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HC Verma Solutions: Chapter 45 - Semiconductors & Semiconductor Devices | Physics Class 11 - NEET PDF Download

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 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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FAQs on HC Verma Solutions: Chapter 45 - Semiconductors & Semiconductor Devices - Physics Class 11 - NEET

1. What are semiconductors and how do they work?
Ans. Semiconductors are materials that have electrical conductivity between that of conductors and insulators. They have the ability to conduct electric current under certain conditions. Semiconductors work by manipulating the movement of electrons in their atomic structure. The addition of impurities, also known as doping, can enhance or control their conductivity.
2. What is the difference between a conductor, an insulator, and a semiconductor?
Ans. Conductors are materials that easily allow the flow of electric current, while insulators have very low conductivity and do not allow the flow of electric current. Semiconductors, on the other hand, have intermediate conductivity, which can be controlled by factors such as temperature, impurities, and applied voltage.
3. How are semiconductors used in electronic devices?
Ans. Semiconductors are used in electronic devices to control the flow of electric current. They form the basis of various electronic components such as diodes, transistors, and integrated circuits. These components are essential for the functioning of devices like computers, smartphones, and televisions.
4. What is the concept of doping in semiconductors?
Ans. Doping is the process of intentionally adding impurities to a semiconductor material. It is done to alter its electrical properties and enhance its conductivity. Doping can introduce either extra electrons (n-type doping) or holes (p-type doping) into the semiconductor, which helps in creating different types of semiconductor devices.
5. What is the role of a diode in a circuit?
Ans. A diode is a semiconductor device that allows current to flow in only one direction. It acts as a one-way valve for electric current. Diodes are commonly used in circuits to protect sensitive components from reverse current, convert AC to DC, and regulate voltage levels. They play a crucial role in rectifying, switching, and controlling electrical currents.
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HC Verma Solutions: Chapter 45 - Semiconductors & Semiconductor Devices Notes

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HC Verma Solutions: Chapter 45 - Semiconductors & Semiconductor Devices NEET Questions

The "HC Verma Solutions: Chapter 45 - Semiconductors & Semiconductor Devices NEET Questions" guide is a valuable resource for all aspiring students preparing for the NEET exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

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