Semiconductor Questions — JEE Physics

Q1

The energy band gap is maximum in

A metals

B superconductors

C insulator

D semiconductor

Correct Answer

Option C

Solution

The energy band gap is maximum in insulators.

Q2

If the ratio of the concentration of electrons to that of holes in a semiconductor is

{7 \over 5}

and the ratio of currents is

{7 \over 4},

then what is the ratio of their drift velocities?

A

{5 \over 8}

B

{4 \over 5}

C

{5 \over 4}

D

{4 \over 7}

Correct Answer

Option C

Solution

{{{I_e}} \over {{I_h}}} = {{n{}_eeA{v_e}} \over {n{}_heA{v_h}}}

\Rightarrow {7 \over 4} = {7 \over 5} \times {{{v_e}} \over {{v_h}}}

\Rightarrow {{{v_e}} \over {{v_h}}} = {5 \over 4}

Q3

The photodiode is used to detect the optical signals. These diodes are preferably operated in reverse biased mode because :

A fractional change in majority carriers produce higher forward bias current

B fractional change in majority carriers produce higher reverse bias current

C fractional change in minority carriers produce higher forward bias current

D fractional change in minority carriers produce higher reverse bias current

Correct Answer

Option D

Solution

A photodiode is reverse biased.

When light falling on it produces charge carriers, the fractional change, in minority carriers is high since the original current is very small.

Q4

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than

2480

nm

is incident on it. The band gap in

(eV)

for the semiconductor is

A

2.5

eV

B

1.1

eV

C

0.7

eV

D

0.5

eV

Correct Answer

Option D

Solution

Band gap $=$ energy of photon of wavelength

2480

nm.

So,

\Delta E = {{hc} \over \lambda }

= \left( {{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {2480 \times {{10}^{ - 9}}}}} \right) \times {1 \over {1.6 \times {{10}^{ - 19}}}}eV

= 0.5\,eV

Q5

The part of a transistor which is most heavily doped to produce large number of majority carriers is

A emmiter

B base

C collector

D can be any of the above three

Correct Answer

Option A

Solution

Emitter sends the majority charge carries towards the collector. Therefore emitter is most heavily doped.

Q6

The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the

A crystal structure

B variation of the number of charge carries with temperature

C type of bonding

D variation of scattering mechanism with temperature

Correct Answer

Option B

Solution

When the temperature increases, certain bounded electrons become free which tend to promote conductivity.

Simultaneously number of collisions between electrons and positive kernels increases

Q7

In the middle of the depletion layer of a reverse- biased

p

-

n

junction, the

A electric field is zero

B potential is maximum

C electric field is maximum

D potential is zero

Correct Answer

Option A

Solution

As in reverse bias, the current through the

0000

is zero through the electric field is also zero.

Q8

In a full wave rectifier circuit operating from

50

Hz

mains frequency, the fundamental frequency in the ripple would be

A

25

Hz

B

50

Hz

C

70.7

Hz

D

100

Hz

Correct Answer

Option D

Solution

Input frequency,

f = 50\,Hz \Rightarrow T = {1 \over {50}}

For full wave rectifier,

{T_1} = {T \over 2} = {1 \over {100}} \Rightarrow {f_1} = 100\,Hz.

Q9

In a common base amplifier, the phase difference between the input signal voltage and output voltage is

A

\pi

B

{\pi \over 4}

C

{\pi \over 2}

D

0

Correct Answer

Option D

Solution

Zero; In common base amplifier circuit, input and output voltage are in the same phase.

Q10

In a common base mode of a transistor, the collector current is

5.488

mA

for an emitter current of

5.60mA.

The value of the base current amplification factor

\left( \beta \right)

will be

A

49

B

50

C

51

D

48

Correct Answer

Option A

Solution

{I_C} = 5.488\,mA,\,\,{I_e} = 5.6\,mA,\,{I_B} = {I_E} - {I_C}

\beta = {{{I_c}} \over {{I_B}}} = {{5.488} \over {5.6 - 5.485}} = 49