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For a transistor in common emitter confi...

For a transistor in common emitter configuration, the voltage drop across the load of 1000 `Omega` is 0.5 V. If the value of `alpha ` for the transistor is 0. 98, then the base current will be approximately equal to

A

5 `muA`

B

8`muA`

C

10`muA`

D

15`muA`

Text Solution

Verified by Experts

The correct Answer is:
C
`beta=0.98/(1-0.98)=98/2 = 49`
`I_C=V/R = 0.5/1000 = 5xx10^(-4) A`
`therefore beta=I_C/I_B`
`therefore I_B=I_C/beta=(5xx10^(-4))/49 = (500xx10^(-6))/49=10 muA`
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