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

Given : \(V_Z = 8\,V,\; I_{ZM} = 10\,mA = 0.01\,A,\; V_i = 10\,V,\; R = 100\,\Omega\)

Total current through series resistor

\(\begin{array}{l} I = \frac{V_i - V_Z}{R} = \frac{10-8}{100} = 0.02\,A = 20\,mA \end{array}\)

This current splits as

\(\begin{array}{l} I = I_Z + I_L \end{array}\)

Maximum \(R_L\) (minimum load current)

For regulation, minimum \(I_Z \approx 0\)

\(\begin{array}{l} I_L = 20\,mA \end{array}\)

\(\begin{array}{l} R_{L(max)} = \frac{V_Z}{I_L} = \frac{8}{0.02} = 400\,\Omega \end{array}\)

Minimum \(R_L\) (maximum Zener current)

\(\begin{array}{l} I_Z = I_{ZM} = 10\,mA \end{array}\)

\(\begin{array}{l} I_L = 20 - 10 = 10\,mA = 0.01\,A \end{array}\)

\(\begin{array}{l} R_{L(min)} = \frac{8}{0.01} = 800\,\Omega \end{array}\)

Required ratio

\(\begin{array}{l} \frac{R_{L(max)}}{R_{L(min)}} = \frac{400}{800} = \frac{1}{2} \end{array}\)

Hence, ratio = \(\frac{1}{2}\)