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Find resistance for the circuit shown to obtain $V_{CE}=5V$ and $IC=2mA$
1 Answer
written 3.5 years ago by |
Applying KVL to the Base - Emitter Loop
$V_C-I_BR_B-V_{BE}=0$
where, $I_B=\dfrac{I_C}\beta$
$V_C-\dfrac{I_C}\beta R_B-V_{BE}=0$
$V_C-V_{BE}=\dfrac{I_C}\beta R_B$
$R_B=\dfrac{V_C-V_{BE}}{\dfrac{I_C}\beta}$
$R_B=\dfrac{(V_C-V_{BE})\times\beta}{I_C}$
$R_B=\dfrac{(10-0.7)\times100}{2\times10^{-3}}$
$\therefore \underline{\underline{R_B=465\ K\Omega}}$
Applying KVL to the Collector - Emitter Loop
$V_C-I_CR_C-V_{CE}=0$
$V_C-V_{CE}=I_CR_C$
$R_C=\dfrac{V_C-V_{CE}}{I_C}$
$R_C=\dfrac{10-5}{2\times10^{-3}}$
$\therefore \underline{\underline{R_C=2.5\ K\Omega}}$