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Semiconductor Electronics: Materials, Devices and Simple Circuits MCQs for NEET — Physics Questions with Answers

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The relation between $α$ and $β$ parameters of current gains for a transistors is given by

$α = \frac{β}{1 - β}$

$α = \frac{β}{1 + β}$

$α = \frac{1 - β}{β}$

$α = \frac{1 + β}{β}$

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Explanation

(b) $i_{e} = i_{b} + i_{c} \Rightarrow \frac{i_{e}}{i_{c}} = \frac{i_{b}}{i_{c}} + 1 \Rightarrow \frac{1}{α} = \frac{1}{β} + 1 \Rightarrow α = \frac{β}{(1 + β)}$

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In the CB mode of a transistor, when the collector voltage is changed by 0.5 volt. The collector current changes by 0.05 mA. The output resistance will be

10 k $Ω$

20 k $Ω$

5 k $Ω$

2.5 k $Ω$

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Explanation

(a) Here $∆ V_{c} = 0.5 V, ∆ i_{c} = 0.05 mA = 0.05 \times 10^{- 3} A$

Output resistance is given by

$R_{out} = \frac{∆ V_{c}}{∆ i_{c}} = \frac{0.5}{0.05 \times 10^{- 3}} = 10^{4} Ω = 10 kΩ$

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Consider an NPN transistor amplifier in the common-emitter configuration. The current gain of the transistor is 100. If the collector current changes by 1 mA, what will be the change in emitter current?

1.1 mA

1.01 mA

0.01 mA

10 mA

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Explanation

(b) Current gain $β = \frac{∆ i_{c}}{∆ i_{b}} \Rightarrow ∆ i_{b} = \frac{1 \times 10^{- 3}}{100} = 10^{- 5} A = 0.01 mA$

By using, $∆ i_{e} = ∆ i_{b} + ∆ i_{c} \Rightarrow ∆ i_{e} = 1.01 + 1 = 1.01 mA$

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In a common base amplifier the phase difference between the input signal voltage and the output voltage is

$π / 4$

$π / 2$

$π$

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Explanation

(a) In CB amplifier Input and output voltage signal are in same phase.

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In NPN transistor the collector current is 10 mA. If 90% of electrons emitted reach the collector, then

Emitter current will be 9 mA

Emitter current will be 11.1 mA

Base current will be 0.1 mA

Base current will be 0.01 mA

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Explanation

(b) As 90% of the electrons reach the collector, then

$\Rightarrow$ 90% of emitter current = 10 mA

$\Rightarrow$ emitter current = $\frac{10}{90} \times 100 = 11.1 m A$

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In a transistor in CE configuration, the ratio of power gain to voltage gain is

$α$

$β / α$

$βα$

$β$

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Explanation

(d) For CE configuration voltage gain $= β \times R_{L} / R_{i}$

Power gain = $β^{2} \times R_{L} / R_{i} \Rightarrow \frac{Power gain}{Voltage gain} = β$

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In a crystal, the atoms are located at the position of

Maximum potential energy

Minimum potential energy

Zero potential energy

Infinite potential energy

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Explanation

(b) Atoms are leaked at minimum potential energy to be in stable equilibrium.

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The distance between the body centred atom and a corner atom in sodium (a = 4.225 Å) is

3.66 Å

3.17 Å

2.99 Å

2.54 Å

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In good conductors of electricity, the type of bonding that exists is

Ionic

Vander Waals

Covalent

Metallic

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Explanation

(d) Metallic bonding exists in good conductors.

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Bonding in a germanium crystal (semi- conductor) is

Metallic

Ionic

Vander Waal's type

Covalent

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Explanation

(d) Covalent bonding exists in semi-conductor.

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