A transistor connected in common emitter mode | vedclass

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Question

(A) The voltage gain $(A_v)$ of a common emitter transistor amplifier is given by the formula: $A_v = \beta 	imes \frac{R_L}{R_{in}}$, where $\beta$

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(A) The voltage gain $(A_v)$ of a common emitter transistor amplifier is given by the formula: $A_v = \beta 	imes \frac{R_L}{R_{in}}$, where $\beta$ is the current gain, $R_L$ is the load resistance, and $R_{in}$ is the input resistance.Given: $\beta = 50$, $R_L = 5 \, k\Omega = 5000 \, \Omega$, and input peak voltage $V_{in} = 5 \, mV = 0.005 \, V$.Assuming the input resistance $R_{in}$ is $1 \, k\Omega$ (as implied by the ratio in the provided context), the voltage gain is calculated as:$A_v = 50 	imes \frac{5 \, k\Omega}{1 \, k\Omega} = 50 	imes 5 = 250$.

$A$ transistor connected in common emitter mode has a load resistance of $5 \, k\Omega$. If the input peak voltage is $5 \, mV$ and the current gain $(\beta)$ is $50$, find the voltage gain.

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