JEE Main 2017 · Physics · Semiconductors and Communication · easy

A 6 V Zener diode is connected in reverse bias across a load of 1 kΩ, in series with a 200 Ω resistor, to a 12 V supply. Assuming the diode is in its…

Single correct option

6\ \mathrm{V}

Zener diode is connected in reverse bias across a load of

1\ \mathrm{k\Omega}

, in series with a

200\ \Omega

resistor, to a

12\ \mathrm{V}

supply. Assuming the diode is in its breakdown region, the current through the Zener diode is

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Answer: (A) $24\ \mathrm{mA}$

The Zener holds the voltage across the load at

6\ \mathrm{V}

. The drop across the series resistor is

12 - 6 = 6\ \mathrm{V}

, so the total current is

I = 6/200 = 30\ \mathrm{mA}

. The load current is

6\ \mathrm{V}/1\ \mathrm{k\Omega} = 6\ \mathrm{mA}

. By Kirchhoff's junction rule the diode current is

I_Z = 30 - 6 = 24\ \mathrm{mA}

. The value

30\ \mathrm{mA}

ignores the current taken by the load, and

6\ \mathrm{mA}

is the load current itself.

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