Question

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`

Answer 1

Correct Answer - 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`