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Question: In a common emitter transistor amplifier an increase of 50 \(\mu A\) in the base current causes an i...

In a common emitter transistor amplifier an increase of 50 μA\mu A in the base current causes an increase of 1 mA in the collector current. The change in emitter current is

A

1050 mA

B

1050 μA\mu A

C

5025 mA

D

5025 μA\mu A

Answer

1050 μA\mu A

Explanation

Solution

: the change in emitter current is

ΔIE=ΔIB+ΔIC\Delta I_{E} = \Delta I_{B} + \Delta I_{C}

=50×10−6+1×10−3= 50 \times 10^{- 6} + 1 \times 10^{- 3}

1050×10−6A=1050μA1050 \times 10^{- 6}A = 1050\mu A