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1

### AIPMT 2012 Prelims

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

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
A
0.1 V
B
1.0 V
C
1 mV
D
10 mV

## Explanation

Here, RC = 2 k$$\Omega$$ = 2 × 103 $$\Omega$$

V0 = 2 V, RB = 1 k$$\Omega$$ = 1 × 103 $$\Omega$$, $$\beta$$ = 100

The output voltage, across the load RC

V0 = ICRC = 2

The collector current (IC)

IC = $${2 \over {2 \times {{10}^3}}}$$ = 10-3 A = 1 mA

Current gain($$\beta$$) = $${{{I_C}} \over {{I_B}}}$$ = 100

$$\Rightarrow$$ IB = $${{{I_C}} \over {100}}$$ = $${{{{10}^{ - 3}}} \over {100}}$$ = 10-5 A

Input voltage, Vi

= IBRB = (10–5 A) (1 × 103 $$\Omega$$) = 10–2 V = 10 mV
3

### AIPMT 2012 Prelims

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

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

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