C and Si both have same lattice structure; having 4 bonding electrons in each. However, C is insulator where as Si is intrinsic semiconductor. This is because
A
In case of C the valence band is not completely filled at absolute zero temperature.
B
In case of C the conduction band is partly filled even at absolute zero temperature.
C
The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third.
D
The four bonding electrons in the case of C lie in the third orbit , whereas for Si they lie in the fourth orbit.
Explanation
Electronic configuration of 6C
6C = 1s2, 2s2 2p2
The electronic configuration of 14Si
14Si = 1s2, 2s2 2p6, 3s2 3p2
As they are away from Nucleus, so effect of
nucleus is low for Si even for Sn and Pb are
almost mettalic.
2
AIPMT 2012 Prelims
MCQ (Single Correct Answer)
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
3
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
4
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
Questions Asked from Semiconductor Electronics
On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions