The voltage-current characteristic of a diode during forward bias is given by `I = 7.8 xx 10^(-5) e^(6.9V_(D))`, where I is the current in mA and `V_(D)` is the diode voltage in V. Find the dynamic resistance of the diode in `Omega`, when the current is 4mA.

Options

A. 18.6
B. 21.7
C. 28.2
D. 36.2

Answer

D

Explanation

Differentiating `I_(D)` w.r.t `V_(D)`, we get
`(d)/(dV_(D)) (I_(D)) = (d)/(dV_(D)) (7.8 xx 10^(-5) e^(6.9V_(D)))`
`= 7.8 xx 10^(-5) xx 6.9 xx e^(6.9 V_(D))`
`= I_(D) xx 6.9`
Dynamic resistance is `(d)/(dI_(D)) (V_(D))`
`rArr (dV_(D))/(di_(D)) = (1)/(6.9 xx I_(D))`
When `I_(D) = 4mA = 4 xx 10^(-3)A`
Then, `(dV_(D))/(dI_(D)) = (1)/(6.9 xx 4 xx 10^(-3))`
`= 36.23 Omega`
`:.` Dynamic resistance `= 36.23 Omega`