Question

For an $n-p-n$, transistor circuit, the collector current is $10mA$. If $90\%$ of the electrons emitted reach the collector then :

• A. the base current will be 1 mA
• B. the base current will be 10 mA
• C. the emitter current will be 1.1 mA
• D. the emitter current will be 9 mA

Answer 1

Answer: (A)

Emitter current is sum of base current and collector current.
We know that amplification factor $\alpha =\frac{\Delta I_C}{\Delta I_E}$
$\therefore \Delta I_E=\frac{\Delta I_C}{\alpha }=\frac{10}{0.9}=11mA$
Also, $\Delta I_E=\Delta I_B+\Delta I_C$
$\Rightarrow \Delta I_B=\Delta I_E-\Delta I_C$
$=11-10$
$=1mA$
Alternative : The collector current
$I_C=10mA$
As given,
$90\%$ of emitter current = collector current
or $\frac{90}{100}{I}_E=I_C$
$I_E=\frac{100}{90}{I}_C$
$=\frac{100}{90}\times 10$
$\simeq 11mA$
Thus, $I_B=I_E-I_C$
$=11-10=1mA$