[I_{B}=40\mu A, I_{C} = 5...] In the circuit shown in the figure, the input voltage V_{i} is 20 V, V_{BE}= 0 and V

For Competitive Exams

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

$I_{B}=20\mu A, I_{C} = 5mA, \beta=250$

$I_{B}=25\mu A, I_{C} = 5mA, \beta=200$

$I_{B}=40\mu A, I_{C} = 10mA, \beta=250$

$I_{B}=40\mu A, I_{C} = 5mA, \beta=125$

Explanation:

$V_{BE}=0$
$V_{CE}=0$
$V_{b}=0$

I_{c}= \frac{20-0}{4*10^{3}}$
$I_{c}=5*10^{-3}=5 mA$
$V_{i}=V_{BE}+I_{B}R_{B}$
$V_{i}=0+I_{B}R_{B}$
$20 =I_{B}*500*10^{3}$
$I_{B}= \dfrac{20}{500* 10^{3}}= 40 \mu A$
$\beta=\dfrac{I_{C}}{I_{B}}=\dfrac{25*10^{-3}}{40*10^{-6}}=125$

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