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Physics Question on Semiconductor electronics: materials, devices and simple circuits

In a common emitter transistor amplifier an increase of 50 μA50 \,\mu A in the base current causes an increase of 1 mA1 \,mA in the collector current. The change in emitter current is

A

1050 mA1050\,mA

B

1050 μA1050\,\mu A

C

5025 mA5025\,mA

D

5025 μA5025\,\mu A

Answer

1050 μA1050\,\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= 1050 \times 10^{-6} A =1050 μA = 1050\, \mu A