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Question:

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?

0-9V

0-12V

0-52V

0-64V

Solution:

The circuit diode for a zener diode with a protective resistance is shown. Applying KVL, V = iR + V_D. V_D = V - iR = V - 100i. Maximum power dissipated across the zener diode is given by: P = V_D i = (V - 100i)i = 1. ∴ 100i² - Vi + 1 ≥ 0. Current should be real, hence determinant is greater than zero. ∴ V² > 400. V > 20V