Chapter 1
Exercise Solutions EX1.1
3 / 2exp
2
g i
E
n BT
kT
−
3 / 2 14
6
1.4
i
or n i =1.8 10 × 6 cm−3
Ge: =( × ) ( ) ⎜⎜⎛2(86×−10− ) ( )300 ⎟⎟⎞
66 0 exp
300 10 66
i
EX1.2
(a) (i) n o =N d =2×1016 cm−3
16
2 10 2
10 125 1 10 2
10 5
×
×
=
=
o
i o n
n
(ii) p o=N a=1015 cm−3
15
2 10 2
10 25 2 10
10 5
=
=
o
i o p
n
(b) (i) n o=N d =2×1016 cm−3
16
2 6 2
10 62 1 10 2
10 8
×
×
=
=
o
i o n
n
(ii) p o=N a=1015 cm−3
15
2 6 2
10 24 3 10
10 8
=
=
o
i o p
n
EX1.3
(a) For n-type;
10 2 6800 10
6 1
1 1
16
×
×
=
d
n N
eμ
(b) J= 1⋅Ε⇒Ε=ρJ=(0.046)( )175 =8.04
-
EX1.4
Diffusion current density due to holes:
( )16 1
p
dp
J eD
dx
x eD
= −
⎞
⎟⎟
⎠
Trang 2(a) At x=0
( 19) ( ) ( )16
2 3
16 / 10
p
−
−
×
(b) At x=10 −3 cm
3
2 3
10
10
p
−
−
EX1.5
(a) ( ) ( )( )
(1.8 10 ) 1.23
10 10 ln 026
6
17 16
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
×
=
bi
(b) ( ) ( )( )
(2.4 10 ) 0.374
10 10 ln 026
13
17 16
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
×
=
bi
EX1.6
1/ 2
bi
V
V
−
and
2
2 10
ln
1.5 10
a d
i
N N
V V
n
V
0.757
−
−
or
C jo =2.21 p F
EX1.7
⎠
⎞
⎜⎜
⎝
⎛
=
S
D T D
I
I V
10 2
10 50 ln 026
6
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
×
D
10 2
10 ln 026
3
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
D
10 2
10 50 ln 026
6
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
×
D
10 2
10 ln 026
3
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
D
Trang 3
EX1.8
T
V
V
3
4
4 10
D
V
× and
(1012)exp
0.026
D D
V
By trial and error, we find I D ≅0.866mA and V D ≅0.535V
EX1.9
20 1
7 0
=
⇒
−
R
V V
5 3
7 0 4
=
−
=
D
P D =I D V D =(0.9429)( )0.7 =0.66 mW
EX1.10
PSpice Analysis
EX1.11
20
7 0 8
=
−
=
D
= = ⇒71.2Ω
365 0
026 0
D
T d I
V r
i d ωt 12.5sinωt
0712 0 20
sin 25 0
⇒ +
10
7 0 8
=
−
=
D
= ⇒35.6Ω
73 0
026 0
d
r
i d ωt 24.9sinωt
0356 0 10
sin 25 0
⇒ +
EX1.12
For the pn junction diode, ln (0.026 ln) 1.2 10153
4 10
D
S
I
V V
I
−
−
×
The Schottky diode voltage will be smaller, so V D =0.6871 0.265 0.4221 − = V
T
V
I I
V
Trang 4or
3
10
0.4221 exp
0.026
−
−
×
A
EX1.13
P= ⋅I V Z⇒10=I( )5.6 ⇒ =I 1.79 mA
Also I 10 5.6 1.79 R 2.46 Ωk
R
−
Test Your Understanding Solutions
TYU1.1
(a) T = 400K
Si: 3/ 2exp
2
g i
E
n BT
kT
−
3/ 2 15
6
1.1
i
×
or
n i =4.76 10 × 12 c m−3
6
0.66
i
×
m
or
n i =9.06 10 × 14 c −3
GaAs:
3 / 2 14
6
1.4
×
m
n
or
n i =2.44 10 × 9 c −3
(b) T = 250 K
6
1.1
i
×
m
or
n i =1.61 10 × 8 c −3
6
0.66
i
×
m
or
n i =1.42 10 × 12 c −3
Trang 5( ) ( ) ( ) ( )
6
1.4
i
×
m
or
n i =6.02 10 × 3 c −3
TYU1.2
(a) = =(1.6×10− 19) ( )480(2×1015)=0.154
a
p N
eμ
= = =6.51Ω
1536 0
1 1
σ
(b) = =(1.6×10− 19) (1350) (2×1017)=43.2
d
n N
eμ
= = =0.0231Ω
2 43
1 1
σ
TYU1.3
(a) J=σΕ=(0.154)( )4 =0.616 A/cm 2 (b) J=σΕ=(43.2)( )4 =172.8 A/cm 2
TYU1.4
(a) J n eD n dn eD n n
Δ
4
0 2.5 10
n
or
J n =202 /A cm2
(b) J p eD p dp eD p p
Δ
4
0 4 10
p
− ×
or
J p = −24.5 /A cm2
TYU1.5
(a)
2 10 2
15
8 10 1.5 10
2.81 10
8 10
i o o
n
n
−
m−
×
×
(b) n=n o+δn= ×8 1015+0.1 10× 15
or
n=8.1 10 × 15c m−3
4
p= p o+δp=2.81 10× 4+101
or
p≅10 14 cm−3
Trang 6
TYU1.6
(1.5 10 ) 0.679
10 5 10 ln 026 0
10
16 15
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
×
×
=
⎟
⎠
⎞
⎜
⎝
⎛
=
i
d a T bi
n
N N V
(1.8 10 ) 1.15
10 5 10 ln 026
6
16 15
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
×
×
=
bi
(2.4 10 ) 0.296
10 5 10 ln 026
13
16 15
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
×
×
=
bi
TYU1.7
026 0
55 0 exp 10
⎠
⎞
⎜
⎝
⎛
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
T
D S
D
V
V I
(ii) ( ) 7.20μ
026 0
65 0 exp
⎠
⎞
⎜
⎝
⎛
D
026 0
75 0 exp
⎠
⎞
⎜
⎝
⎛
D
(b) (i) I D =−10− 16 A (ii) I D =−10−16 A
TYU1.8
Δ =T 100C so ΔV D ≅ ×2 100 200 = m V
V
Then V D =0 650 0 20 0 450 . − . = .
TYU1.9
Trang 7
TYU1.10
(a) I D =0
4
7 0
=
D
4
7 0
=
D
(d) I D =0
(e) I D =0
TYU1.11
P=I V D D⇒1.05=I D(0.7) so I D =1 5 mA
1.5
PS D
V V
I
γ
TYU1.12
0.026
D d T
I
V
TYU1.13
010 0
026
=
=
D
T d I
V
= ⇒260Ω
10 0
026 0
d
r
= ⇒26Ω
1
026 0
d r
-
TYU1.14
50
T
V
or
I D =0.52 mA
TYU1.15
For the pn junction diode,
4 0.7 0.825
4
D
For the Schottky diode, 4 0.3 0.925
4
D
Trang 8
TYU1.16
V z =V zo+I r z z⇒V zo =V z −I r z z so V zo =5.20−( )10−3 ( )20 =5.18 V
Then V z =5.18+(10 10× −3) ( )20 ⇒V z =5.38 V
TYU1.17
6 3
5
6 =
=
=
⇒
=
Z Z Z Z
V
P I V I
V PS =I Z R+V Z =( )( )1.81 4 +3.6=10.8 V