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

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If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be

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Explanation

(c) In full wave rectifier, the fundamental frequency in ripple is twice that of input frequency.

A diode having potential difference 0.5 V across its junction which does not depend on current, is connected in series with resistance of 20 Ω across source. If 0.1 A passes through resistance then what is the voltage of the source

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Explanation

The given information: Potential drop across the diode = 0.5 V, Resistance = 20 Ω, Current through resistance = 0.1 A. Voltage drop across the resistance = IR = 0.1 A × 20 Ω = 2 V. Total voltage of the source = Voltage drop across diode + Voltage drop across resistance = 0.5 V + 2 V = 2.5 V.

The emitter-base junction of a transistor is …… biased while the collector-base junction is ……. biased

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Explanation

(d) The emitter base junction is forward biased while collector base junction is reversed biased.

In a PNP transistor the base is the N-region. Its width relative to the P-region is 

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Explanation

(a) The base is always thin

A common emitter amplifier is designed with NPN transistor (α = 0.99). The input impedance is 1 KΩ and load is 10 KΩ. The voltage gain will be 

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Explanation

(c) Voltage gain = β× Resistance gain

β=α1-α=0.99(1-0.99)=99

Resistance gain = 10×103103=10

 Voltage gain = 99×10=990.

The most commonly used material for making transistor is

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Explanation

(b) Silicon is commonly used in transistor as it is a cheap semi-conductor.

The part of a transistor which is heavily doped to produce a large number of majority carriers is-

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Explanation

(b) Emitter is heavily doped

For a transistor, the current amplification factor is 0.8. The transistor is connected in common emitter configuration. The change in the collector current when the base current changes by 6 mA is 

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Explanation

(c) 

α=0.8β=0.81-0.8=4Also β=icibic=β×ib=4×6=24mA

In a common base amplifier circuit, calculate the change in base current if that in the emitter current is 2 mA and α = 0.98

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Explanation

(a) 

ic=αie=0.98×2=1.96 mA ib=ie-ic=2-1.96=0.04 mA

For a transistor, in a common emitter arrangement, the alternating current gain β is given by

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Explanation

(a) For common emitter transistor : β=IcIBVC

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