2.1 Semiconductor Physics In the problems assume the parameter given in

2.1
Semiconductor Physics
In the problems assume the parameter given in
eV and material B has a bandgap energy of 1.2 eV.
following table. Use the temperature T = 300 K unless
The ratio of intrinsic concentration of material A to
otherwise stated.
that of material B is
Property
Si
Ga As
Bandgap Energy
1.12
1.42
0.66
Dielectric Constant
11.7
13.1
16.0
Effective density of
states in conduction
band Nc (cm-3)
2.8 ´ 10
Effective density of
states in valence
band Nv (cm-3)
1.04 ´ 1019
Intrinsic carrier
concertration
ni (cm-3)
Mobility
Electron
Hole
Ge
(A) 2016
(B) 47.5
(C) 58.23
(D) 1048
4. In silicon at T = 300 K the thermal-equilibrium
concentration of electron is n0 = 5 ´ 10 4 cm -3. The hole
concentration is
19
4.7 ´ 10
17
7.0 ´ 1018
1.04 ´ 10
19
6.0 ´ 1018
(A) 4.5 ´ 1015 cm -3
(B) 4.5 ´ 1015 m -3
(C) 0.3 ´ 10 -6 cm -3
(D) 0.3 ´ 10 -6 m -3
5. In silicon at T = 300 K if the Fermi energy is 0.22 eV
above the valence band energy, the value of p0 is
1.5 ´ 10
1.8 ´ 10
2.4 ´ 10
1350
480
8500
400
3900
1900
10
6
18
1. In germanium semiconductor material at T = 400 K
the intrinsic concentration is
(A) 26.8 ´ 1014 cm -3
(B) 18.4 ´ 1014 cm -3
(C) 8.5 ´ 1014 cm -3
(D) 3.6 ´ 1014 cm -3
(A) 2 ´ 1015 cm -3
(B) 1015 cm -3
(C) 3 ´ 1015 cm -3
(D) 4 ´ 1015 cm -3
6. The thermal-equilibrium concentration of hole p0 in
silicon at T = 300 K is 1015 cm -3. The value of n0 is
(A) 3.8 ´ 108 cm -3
(B) 4.4 ´ 10 4 cm -3
(C) 2.6 ´ 10 4 cm -3
(D) 4.3 ´ 108 cm -3
7. In germanium semiconductor at T = 300 K, the
acceptor concentrations is N a = 1013 cm -3 and donor
concentration is N d = 0. The thermal equilibrium
concentration p0 is
2. The intrinsic carrier concentration in silicon is to be
no greater than ni = 1 ´ 1012 cm -3. The maximum
temperature allowed for the silicon is (Assume
(A) 2.97 ´ 10 9 cm -3
(B) 2.68 ´ 1012 cm -3
(C) 2.95 ´ 1013 cm -3
(D) 2.4 cm -3
Statement for Q.8-9:
In germanium semiconductor at T = 300 K, the
E g = 112
. eV)
(A) 300 K
(B) 360 K
(C) 382 K
(D) 364 K
impurity concentration are
N d = 5 ´ 1015 cm -3 and N a = 0
3. Two semiconductor material have exactly the same
8. The thermal equilibrium electron concentration n0
properties except that material A has a bandgap of 1.0
is
www.nodia.co.in
108
Semiconductor Physics
(A) 5 ´ 1015 cm -3
(C) 115
. ´ 10 cm
9
(B) 115
. ´ 1011 cm -3
-3
(D) 5 ´ 10 cm
6
position of Fermi level with respect to the center of the
bandgap is
-3
9. The thermal equilibrium hole concentration p0 is
(A) 396
. ´ 1013
(B) 195
. ´ 1013 cm -3
(C) 4.36 ´ 1012 cm -3
(D) 396
. ´ 1013 cm -3
boron at a concentration of 2.5 ´ 10
cm
1 ´ 10
-3
13
arsenic at a concentration of
(A) +0.045 eV
(B) - 0.046 eV
(C) +0.039 eV
(D) - 0.039 eV
16. A silicon sample contains acceptor atoms at a
concentration of N a = 5 ´ 1015 cm -3. Donor atoms are
10. A sample of silicon at T = 300 K is doped with
13
Chap 2.1
added
forming
and
n - type
compensated
and with
semiconductor such that the Fermi level is 0.215 eV
cm -3. The
below the conduction band edge. The concentration of
material is
donors atoms added are
(A) p - type with p0 = 15
. ´ 1013 cm -3
(A) 12
. ´ 1016 cm -3
(B) 4.6 ´ 1016 cm -3
(B) p - type with p0 = 15
. ´ 10 7 cm -3
(C) 39
. ´ 1012 cm -3
(D) 2.4 ´ 1012 cm -3
(C) n - type with n0 = 15
. ´ 1013 cm -3
17. A silicon semiconductor sample at T = 300 K is
(D) n - type with n0 = 15
. ´ 10 7 cm -3
doped with phosphorus atoms at a concentrations of
11. In a sample of gallium arsenide at T = 200 K,
n0 = 5 p0 and N a = 0. The value of n0 is
(A) 9.86 ´ 10 9 cm -3
(B) 7 cm -3
(C) 4.86 ´ 10 3 cm -3
(D) 3 cm -3
12. Germanium at T = 300 K is uniformly doped with
-3
an acceptor concentration of N a = 10 cm and a donor
15
concentration of N d = 0. The position of fermi energy
with respect to intrinsic Fermi level is
1015 cm -3. The position of the Fermi level with respect
to the intrinsic Fermi level is
(A) 0.3 eV
(B) 0.2 eV
(C) 0.1 eV
(D) 0.4 eV
18. A silicon crystal having a cross-sectional area of
0.001 cm 2 and a length of 20 mm is connected to its
ends to a 20 V battery. At T = 300 K, we want a
current of 100
mA in crystal. The concentration of
donor atoms to be added is
(A) 0.02 eV
(B) 0.04 eV
(A) 2.4 ´ 1013 cm -3
(B) 4.6 ´ 1013 cm -3
(C) 0.06 eV
(D)0.08 eV
(C) 7.8 ´ 1014 cm -3
(D) 8.4 ´ 1014 cm -3
13.
In
germanium
T = 300
at
concentration are N d = 10
14
cm
-3
K,
the
donor
N a = 0. The
and
Fermi energy level with respect to intrinsic Fermi
level is
(A) 0.04 eV
(B) 0.08 eV
(C) 0.42 eV
(D) 0.86 eV
14. A GaAs device is doped with a donor concentration
19. The cross sectional area of silicon bar is 100 mm 2 .
The length of bar is 1 mm. The bar is doped with
arsenic atoms. The resistance of bar is
(A) 2.58 mW
(B) 11.36 kW
(C) 1.36 mW
(D) 24.8 kW
20. A thin film resistor is to be made from a GaAs film
doped n - type. The resistor is to have a value of 2 kW.
of 3 ´ 1015 cm -3. For the device to operate properly, the
The resistor length is to be 200 mm and area is to be
intrinsic carrier concentration must remain less than
10 -6 cm 2 . The doping efficiency is known to be 90%.
5%
The mobility of electrons is 8000
of
the
total
concentration.
The
maximum
cm 2 V - s. The
temperature on that the device may operate is
doping needed is
(A) 763 K
(B) 942 K
(A) 8.7 ´ 1015 cm -3
(B) 8.7 ´ 10 21 cm -3
(C) 486 K
(D) 243 K
(C) 4.6 ´ 1015 cm -3
(D) 4.6 ´ 10 21 cm -3
15. For a particular semiconductor at T = 300 K
E g = 15
. eV, m = 10 m
*
p
*
n
and ni = 1 ´ 10
15
-3
cm . The
21. A silicon sample doped n - type at 1018 cm -3 have a
resistance of 10 W . The sample has an area of 10 -6
www.nodia.co.in
Chap 2.1
Semiconductor Physics
109
cm 2 and a length of 10 mm . The doping efficiency of
27. For a sample of GaAs scattering time is t sc = 10 -13 s
the sample is (m n = 800 cm 2 V - s)
and electron’s effective mass is me* = 0.067 mo . If an
(A) 43.2%
(B) 78.1%
electric field of 1 kV cm is applied, the drift velocity
(C) 96.3%
(D) 54.3%
produced is
22. Six volts is applied across a 2 cm long
(A) 2.6 ´ 106 cm s
(B) 263 cm s
(C) 14.8 ´ 10
(D) 482
6
cm s
semiconductor bar. The average drift velocity is 10 4
28. A gallium arsenide semiconductor at T = 300 K is
cm s. The electron mobility is
(A) 4396 cm V - s
(B) 3 ´ 10 cm V - s
doped with impurity concentration N d = 1016 cm -3. The
(C) 6 ´ 10 4 cm 2 V - s
(D) 3333 cm 2 V - s
mobility m n is 7500 cm 2 V - s. For an applied field of
2
4
2
10 V cm the drift current density is
23. For a particular semiconductor material following
(A) 120 A cm 2
(B) 120 A cm 2
parameters are observed:
(C) 12 ´ 10 4 A cm 2
(D) 12 ´ 10 4 A cm 2
m n = 1000 cm 2 V - s ,
29. In a particular semiconductor the donor impurity
m p = 600 cm 2 V - s ,
concentration is N d = 1014 cm -3. Assume the following
N c = N v = 1019 cm -3
These
parameters,
parameters
temperature.
The
are
measured
independent
conductivity
-6
intrinsic material is s = 10 (W - cm)
-1
m n = 1000 cm 2 V - s,
of
of
the
æ T ö
N c = 2 ´ 1019ç
÷
è 300 ø
at T = 300 K.
The conductivity at T = 500 K is
(A) 2 ´ 10 -4 (W - cm) -1
(B) 4 ´ 10 -5 (W - cm) -1
(C) 2 ´ 10 -5 (W - cm) -1
(D) 6 ´ 10 -3 (W - cm) -1
æ T ö
N v = 1 ´ 1019ç
÷
è 300 ø
32
cm -3,
32
cm -3,
. eV.
E g = 11
24. An n - type silicon sample has a resistivity of 5
W - cm at T = 300 K. The mobility is m n = 1350
cm 2 V - s. The donor impurity concentration is
(A) 2.86 ´ 10 -14 cm -3
(C) 11.46 ´ 10 cm
15
25.
(B) 9.25 ´ 1014 cm -3
-3
(D) 11
. ´ 10
-15
cm
-3
An electric field of E = 10 V cm is applied. The
electric current density at 300 K is
(A) 2.3 A cm 2
(B) 1.6 A cm 2
(C) 9.6 A cm 2
(D) 3.4 A cm 2
Statement for Q.30-31:
In a silicon sample the electron concentration
drops linearly from 1018 cm -3 to 1016 cm -3 over a length
A semiconductor has following parameter
of 2.0 mm. The current density due to the electron
m n = 7500 cm 2 V - s,
diffusion current is ( Dn = 35 cm 2 s).
m p = 300 cm 2 V - s,
(A) 9.3 ´ 10 A cm
4
(B) 2.8 ´ 10 A cm
2
4
(C) 9.3 ´ 10 9A cm 2
2
ni = 3.6 ´ 1012 cm -3
(D) 2.8 ´ 10 9 A cm 2
30.
26. In a GaAs sample the electrons are moving under
an
electric
field
of
5
kV cm and
16
concentration is uniform at 10
the
carrier
-3
cm . The electron
7
When
conductivity
is
minimum,
the
hole
concentration is
(A) 7.2 ´ 1011 cm -3
(B) 1.8 ´ 1013 cm -3
(C) 1.44 ´ 1011 cm -3
(D) 9 ´ 1013 cm -3
velocity is the saturated velocity of 10 cm s. The drift
31. The minimum conductivity is
current density is
(A) 1.6 ´ 10 A cm
4
(C) 1.6 ´ 108 A cm 2
2
2
(A) 0.6 ´ 10 -3 (W - cm) -1
(B) 17
. ´ 10 -3 (W - cm) -1
(D) 2.4 ´ 108 A cm 2
(C) 2.4 ´ 10 -3 (W - cm) -1
(D) 6.8 ´ 10 -3 (W - cm) -1
(B) 2.4 ´ 10 A cm
4
www.nodia.co.in
110
Semiconductor Physics
32. A particular intrinsic semiconductor has a
Chap 2.1
æ -x ö
ç
÷
ç Lp ÷
ø
Hole concentration p0 = 1015 e è
resistivity of 50 (W - cm) at T = 300 K and 5 (W - cm) at
T = 330 K. If change in mobility with temperature is
cm -3,x ³ 0
æ -x ö
÷
ç
neglected, the bandgap energy of the semiconductor is
Electron concentration n0 = 5 ´ 1014 e è L n ø cm -3,x £ 0
(A) 1.9 eV
(B) 1.3 eV
Hole diffusion coefficient Dp = 10 cm 2 s
(C) 2.6 eV
(D) 0.64 eV
Electron diffusion coefficients Dn = 25 cm 2 s
a
Hole diffusion length Lp = 5 ´ 10 -4 cm,
semiconductor. If only the first mechanism were
Electron diffusion length Ln = 10 -3 cm
33.
Three
scattering
mechanism
exist
in
present, the mobility would be 500 cm V - s. If only
2
the second mechanism were present, the mobility
The total current density at x = 0 is
would be 750 cm 2 V - s. If only third mechanism were
(A) 1.2 A cm 2
(B) 5.2 A cm 2
present, the mobility would be 1500 cm 2 V - s. The net
(C) 3.8 A cm 2
(D) 2 A cm 2
mobility is
(A) 2750 cm 2 V - s
(B) 1114 cm 2 V - s
(C) 818 cm 2 V - s
(D) 250 cm 2 V - s
37. A germanium Hall device is doped with 5 ´ 1015
donor atoms per cm 3 at T = 300 K. The device has the
geometry d = 5 ´ 10 -3 cm, W = 2 ´ 10 -2 cm and L = 0.1
34. In a sample of silicon at T = 300 K, the electron
cm. The current is I x = 250 mA, the applied voltage is
mV, and the magnetic flux is Bz = 5 ´ 10 -2
concentration varies linearly with distance, as shown
Vx = 100
in fig. P2.1.34. The diffusion current density is found
tesla. The Hall voltage is
to be J n = 0.19 A cm . If the electron diffusion
(A) -0.31mV
(B) 0.31 mV
coefficient is Dn = 25 cm s, The electron concentration
(C) 3.26 mV
(D) -3.26 mV
2
2
at is
Statement for Q.38-39:
A silicon Hall device at T = 300 K has the
n(cm-3)
5´10
14
geometry d = 10 -3 cm , W = 10 -2 cm, L = 10 -1 cm. The
following parameters are measured: I x = 0.75 mA,
Vx = 15 V, V H = +5.8 mV, tesla
n(0)
0.010
0
x(cm)
38. The majority carrier concentration is
Fig. P2.1.34
(A) 4.86 ´ 10 cm
8
-3
(C) 9.8 ´ 10 26 cm -3
(A) 8 ´ 1015 cm -3, n - type
-3
(B) 8 ´ 1015 cm -3, p - type
(D) 5.4 ´ 1015 cm -3
(C) 4 ´ 1015 cm -3, n - type
(B) 2.5 ´ 10 cm
13
35. The hole concentration in p - type GaAs is given by
39. The majority carrier mobility is
xö
æ
p = 1016 ç 1 - ÷ cm -3 for 0 £ x £ L
Lø
è
where L = 10 mm. The hole diffusion coefficient is
10 cm 2 s. The hole diffusion current density at
x = 5 mm is
(A) 20 A cm 2
(B) 16 A cm 2
(C) 24 A cm 2
(D) 30 A cm 2
(A) 430 cm 2 V - s
(B) 215 cm 2 V - s
(C) 390 cm 2 V - s
(D) 195 cm 2 V - s
40. In a semiconductor n0 = 1015 cm -3 and ni = 1010
cm -3. The excess-carrier life time is 10 -6 s. The excess
hole
concentration
is
dp = 4 ´ 1013
cm -3.
electron-hole recombination rate is
36. For a particular semiconductor sample consider
following parameters:
(D) 4 ´ 1015 cm -3, p - type
(A) 4 ´ 1019 cm -3s -1
(B) 4 ´ 1014 cm -3s -1
(C) 4 ´ 10 24 cm -3s -1
(D) 4 ´ 1011 cm -3s -1
www.nodia.co.in
The
Chap 2.1
Semiconductor Physics
111
41. A semiconductor in thermal equilibrium, has a
Solutions
hole concentration of p0 = 1016 cm -3 and an intrinsic
concentration of ni = 1010 cm -3. The minority carrier
life
time
4 ´ 10 -7s.
is
The
thermal
equilibrium
recombination rate of electrons is
1. (D) n = N c N v e
2
i
æ -Eg ö
-ç
ç kT ÷÷
è
ø
(A) 2.5 ´ 10 22 cm -3 s -1
(B) 5 ´ 1010 cm -3 s -1
æ 400 ö
Vt = 0.0259ç
÷ = 0.0345
è 300 ø
(C) 2.5 ´ 1010 cm -3 s -1
(D) 5 ´ 10 22 cm -3 s -1
For Ge at 300 K,
N c = 104
. ´ 1019, N v = 6.0 ´ 1018 , E g = 0.66 eV
Statement for Q.42-43:
æ 0 .66 ö
3
concentration of N d = 106 cm -3. The minority carrier
-ç
÷
æ 400 ö
0 .0345 ø
ni2 = 104
. ´ 1019 ´ 6.0 ´ 1018 ´ ç
÷ ´e è
è 300 ø
hole lifetime is t p 0 = 10 m s.
Þ
A n-type
silicon
sample
contains
a
donor
42. The thermal equilibrium generation rate of hole is
ni = 8.5 ´ 1014 cm -3
æ -Eg ö
-ç
ç kT ÷÷
è
ø
(A) 5 ´ 108 cm -3 s -1
(B) 10 4 cm -3 s -1
2. (C) n = N c N v e
(C) 2. 25 ´ 10 9 cm -3 s -1
(D) 10 3 cm -3 s -1
æ T ö - çè
(1012 ) 2 = 2.8 ´ 1019 ´ 104
. ´ 1019ç
÷ e
è 300 ø
2
i
3
43. The thermal equilibrium generation rate for
3
T e
electron is
(A) 1125
.
´ 10 9 cm -3 s -1
(B) 2.25 ´ 10 9 cm -3 s -1
(C) 8.9 ´ 10 -10 cm -3 s -1
(D) 4 ´ 10 9 cm -3 s -1
-13´10 3
T
= 9.28 ´ 10 -8 , By trial T = 382 K
-
44. A n -type silicon sample contains a donor
concentration of N d = 1016 cm -3. The minority carrier
hole lifetime is t p 0 = 20 m s. The lifetime of the majority
(B) 8.9 ´ 10 -6 s
(C) 4.5 ´ 10
-17
s
(D) 113
. ´ 10
-7
E gA
4. (A) p0 =
45. In a silicon semiconductor material the doping
concentration are N a = 1016 cm -3 and N a = 0. The
= 2257.5
Þ
-
6. (B) p0 = N v e
( EF - Ev )
-
kT
-0 .22
= 104
. ´ 1019 e 0 .0259 = 2 ´ 1015 cm -3
( EF - Ev )
kT
Þ
æN ö
EF - Ev = kT ln ç v ÷
è p0 ø
equilibrium recombination rate is Rp 0 = 1011 cm -3s -1 . A
At 300 K, N v = 10
. ´ 1019 cm -3
uniform generation rate produces an excess- carrier
æ 104
. ´ 1019 ö
EF - Ev = 0.0259 lnç
÷ = 0. 239 eV
1015
è
ø
concentration of dn = dp = 1014 cm -3. The factor, by
which the total recombination rate increase is
(A) 2.3 ´ 1013
(B) 4.4 ´ 1013
(C) 2.3 ´ 10 9
(D) 4.4 ´ 10 9
niA
= 47.5
niB
ni2 (15
. ´ 1010 ) 2
=
= 4.5 ´ 1015 cm -3
n0
5 ´ 10 4
5. (A) p0 = N v e
s
æ E gA - E gB ö
÷÷
kT
ø
-ç
ç
n2
e kT
3. (B) iA
= E gB = e è
2
niB
e kT
carrier is ( ni = 15
. ´ 1010 cm -3)
(A) 8.9 ´ 106 s
æ 1 .12 e ö
÷
kT ø
( Ec - EF )
n0 = N c e
kT
At 300 K, N c = 2.8 ´ 1019 cm -3
Ec - EF = 112
. - 0. 239 = 0.881 eV
n0 = 4.4 ´ 10 4 cm -3
***********
2
7. (C) p0 =
Na - Nd
N ö
æ
+ ç N a - d ÷ + ni2
2
2 ø
è
For Ge ni = 2.4 ´ 10 3
2
p0 =
www.nodia.co.in
æ 1013 ö
1013
+ ç
÷ + (2.4 ´ 1013) 2 = 2.95 ´ 1013 cm -3
2
2
è
ø
112
Semiconductor Physics
2
8. (A) n0 =
2
Nd - Na
N ö
æ
+ ç N d - a ÷ + ni2
2
2 ø
è
n0 =
2
=
Chap 2.1
æ 5 ´ 1015 ö
5 ´ 1015
+ ç
÷ + (2.4 ´ 1013) 2 = 5 ´ 1015 cm -3
2
2
è
ø
æ 1014 ö
1014
+ ç
.
´ 1014 cm -3
÷ + (2.4 ´ 1013) 2 = 1055
2
2
è
ø
æn ö
æ 1055
.
´ 1014 ö
EF - EFi = kT ln ç 0 ÷ = 0.0259 ln ç
÷
13
è ni ø
è 2.4 ´ 10 ø
= 0.0383 eV
n2 (2.4 ´ 1013) 2
9. (B) p0 = i =
= 195
. ´ 1013 cm -3
p0
2.95 ´ 1013
2
14. (A) n0 =
10. (A) Since N a > N d , thus material is p-type ,
p0 = N a - N d = 2.5 ´ 10 - 1 ´ 10 = 15
. ´ 10 cm
3
3
13
ni = 0.05 n0
-3
æ 200 ö
11. (D) kT = 0.0259ç
÷ = 0.0173 eV
è 300 ø
æ 1 .42 ö
n02
5
15. (A) EFi - Emidgap =
. ´ 10 cm
N c = 104
-3
, N v = 6 ´ 10
18
3
æ 400 ö
ni2 = 104
. ´ 1019 ´ 6 ´ 1018 ç
÷ e
è 300 ø
cm
0 .66 ö
- æç
÷
è 0 .0345 ø
-3
, E g = 0.66 eV
= 7. 274 ´ 10 29
Ec - EF ö
÷
kT ø
N d = 5 ´ 1015 + 2.8 ´ 1019 e
0 .215 ö
- æç
÷
è 0 .0259 ø
= 119
. ´ 1016 cm -3
æN ö
17. (A) EF - EFi = kT ln ç d ÷
è Ni ø
æ 1015 ö
= 0.0259 ln ç
÷ = 0.287 eV
. ´ 1010 ø
è 15
ni = 8.528 ´ 1014 cm -3
2
Na
æN ö
+ ç a ÷ + ni2
2
è 2 ø
2
æ 10 ö
10
+ ç
÷ + 7. 274 ´ 10 29
2
è 2 ø
p0 = 1.489 ´ 1015 cm -3
æ p ö
æ 1.489 ´ 1015 ö
EFi - EF = kT ln ç 0 ÷ = 0.0345 ln ç
÷
14
è n0 ø
è 8.528 ´ 10 ø
= 0.019 eV
18. (D) R =
s=
Nd
æN ö
+ ç d ÷ + ni2
2
è 2 ø
For Ge at T = 300 K, ni = 2.4 ´ 1013 cm -3
V
20
=
= 200 W
I 100m
L
2 ´ 10 -3
=
= 0.01( W - cm) -1
RA (200) (0.001)
s » em n n0 , For Si, m n = 1350.
Þ
0.01 = (1.6 ´ 10 -19)(1350)n0
n0 = 4.6 ´ 1013 cm -3,
n0 >> ni
2
13. (A) n0 =
- æç
è
For Si, N c = 2.8 ´ 1019 cm -3
19
Þ
æ 1 .42 ´300 ö
æ mp* ö
3
kT ln ç * ÷ = 0.0446 eV
çm ÷
4
è pø
16. (A) n0 = N d - N a = N c e
For Ge at 300 K,
=
E g = 1.42 eV
N v = 7 ´ 1018 ,
By trial and error T = 763 K
æ -Eg ö
-ç
ç kT ÷÷
è
ø
15
For GaAs at T = 300 K,
æ T ö - çè 0 .0259T ÷ø
(1504
.
) 2 = 4.7 ´ 1017 ´ 7 ´ 1018 ç
÷ e
è 300 ø
æ 400 ö
12. (A) kT = 0.0259ç
÷ = 0.0345 eV
è 300 ø
15
æ -Eg ö
-ç
ç kT ÷÷
è
ø
3
n0 = 5 ni = 3.3 cm -3
p0 =
.
n0 = 30075
´ 1015 cm -3
N c = 4.7 ´ 1017,
ni = 1.48 cm -3
ni2 = N c N v e
Þ
n = Nc Nv e
E g = 1.42 eV
N v = 7.0 ´ 1018 ,
3
ni2 = n0 p0 = 5 p02 =
n0 = 15
. ´ 1015 + (15
. ´ 1015) 2 + (0.05 n0 ) 2
2
i
æ 200 ö - çè 0 .0173 ÷ø
ni2 = 4.7 ´ 1017 ´ 7.0 ´ 1018 ç
÷ e
è 300 ø
Þ
Þ
ni = 1504
.
´ 1014 cm -3
For GaAs at 300 K,
N c = 4.7 ´ 1017 cm -3,
Nd
æN ö
+ ç d ÷ + ni2
2
è 2 ø
Þ
n0 = N d
19. (B) N d >> ni
R=
www.nodia.co.in
Þ
L
L
,
=
s A em n N d A
n0 = N d , s » e m n n0 ,
Chap 2.1
=
Semiconductor Physics
(1.6 ´ 10
0.1
= 11.36 kW
)(1100)(5 ´ 1016 )(100 ´ 10 -8 )
-19
L
,
sA
20. (A) R =
Þ n0 =
s » e m n n0 ,
R=
L
em n n0 A
et sc E (1.6 ´ 10 -19)(10 -13)(10 5)
=
(0.067)(9.1 ´ 10 -31 )
me*
= 26.2 ´ 10 3 m s = 2.6 ´ 106 cm s
L
em n AR
28. (A) N d >> ni
-4
20 ´ 10
= 8.7 ´ 1015 cm -3
(0.9)(1.6 ´ 10 )( 8000)(10 -6 )(2 ´ 10 3)
Þ
J = em n n0 E = (1.6 ´ 10
-19
21. (B) s » e m n n0 , R =
=
26. (A) J = evn = (1.6 ´ 10 -19)(10 7)(1016 ) = 1.6 A cm 2
27. (A) vd =
n0 = 0.9 N d
=
113
L
L
, n0 =
sA
em n AR
10 ´ 10 -4
= 7.81 ´ 1017 cm -3
-19
(1.6 ´ 10 )( 800)(10 -6 )(10)
n0
7.8 ´ 10
´ 100 =
´ 100 = 78.1 %
Nd
1018
17
Efficiency =
V 6
22. (D) E =
= = 3 V/cm, vd = m n E,
L 2
vd 10 4
= 3333 cm 2 V - s
=
E
3
mn =
29. (D) n = N c N v e
2
i
At T = 300 K, ni = 391
. ´ 10 cm
-3
æN N ö
Þ E g = kT ln ç c 2 v ÷
è ni ø
æ 500 ö
At T = 500 K , kT = 0.0259ç
÷ = 0.0432 eV,
è 300 ø
Þ
= (1.6 ´ 10 -19)(1000)(1014 )(100) = 1.6 A cm 2
30. (A) s = em n n0 + em p p0 and n0 =
= (1.6 ´ 10 -19)(2.29 ´ 1013)(1000 + 600)
= 5.86 ´ 10 (W - cm)
24. (B) r =
Nd =
-1
1
1
=
s em n N d
1
1
=
= 9.25 ´ 1014 cm -3
5(1.6 ´ 10 -19)(1350)
re m n
25. (B) J n = eDn
s = em n
ni2
p0
ni2
+ em p p0 ,
p0
Þ
æm
p0 = ni ç n
çm
è p
dn
dx
æ 1018 - 1016 ö
= (1.6 ´ 10 -19)( 35)ç
÷ = 2.8 ´ 10 4 A cm 2
-4
2
´
10
è
ø
1
ö2
2
÷ = 3.6 ´ 1012 æç 7500 ö÷
÷
è 300 ø
ø
= 7. 2 ´ 1011 cm -3
31. (B) smin =
2 s i m pm n
mp + mn
= 2 en i m pm n
= 2 ´ 1.6 ´ 10 -19( 3.6 ´ 1012 ) (7500)( 300)
. ´ 10 -3(W - cm) -1
= 17
cm -3,
ni = 2.29 ´ 1013 cm -3
-3
N d = n0
1
æ 1019
ö
eV
Þ E g = 2(0.0259) ln ç
.
÷ = 1122
9
391
.
´
10
è
ø
ni2 = (1019) 2 e
Þ
( -1) em n ni2
ds
=0 =
+ em p
dp0
p02
9
1 .122 ö
- æç
÷
è 0 .0432 ø
= 7.18 ´ 1019
ni = 8.47 ´ 10 9 cm -3
N d >> ni
10 -6 = (1.6 ´ 10 -19)(1000 + 600)ni
ni2 = N c N v e
1 .1 ö
- æç
÷
è 0 .0259 ø
J = sE = em n n0 E
Þ
23. (D) s1 = eni (m n + m p )
æ Eg ö
-ç
ç kT ÷÷
è
ø
)(7500)(1016 )(10) = 120 A cm 2
æ -Eg ö
-ç
ç kT ÷÷
è
ø
= (2 ´ 1019)(1 ´ 1019) e
Þ
n0 = N d
-19
32. (B) s =
1
= emni ,
r
Eg
1
2 kT1
n
e
r1
= i1 =
Eg
1
ni 2
2 kT2
e
r2
0.1 = e
-
Eg æ 1
1 ö
÷
ç
2 k è T1
T2 ø
E g æ 330 - 300 ö
ç
÷ = ln 10
2 k è 330 ´ 300 ø
E g = 22( k300) ln 10 = 1.31 eV
33. (D)
www.nodia.co.in
1 1
1
1
=
+
+
m m1 m 2 m 3
114
Semiconductor Physics
1
1
1
1
=
+
+
m 500 750 1500
Þ
m = 250 cm 2 V - s
41. (C) n0 =
æ 5 ´ 1014 - n(0) ö
0.19 = (1.6 ´ 10 -19)(25)ç
÷,
0.010
è
ø
=
p0 =
16
x ö ö e10 Dp
æ
1
=
÷
ç
÷
L øø
L
è
(1.6 ´ 10 -19)(1016 )(10)
10 ´ 10 -4
dn0
dx
=
x =0
J = Jp + Jn =
=
1015 eDp
x =0
Lp
5 ´ 1014 eDn
Ln
1015 eDp
Lp
+
5 ´ 1014 eDn
Ln
æ 1015(10)
5 ´ 1014 (25) ö
= 1.6 ´ 10 -19ç
+
÷ = 5. 2 A cm 2
-4
-3
15
´
10
10
è
ø
=
- I x Bz
ned
38. (B) V H is positive p-type
p=
I x Bz
Þ
epd
p=
I x Bz
eV H d
(0.75 ´ 10 -3)(10 -1 )
(1.6 ´ 10 -19)(5.8 ´ 10 -3)(10 -5)
= 8.08 ´ 10 21 m -3 = 8.08 ´ 1015 cm -3
39. (C) up =
Ix L
epVxWd
=
(0.75 ´ 10 -3)(10 -3)
)( 8.08 ´ 10 21 )(15)(10 -4 )(10 -5)
(1.6 ´ 10
Recombination rate.
G = Rn 0 = Rp 0 = 2.25 ´ 10 9 cm -3s -1
44. (A) Recombination rates are equal
N d >> ni
n0 = N d , p0 =
ni2
n0
n0
n2
t p 0 = 02 t p 0
p0
ni
tn 0 =
æ 1016 ö
=ç
÷ ´ 20 ´ 10 -6 = 8.9 ´ 106 s
10
´
15
10
.
è
ø
45. (D) N d >> ni
p0 =
Rp =
Rp
Rp 0
-19
Þ
n0 = N d
n
(15
. ´ 10 )
=
= 2. 25 ´ 10 4 cm -3
n0
1016
2
i
Rp 0 =
p0
tp 0
10 2
Þ
tp 0 =
p0
2. 25 ´ 10 4
=
= 2. 25 ´ 10 7 s
Rp 0
1011
dp
1014
=
= 4.44 ´ 10 20 cm -3s -1
t p 0 2.25 ´ 10 -7
=
4.44 ´ 10 20
= 4.44 ´ 10 9
1011
***********
m p = 39
. ´ 10 -2 m 2 V - s = 390 cm 2 V - s
40. (A) n-type semiconductor
R=
n0
p
= 0,
tn 0 tp 0
2
(250 ´ 10 -6 )(5 ´ 10 -2 )
= -0.313 mV
(5 ´ 10 21 )(1.6 ´ 10 -19)(5 ´ 10 -5)
VH =
2.25 ´ 10 4
= 2.25 ´ 10 9 cm -3s -1
10 ´ 10 -6
Rn 0 =
carrier are equal . The generation rate is equal to
dp
36. (B) J p = -eDp 0
dx
37. (A) V H =
(15
. ´ 1010 ) 2
= 2.25 ´ 10 4 cm -3
1016
43. (B) Recombination rate for minority and majority
J = 16 A cm 2
J n = e Dn
n2
p0
, p0 = i , n0 = N d = 106 cm -3
n0
tp 0
42. (C) Rn 0 =
n(0) = 2.5 ´ 1013 cm -3
dp
d æ 16
= - eDp
ç 10
dx
dx è
ni2 (1010 ) 2
=
= 10 4 cm -3
p0
1016
n0
10 4
=
= 2.5 ´ 1010 cm -3s -1
t n 0 4 ´ 10 -7
Rn 0 =
dn
34. (B) J n = eDn
dx
35. (B) J = -eDp
Chap 2.1
dp 4 ´ 1013
=
= 4 ´ 1019 cm -3s -1
tp 0
10 -6
www.nodia.co.in