Question

An $N-P-N$ transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of $4 \, mA$ . The terminal of a $8 \, V$ battery is connected to the collector through a load resistance $R_{L}$ and to the base through a resistance $R_{B}$ . The collector-emitter voltage $V_{CE}=4V$ base-emitter voltage $V_{BE}=0.6 \, V$ and base current amplification factor $\beta _{d . c .}=100$ . Find the values of $R_{L}$ and $R_{B}$ .

• A. $R_{L}=1k\Omega $ , $R_{B}=185 \, k\Omega $
• B. $R_{L}=2k\Omega , \, R_{B}=150 \, k\Omega $
• C. $R_{L}=1 \, k\Omega , \, R_{B}=240 \, k\Omega $
• D. $R_{L}=3k\Omega , \, R_{B}=185k\Omega $

Answer 1

Answer: (A)

See figure. Potential difference across $R_{L}$

$I_{c}R_{L}=8V-V_{CE}$
$=8V-4V=4V$
Now $I_{C}R_{L}=4V$
$R_{L}=\frac{4}{4 \times 10^{- 3}}=10^{3}\Omega =1k\Omega $
Further, for base-emitter equation,
$V_{CC}=I_{B}R_{B}+V_{BE}$
or $I_{B}R_{B}=$ Potential difference across $R_{B}$
$=V_{CC}-V_{BE}=8-0.6=7.4V$
Again $I_{B}=\frac{I_{C}}{\beta }=\frac{4 \times 10^{- 3}}{100}=4\times 10^{- 5}A$
$R_{B}=\frac{7 . 4}{4 \times 10^{- 5}}=1.85\times 10^{5}\Omega =185 \, k\Omega $