A BJT has its base current as 0.02 mA, and the current amplification factor as 0.9. Determine the value of the emitter current if the I_CBO is found to be 30 μA.

Question

Question

This question was previously asked in

PGCIL Diploma Trainee EE Official Paper held on 27 October 2018

Answer 1

Concept:

In a common base connection,

I_E = I_B + I_C

And I_C = α I_E + I_CBO

Where α is the current amplification factor

I_E = I_B + α I_E + I_CBO

⇒ I_E (1 – α) = I_B + I_CBO

\( \Rightarrow {I_E} = \frac{1}{{\left( {1 - \alpha } \right)}}\left( {{I_B} + {I_{CBO}}} \right)\)

Calculation:

Given that, I_B = 0.02 mA

Current amplification factor (α) = 0.9

I_CBO = 30 μA = 0.03 mA

The emitter current is,

\({I_E} = \frac{1}{{1 - 0.9}}\left( {0.02 + 0.03} \right) = 0.5\;mA\)