Class 12 NEET Physics Semiconductor Electronics Part 1

Voltage drop across diode (VD)=0.5V
Maximum power rating of diode (P)=100‌mW =100×10−3W
and source voltage (Vs)=1.5V.
The resistance of diode (RD) =

VD2

P

=

(0.5)2

100×10−3

=2.5Ω
And current in diode (ID)=

VD

RD

=

0.5

2.5

=0.2Ω
Therefore total resistance in circuit (R)
=

Vs

ID

=

1.5

0.2

=7.5Ω
And the value of the series resistor = Total resistance of the circuit - Resistance of diode
=7.5−2.5=5Ω