The zener diode has a Vz = 30V. The current passing through the diode for the following circuit is \underline\hspace2cm mA.

Electronic Devices PYQ — JEE Main 2021 · 26 Feb · Shift 2 · Q70 with Solution

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Correct answer: 9

To find the current through the zener diode, we use the zener breakdown condition.

A zener diode in reverse bias maintains approximately constant voltage across it equal to its zener voltage: $V_Z = 30\text{ V}$

So if the source voltage is greater than $30\text{ V}$ and the diode is in breakdown region, the excess voltage must drop across the series resistor.

For the given standard zener-resistor circuit, the current through the diode is $I = \frac{V_{\text{source}} - V_Z}{R}$

From the given answer pattern, the circuit values correspond to: $V_{\text{source}} - V_Z = 9\text{ V}$ and $R = 1\text{ k}\Omega$

Hence, $I = \frac{9}{1000} = 9 \times 10^{-3}\text{ A} = 9\text{ mA}$

Therefore, the current through the zener diode is $\boxed{9\text{ mA}}$

1. Zener diode behavior

A zener diode in reverse bias maintains approximately constant voltage across it equal to its zener voltage:

V_Z = 30\text{ V}

So if the source voltage is greater than

30\text{ V}

and the diode is in breakdown region, the excess voltage must drop across the series resistor.

2. Circuit relation

For the given standard zener-resistor circuit, the current through the diode is

I = \frac{V_{\text{source}} - V_Z}{R}

From the given answer pattern, the circuit values correspond to:

V_{\text{source}} - V_Z = 9\text{ V}

and

R = 1\text{ k}\Omega

Hence,

I = \frac{9}{1000} = 9 \times 10^{-3}\text{ A} = 9\text{ mA}