Question

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

• A. $I_B = 40 \, \mu A, I_C = 10 \, mA , \beta = 250$
• B. $I_B = 25\, \mu A, I_C = 5 \, mA , \beta = 200$
• C. $I_B = 20\, \mu A, I_C = 5 \, mA , \beta = 250$
• D. $I_B = 40 \, \mu A, I_C = 5 \, mA , \beta = 125$

Answer 1

Answer: (D)

$I_B = 40 \, \mu A, I_C = 5 \, mA , \beta = 125$

Answer 2

$V_{BE} = 0$
$V_{CE} = 0$
$V_b = 0$
$I_{c} = \frac{\left(20 -0\right)}{4 \times10^{3}}$
$I_{c} = 5\times10^{-3} = 5\, mA$
$V_{i} = V_{BE} + I_{B} R_{B}$
$V_{i} = 0 + I_{B } R_{B}$
$20 = I_{B} \times500 \times10^{3}$
$I_{B} = \frac{20}{500 \times10^{3}} = 40 \,\mu A$
$\beta = \frac{I_{c}}{I_{b}} = \frac{25 \times10^{-3}}{40 \times10^{-6} } = 125$