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Q.

A pnp transistor working in a common base configuration sends 1.96 mA current as the output with α=0.98. If the device has a leakage current of 5 μA then the base current is :

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a

40 μA

b

45 μA

c

None

d

35 μA

answer is A.

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Detailed Solution

IC=1.96mA

IE=ICα=1.960.98=2mA

IE=IB+IC+Ileakage 

2=IB+1.96+0.005

IB=0.035mA=35μA

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