Find the maximum zener current for the zener diode as shown in figure. Given, $V_{Z} = 6 V, R_{Z} = 1.5 Ω, R = 400 Ω$

35.00 m A

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30.87 m A

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34.87 m A

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34.87 A

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Solution

The correct option is C $34.87 m A$
Here, $I_{T} = I_{Z} + I_{L}$

Thus, the zener current will be maximum when $I_{L} = 0$ for infinite load resistance. So, $I_{T} = I_{Z} (m a x)$

By Kirchhoff's voltage law, $V_{i n} - V_{Z} = (R + R_{Z}) I_{T}$

Thus, $I_{Z} (m a x) = I_{T} = \frac{V_{i n} - V_{Z}}{R + R_{z}} = \frac{20 - 6}{400 + 1.5} = 34.87 m A$

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