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

The current gain of a transistor in common emitter mode is 49.49 . The change in collector current and emitter current corresponding to the change in base current by 5.0 μA5.0\, \mu A are :

A

ΔiC=245 μA,ΔiE=250 μA\Delta i_{C}=245\, \mu A , \Delta i_{E}=250\, \mu A

B

ΔiC=252 μA,ΔiE=145 μA\Delta i_{C}=252\, \mu A , \Delta i_{E}=145\, \mu A

C

ΔiC=125 μA,ΔiE=250 μA\Delta i_{C}=125\, \mu A , \Delta i_{E}=250\, \mu A

D

ΔiC=252 μA,ΔiE=230 μA\Delta i_{C}=252\, \mu A , \Delta i_{E}=230\, \mu A

Answer

ΔiC=245 μA,ΔiE=250 μA\Delta i_{C}=245\, \mu A , \Delta i_{E}=250\, \mu A

Explanation

Solution

Current gain in common emitter mode of transistor
β=ΔiCΔiB\beta=\frac{\Delta i_{C}}{\Delta i_{B}}
or ΔiC=βΔiB\Delta i_{C}=\beta \Delta i_{B}
Given, β=49,ΔiB=5.0 μA\beta=49, \Delta i_{B}=5.0\, \mu A
∴ΔiC=49×5.0=245 μA\therefore \Delta i_{C}=49 \times 5.0=245\, \mu A
For a transistor, emitter current is the sum of base current and collector current.
i.e., iE=iC+iBi_{E}=i_{C}+i_{B}
⇒ΔiE=ΔiC+ΔiB\Rightarrow \Delta i_{E}=\Delta i_{C}+\Delta i_{B}
∴ΔiE=245+5.0=250 μA\therefore \Delta i_{E}=245+5.0=250\, \mu A