An experiment is performed to determine the I−V characteristics of a Zener diode, which has a protective resistance of R=100Ω, and a maximum power of dissipation rating of 1W. The minimum voltage range of the DC source in the circuit is
A
0−24V
B
0−5V
C
0−12V
D
0−8V
Hard
JEE Mains
Open in App
Updated on : 2022-09-05
Solution
Verified by Toppr
Correct option is A)
The circuit diode for a zener diode with a protective resistance is shown.
Applying KVL,
V=iR+VD
VD=V−iR=V−100i
Maximum power dissipated across the zener diode is given by:
P=VDi=(V−100i)i=1
∴100i2−Vi+1≥0
Current should be real, hence determinant is greater than zero.
∴V2>400
V>20V
Solve any question of Semiconductor Electronics: Materials, Devices And Simple Circuits with:-