Ch01p

Chapter 1
Exercise Problems
EX1.1
⎛ − Eg ⎞
ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
GaAs: ni = ( 2.1× 1014 ) ( 300 )
Ge: ni = (1.66 × 1013 ) ( 300 )
3/ 2
3/ 2
⎛
⎞
−1.4
⎟ or ni = 1.8 × 106 cm −3
exp ⎜
⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟
⎝
⎠
⎛
⎞
−0.66
⎟ or ni = 2.40 × 1013 cm −3
exp ⎜
⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟
⎝
⎠
EX1.2
(a) majority carrier: holes, po = 1017 cm −3 minority carrier: electrons,
n 2 (1.5 × 10
no = i =
1017
po
)
10 2
= 2.25 × 103 cm −3
(b) majority carrier: electrons, no = 5 × 1015 cm −3 minority carrier: holes,
n 2 (1.5 × 10 )
= 4.5 × 104 cm −3
po = i =
5 × 1015
no
10 2
EX1.3
For n-type, drift current density J ≅ eμn nE or 200 = (1.6 × 10−19 ) ( 7000 ) (1016 ) E which yields
E = 17.9 V / cm
EX1.4
Diffusion current density due to holes:
dp
J p = −eD p
dx
⎛ −1 ⎞
⎛ −x ⎞
= −eD p (1016 ) ⎜ ⎟ exp ⎜ ⎟
⎜L ⎟
⎜L ⎟
⎝ p⎠
⎝ p⎠
(a) At x = 0
(1.6 ×10 ) (10 ) (10 ) = 16 A / cm
=
−19
Jp
16
2
10−3
−3
(b) At x = 10 cm
⎛ −10−3 ⎞
J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm 2
⎝ 10 ⎠
EX1.5
⎡N N
Vbi = VT ln ⎢ a 2 d
⎣ ni
⎡ (1016 )(1017 ) ⎤
⎤
⎢
⎥ or Vbi = 1.23 V
=
0.026
ln
(
)
⎥
6 2 ⎥
⎢
⎦
⎣ (1.8 × 10 ) ⎦
EX1.6
⎛ V ⎞
C j = C jo ⎜1 + R ⎟
⎝ Vbi ⎠
and
−1/ 2
⎡N N ⎤
Vbi = VT ln ⎢ a 2 d ⎥
⎣ ni ⎦
⎡ (1017 )(1016 ) ⎤
⎥ = 0.757 V
= ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣
⎦
5 ⎞
⎛
Then 0.8 = C jo ⎜ 1 +
⎟
⎝ 0.757 ⎠
or
C jo = 2.21 pF
−1/ 2
= C jo ( 7.61)
−1/ 2
EX1.7
⎡
⎛v ⎞ ⎤
iD = I S ⎢exp ⎜ D ⎟ − 1⎥
⎝ VT ⎠ ⎥⎦
⎣⎢
⎡
⎛ v ⎞ ⎤
so 10−3 = (10−13 ) ⎢ exp ⎜ D ⎟ − 1⎥
⎝ 0.026 ⎠ ⎦
⎣
⎡ 10−3
⎤
Solving for the diode voltage, we find vD = ( 0.026 ) ln ⎢ −13 + 1⎥
⎣10
⎦
or
vD ≅ ( 0.026 ) ln (1010 )
which yields
vD = 0.599 V
EX1.8
⎛V ⎞
VPS = I D R + VD and I D ≅ I S exp ⎜ D ⎟
⎝ VT ⎠
( 4 − VD )
so 4 = I D ( 4 ×103 ) + VD ⇒ I D =
4 ×103
and
⎛ V ⎞
I D = (10 −12 ) exp ⎜ D ⎟
⎝ 0.026 ⎠
By trial and error, we find I D ≅ 0.864 mA and VD ≅ 0.535 V
EX1.9
(a)
ID =
(b)
ID =
Then R =
(c)
VPS − Vγ
R
VPS − Vγ
R
5 − 0.7
⇒ I D = 1.08 mA
4
VPS − Vγ
⇒R=
ID
=
8 − 0.7
= 6.79 kΩ
1.075
ID(mA)
Diode curve
1.25
1.08
Load lines
(b)
(a)
0
0.7
2
4
VD(v)
6
8
EX1.10
PSpice analysis
EX1.11
Quiescent diode current I DQ =
VPS − Vγ
=
10 − 0.7
= 0.465 mA
20
R
Time-varying diode current:
V
0.026
We find that rd = T =
= 0.0559 kΩ
I DQ 0.465
Then id =
vI
0.2sin ω t (V )
=
⋅
or id = 9.97sin ω t ( μ A)
rd + R 0.0559 + 20 ( kΩ )
EX1.12
⎛I ⎞
⎛ 1.2 × 10−3 ⎞
or VD = 0.6871 V
For the pn junction diode, VD ≅ VT ln ⎜ D ⎟ = ( 0.026 ) ln ⎜
−15 ⎟
⎝ 4 × 10 ⎠
⎝ IS ⎠
The Schottky diode voltage will be smaller, so VD = 0.6871 − 0.265 = 0.4221 V
⎛V ⎞
Now I D ≅ I S exp ⎜ D ⎟
⎝ VT ⎠
or
1.2 × 10−3
IS =
⇒ I S = 1.07 × 10−10 A
0.4221
⎛
⎞
exp ⎜
⎟
⎝ 0.026 ⎠
EX1.13
P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA
Also I =
10 − 5.6
= 1.79 ⇒ R = 2.46 kΩ
R
Test Your Understanding Exercises
TYU1.1
(a) T = 400K
⎛ − Eg ⎞
Si: ni = BT 3 / 2 exp ⎜
⎟
⎝ 2kT ⎠
ni = ( 5.23 × 1015 ) ( 400 )
or
ni = 4.76 × 1012 cm −3
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦
Ge: ni = (1.66 × 1015 ) ( 400 )
3/ 2
⎡
⎤
−0.66
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦
or
ni = 9.06 × 1014 cm −3
GaAs:
ni = ( 2.1× 1014 ) ( 400 )
3/ 2
⎡
⎤
−1.4
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦
or
ni = 2.44 × 109 cm −3
(b) T = 250 K
Si: ni = ( 5.23 × 1015 ) ( 250 )
3/ 2
⎡
⎤
−1.1
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
or
ni = 1.61× 108 cm −3
Ge: ni = (1.66 × 1015 ) ( 250 )
3/ 2
⎡
⎤
−0.66
⎥
exp ⎢
−6
⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦
or
ni = 1.42 × 1012 cm −3
GaAs: ni = ( 2.10 × 1014 ) ( 250 )
3/ 2
⎡
⎤
−1.4
⎥
exp ⎢
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
or
ni = 6.02 × 103 cm −3
TYU1.2
(a)
n = 5 × 1016 cm −3 , p <<< n, so σ ≅ eμ n n = (1.6 × 10 −19 ) (1350 ) ( 5 × 1016 )
or
σ = 10.8 ( Ω − cm )
(b)
−1
p = 5 × 1016 cm −3 , n <<< p, so σ ≅ eμ p p = (1.6 × 10−19 ) ( 480 ) ( 5 × 1016 )
or
σ = 3.84 ( Ω − cm )
−1
TYU1.3
J = σ E = (10 )(15 ) or J = 150 A / cm2
TYU1.4
(a)
J n = eDn
⎛ 1015 − 1016 ⎞
dn
Δn
so J n = 1.6 × 10−19 ( 35 ) ⎜
= eDn
−4 ⎟
dx
Δx
⎝ 0 − 2.5 × 10 ⎠
(
)
or
J n = 202 A / cm 2
(b)
J p = −eD p
or
J p = −24.5 A / cm2
TYU1.5
⎛ 1014 − 5 × 1015 ⎞
dp
Δp
so J p = − 1.6 × 10−19 (12.5 ) ⎜
= −eD p
−4 ⎟
dx
Δx
⎝ 0 − 4 × 10 ⎠
(
)
no = N d = 8 × 1015 cm −3
(a)
10
n 2 (1.5 × 10 )
po = i =
= 2.81× 10 4 cm −3
no
8 × 1015
2
(b) n = no + δ n = 8 × 1015 + 0.1× 1015
or
n = 8.1×1015 cm−3
p = po + δ p = 2.81 × 10 4 + 1014
or
p ≅ 1014 cm −3
TYU1.6
(a)
⎡ (1015 )(1017 ) ⎤
⎡N N ⎤
⎥ = 0.697 V
Vbi = VT ln ⎢ a 2 d ⎥ so Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣ ni ⎦
⎣
⎦
(b)
⎡ (1017 )(1017 ) ⎤
⎥ = 0.817 V
Vbi = ( 0.026 ) ln ⎢
⎢ (1.5 × 1010 )2 ⎥
⎣
⎦
TYU1.7
⎡
⎛V
I D = I S ⎢exp ⎜ D
⎝ VT
⎣⎢
(a)
⎞ ⎤
⎟ − 1⎥
⎠ ⎦⎥
⎛ 0.5 ⎞
I D ≅ 10−14 exp ⎜
⎟
⎝ 0.026 ⎠
Then, for
VD = 0.5 V, I D = 2.25 μ A
VD = 0.6 V, I D = 0.105 mA
VD = 0.7 V, I D = 4.93 mA
(b)
I D ≅ − I S = −10 −14 A
for both cases.
TYU1.8
ΔT = 100C so ΔVD ≅ 2 × 100 = 200 mV
Then VD = 0.650 − 0.20 = 0.450 V
TYU1.9
ID(mA)
Diode
1.0
艐0.87
Load line
0
1
艐0.54v
2
VD(v)
VD
ID
0.45
0.50
0.55
0.033
0.225
1.54
3
TYU1.10
P = I DVD ⇒ 1.05 = I D ( 0.7 ) so I D = 1.5 mA
4
Now R =
VPS − Vγ
ID
=
10 − 0.7
⇒ R = 6.2 kΩ
1.5
TYU1.11
I
0.8
gd = D =
= 30.8 mS
VT 0.026
TYU1.12
V
0.026
0.026
rd = T ⇒ 50 =
⇒ ID =
ID
ID
50
or
I D = 0.52 mA
TYU1.13
For the pn junction diode,
4 − 0.7
ID =
= 0.825 mA
4
For the Schottky diode, I D =
4 − 0.3
= 0.925 mA
4
TYU1.14
Vz = Vzo + I z rz ⇒ Vzo = Vz − I z rz so Vzo = 5.20 − (10 −3 ) ( 20 ) = 5.18 V
Then Vz = 5.18 + (10 × 10−3 ) ( 20 ) ⇒ Vz = 5.38 V