A common emitter amplifier circuit,built usin | vedclass

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(B) At the saturation state,the collector-emitter voltage $V_{CE}$ becomes zero.Applying Kirchhoff's voltage law to the output loop:$V_{CC} - i_C R_C

Vedclass · Accepted Answer

$40$

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(B) At the saturation state,the collector-emitter voltage $V_{CE}$ becomes zero.Applying Kirchhoff's voltage law to the output loop:$V_{CC} - i_C R_C - V_{CE} = 0$Since $V_{CE} = 0$ at saturation:$i_C = \frac{V_{CC}}{R_C} = \frac{10\,V}{1000\,\Omega} = 10\,mA = 10 	imes 10^{-3}\,A$.The current gain $\beta$ is defined as the ratio of collector current to base current:$\beta = \frac{i_C}{i_B}$Given $\beta = 250$,we can find the minimum base current $i_B$ required for saturation:$i_B = \frac{i_C}{\beta} = \frac{10\,mA}{250} = \frac{10 	imes 10^{-3}\,A}{250} = 0.04 	imes 10^{-3}\,A = 40 	imes 10^{-6}\,A = 40\,\mu A$.

$A$ common emitter amplifier circuit,built using an $npn$ transistor,is shown in the figure. Its $dc$ current gain is $250$,$R_C = 1\,k\Omega$ and $V_{CC} = 10\,V$. What is the minimum base current for $V_{CE}$ to reach saturation? (in $\mu A$)

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