Conceptual doubt about voltage drop in a diode In the given circuit, I am not sure how to solve it, I know that due to the current the diode is in reverse bias, hence no current should flow through it, But the answer is not given that and a finite value of I2 is given. How is this happening also when a diode is in reverse bias does the voltage across it still drop or is it only in forward bias? For example everyone knows voltage drop for Silicon diode is 0.7V. So this will drop when its in forward bias or reverse bias. Can someone explain?

 A: TL;DR The diode in your schematic is called a Zener diode. Without the diode, the voltage drop on the $5 \text{ k}\Omega$ resistor would be $80 \text{ V}$. However, the Zener diode clamps that voltage to $60 \text{ V}$ by shunting as much current as necessary in the reverse-bias direction. From this information you can calculate all currents in your circuit.

A Zener diode is a type of diode designed to allow current to flow in the opposite direction, i.e. when the diode is reverse-biased. When forward-biased the diode's voltage drop is standard $0.7 \text{ V}$, and when reverse-biased the diode's maximum voltage drop is as indicated on the schematic. Ideally, Zener diodes behave as follows:

*

*when the reverse-bias voltage is lower than nominal voltage (in your case $60 \text{ V}$), no current flows in the opposite direction;

*when the reverse-bias voltage equals nominal voltage, the Zener diode starts shunting current such that its reverse-bias voltage does not exceed the nominal voltage.

In reality this voltage drop depends on the reverse current magnitude; graphs are usually provided in the diode datasheet. Zener diodes are often used as (cheap) voltage references or voltage regulators where only small current is needed.

Figure: Zener diode I-V characteristics (Source)
