A transistor is used in common-emitter mode in an amplifier circuit- When a signal of 20 mV is added to the base emitter voltage- the base current changesby 20 - A and the collector current changes by 2 mA- The load resistance is 5 k - The trans conductance and the voltage gain are A- 0-1 mho - 500B- 0-2 mho - 500C- 0-3 mho - 1000D- 0-2 mho - 100

The correct options are A 0.1 mho, 500Transistor conductance = Δ l_C/Δ V_BE = 2mA/20mA = 0.1mho The change in output voltage is R_L, Δ l_C = (5 k Ω)(2mA) = 10V ...

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Byju's Answer

Standard XII

Physics

Transistor: Working

A transistor ...

Question

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 20 mV is added to the base emitter voltage, the base current changes by 20 $μ$ A and the collector current changes by 2mA. The load resistance is 5k $Ω$ . The trans conductance and the voltage gain are

0.1 mho, 500

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0.2 mho, 500

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0.3 mho, 1000

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0.2 mho, 100

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Solution

The correct option is A 0.1 mho, 500
Transistor conductance = $\frac{Δ l_{C}}{Δ V_{B E}}$
$= \frac{2 m A}{20 m A} = 0.1$ mho

The change in output voltage is $R_{L}, Δ l_{C} = (5 k Ω) (2 m A) = 10 V$

The applied signal voltage = 20 mV
Thus, the voltage gain is $\frac{10 V}{20 m V} = 500$

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A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 20 mV is added to the base emitter voltage, the base current changes by 20 μA and the collector current changes by 2mA. The load resistance is 5kΩ. The trans conductance and the voltage gain are

The correct option is A 0.1 mho, 500Transistor conductance = ΔlCΔVBE =2mA20mA=0.1mho The change in output voltage is RL, ΔlC=(5kΩ)(2mA)=10V The applied signal voltage = 20 mV Thus, the voltage gain is 10V20mV=500

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 20 mV is added to the base emitter voltage, the base current changes by 20 $μ$ A and the collector current changes by 2mA. The load resistance is 5k $Ω$ . The trans conductance and the voltage gain are

The correct option is A 0.1 mho, 500
Transistor conductance = $\frac{Δ l_{C}}{Δ V_{B E}}$
$= \frac{2 m A}{20 m A} = 0.1$ mho

The change in output voltage is $R_{L}, Δ l_{C} = (5 k Ω) (2 m A) = 10 V$

The applied signal voltage = 20 mV
Thus, the voltage gain is $\frac{10 V}{20 m V} = 500$