(D) The voltage gain $A_v$ of a $CE$ amplifier is given by the formula: $A_v = \beta \times \frac{R_C}{R_{in}}$ Given: $\beta = 50$ $R_C = 5 \, k\Omeg

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(D) The voltage gain $A_v$ of a $CE$ amplifier is given by the formula: $A_v = \beta \times \frac{R_C}{R_{in}}$ Given: $\beta = 50$ $R_C = 5 \, k\Omega$ $R_{in} = 1 \, k\Omega$ $V_{in} = 0.01 \, V$ Substituting the values: $A_v = 50 \times \frac{5 \, k\Omega}{1 \, k\Omega} = 250$ The output voltage $V_{out}$ is: $V_{out} = A_v \times V_{in} = 250 \times 0.01 \, V = 2.5 \, V$

Class 12 Physics Semiconductor Electronics Junction Transistor

Medium

The current amplification factor of a transistor is $50$. In a $CE$ amplifier circuit, the collector resistance is $5 \, k\Omega$ and the input resistance is $1 \, k\Omega$. If the input voltage is $0.01 \, V$, the output voltage is ........ $V$.

A
$1.7$
B
$0.4$
C
$4.7$
D
$2.5$

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Junction Transistor

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The current amplification factor of a transistor is $50$. In a $CE$ amplifier circuit, the collector resistance is $5 \, k\Omega$ and the input resistance is $1 \, k\Omega$. If the input voltage is $0.01 \, V$, the output voltage is ........ $V$.

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In a common base amplifier,the phase difference between the input signal voltage and the output voltage is

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In an $n-p-n$ transistor,$10^{8}$ electrons enter the emitter in $10^{-8} \,s$. If $1\%$ of electrons are lost in the base,the fraction of current that enters the collector and the current amplification factor $\beta$ are respectively:

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