Class 12-science H C VERMA Solutions Physics Chapter 23 - Semiconductors and Semiconducting Devices
Semiconductors and Semiconducting Devices Exercise 419
Solution 1
We know ,
And no. of atoms is given as :
Total no. of states = 2n
Total no. of empty
states =
Solution 2
In case of pure semiconductor,
No. of electrons = No. of holes
Also,
And from 1 we know
Solution 3
ABC
Solution 4
a) 
b)
T = 3000.47k
Solution 5
Energy gap between two gaps is:
E=2kT
E = 50meV
Solution 6
Band gap is given as :
Solution 7
We know,
Solution 8
We know,
Solution 9
We know
And energy required to overcome the gap is band gap.
Thus ,
E=2eV
Solution 10
As
given : 
And
For
diamond ΔE is more 
the
conduction electrons is apporxinately equal zero.
Solution 11
As given :
And
Taking ratio or equation 1 and 2
Solution 12
Total
no. Of charge carriers = 
Also we know,
Solving we get
As
value is more after doping 
And
no. Of atoms of boronadded 
Solution 13
Initially ,
After doping concentration becomes,
Product of concentration remains constant. Thus,
Solution 14
As given ,
And,
Taking log on both sides and solving we get;
Solution 15
Net
band gap is: 
Also,
Now
Range is 23.2k to 231.8k
Solution 16
a) Potential barrier is :
V =Ed
=0.2V
b) kinetic energy is :
=0.2eV
Solution 17
a)
=0.2eV (No bias condition)
b) When biasing is forward.
K.E. + Ve = 0.2e
K.E.=0.2e-Ve
=(0.2-0.1)e
=0.1e
c) When bias is reversed
K.E. - Ve = 0.2e
K.E.=0.2e+Ve
=(0.2+0.1)e
=0.3e
Solution 18
a) Kinetic energy of hole is decreased if junction is forward biased. Thus,
K.E. = 300-250
=50meV
b) Kinetic energy of hole is increased if junction is reverse biased. Thus,
K.E. = 300+250
=550meV
Solution 19
a) unbiased condition:
diffusion current = drift current
Diffusion current 
b) Reverse bias condition :
Diffusion current =OA
c) For forward bias condition :
Semiconductors and Semiconducting Devices Exercise 420
Solution 20
No. Of electrons is given as :
Solution 21
a) We know,
b) We know,
And 
Differentiating w.r.t V
c) From above part
V = 0.25V
Solution 22
a) According to question :
i=2A
b)
Solving we get
V = 318mV
Solution 23
a)
b)
Solution 24
There is no current in diode by wheatstone bridge's principle
Now,
net resistance is 
R=20Ω
Solution 25
a)
According to Ist figure both diodes are in forward bias 
and current is given as :
b) In this case (according to IInd figure) one diode is in forward bias while other is in reverse bias.
Thus,
And current is given as :
I = OA
c) In this case (figure III) both diodes are forward biased
Thus,
and current is given as :
d) In this case (figure IV) one diode is forward bias and other is reverse bias but they are in parallel and thus current passes through diode which forward bias.
Solution 26
a) In A1 current is o because diode is reverse bias and thus resistance is infinite.
b) In A2 current is given as :
Solution 27
a)
According to 
figure , both diodes are forward bias
∴ Net resistance offered by diodes is O.
Now 
Now,
b) According to IInd figure one diode is forward bias and other diode is reverse bias thus current will pass through diode which is forward biased.
Thus,
Solution 28
a)
I= 0.42A
b)
I= 0.16A
Solution 29
a)
b)
Solution 30
There are two cases:
a)
If 
,
then diode is forward biased and resistance will be :
b)
If 
,
then diode is reverse biased and resistance will be:
Solution 31
We know,
Solution 32
Now,
a)
change 
Change in V = 8000v
b)
change in input voltage 
=25mV
c)
Power gain ,
Solution 33
a) X=1 ( if A = 1, B = 0, C=1)
b) X=0 (if A=B=C=1)
c) X=0 (if A=B=C=0)
Solution 34
Circuit
diagram for 
is shown below :
Solution 35
To prove 
a) If A=0
B=0
=0+1
=1
b) If A=0 B=1 or A=1 B=0
=0+1
=1
c) If A=1 B=1
=1+0
=1
