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Standard XII

Physics

Transistor

The input res...

Question

The input resistance of a silicon transistor is $100 Ω$ . Base current is changed by $40 μ A$ , which results in a change in collector current by 2 $m A$ . This transistor is used as a common-emitter amplifier with a load of $4 k Ω$ . The voltage gain of the amplifier is

2000

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3000

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4000

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1000

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Solution

The correct option is A 2000
$V_{g} = β \frac{R_{L}}{R_{I}} = \frac{Δ I_{C}}{Δ I_{B}} \frac{R_{L}}{R_{I}} (S i n c e, β = \frac{Δ I_{C}}{Δ I_{B}}) P u t t i n g v a l u e s V_{g} = \frac{2 \times 10^{- 3}}{40 \times 10^{- 6}} \times \frac{4 \times 10^{3}}{100} V_{g} = 2000$

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The input resistance of a silicon transistor is 100 Ω. Base current is changed by 40 μA , which results in a change in collector current by 2 mA. This transistor is used as a common-emitter amplifier with a load of 4 kΩ. The voltage gain of the amplifier is

The correct option is A 2000 Vg=βRLRI =ΔICΔIBRLRI ( Since, β=ΔICΔIB)Putting values Vg=2×10−340×10−6×4×103100Vg=2000

The input resistance of a silicon transistor is $100 Ω$ . Base current is changed by $40 μ A$ , which results in a change in collector current by 2 $m A$ . This transistor is used as a common-emitter amplifier with a load of $4 k Ω$ . The voltage gain of the amplifier is

The correct option is A 2000
$V_{g} = β \frac{R_{L}}{R_{I}} = \frac{Δ I_{C}}{Δ I_{B}} \frac{R_{L}}{R_{I}} (S i n c e, β = \frac{Δ I_{C}}{Δ I_{B}}) P u t t i n g v a l u e s V_{g} = \frac{2 \times 10^{- 3}}{40 \times 10^{- 6}} \times \frac{4 \times 10^{3}}{100} V_{g} = 2000$