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
Questions Asked from Semiconductor Electronics
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