1
MCQ (Single Correct Answer)

AIPMT 2011 Mains

Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 $$ \times $$ 1016 m$$-$$3. Doping by indium increases nh to 4.5 $$ \times $$ 1022 m$$-$$3. The doped semiconductor is of
A
p-type having electron concentration ne = 5 $$ \times $$ 109 m$$-$$3
B
n-type with electron concentration ne = 5 $$ \times $$ 1022 m$$-$$3
C
p-type with electron concentration ne = 2.5 $$ \times $$ 1010 m$$-$$3
D
n-type with electron concentration ne = 2.5 $$ \times $$ 1023 m$$-$$3

Explanation

(ni)2 = ne × nh

(1.5 × 1016)2 = ne (4.5 × 1022)

So ne = 5 × 109

Now nh = 4.5 × 1022

$$ \Rightarrow $$ nh $$ >> $$ ne

Hence, semiconductor is p-type

and ne = 5 × 109 m–3
2
MCQ (Single Correct Answer)

AIPMT 2011 Mains

In the following figure, the diodes which are forward biased, are

A
(A), (B) and (D)
B
(C) only
C
(C) and (A)
D
(B) and (D)

Explanation

p-n junction is said to be forward biased when p side is at high potential than n side. It is for circuit (A) and (C).
3
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

In forward biasing of the p-n junction
A
the positive terminal of the battery is connected to p-side and the depletion region becomes thick.
B
the positive terminal of the battery is connected to n-side and the depletion region becomes thin.
C
the positive terminal of the battery is connected to n-side and the depletion region becomes thick.
D
the positive terminal of the battery is connected to p-side and the depletion region becomes thin.

Explanation

In forward biasing, the positive terminal of the battery is connected to p-side and the negative terminal to n-side of p-n junction. The forward bias voltage opposes the potential barrier. Due to it, the depletion region becomes thin.
4
MCQ (Single Correct Answer)

AIPMT 2011 Mains

A Zener diode, having breakdown voltage equal to 15 V, is used in a voltage regulator circuit shown in figure. The current through the diode is

A
5 mA
B
10 mA
C
15 mA
D
20 mA

Explanation

Voltage across 250 Ω resistance = 20 V – 15 V = 5 V

Now current through 250 $$\Omega $$ resistance: 5/250 = 20 mA

If voltage across load resistance 1 k$$\Omega $$ is 15 V, then

current through 1 kΩ is 15/1000 = 15 mA

The current through the zener diode is

= Current through 250 $$\Omega $$ resistance – Current through 1 k$$\Omega $$ resistance.

= 20 - 15

= 5 mA

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Medical

NEET

Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI