1

### 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
2

### 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
3

### 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).
4

### 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

#### Graduate Aptitude Test in Engineering

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

NEET

Class 12