Physics - Semiconductors Question with Solution | TestHub

When switched on,

$V_{CE}=0$

$V_{CC}-R_C i_C=0$

$i_c=\frac{V_{CC}}{R_C}=\frac{5}{1\times {10}^3}=5\times {10}^{-3} \mathrm{A}$

$i_c=\beta i_B$

$i_B=\frac{i_C}{\beta }=\frac{5\times {10}^{-3}}{200}=2.5\times {10}^{-6} \mathrm{A} =\text{2.5} \mathrm{\mu A}$

Using KVL at input side,

$V_{BB}-i_B R_B-V_{BE}=0$

$V_{BB}=V_{BE}+i_B{R}_B$

$=1+100\times {10}^3\times 25\times {10}^{-6}$

$=1+2.5$

$=3.5\mathit{ }\mathrm{V}$