In a silicon transistor, the base current is changed by 20 . This… - Vidyadip

Question

In a silicon transistor, the base current is changed by $20\mu\text{A}.$ This results in a change of 0.02V in base to emitter voltage and a change of 2mA in the collector current.

Find the input resistance, $\beta-$ac and trans conductance of the transistor.
This transistor is used as an amplifier in CE configuration with a load resistance $5\text{k}\Omega$ What is the voltage gain of the amplifier?

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Answer

Given $\Delta\text{I}_\text{B} = 20\mu\text{A} = 20 \times 10^{-3}\text{mA} = 0.020\text{mA,}$
$\Delta\text{V}_\text{BE} = 0.02\text{V}, \Delta\text{I}_\text{C} = 2\text{mA}$
Input resistance, $\text{R}_\text{i}=\frac{\Delta\text{V}_\text{BE}}{\Delta\text{I}_\text{B}}=\frac{ 0.02}{20\times10^{-6}}\Omega=10^3\Omega=1\text{k}\Omega$
Current gain, $\beta_\text{ac}=\frac{\Delta\text{I}_\text{C}}{\Delta\text{I}_\text{B}}=\frac{\text{2mA}}{0.020\text{mA}}=100$
Trans conductance of a transistor is defined as the ratio of change in collector current to the change in base to emitter voltage at constant collector to emitter voltage, i.e.,
$\text{gm}=\Big(\frac{\Delta\text{I}_\text{C}}{\Delta\text{V}_\text{BE}}\Big)_{\text{V}_\text{CE} =\text{cons}\tan}\text{t}=\frac{2\times10^{-3}}{0.02}=0.1\text{W}^{-1}$
Voltage gain, $\text{A}_\text{v}=\frac{\text{R}_\text{L}}{\text{R}_\text{i}}\times\beta$
Given $\text{RL}=5\text{k}\Omega=5\times103\Omega,$
$\therefore\text{A}_\text{v}=\frac{5\times10^3}{1000}\times1000=500$
As CE amplifier causes a phase shift of 180° between input and output voltages, so voltage gain, $A_ν = -500$.

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