Kelvi

Kelvi

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Kelvi

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In the circuit shown in the figure, the input voltage $V_i$ is $20V, V_{BE} = 0$ and $V_{CE} = 0.$ The values of $I_B, I_C$ and $\beta$ are given by

hard

Current Electricity

2018

physics

$I_B = 20 \mu A, I_C = 5 mA, \beta = 250$

Explanation

To solve this problem, we need to analyze the given transistor circuit. $\\$ Given: $\\ V_i = 20 \,$ V $, \, V_{BE} = 0, \, V_{CE} = 0 \\$

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$I_B = 25 \mu A, I_C = 5 mA, \beta = 200$

$I_B = 40 \mu A, I_C = 10 mA, \beta = 250$

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

The transistor is in saturation, so

V_{CE} = 0. \\

Let's calculate the base current

I_B: \\

The voltage across

R_B

is

V_i - V_{BE} = 20 \,

V.

\\ I_B = \frac{V_i - V_{BE}}{R_B} = \frac{20 \, V}{500 \, k \Omega} \\ I_B = \frac{20}{500 \times 10^3} \\ I_B = 40 \, \mu A \\

Next, calculate the collector current

I_C: \\

Since the transistor is in saturation,

V_{CE} = 0. \\

The voltage across

R_C

is

20 \,

V

. \\ I_C = \frac{20 \, V}{4 \, k \Omega} \\ I_C = \frac{20}{4 \times 10^3} \\ I_C = 5 \, mA \newline

Finally, calculate

\beta: \\ \beta = \frac{I_C}{I_B} = \frac{5 \, mA}{40 \, \mu A} \\ \beta = \frac{5 \times 10^{-3}}{40 \times 10^{-6}} \\ \beta = 125 \\

Therefore, the values are

I_B = 40 \, \mu

A

, \, I_C = 5 \,

mA

, \, \beta = 125. \\

This corresponds to Option 4.