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 :-
Answers
Answered by
4
1.5 =5 x10^-3 R
R=300ohms
Answered by
9
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.
Similar questions