Question

A zener of breakdown voltage V_Z = 8 V and maximum Zener current, I_ZM = 10 mA is subjected to an input voltage V_i = 10 V with series resistance R = 100 Ω. In the given circuit R_L represents the variable load resistance. The ratio of maximum and minimum value of R_L is __________.

Answer 1

The correct answer is 2

Fig.

Minimum value of R_L for which the diode is shorted is
\(\frac{R_L}{R_L}\)+100×10=8 ⇒R_L=400 Ω
For maximum value of R_L, current through diode is 10 mA.
So
\(i_R= i_{R_L} +I_{ZM}\)
\(\frac{2}{100}\)=\(\frac{8}{R_L}\)+10×10−3
10×10⁻³=\(\frac{8}{R_L}\)
R_L = 800 Ω
So
\(\frac{R_{Lmax}}{R_{Lmin}}=\frac{800}{400}=2\)
So, ratio of maximum and minimum value of R_L is 2