Question 18.31: Design a Zener shunt voltage regulator with the following sp...

Refer to Fig. 18.19.
Selection of zener diode

V_{z} = V_{o} = 10 V
I_{z\,max} = 40 mA
P_{z} = V_{z} × I_{z\,max} = 10 × 40 × 10^{– 3} = 0.4 W

Hence a 0.5WZ 10 zener can be selected
Value of load resistance, R_{L}

R_{L\,min} =\frac{V_{o}}{I_{L\,max}}=\frac{10}{50 × 10^{– 3}} = 200 Ω

R_{L\,max} =\frac{V_{o}}{I_{L\,min}}=\frac{10}{30 × 10^{–3}} = 333 Ω

Value of input resistance, R

R_{max} =\frac{V_{in(max)} – V_{o}}{I_{Lmin} + I_{z(max)}}
= \frac{30 – 10}{(30 + 40) × 10^{– 3}} = 286 Ω

R_{min} =\frac{V_{in (min)} – V_{o}}{I_{L\,max} + I_{z (min)}}
= \frac{20 – 10}{(50 + 20) × 10^{– 3}} = 143 Ω

Therefore, R = \frac{R_{max} + R_{min}}{2} = 215 Ω