Question

In an npn transistor, $108$ electrons enter the emitter in $10^{- 8}$ second. If $1\%$ electrons entering to emitter are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively:

• A. $0.8$ and $49$
• B. $0.9$ and $90$
• C. $0.88$ and $88$
• D. $0.99$ and $99$

Answer 1

Answer: (D)

$108$ electron enter the emitter in $10^{- 8}$ second
$\therefore I_{E}=\frac{108 \times 1 . 6 \times 10^{- 19}}{10^{- 8}}=172.8\times 10^{- 11}A$
$\therefore 1\%$ of $I_{E}$ is lost in base $,I_{B}=\frac{I_{E}}{100}$
$\Rightarrow 99\%$ of $I_{E}$ i.e., $\frac{99}{100}I_{E}$ enters the collector
$\Rightarrow I_{C}=0.99I_{E}$
Now, $\beta =\frac{I_{C}}{I_{B}}=\frac{0 . 99 I_{E}}{0 . 01 I_{E}}=99$