Question

The diagram shows the biasing of an n-p-n transistor in common emitter configuration used in an amplifier. The design of the transistor is such that $98\%$ of the charge carriers from emitter reach the collector. If base current changes from $50\mu A$ to $100\mu A$, then the corresponding change in the voltage across the load resistance $R_L$ will be

• A. $0.25V$
• B. $0.25V$
• C. $24.5V$
• D. $49.0V$

Answer 1

The correct option is C $24.5V$


From the given data
$I_C=0.98I_E$
$\Delta I_B=50\times {10}^{-6}A$

voltage across the $R_l=R_L\times \Delta I_C$
where, $\beta =\frac{\Delta I_C}{\Delta I_B}\Delta I_C=\beta \times I_B$
here, $\beta =\frac{I_C}{I_B}$

For CE-configuration,
$I_E=0.98I_E$
$I_E=I_B+I_C$
so, $I_B=0.02I_E$

so $\beta =\frac{0.987I_E}{0.02I_C}=49$
form eq.(1)
$\Delta I_C=49\times 50\times {10}^{-6}$

So the voltage across $R_L=10\times {10}^3\times 49\times 50\times {10}^{-6}\Rightarrow R_L=24.5V$