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When the voltage drop across a p.n junct...

When the voltage drop across a `p.n` junction diode is increased from `0.65V` to `0.70V`, the change in the diode current is `5mA`. What is the dynamic resistance of the diode?

A

`5 Omega`

B

`10 Omega`

C

`20 Omega`

D

`25 Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
Dynamic resistance is `r_d=(DeltaV)/(DeltaI)`
Here, `DeltaV` = 0.7 - 0.65 V = 0.05 V ,
`DeltaI=5mA = 5xx 10^(-3) A`
`therefore r_d=0.05/(5xx10^(3))=10 Omega`
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