Ncert Exercises & Solutions

For the transistor circuit shown in figure, evaluate $V_{E}, R_{B}, R_{E},$ given $I_{C} = 1 mA, V_{CE} = 3 V, V_{BE} = 0.5 V and V_{CC} = 12 V, β = 100 .$

Hint:

I_{C} (R_{C} + R_{E}) + V_{CE} = V_{CC}

and

V_{B} = V_{E} + V_{BE}

Step 1: Find the emitter resistance.

Consider the fig. (b) given here to solve this problem

I_{C} \approx I_{E} [As base current is very small .]

R = 7.8 kΩ

From the figure,

I_{C} (R_{C} + R_{E}) + V_{CE} = 12

(R_{E} + R_{C}) \times 1 \times 10^{- 3} + 3 = 12

Step 3: Find the emitter voltage.

R_{E} + R_{C} = 9 \times 10^{3} = 9 kΩ

R = 9 - 7.8 = 1.2 kΩ

V_{E} = I_{E} \times R

= 1 \times 10^{- 3} \times 1.2 \times 10^{3} = 1.2 V

Step 3: Find the base voltage.

Voltage

V_{B} = V_{E} + V_{BE} = 1.2 + 0.5 = 1.7 V

Current I = \frac{V_{B}}{20 \times 10^{3}} = \frac{1.7}{20 \times 10^{3}} = 0.085 mA

Step 4: Find the base resistance.

Resistance R_{B} = \frac{12 - 1.7}{\frac{I_{C}}{β} + 0.085} = \frac{10.3}{0.01 + 0.085} [Given, β = 100]

= 108 kΩ

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