For CE transistor amplifier- the audio signal voltage across the collector resistance of 2 k- is 4V- If the current amplification factor of the transistor is 100 and the base resistance is 1 k- then the input signal voltage is

Output current, I_C = V_C/R_C = 4/2000 = 2 mA Now, base current, I_B = I_C/β = 2 mA/100 = 20 μ A So, input voltage, V_B = I_BR_B = 20 μ A × 1 kΩ = 20 mV ...

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Standard XII

Physics

Transistor

For CE transi...

Question

For CE transistor amplifier, the audio signal voltage across the collector resistance of $2 k Ω$ is 4V. If the current amplification factor of the transistor is 100 and the base resistance is $1 k Ω$ , then the input signal voltage is

10 mV

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20 mV

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30 mV

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15 mV

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Solution

The correct option is B
20 mV

Output current, $I_{C} = \frac{V_{C}}{R_{C}} = \frac{4}{2000} = 2 m A$
Now, base current, $I_{B} = \frac{I_{C}}{β} = \frac{2 m A}{100} = 20 μ A$
So, input voltage, $V_{B} = I_{B} R_{B} = 20 μ A \times 1 k Ω = 20 m V$

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For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is

The correct option is B 20 mV Output current, IC=VCRC=42000=2 mA Now, base current, IB=ICβ=2 mA100=20 μA So, input voltage, VB=IBRB=20 μA×1 kΩ=20mV

For CE transistor amplifier, the audio signal voltage across the collector resistance of $2 k Ω$ is 4V. If the current amplification factor of the transistor is 100 and the base resistance is $1 k Ω$ , then the input signal voltage is

The correct option is B
20 mV

Output current, $I_{C} = \frac{V_{C}}{R_{C}} = \frac{4}{2000} = 2 m A$
Now, base current, $I_{B} = \frac{I_{C}}{β} = \frac{2 m A}{100} = 20 μ A$
So, input voltage, $V_{B} = I_{B} R_{B} = 20 μ A \times 1 k Ω = 20 m V$