Question

A 2 V battery forward biases a diode however there is a drop of 0.5 V across the diode which isindependent of current. Also a current greater then10 mA produces large joule loss and damagesdiode. If diode is to be operated at 5 mA, the seriesresistance to be put is :-

Answer 1

1.5 =5 x10^-3 R

R=300ohms

Answer 2

The resistance required is 300Ω.

Energy of the battery = 2V (Given)

Drop in energy = 0.5V (Given)

Current in the diode = 5mA (Given)

According to Kirchhoff's voltage law -

2 = 0.5+IR

= 0.5 + 5 × 10³R

R = 15 / 5 × 10³R

R = 1500/5

R = 300

Thus, the resistance to be put is 300 ohms.