Chapter 45 : Semiconductors and Semiconductor Devices - HC Verma Solution, Physics Class 11 Notes | EduRev

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JEE : Chapter 45 : Semiconductors and Semiconductor Devices - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


1
CHAPTER - 45
SEMICONDUCTOR AND SEMICONDUCTOR DEVICES
1. f = 1013 kg/m
3
, V = 1 m
3
m = fV = 1013 ? 1 = 1013 kg
No.of atoms = 
3 23
1013 10 6 10
23
? ? ?
= 264.26 ? 10
26
.
a) Total no.of states = 2 N = 2 ? 264.26 ? 10
26 
= 528.52 = 5.3 ? 10
28
? 10
26
b) Total no.of unoccupied states = 2.65 ? 10
26
.
2. In a pure semiconductor, the no.of conduction electrons = no.of holes
Given volume = 1 cm ? 1 cm ? 1 mm
= 1 ? 10
–2 
? 1 ? 10
–2
? 1 ? 10
–3
= 10
–7
m
3
No.of electrons = 6 ? 10
19
? 10
–7
= 6 ? 10
12
.
Hence no.of holes = 6 ? 10
12
.
3. E = 0.23 eV, K = 1.38 ? 10
–23
KT = E
? 1.38 ? 10
–23
? T = 0.23 ? 1.6 ? 10
–19
? T = 
19 4
23
0.23 1.6 10 0.23 1.6 10
1.38 1.38 10
?
?
? ? ? ?
?
?
= 0.2676 ? 10
4
= 2670.
4. Bandgap = 1.1 eV, T = 300 K
a) Ratio = 
5 2
1.1 1.1
KT 8.62 10 3 10
?
?
? ? ?
= 42.53= 43
b) 4.253 ? = 
5
1.1
8.62 10 T
?
? ?
or T = 
5
1.1 10
4.253 8.62
?
?
= 3000.47 K.
5. 2KT = Energy gap between acceptor band and valency band
? 2 ? 1.38 ? 10
–23
? 300 
? E = (2 ? 1.38 ? 3) ? 10
–21
J = 
21
2
19
6 1.38 10 6 1.38
eV 10 eV
1.6 1.6 10
?
?
?
? ? ? ?
? ? ?
? ?
? ?
= 5.175 ? 10
–2
eV = 51.75 meV = 50 meV.
6. Given :
Band gap = 3.2 eV,
E = hc / ? = 1242 / ? = 3.2 or ? = 388.1 nm. ?
7. ? = 820 nm
E = hc / ? = 1242/820 = 1.5 eV ?
8. Band Gap = 0.65 eV, ? =?
E = hc / ? = 1242 / 0.65 = 1910.7 ? 10
–9
m = 1.9 ? 10
–5
m.
?
9. Band gap = Energy need to over come the gap
hc 1242eV nm
620nm
?
?
?
= 2.0 eV.
10. Given n = 
E / 2KT
e
? ?
, ?E = Diamon ? 6 eV ; ?E Si ?  1.1 eV
Now, n
1
= 
5
1
6
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
n
2
= 
5
2
1.1
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
51
1
10
2
n 4.14772 10
n 5.7978 10
?
?
?
?
?
= 7.15 ? 10
–42
.
Due to more ?E, the conduction electrons per cubic metre in diamond is almost zero. ?
Page 2


1
CHAPTER - 45
SEMICONDUCTOR AND SEMICONDUCTOR DEVICES
1. f = 1013 kg/m
3
, V = 1 m
3
m = fV = 1013 ? 1 = 1013 kg
No.of atoms = 
3 23
1013 10 6 10
23
? ? ?
= 264.26 ? 10
26
.
a) Total no.of states = 2 N = 2 ? 264.26 ? 10
26 
= 528.52 = 5.3 ? 10
28
? 10
26
b) Total no.of unoccupied states = 2.65 ? 10
26
.
2. In a pure semiconductor, the no.of conduction electrons = no.of holes
Given volume = 1 cm ? 1 cm ? 1 mm
= 1 ? 10
–2 
? 1 ? 10
–2
? 1 ? 10
–3
= 10
–7
m
3
No.of electrons = 6 ? 10
19
? 10
–7
= 6 ? 10
12
.
Hence no.of holes = 6 ? 10
12
.
3. E = 0.23 eV, K = 1.38 ? 10
–23
KT = E
? 1.38 ? 10
–23
? T = 0.23 ? 1.6 ? 10
–19
? T = 
19 4
23
0.23 1.6 10 0.23 1.6 10
1.38 1.38 10
?
?
? ? ? ?
?
?
= 0.2676 ? 10
4
= 2670.
4. Bandgap = 1.1 eV, T = 300 K
a) Ratio = 
5 2
1.1 1.1
KT 8.62 10 3 10
?
?
? ? ?
= 42.53= 43
b) 4.253 ? = 
5
1.1
8.62 10 T
?
? ?
or T = 
5
1.1 10
4.253 8.62
?
?
= 3000.47 K.
5. 2KT = Energy gap between acceptor band and valency band
? 2 ? 1.38 ? 10
–23
? 300 
? E = (2 ? 1.38 ? 3) ? 10
–21
J = 
21
2
19
6 1.38 10 6 1.38
eV 10 eV
1.6 1.6 10
?
?
?
? ? ? ?
? ? ?
? ?
? ?
= 5.175 ? 10
–2
eV = 51.75 meV = 50 meV.
6. Given :
Band gap = 3.2 eV,
E = hc / ? = 1242 / ? = 3.2 or ? = 388.1 nm. ?
7. ? = 820 nm
E = hc / ? = 1242/820 = 1.5 eV ?
8. Band Gap = 0.65 eV, ? =?
E = hc / ? = 1242 / 0.65 = 1910.7 ? 10
–9
m = 1.9 ? 10
–5
m.
?
9. Band gap = Energy need to over come the gap
hc 1242eV nm
620nm
?
?
?
= 2.0 eV.
10. Given n = 
E / 2KT
e
? ?
, ?E = Diamon ? 6 eV ; ?E Si ?  1.1 eV
Now, n
1
= 
5
1
6
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
n
2
= 
5
2
1.1
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
51
1
10
2
n 4.14772 10
n 5.7978 10
?
?
?
?
?
= 7.15 ? 10
–42
.
Due to more ?E, the conduction electrons per cubic metre in diamond is almost zero. ?
Semiconductor devices
2
11. ? = T
3/2 E / 2KT
e
? ?
at 4°K
? = 
5
0.74
3 / 2
2 8.62 10 4
4 e
?
?
? ? ?
? = 8 ? e
–1073.08
.
At 300 K, 
? = 
5
0.67
3 / 2 12.95
2 8.62 10 300
3 1730
300 e e
8
?
?
?
? ? ?
?
? .
Ratio = 
1073.08
12.95
8 e
[(3 1730)/ 8] e
?
?
?
? ?
= 
1060.13
64
e
3 1730
?
?
. ?
12. Total no.of charge carriers initially = 2 ? 7 ? 10
15
= 14 ? 10
15
/Cubic meter
Finally the total no.of charge carriers = 14 ? 10
17
/ m
3
We know :
The product of the concentrations of holes and conduction electrons remains, almost the same.
Let x be the no.of holes.
So, (7 ? 10
15
) ?  (7 ? 10
15
) = x ? (14 ? 10
17
– x)
? 14x ? 10
17
– x
2
= 79 ? 10
30
? x
2
– 14x ? 10
17
– 49 ? 10
30
= 0
x = 
17 2 34 30
14 10 14 10 4 49 10
2
? ? ? ? ? ?
= 14.00035 ? 10
17
.
= Increased in no.of holes or the no.of atoms of Boron added.
? 1 atom of Boron is added per 
28
15
5 10
1386.035 10
?
?
= 3.607 ? 10
–3
? 10
13 
= 3.607 ? 10
10
.
13. (No. of holes) (No.of conduction electrons) = constant.
At first :
No. of conduction electrons = 6 ? 10
19
No.of holes = 6 ? 10
19
After doping
No.of conduction electrons = 2 ? 10
23
No. of holes = x.
(6 ? 10
19
) (6 ? 10
19
) = (2 ? 10
23
)x
?
19 19
23
6 6 10
2 10
?
? ?
?
= x
? x = 18 ? 10
15
= 1.8 ? 10
16
.
14. ? = 
E / 2KT
0
e
? ?
?
?E = 0.650 eV, T = 300 K
According to question, K = 8.62 ? 10
–5
eV
E
E / 2KT
2 K 300
0 0
e 2 e
? ?
? ?
? ?
? ? ? ?
?
5
0.65
2 8.62 10 T
e
?
?
? ? ?
= 6.96561 ? 10
–5
Taking in on both sides,
We get, 
5
0.65
2 8.62 10 T'
?
?
? ? ?
= –11.874525
?
5
1 11.574525 2 8.62 10
T' 0.65
?
? ? ?
?
? T’ = 317.51178 = 318 K.
Page 3


1
CHAPTER - 45
SEMICONDUCTOR AND SEMICONDUCTOR DEVICES
1. f = 1013 kg/m
3
, V = 1 m
3
m = fV = 1013 ? 1 = 1013 kg
No.of atoms = 
3 23
1013 10 6 10
23
? ? ?
= 264.26 ? 10
26
.
a) Total no.of states = 2 N = 2 ? 264.26 ? 10
26 
= 528.52 = 5.3 ? 10
28
? 10
26
b) Total no.of unoccupied states = 2.65 ? 10
26
.
2. In a pure semiconductor, the no.of conduction electrons = no.of holes
Given volume = 1 cm ? 1 cm ? 1 mm
= 1 ? 10
–2 
? 1 ? 10
–2
? 1 ? 10
–3
= 10
–7
m
3
No.of electrons = 6 ? 10
19
? 10
–7
= 6 ? 10
12
.
Hence no.of holes = 6 ? 10
12
.
3. E = 0.23 eV, K = 1.38 ? 10
–23
KT = E
? 1.38 ? 10
–23
? T = 0.23 ? 1.6 ? 10
–19
? T = 
19 4
23
0.23 1.6 10 0.23 1.6 10
1.38 1.38 10
?
?
? ? ? ?
?
?
= 0.2676 ? 10
4
= 2670.
4. Bandgap = 1.1 eV, T = 300 K
a) Ratio = 
5 2
1.1 1.1
KT 8.62 10 3 10
?
?
? ? ?
= 42.53= 43
b) 4.253 ? = 
5
1.1
8.62 10 T
?
? ?
or T = 
5
1.1 10
4.253 8.62
?
?
= 3000.47 K.
5. 2KT = Energy gap between acceptor band and valency band
? 2 ? 1.38 ? 10
–23
? 300 
? E = (2 ? 1.38 ? 3) ? 10
–21
J = 
21
2
19
6 1.38 10 6 1.38
eV 10 eV
1.6 1.6 10
?
?
?
? ? ? ?
? ? ?
? ?
? ?
= 5.175 ? 10
–2
eV = 51.75 meV = 50 meV.
6. Given :
Band gap = 3.2 eV,
E = hc / ? = 1242 / ? = 3.2 or ? = 388.1 nm. ?
7. ? = 820 nm
E = hc / ? = 1242/820 = 1.5 eV ?
8. Band Gap = 0.65 eV, ? =?
E = hc / ? = 1242 / 0.65 = 1910.7 ? 10
–9
m = 1.9 ? 10
–5
m.
?
9. Band gap = Energy need to over come the gap
hc 1242eV nm
620nm
?
?
?
= 2.0 eV.
10. Given n = 
E / 2KT
e
? ?
, ?E = Diamon ? 6 eV ; ?E Si ?  1.1 eV
Now, n
1
= 
5
1
6
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
n
2
= 
5
2
1.1
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
51
1
10
2
n 4.14772 10
n 5.7978 10
?
?
?
?
?
= 7.15 ? 10
–42
.
Due to more ?E, the conduction electrons per cubic metre in diamond is almost zero. ?
Semiconductor devices
2
11. ? = T
3/2 E / 2KT
e
? ?
at 4°K
? = 
5
0.74
3 / 2
2 8.62 10 4
4 e
?
?
? ? ?
? = 8 ? e
–1073.08
.
At 300 K, 
? = 
5
0.67
3 / 2 12.95
2 8.62 10 300
3 1730
300 e e
8
?
?
?
? ? ?
?
? .
Ratio = 
1073.08
12.95
8 e
[(3 1730)/ 8] e
?
?
?
? ?
= 
1060.13
64
e
3 1730
?
?
. ?
12. Total no.of charge carriers initially = 2 ? 7 ? 10
15
= 14 ? 10
15
/Cubic meter
Finally the total no.of charge carriers = 14 ? 10
17
/ m
3
We know :
The product of the concentrations of holes and conduction electrons remains, almost the same.
Let x be the no.of holes.
So, (7 ? 10
15
) ?  (7 ? 10
15
) = x ? (14 ? 10
17
– x)
? 14x ? 10
17
– x
2
= 79 ? 10
30
? x
2
– 14x ? 10
17
– 49 ? 10
30
= 0
x = 
17 2 34 30
14 10 14 10 4 49 10
2
? ? ? ? ? ?
= 14.00035 ? 10
17
.
= Increased in no.of holes or the no.of atoms of Boron added.
? 1 atom of Boron is added per 
28
15
5 10
1386.035 10
?
?
= 3.607 ? 10
–3
? 10
13 
= 3.607 ? 10
10
.
13. (No. of holes) (No.of conduction electrons) = constant.
At first :
No. of conduction electrons = 6 ? 10
19
No.of holes = 6 ? 10
19
After doping
No.of conduction electrons = 2 ? 10
23
No. of holes = x.
(6 ? 10
19
) (6 ? 10
19
) = (2 ? 10
23
)x
?
19 19
23
6 6 10
2 10
?
? ?
?
= x
? x = 18 ? 10
15
= 1.8 ? 10
16
.
14. ? = 
E / 2KT
0
e
? ?
?
?E = 0.650 eV, T = 300 K
According to question, K = 8.62 ? 10
–5
eV
E
E / 2KT
2 K 300
0 0
e 2 e
? ?
? ?
? ?
? ? ? ?
?
5
0.65
2 8.62 10 T
e
?
?
? ? ?
= 6.96561 ? 10
–5
Taking in on both sides,
We get, 
5
0.65
2 8.62 10 T'
?
?
? ? ?
= –11.874525
?
5
1 11.574525 2 8.62 10
T' 0.65
?
? ? ?
?
? T’ = 317.51178 = 318 K.
Semiconductor devices
3
15. Given band gap = 1 eV
Net band gap after doping = (1 – 10
–3
)eV = 0.999 eV
According to the question, KT
1
= 0.999/50
? T
1
= 231.78 = 231.8
For the maximum limit KT
2
= 2 ? 0.999
? T
2
= 
3
2
5
2 1 10 2
10 23.2
8.62 8.62 10
?
?
? ?
? ? ?
?
.
Temperature range is (23.2 – 231.8).
16. Depletion region ‘d’ = 400 nm = 4 ? 10
–7
m
Electric field E = 5 ? 10
5
V/m
a) Potential barrier V = E ? d = 0.2 V
b) Kinetic energy required = Potential barrier ? e = 0.2 eV [Where e = Charge of electron]
17. Potential barrier = 0.2 Volt
a) K.E. = (Potential difference) ? e = 0.2 eV (in unbiased cond
n
)
b) In forward biasing
KE + Ve = 0.2e
? KE = 0.2e – 0.1e = 0.1e.
c) In reverse biasing
KE – Ve = 0.2 e
? KE = 0.2e + 0.1e = 0.3e.
18. Potential barrier ‘d’ = 250 meV
Initial KE of hole = 300 meV
We know : KE of the hole decreases when the junction is forward biased and increases when reverse 
blased in the given ‘Pn’ diode.
So,
a) Final KE = (300 – 250) meV = 50 meV
b) Initial KE = (300 + 250) meV = 550 meV
19. i
1
= 25 ?A, V = 200 mV, i
2
= 75 ?A
a) When in unbiased condition drift current = diffusion current
? Diffusion current = 25 ?A.
b) On reverse biasing the diffusion current becomes ‘O’.
c) On forward biasing the actual current be x.
x – Drift current = Forward biasing current
? x – 25 ?A = 75 ?A
? x = (75 + 25) ?A = 100 ?A. ?
20. Drift current = 20 ?A = 20 ? 10
–6
A.
Both holes and electrons are moving
So, no.of electrons = 
6
19
20 10
2 1.6 10
?
?
?
? ?
= 6.25 ? 10
13
.
21. a) e
aV/KT
= 100
?
5
V
8.62 10 300
e
?
? ?
= 100
?
5
V
8.62 10 300
?
? ?
= 4.605 ? V = 4.605 ? 8.62 ? 3 ? 10
–3
= 119.08 ? 10
–3
R = 
ev / KT 1
0
V V
I I (e )
?
? = 
3 3
6 5
119.08 10 119.08 10
10 10 (100 1) 99 10
? ?
? ?
? ?
?
? ? ? ?
= 1.2 ? 10
2
.
V
0
= I
0
R
? 10 ?  10
-6
? 1.2 ? 10
2
= 1.2 ? 10
–3
= 0.0012 V.
Page 4


1
CHAPTER - 45
SEMICONDUCTOR AND SEMICONDUCTOR DEVICES
1. f = 1013 kg/m
3
, V = 1 m
3
m = fV = 1013 ? 1 = 1013 kg
No.of atoms = 
3 23
1013 10 6 10
23
? ? ?
= 264.26 ? 10
26
.
a) Total no.of states = 2 N = 2 ? 264.26 ? 10
26 
= 528.52 = 5.3 ? 10
28
? 10
26
b) Total no.of unoccupied states = 2.65 ? 10
26
.
2. In a pure semiconductor, the no.of conduction electrons = no.of holes
Given volume = 1 cm ? 1 cm ? 1 mm
= 1 ? 10
–2 
? 1 ? 10
–2
? 1 ? 10
–3
= 10
–7
m
3
No.of electrons = 6 ? 10
19
? 10
–7
= 6 ? 10
12
.
Hence no.of holes = 6 ? 10
12
.
3. E = 0.23 eV, K = 1.38 ? 10
–23
KT = E
? 1.38 ? 10
–23
? T = 0.23 ? 1.6 ? 10
–19
? T = 
19 4
23
0.23 1.6 10 0.23 1.6 10
1.38 1.38 10
?
?
? ? ? ?
?
?
= 0.2676 ? 10
4
= 2670.
4. Bandgap = 1.1 eV, T = 300 K
a) Ratio = 
5 2
1.1 1.1
KT 8.62 10 3 10
?
?
? ? ?
= 42.53= 43
b) 4.253 ? = 
5
1.1
8.62 10 T
?
? ?
or T = 
5
1.1 10
4.253 8.62
?
?
= 3000.47 K.
5. 2KT = Energy gap between acceptor band and valency band
? 2 ? 1.38 ? 10
–23
? 300 
? E = (2 ? 1.38 ? 3) ? 10
–21
J = 
21
2
19
6 1.38 10 6 1.38
eV 10 eV
1.6 1.6 10
?
?
?
? ? ? ?
? ? ?
? ?
? ?
= 5.175 ? 10
–2
eV = 51.75 meV = 50 meV.
6. Given :
Band gap = 3.2 eV,
E = hc / ? = 1242 / ? = 3.2 or ? = 388.1 nm. ?
7. ? = 820 nm
E = hc / ? = 1242/820 = 1.5 eV ?
8. Band Gap = 0.65 eV, ? =?
E = hc / ? = 1242 / 0.65 = 1910.7 ? 10
–9
m = 1.9 ? 10
–5
m.
?
9. Band gap = Energy need to over come the gap
hc 1242eV nm
620nm
?
?
?
= 2.0 eV.
10. Given n = 
E / 2KT
e
? ?
, ?E = Diamon ? 6 eV ; ?E Si ?  1.1 eV
Now, n
1
= 
5
1
6
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
n
2
= 
5
2
1.1
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
51
1
10
2
n 4.14772 10
n 5.7978 10
?
?
?
?
?
= 7.15 ? 10
–42
.
Due to more ?E, the conduction electrons per cubic metre in diamond is almost zero. ?
Semiconductor devices
2
11. ? = T
3/2 E / 2KT
e
? ?
at 4°K
? = 
5
0.74
3 / 2
2 8.62 10 4
4 e
?
?
? ? ?
? = 8 ? e
–1073.08
.
At 300 K, 
? = 
5
0.67
3 / 2 12.95
2 8.62 10 300
3 1730
300 e e
8
?
?
?
? ? ?
?
? .
Ratio = 
1073.08
12.95
8 e
[(3 1730)/ 8] e
?
?
?
? ?
= 
1060.13
64
e
3 1730
?
?
. ?
12. Total no.of charge carriers initially = 2 ? 7 ? 10
15
= 14 ? 10
15
/Cubic meter
Finally the total no.of charge carriers = 14 ? 10
17
/ m
3
We know :
The product of the concentrations of holes and conduction electrons remains, almost the same.
Let x be the no.of holes.
So, (7 ? 10
15
) ?  (7 ? 10
15
) = x ? (14 ? 10
17
– x)
? 14x ? 10
17
– x
2
= 79 ? 10
30
? x
2
– 14x ? 10
17
– 49 ? 10
30
= 0
x = 
17 2 34 30
14 10 14 10 4 49 10
2
? ? ? ? ? ?
= 14.00035 ? 10
17
.
= Increased in no.of holes or the no.of atoms of Boron added.
? 1 atom of Boron is added per 
28
15
5 10
1386.035 10
?
?
= 3.607 ? 10
–3
? 10
13 
= 3.607 ? 10
10
.
13. (No. of holes) (No.of conduction electrons) = constant.
At first :
No. of conduction electrons = 6 ? 10
19
No.of holes = 6 ? 10
19
After doping
No.of conduction electrons = 2 ? 10
23
No. of holes = x.
(6 ? 10
19
) (6 ? 10
19
) = (2 ? 10
23
)x
?
19 19
23
6 6 10
2 10
?
? ?
?
= x
? x = 18 ? 10
15
= 1.8 ? 10
16
.
14. ? = 
E / 2KT
0
e
? ?
?
?E = 0.650 eV, T = 300 K
According to question, K = 8.62 ? 10
–5
eV
E
E / 2KT
2 K 300
0 0
e 2 e
? ?
? ?
? ?
? ? ? ?
?
5
0.65
2 8.62 10 T
e
?
?
? ? ?
= 6.96561 ? 10
–5
Taking in on both sides,
We get, 
5
0.65
2 8.62 10 T'
?
?
? ? ?
= –11.874525
?
5
1 11.574525 2 8.62 10
T' 0.65
?
? ? ?
?
? T’ = 317.51178 = 318 K.
Semiconductor devices
3
15. Given band gap = 1 eV
Net band gap after doping = (1 – 10
–3
)eV = 0.999 eV
According to the question, KT
1
= 0.999/50
? T
1
= 231.78 = 231.8
For the maximum limit KT
2
= 2 ? 0.999
? T
2
= 
3
2
5
2 1 10 2
10 23.2
8.62 8.62 10
?
?
? ?
? ? ?
?
.
Temperature range is (23.2 – 231.8).
16. Depletion region ‘d’ = 400 nm = 4 ? 10
–7
m
Electric field E = 5 ? 10
5
V/m
a) Potential barrier V = E ? d = 0.2 V
b) Kinetic energy required = Potential barrier ? e = 0.2 eV [Where e = Charge of electron]
17. Potential barrier = 0.2 Volt
a) K.E. = (Potential difference) ? e = 0.2 eV (in unbiased cond
n
)
b) In forward biasing
KE + Ve = 0.2e
? KE = 0.2e – 0.1e = 0.1e.
c) In reverse biasing
KE – Ve = 0.2 e
? KE = 0.2e + 0.1e = 0.3e.
18. Potential barrier ‘d’ = 250 meV
Initial KE of hole = 300 meV
We know : KE of the hole decreases when the junction is forward biased and increases when reverse 
blased in the given ‘Pn’ diode.
So,
a) Final KE = (300 – 250) meV = 50 meV
b) Initial KE = (300 + 250) meV = 550 meV
19. i
1
= 25 ?A, V = 200 mV, i
2
= 75 ?A
a) When in unbiased condition drift current = diffusion current
? Diffusion current = 25 ?A.
b) On reverse biasing the diffusion current becomes ‘O’.
c) On forward biasing the actual current be x.
x – Drift current = Forward biasing current
? x – 25 ?A = 75 ?A
? x = (75 + 25) ?A = 100 ?A. ?
20. Drift current = 20 ?A = 20 ? 10
–6
A.
Both holes and electrons are moving
So, no.of electrons = 
6
19
20 10
2 1.6 10
?
?
?
? ?
= 6.25 ? 10
13
.
21. a) e
aV/KT
= 100
?
5
V
8.62 10 300
e
?
? ?
= 100
?
5
V
8.62 10 300
?
? ?
= 4.605 ? V = 4.605 ? 8.62 ? 3 ? 10
–3
= 119.08 ? 10
–3
R = 
ev / KT 1
0
V V
I I (e )
?
? = 
3 3
6 5
119.08 10 119.08 10
10 10 (100 1) 99 10
? ?
? ?
? ?
?
? ? ? ?
= 1.2 ? 10
2
.
V
0
= I
0
R
? 10 ?  10
-6
? 1.2 ? 10
2
= 1.2 ? 10
–3
= 0.0012 V.
Semiconductor devices
4
c) 0.2 = 
eV / KT
0
KT
e
ei
?
K = 8.62 ? 10
–5
eV/K, T = 300 K
i
0
= 10 ? 10
–5
A.
Substituting the values in the equation and solving 
We get V = 0.25
22. a) i
0
= 20 ? 10
–6
A, T = 300 K, V = 300 mV
i = 
ev
1
KT
0
i e
?
= 20 ?
100
6
8.62
10 (e 1)
?
? = 2.18 A = 2 A.
b) 4 = 
2
V
6
8.62 3 10
20 10 (e 1)
?
?
? ?
? ? ?
3
V 10
6
8.62 3
4 10
e 1
20
?
?
?
? ?
?
3
V 10
8.62 3
e 200001
?
?
? ?
3
V 10
12.2060
8.62 3
?
?
?
? V = 315 mV = 318 mV.
23. a) Current in the circuit = Drift current
(Since, the diode is reverse biased = 20 ?A)
b) Voltage across the diode = 5 – (20 ? 20 ? 10
–6
) 
= 5 – (4 ? 10
-4
) = 5 V. ?
24. From the figure :
According to wheat stone bridge principle, there is no current through the 
diode.
Hence net resistance of the circuit is 
40
2
= 20 ?.
25. a) Since both the diodes are forward biased net resistance = 0
i = 
2V
2 ?
= 1 A
b) One of the diodes is forward biased and other is reverse biase.
Thus the resistance of one becomes ?.
i = 
2
2 ? ?
= 0 A.
Both are forward biased.
Thus the resistance is 0.
i = 
2
2
= 1 A.
One is forward biased and other is reverse biased.
Thus the current passes through the forward biased diode.
? i = 
2
2
= 1 A.
26. The diode is reverse biased. Hence the resistance is infinite. So, current through 
A
1
is zero.
For A
2
, current = 
2
10
= 0.2 Amp.
5 ? 20 
A ?
20 ? ? 20 ? ?
B ?
20 ? ? 20 ? ?
i ?
2V 2 ? ?
2V 2 ? ?
2V 2 ? ?
2V 2 ? ?
2V 
10 ? ?
A 1
A 2 ?
Page 5


1
CHAPTER - 45
SEMICONDUCTOR AND SEMICONDUCTOR DEVICES
1. f = 1013 kg/m
3
, V = 1 m
3
m = fV = 1013 ? 1 = 1013 kg
No.of atoms = 
3 23
1013 10 6 10
23
? ? ?
= 264.26 ? 10
26
.
a) Total no.of states = 2 N = 2 ? 264.26 ? 10
26 
= 528.52 = 5.3 ? 10
28
? 10
26
b) Total no.of unoccupied states = 2.65 ? 10
26
.
2. In a pure semiconductor, the no.of conduction electrons = no.of holes
Given volume = 1 cm ? 1 cm ? 1 mm
= 1 ? 10
–2 
? 1 ? 10
–2
? 1 ? 10
–3
= 10
–7
m
3
No.of electrons = 6 ? 10
19
? 10
–7
= 6 ? 10
12
.
Hence no.of holes = 6 ? 10
12
.
3. E = 0.23 eV, K = 1.38 ? 10
–23
KT = E
? 1.38 ? 10
–23
? T = 0.23 ? 1.6 ? 10
–19
? T = 
19 4
23
0.23 1.6 10 0.23 1.6 10
1.38 1.38 10
?
?
? ? ? ?
?
?
= 0.2676 ? 10
4
= 2670.
4. Bandgap = 1.1 eV, T = 300 K
a) Ratio = 
5 2
1.1 1.1
KT 8.62 10 3 10
?
?
? ? ?
= 42.53= 43
b) 4.253 ? = 
5
1.1
8.62 10 T
?
? ?
or T = 
5
1.1 10
4.253 8.62
?
?
= 3000.47 K.
5. 2KT = Energy gap between acceptor band and valency band
? 2 ? 1.38 ? 10
–23
? 300 
? E = (2 ? 1.38 ? 3) ? 10
–21
J = 
21
2
19
6 1.38 10 6 1.38
eV 10 eV
1.6 1.6 10
?
?
?
? ? ? ?
? ? ?
? ?
? ?
= 5.175 ? 10
–2
eV = 51.75 meV = 50 meV.
6. Given :
Band gap = 3.2 eV,
E = hc / ? = 1242 / ? = 3.2 or ? = 388.1 nm. ?
7. ? = 820 nm
E = hc / ? = 1242/820 = 1.5 eV ?
8. Band Gap = 0.65 eV, ? =?
E = hc / ? = 1242 / 0.65 = 1910.7 ? 10
–9
m = 1.9 ? 10
–5
m.
?
9. Band gap = Energy need to over come the gap
hc 1242eV nm
620nm
?
?
?
= 2.0 eV.
10. Given n = 
E / 2KT
e
? ?
, ?E = Diamon ? 6 eV ; ?E Si ?  1.1 eV
Now, n
1
= 
5
1
6
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
n
2
= 
5
2
1.1
E / 2KT
2 300 8.62 10
e e
?
?
? ?
? ? ?
?
51
1
10
2
n 4.14772 10
n 5.7978 10
?
?
?
?
?
= 7.15 ? 10
–42
.
Due to more ?E, the conduction electrons per cubic metre in diamond is almost zero. ?
Semiconductor devices
2
11. ? = T
3/2 E / 2KT
e
? ?
at 4°K
? = 
5
0.74
3 / 2
2 8.62 10 4
4 e
?
?
? ? ?
? = 8 ? e
–1073.08
.
At 300 K, 
? = 
5
0.67
3 / 2 12.95
2 8.62 10 300
3 1730
300 e e
8
?
?
?
? ? ?
?
? .
Ratio = 
1073.08
12.95
8 e
[(3 1730)/ 8] e
?
?
?
? ?
= 
1060.13
64
e
3 1730
?
?
. ?
12. Total no.of charge carriers initially = 2 ? 7 ? 10
15
= 14 ? 10
15
/Cubic meter
Finally the total no.of charge carriers = 14 ? 10
17
/ m
3
We know :
The product of the concentrations of holes and conduction electrons remains, almost the same.
Let x be the no.of holes.
So, (7 ? 10
15
) ?  (7 ? 10
15
) = x ? (14 ? 10
17
– x)
? 14x ? 10
17
– x
2
= 79 ? 10
30
? x
2
– 14x ? 10
17
– 49 ? 10
30
= 0
x = 
17 2 34 30
14 10 14 10 4 49 10
2
? ? ? ? ? ?
= 14.00035 ? 10
17
.
= Increased in no.of holes or the no.of atoms of Boron added.
? 1 atom of Boron is added per 
28
15
5 10
1386.035 10
?
?
= 3.607 ? 10
–3
? 10
13 
= 3.607 ? 10
10
.
13. (No. of holes) (No.of conduction electrons) = constant.
At first :
No. of conduction electrons = 6 ? 10
19
No.of holes = 6 ? 10
19
After doping
No.of conduction electrons = 2 ? 10
23
No. of holes = x.
(6 ? 10
19
) (6 ? 10
19
) = (2 ? 10
23
)x
?
19 19
23
6 6 10
2 10
?
? ?
?
= x
? x = 18 ? 10
15
= 1.8 ? 10
16
.
14. ? = 
E / 2KT
0
e
? ?
?
?E = 0.650 eV, T = 300 K
According to question, K = 8.62 ? 10
–5
eV
E
E / 2KT
2 K 300
0 0
e 2 e
? ?
? ?
? ?
? ? ? ?
?
5
0.65
2 8.62 10 T
e
?
?
? ? ?
= 6.96561 ? 10
–5
Taking in on both sides,
We get, 
5
0.65
2 8.62 10 T'
?
?
? ? ?
= –11.874525
?
5
1 11.574525 2 8.62 10
T' 0.65
?
? ? ?
?
? T’ = 317.51178 = 318 K.
Semiconductor devices
3
15. Given band gap = 1 eV
Net band gap after doping = (1 – 10
–3
)eV = 0.999 eV
According to the question, KT
1
= 0.999/50
? T
1
= 231.78 = 231.8
For the maximum limit KT
2
= 2 ? 0.999
? T
2
= 
3
2
5
2 1 10 2
10 23.2
8.62 8.62 10
?
?
? ?
? ? ?
?
.
Temperature range is (23.2 – 231.8).
16. Depletion region ‘d’ = 400 nm = 4 ? 10
–7
m
Electric field E = 5 ? 10
5
V/m
a) Potential barrier V = E ? d = 0.2 V
b) Kinetic energy required = Potential barrier ? e = 0.2 eV [Where e = Charge of electron]
17. Potential barrier = 0.2 Volt
a) K.E. = (Potential difference) ? e = 0.2 eV (in unbiased cond
n
)
b) In forward biasing
KE + Ve = 0.2e
? KE = 0.2e – 0.1e = 0.1e.
c) In reverse biasing
KE – Ve = 0.2 e
? KE = 0.2e + 0.1e = 0.3e.
18. Potential barrier ‘d’ = 250 meV
Initial KE of hole = 300 meV
We know : KE of the hole decreases when the junction is forward biased and increases when reverse 
blased in the given ‘Pn’ diode.
So,
a) Final KE = (300 – 250) meV = 50 meV
b) Initial KE = (300 + 250) meV = 550 meV
19. i
1
= 25 ?A, V = 200 mV, i
2
= 75 ?A
a) When in unbiased condition drift current = diffusion current
? Diffusion current = 25 ?A.
b) On reverse biasing the diffusion current becomes ‘O’.
c) On forward biasing the actual current be x.
x – Drift current = Forward biasing current
? x – 25 ?A = 75 ?A
? x = (75 + 25) ?A = 100 ?A. ?
20. Drift current = 20 ?A = 20 ? 10
–6
A.
Both holes and electrons are moving
So, no.of electrons = 
6
19
20 10
2 1.6 10
?
?
?
? ?
= 6.25 ? 10
13
.
21. a) e
aV/KT
= 100
?
5
V
8.62 10 300
e
?
? ?
= 100
?
5
V
8.62 10 300
?
? ?
= 4.605 ? V = 4.605 ? 8.62 ? 3 ? 10
–3
= 119.08 ? 10
–3
R = 
ev / KT 1
0
V V
I I (e )
?
? = 
3 3
6 5
119.08 10 119.08 10
10 10 (100 1) 99 10
? ?
? ?
? ?
?
? ? ? ?
= 1.2 ? 10
2
.
V
0
= I
0
R
? 10 ?  10
-6
? 1.2 ? 10
2
= 1.2 ? 10
–3
= 0.0012 V.
Semiconductor devices
4
c) 0.2 = 
eV / KT
0
KT
e
ei
?
K = 8.62 ? 10
–5
eV/K, T = 300 K
i
0
= 10 ? 10
–5
A.
Substituting the values in the equation and solving 
We get V = 0.25
22. a) i
0
= 20 ? 10
–6
A, T = 300 K, V = 300 mV
i = 
ev
1
KT
0
i e
?
= 20 ?
100
6
8.62
10 (e 1)
?
? = 2.18 A = 2 A.
b) 4 = 
2
V
6
8.62 3 10
20 10 (e 1)
?
?
? ?
? ? ?
3
V 10
6
8.62 3
4 10
e 1
20
?
?
?
? ?
?
3
V 10
8.62 3
e 200001
?
?
? ?
3
V 10
12.2060
8.62 3
?
?
?
? V = 315 mV = 318 mV.
23. a) Current in the circuit = Drift current
(Since, the diode is reverse biased = 20 ?A)
b) Voltage across the diode = 5 – (20 ? 20 ? 10
–6
) 
= 5 – (4 ? 10
-4
) = 5 V. ?
24. From the figure :
According to wheat stone bridge principle, there is no current through the 
diode.
Hence net resistance of the circuit is 
40
2
= 20 ?.
25. a) Since both the diodes are forward biased net resistance = 0
i = 
2V
2 ?
= 1 A
b) One of the diodes is forward biased and other is reverse biase.
Thus the resistance of one becomes ?.
i = 
2
2 ? ?
= 0 A.
Both are forward biased.
Thus the resistance is 0.
i = 
2
2
= 1 A.
One is forward biased and other is reverse biased.
Thus the current passes through the forward biased diode.
? i = 
2
2
= 1 A.
26. The diode is reverse biased. Hence the resistance is infinite. So, current through 
A
1
is zero.
For A
2
, current = 
2
10
= 0.2 Amp.
5 ? 20 
A ?
20 ? ? 20 ? ?
B ?
20 ? ? 20 ? ?
i ?
2V 2 ? ?
2V 2 ? ?
2V 2 ? ?
2V 2 ? ?
2V 
10 ? ?
A 1
A 2 ?
Semiconductor devices
5
27. Both diodes are forward biased. Thus the net diode resistance is 0.
i = 
5 5
(10 10)/10.10 5
?
?
= 1 A.
One diode is forward biased and other is reverse biased.
Current passes through the forward biased diode only.
i = 
net
V 5
R 10 0
?
?
= 1/2 = 0.5 A.
28. a) When R = 12 ?
The wire EF becomes ineffective due to the net (–)ve voltage.
Hence, current through R = 10/24 = 0.4166 = 0.42 A.
b) Similarly for R = 48 ?.
? i = 
10
(48 12) ?
= 10/60 = 0.16 A. ?
29.
Since the diode 2 is reverse biased no current will pass through it.
30. Let the potentials at A and B be V
A
and V
B
respectively.
i) If V
A
> V
B
Then current flows from A to B and the diode is in forward biased.
Eq. Resistance = 10/2 = 5 ?.
ii) If V
A
< V
B
? Then current flows from B to A and the diode is reverse biased.
Hence Eq.Resistance = 10 ?. ?
31. ?I
b
= 80 ?A – 30 ?A = 50 ?A = 50 ? 10
–6
A
?I
c
= 3.5 mA – 1 mA = –2.5 mA = 2.5 ? 10
–3
A
? = 
c
ce
b
I
V
I
? ? ?
? ?
?
? ?
= constant  
?  
3
6
2.5 10 2500
50 50 10
?
?
?
?
?
= 50.
Current gain = 50.
i ? 10 ? ?
5V 
10 ? ?
i ?
5V 
i ?
10 ? ?
10 ? ?
A ?
B ?
12 ? ?
4V 
6V R ?
1 ? ?
E ?
D ? C 
F ?
A ?
B ? 10 ? ?
V ?
I 
X ? V ?
I 
A ? B ?
i ? 1 ?
2 ?
V ?
I 
X ? V ?
I 
A ? B ?
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