Question

In the circuit of $CE$ amplifier, a silicon transistor is used. The value of $V_{C C}=+20\, V, R_{L}=3\, k \Omega$, collector voltage $=5 \,V , \beta=100$. Then the base resistance $R _{ B }$ should be _________$\times 10^{5} \Omega$. (Take input voltage drop across Si transistor = $0.7 \,V$ )

Answer 1

$V _{ CC }= V _{ C }+ I _{ C } R _{ L }$
$\therefore 20 =5+3 \times 10^{3} I _{ C }$
$\therefore I _{ C } =\frac{15}{3 \times 10^{3}} $
$=5 \times 10^{-3} A =5 \,mA ,$
$\beta =\frac{ I _{ C }}{ I _{ B }}$
$ \therefore I_{B}=\frac{I_{C}}{\beta} =\frac{5 \times 10^{-3}}{100} $
$=5 \times 10^{-5} A$
$ V_{C C}=I_{B} R_{B}+$ P.D. across input junction
As silicon transistor is in forward bias,
the P.D. across input junction is $ 0.7 $ volt.
$ \therefore 20=5 \times 10^{-5} R_{B}+0.7 $
$ \therefore R_{B}=\frac{19.3}{5 \times 10^{-5}}$
$ \therefore R_{B}=3.86 \times 10^{5} \Omega$
As silicon transistor is in forward bias, the P.D. across input junction is $0.7$ volt.
$\therefore 20=5 \times 10^{-5} R _{ B }+0.7$
$\therefore R _{ B }=\frac{19.3}{5 \times 10^{-5}} $
$\therefore R _{ B }=3.86 \times 10^{5} \Omega$