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In the circuit shown in the figure, the input voltage $V_i$ is 20 V, $V_{BE} = 0$ and $V_{CE} = 0$. The values of $I_B$, $I_C$ and $\beta$ are given by
Explanation
$I_C = 20/4k$, $I_B = 20/500k$, and $\beta = I_C/I_B$.
Detailed Solution
$V_{BE} = 0$, $V_{CE} = 0$, $V_b = 0$
$I_C = \frac{(20 - 0)}{4\times10^{3}} = 5\times10^{-3}$ A $= 5$ mA
$V_i = V_{BE} + I_BR_B = 0 + I_BR_B$
$20 = I_B \times 500 \times 10^{3}$
$I_B = \frac{20}{500\times10^{3}} = 40\ \mu A$
$\beta = \frac{I_C}{I_B} = \frac{5\times10^{-3}}{40\times10^{-6}} = 125$
So $I_B = 40\ \mu A$, $I_C = 5$ mA and $\beta = 125$.
$I_C = \frac{(20 - 0)}{4\times10^{3}} = 5\times10^{-3}$ A $= 5$ mA
$V_i = V_{BE} + I_BR_B = 0 + I_BR_B$
$20 = I_B \times 500 \times 10^{3}$
$I_B = \frac{20}{500\times10^{3}} = 40\ \mu A$
$\beta = \frac{I_C}{I_B} = \frac{5\times10^{-3}}{40\times10^{-6}} = 125$
So $I_B = 40\ \mu A$, $I_C = 5$ mA and $\beta = 125$.