SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

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Question : 135 of 200

Marks: +1, -0

Calculate the value of emitter current for a transistor with

\α_{ \dc }=0.98, \I _{ \CBO }=5 \µ\A

and

\I _{ \B }=95 \µ\A

4.5 mA
4 mA
3.5 mA
5 mA

Solution:

\α_{ \dc }=0.98, \I _{ \CBO }=5 \µ\A

and

\I _{ \B }=95 \µ\A

\β={\α}/{1-\α}=0.98/{1-0.98}=49

\I _{ \CEO }={ \I _{ \CBO }}/{1-\α}

={5×10^{-6}}/(1-0.98)=250\µ\A

\I _{ \C }={\α\I _{ \B }}/{1-\α}+{ \I _{ \CBO }}/{1-\α}

=49 × (95 × 10^{-6})+250 × 10^{-6}

=4905 ×10^{-6} \A

\I _{ \E }={(4905-5) ×10^{-6}}/{0.98}

={4900 ×10^{-6}}/{0.98}=5 \mA

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