First slide

Transistor

Question

In a common emitter amplifier, using output resistance of 5000 ohm and input resistance of 2000 ohm, if the peak value of input signal voltage is 10 mV and $β$ = 50, then the peak value of output voltage is

Moderate

A

$5 \times 10^{- 6} volt$
B

$2.5 \times 10^{- 4} volt$
C

$1.25 volt$
D

125 volt

Solution

In common emitter mode, the transistor is current amplifier.
$∆ I_{B} = \frac{10 \times 10^{- 3}}{2000} = 5 \times 10^{- 6} A Again, β = \frac{∆ I_{C}}{∆ I_{B}} or ∆ I_{C} = β \times ∆ I_{B} = 50 \times 5 \times 10^{- 6} A = 250 \times 10^{- 6} A = 2.5 \times 10^{- 4} A Peak value of output voltage = 2.5 \times 10^{- 4} \times 5000 volt = 1.25 volt$

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