Physics - Semiconductors Question with Solution | TestHub

$V_{\mathrm{C}\mathrm{C}}=V_{\mathrm{C}}+I_{\mathrm{C}}{R}_{\mathrm{L}}$
$\therefore 20=5+3\times {10}^3{\mathit{\text{I}}}_{\text{C}}$
$\therefore {\mathit{\text{I}}}_{\text{C}}=\frac{15}{3\times {10}^3}$
$=5\times {10}^{-3} \text{A}=5 \text{mA}$,
$\beta =\frac{{\mathit{\text{I}}}_{\mathit{\text{C}}}}{{\mathit{\text{I}}}_{\text{B}}}$
$\therefore {\mathit{\text{I}}}_{\text{B}}=\frac{{\mathit{\text{I}}}_{\text{C}}}{\beta }=\frac{5\times {10}^{-3}}{100}$
$=5\times {10}^{-5} \text{A}$
$V_{\mathrm{C}\mathrm{C}}=I_{\mathrm{B}}{R}_{\mathrm{B}}+\mathrm{P}.\mathrm{D}.$across input junction
As silicon transistor is in forward bias, the$\text{P}.\text{D}.$across input junction is$0.\text{7}$volt.
$\therefore 20=5\times {10}^{-5}{\mathit{\text{R}}}_{\text{B}}+0.7$
$\therefore {\mathit{\text{R}}}_{\text{B}}=\frac{19.3}{5\times {10}^{-5}}$
$\therefore {\mathit{\text{R}}}_{\text{B}}=3.86\times {10}^5\mathrm{\Omega }$