Semiconductors(11th And 12th > Physics ) Questions and answers for exam Preparation

Question 1.

Diode valve is discovered by

Richardson
Dushman
Edison
Fleming

Explanation:-

Answer: Option D. -> Fleming
:
D

Diode valve is discovered by Fleming.

Question 2.

An n-type semiconductor is

negatively charged
positively charged
neutral
negatively or positively charged depending upon the amount of impurity

Explanation:-

Answer: Option C. -> neutral
:
C

N-type semiconductor is having excess of free electrons (unbound electrons) for conduction. The total number of electrons in an atom is equal to total number of protons in the nucleus. Therefore, n –type semiconductor is neutral.

Question 3.

p-type semiconductor are

positively charged
produced when boron is added as an impurity
produced when phosphorus is added as an impurity to silicon
produced when carbon is added as an impurity to germanium.

Explanation:-

Answer: Option B. -> produced when boron is added as an impurity
:
B

Boron has valency three. When boron is doped in a pure semiconductor, then p-type semiconductor is formed.

Question 4.

The relation between number of free electrons (n) is a semiconductor and temperature (T) is given by

n $α$ T
n $α$ $T^{2}$
n $α$ $\sqrt{T}$
n $α$ $T^{\frac{3}{2}}$

Explanation:-

Answer: Option D. -> n

α

T^{\frac{3}{2}}

:
D

For semiconductor, $n = A T^{\frac{3}{2}} e^{\frac{- E_{g}}{2 k T}};$
So, $n α T^{\frac{3}{2}}$

Question 5.

The correct curve between potential (V) and distance (d) near p-n junction is

a
b
c
d

Explanation:-

Answer: Option A. -> a
:
A

V-d curve near the junction will be as shown by curve (a).

Question 6.

Two identical p-n junction may be connected in series with a battery in three ways as shown in the adjoining figure. The potential drop across the p-n junctions are equal in

circuit 1 and circuit 2
circuit 2 and circuit 3
circuit 3 and circuit 1
circuit 1 only

Explanation:-

Answer: Option B. -> circuit 2 and circuit 3
:
B

In circuit 1, N is connected with N, which is not a series combination of p-n junction. In circuit 2, each p-n junction is forward biased, hence same current flows, giving same potential difference across p-n junction. In circuit 2, each p-n junction is reverse biased; same leakage current will flow, giving equal potential difference across each p-n junction diode.

Question 7.

A sinusoidal voltage of peak value 200 volt is connected to a diode and resistor R in the circuit figure, so that halfwave rectification occurs. If the forward resistance of the diode is negligible compared to R, the RMS voltage (in volt) across R is approximately

200
100
$\frac{200}{\sqrt{2}}$
280

Explanation:-

Answer: Option C. ->

\frac{200}{\sqrt{2}}

:
C

If half wave rectifier the output voltage is the RMS voltage
$= \frac{V_{o}}{\sqrt{2}} = \frac{200}{\sqrt{2}}$

Question 8.

The average value of output direct current in a half wave rectifier is

$\frac{I_{0}}{π}$
$\frac{I_{0}}{2}$
$\frac{π I_{0}}{2}$
$\frac{2 I_{0}}{π}$

Explanation:-

Answer: Option A. ->

\frac{I_{0}}{π}

:
A

The average value of output direct current in a half wave rectifier is

= $\frac{(average value of current over a cycle)}{2}$

= $\frac{(\frac{2 I_{0}}{π})}{2} = \frac{I_{0}}{π}$

Question 9.

In the half wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be

25 Hz
50 Hz
70.7 Hz
100 Hz

Explanation:-

Answer: Option B. -> 50 Hz
:
B

In half wave rectifier, we get the output only in one half cycle of input AC therefore, the frequency of the ripple of the output is same as that of input AC ie, 50 Hz.

Question 10.

The value of current in the following diagram will be

zero
10 $^{- 2}$ A
10 A
0.025 A

Explanation:-

Answer: Option B. -> 10

^{- 2}

A
:
B

Here p-n junction as forward biased with voltage
= 5 - 3 = 2 V.
$∴$ Current $I = \frac{2}{200} = \frac{1}{100} = 10^{- 2} A$