1
MCQ (Single Correct Answer)

NEET 2013

In a n-type semiconductor, which of the following statement is true.
A
Holes are minority carries and pentavalent atoms are dopants.
B
Holes are majority carries and trivalent atoms are dopants.
C
Electrons are majority carries and trivalent atoms are dopants.
D
Electrons are minority carriers and pentavalent atoms are dopants.

Explanation

In n-type semiconductor, electrons are majority charge carriers and holes are minority charge carriers and pentavalent atoms are dopants.
2
MCQ (Single Correct Answer)

NEET 2013

In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will be
A
$${1 \over 3}G$$
B
$${5 \over 4}G$$
C
$${2 \over 3}G$$
D
1.5G

Explanation

Formula for trans conductance $$g$$m is

$$g$$m = $${\beta \over {{r_i}}}$$

Formula for voltage gain A,

A = $$\beta $$$${{{R_L}} \over {{r_i}}}$$

$$ \Rightarrow $$ A = $$g$$mRL

Given A = G

$$ \therefore $$ G = $$g$$mRL

$$ \Rightarrow $$ G $$ \propto $$ $$g$$m

$$ \Rightarrow $$ $${{{G_2}} \over {{G_1}}} = {{{g_{{m_2}}}} \over {{g_{{m_1}}}}}$$

$$ \Rightarrow $$ G2 = $${{0.02} \over {0.03}}$$ $$ \times $$ G = $${2 \over 3}G$$
3
MCQ (Single Correct Answer)

AIPMT 2012 Mains

To get an output Y= 1 in given circuit which of the following input will be correct ?

A
A $$ \to $$ 1,  B $$ \to $$ 0,  C $$ \to $$ 0
B
A $$ \to $$ 1,  B $$ \to $$ 0,  C $$ \to $$ 1
C
A $$ \to $$ 1,  B $$ \to $$ 1,  C $$ \to $$ 0
D
A $$ \to $$ 0,  B $$ \to $$ 1,  C $$ \to $$ 0

Explanation



The Boolen expression for the given combination is

output Y = (A + B) ยท C

So it is clear that Y = 1, when A = 1, B = 0 and C = 1
4
MCQ (Single Correct Answer)

AIPMT 2012 Mains

The input resistance of a silicon transistor is 100 $$\Omega $$. Base current is changed by 40 $$\mu $$A which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 k$$\Omega $$. The voltage gain of the amplifier is
A
2000
B
3000
C
4000
D
1000

Explanation

Current gain ($$\beta $$) :

$$\beta $$ = $${{\Delta {I_C}} \over {\Delta {I_B}}}$$

= $${{2 \times {{10}^{ - 3}}} \over {40 \times {{10}^{ - 6}}}}$$ = 50

Voltage gain of the amplifier is

AV = $$\beta {{{R_L}} \over {{R_i}}}$$

= 50 $$ \times $$ $${{4 \times {{10}^3}} \over {100}}$$ = 2000

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