Question

In an $p-n-p$ transistor, ${10}^{10}$ holes enter the emitter in ${10}^{-6}s$. If $2$% of holes is lost in the base, then the current amplification factor is

• A. $49$
• B. $19$
• C. $29$
• D. $39$
• E. $59$

Answer 1

Answer: (A)

$49$

Given, number of holes enter the emitter $={10}^{10}$;
$Time={10}^{-6}s$
Lost of holes in the base $=2$%
We know that,
$I_e=\frac{\text{number of enter holes}\times e}{time}$
$\Rightarrow I_e=\frac{{10}^{10}\times 1.6\times {10}^{-19}}{{10}^{-6}}$
$\Rightarrow I_e=1.6\times {10}^{-3}A$
After the $2$% lost of holes in the base
$I_c=98I_e$
$\Rightarrow \beta =\frac{98}{2}=49$.