In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k$$\Omega $$ is 2 V. If the base resistance is 1 k$$\Omega $$ and the current amplification of the transistor is 100, the input signal voltage is
= IBRB
= (10–5 A) (1 × 103 $$\Omega $$)
= 10–2 V = 10 mV
2
AIPMT 2012 Prelims
MCQ (Single Correct Answer)
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is
A
0.75 A
B
zero
C
0.25 A
D
0.5 A
Explanation
Here D1 is in forward bias and D2 is in
reverse bias so, D1 will conduct and D2 will not
conduct.
So, the current supplied by the battery is
I = $${5 \over {10}}$$ = 0.5 A
3
AIPMT 2011 Mains
MCQ (Single Correct Answer)
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
4
AIPMT 2011 Mains
MCQ (Single Correct Answer)
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).
Questions Asked from Semiconductor Electronics
On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
