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For a BJT, α = 0.97 and collector base junction reverse saturation current is given by 0.4 μA. This BJT is connected in common emitter configuration and operating in active region. If IB = 15 μA then the collector current for the device will be _________
1. 506 μA
2. 498.34 μA
3. 101 mA
4. 1 mA

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Correct Answer - Option 2 : 498.34 μA

Concept: 

ICEO is the reverse leakage current in the common-emitter configuration of BJT when the base is open.

ICBO is the reverse leakage current in the common-base configuration of BJT when the emitter is open.

Also, ICEO > ICBO

And they are related by the relation:

ICEO = (1 + β) ICBO

Total collector current is given by:

IC = βIB + ICBO(β + 1)

Application:

Given that

Common base current gain α = 0.97

So, the common-emitter current gain:

\(β = \frac{α }{{1 - α }} = \frac{{0.97}}{{1 - 0.97}} = 32.33\)

ICO = reverse saturation current = 0.4 μA

IB = Base current = 15 μA

Then collector current is given by  

IC = βIB + (1 + β) ICO

IC = 32.33 × 15 × 10-6 + (1 + 32.33) 0.4 × 10-6 

IC = 498.34 μA

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