 5 33.
15.2
From the junction breakdown curve, for
BV VCBO = 1000 , we need the collector doping
concentration to be N x cmC ≈
−2 1014 3
Depletion width into the base (neglect Vbi )
x
V
e
N
N N Np
BC C
B C B
=
∈
+
F
HG
I
KJ
F
HG
I
KJ
L
NM
O
QP
2 1
1 2/
=
( ) ( )L
NM
−
−
2 11 7 8 85 10 1000
1 6 10
14
19
. .
.
x
x
b g
×
+
F
HG
I
KJ
FH IK
O
QP
2 10
5 10
1
5 10 2 10
14
15 15 14
1 2
x
x x x
/
or
x mp = 316. µ (Minimum base width)
Depletion width into the collector
x
x
xn
=
( ) ( )L
NM
−
−
2 11 7 8 85 10 1000
1 6 10
14
19
. .
.
b g
×
+
F
HG
I
KJ
F
H
I
K
O
QP
5 10
2 10
1
5 10 2 10
15
14 15 14
1 2
x
x x x
/
or
x mn = 78 9. µ (Minimum collector width)
15.3
Compute plot
15.4
(a)
We have β β β β βeff A B A B= + +
Then
180 25 25= + +β βB B
or
155 26= β B
which yields
β B = 5 96.
(b)
We have
β B EA CBi i=
or
β ββB
A
A
CA CBi i
1+
=
F
HG
I
KJ
so
5 96
1 25
25
20.( )FH IK
+
=iCA
which yields
i ACA = 3 23.
15.5
Sketch
15.6
We want
P I
V
IT
CC
C rated C rated= ⋅ ⇒ =
FH IK
1
2 2
1
2
24
2
20, ,
which yields
I AC rated, .= 3 33
Then
R
V
IL
CC
C rated
= =
, .
24
3 33
or
RL = 7.2 Ω
15.7
If V VCC = 25 , then
I
V
R
A IC
CC
L
C ratedmax . ,( ) = = = <
25
100
0 25
The power
P I V I V I RC CE C CC C L= = −a f
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 15
Solutions Manual Problem Solutions
226
To find the maximum power point, set
dP
dI
V I R I
C
CC C L C= = − = − ( )0 2 25 2 100
which yields I AC = 0 125.
So
P max . .( ) ( ) ( )( )= −0 125 25 0 125 100
or
P W PTmax .( ) = <156
So, maximum VCC is V VCC = 25
15.8
Now R
V
Ion
DS
D
=
Power dissipated in transistor
P I V
V
RD DS
DS
on
= =
2
We have
I
V
D
DS
=
−200
100
so we can write
P
V
V
V
R
DS
DS
DS
on
=
−
⋅ =
FH IK
200
100
2
For T C Ron= =25 2, Ω ,
Then
200
100 2
2
−
⋅ =
F
H
I
K
V
V
VDS
DS
DS
which yields
V VDS = 3 92. and
P W=
−
=
FH IK( )
200 3 92
100
3 92 7.69
.
.
We then have
T Ron VDS P
25
50
75
100
2.0
2.33
2.67
3
3.92
4.55
5.20
5.83
7.69
8.89
10.1
11.3
15.9
(a)
We have, for three devices in parallel,
V V V
V
18 2 2.2
5 151 5
.
.+ + = ⇒ =( )
or
V V= 3 31.
Then, I
V
R
= , so that
I A1 1839= .
I A2 1 655= .
I A3 1505= .
Now, P IV= , so
P W1 6 09= .
P W2 5 48= .
P W3 4.98=
(b)
Now
V V
1
18
1
3 6
1
2.2
5 1 288
. .
.+ + = =FH IK ( )
or
V V= 3 88.
Then
I A P W1 12.16 8 38= =, .
I A P W2 21 08 4.19= =. ,
I A P W3 31 77 6 85= =. , .
15.10
For BV V= 200 , from the junction breakdown
curve, we need the drain doping concentration to
be N x cmD ≈
−15 1015 3.
For the channel length (neglect Vbi )
L
V
e
N
N N N
D D
B D B
min
/
( ) FHG
I
KJ
F
HG
I
KJ
L
NM
O
QP=
∈
+
2 1
1 2a f
=
( ) ( )L
NM
−
−
2 117 8 85 10 200
1 6 10
14
19
. .
.
x
x
b g
×
+
F
HG
I
KJ
F
H
I
K
O
QP
15 10
10
1
15 10 10
15
16 15 16
1 2
.
.
/
x
x
or L mmin .( ) = 184 µ
For the drift region
W
x
x
min
. .
.
( ) ( ) ( )L
NM
=
−
−
2 11 7 8 85 10 200
1 6 10
14
19
b g
×
+
F
HG
I
KJ
FH IK
O
QP
10
15 10
1
15 10 10
16
15 15 16
1 2
. .
/
x x
or W mmin( ) = 12.3 µ
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 15
Solutions Manual Problem Solutions
227
15.11
(b) In saturation region,
I K V V VD n GS T GS= − = −a f a f2 20 25 4. and
V V I R IDS DD D D= − = − ( )40 10
We find
V V I A V V satGS D DS DS= = = > ( )5 0 25 37.5, . ,
V V I A V V satGS D DS DS= = = > ( )6 1 30, ,
V V I A V V satGS D DS DS= = = > ( )7 2.25 17.5, ,
For V VGS = 8 and V VGS = 9 , transistor is
biased in the nonsaturation region. For
V VGS = 8 .
I
V
V VD
DS
DS DS=
−
= − −( )40
10
0 25 2 8 4 2.
We find
V V I ADS D= =2.92 3 71, .
For V VGS = 9 ,
I
V
V VD
DS
DS DS=
−
= − −( )40
10
0 25 2 9 4 2.
and we find
V V I ADS D= =188 3 81. , .
Power dissipated in the transistor is P I VT D DS= .
We find
V V P WGS T= =5 9.375,
V V P WGS T= =6 30,
V V P WGS T= =7 39.4,
V V P WGS T= =8 10 8, .
V V P WGS T= =9 7.16,
15.12
T T Pdev amb dev case case ambD− = −− −θ θa f
which can be written as
θ θdev case
dev amb
case amb
T T
PD
− −
=
−
−
=
175 25
10
6 9
−
− = °C W/
Now
P
T T
D rated
j amb
dev case
,
,max
=
−
=
−
−
θ
175 25
9
or
P WD rated, .= 16 7
15.13
P
T T
D rated
j amb
dev case
,
,max
=
−
−
θ
or
θ dev case
j amb
D rated
T T
P−
=
−,max
,
=
−
= °
150 25
50
2.5 C W/
Then
T T Pdev amb D dev case case amb− = +− −θ θa f
so
150 25 2.5− = +
−
PD case ambθa f
or
125 2.5= +
−
PD case ambθa f
15.14
We have
P I V WD D DS= ⋅ = =( )( )4 5 20
Now
T T Pdev amb D dev case case snk snk amb− = + +− − −θ θ θa f
or
Tdev − = + + =( )25 20 1 75 0 8 3 111. .
which yields
T Cdev = °136
Also
T T Pdev case dev caseD− = ⋅ = =− ( )( )θ 20 1 75 35.
so
T Tcase dev= − = −35 136 35
or
T Ccase = °101
And
T T P Ccase snk D case snk− = ⋅ = = °− ( )( )θ 20 0 8 16.
so
T Tsnk case= − = −16 101 16
or
T Csnk = °85
15.15
We have
T T Pdev amb D dev case case amb− = +− −θ θa f
so
200 25 25 3− = +
−
θ case amba f
or
θ case amb C W− = °4 /
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 15
Solutions Manual Problem Solutions
228
15.16
We have
θ dev case
j amb
D rated
T T
P
C W
−
=
−
=
−
= °
,max
,
/
175 25
15
10
Now
P
T T
D
j amb
dev case case snk snk amb
=
−
+ +
− − −
,max
θ θ θ
=
−
+ +
175 25
10 1 4
or
P WD = 10
15.17
We have α α1 2 1+ = . Now
α
β
β1
1
11
=
+
and α
β
β2
2
21
=
+
so
α α
β
β
β
β1 2
1
1
2
21 1
1+ =
+
+
+
=
which can be written as
1
1 1
1 1
1 2 2 1
1 2
=
+ + +
+ +
β β β β
β β
a f a f
a fa f
or
1 1 1 11 2 1 2 2 1+ + = + + +β β β β β βa fa f a f a f
Expanding, we find
1 1 2 1 2+ + +β β β β
= + + +β β β β β β1 1 2 2 1 2
which yields
β β1 2 1=
15.18
The reverse-biased p-well to substrate junction
corresponds to the J2 junction in an SCR. The
photocurrent generated in this junction will be
similar to the avalanche generated current in an
SCR, which can trigger the device.
15.19
Case 1: Terminal 1(+), terminal 2(-), and IG
negative. This triggering was discussed in the
text.
Case 2: Terminal 1(+), terminal 2(-), and IG
positive. Gate current enters the P2 region
directly so that J3 becomes forward biased.
Electrons are injected from N2 and diffuse into
N1, lowering the potential of N1. The junction
J2 becomes more forward biased, and the
increased current triggers the SCR so that
P2N1P1N4 turns on.
Case 3: Terminal 1(-), terminal 2(+), and IG
positive. Gate current enters the P2 region
directly so that the J3 junction becomes more
forward biased. More electrons are injected from
N2 into N1 so that J1 also becomes more
forward biased. The increased current triggers
the P1N1P2N2 device into its conducting state.
Case 4: Terminal 1(-), terminal 2(+), and IG
negative. In this case, the J4 junction becomes
forward biased. Electrons are injected from N3
and diffuse into N1. The potential of N1 is
lowered which increases the forward biased
potential of J1. This increased current then
triggers the P1N1P2N2 device into its
conducting state.
Problem Solutions