The zener diode has a V z = 30V. The current passing through the diode for the following circuit is ________ mA.

Solution

Physics sir · 4 steps

Fix 30 V across the zener and the load, then the zener carries whatever the 5 k resistor does not take.

1. Is it in breakdown. The zener is reverse biased: its cathode faces the positive terminal. Remove it and the divider puts 90 × 5/9 = 50 V across the 5 k resistor. That is more than 30 V, so the zener breaks down and holds the voltage at 30 V.
2. Series resistor current. The 4 k resistor has 90 - 30 = 60 V across it, so I = 60/(4 kΩ) = 15 mA. This is the total current from the battery.
3. Load current. The 5 k resistor is in parallel with the zener, so it also has 30 V across it: I_L = 30/(5 kΩ) = 6 mA.

   I_Z = I - I_L = 15 mA - 6 mA = 9 mA

4. So. The current through the zener diode is 9 mA.

The common slip: 15. Taking the 4 k resistor current as the zener current gives 15 mA. You must subtract the 6 mA that flows through the 5 k load.

The idea to keep. In a zener regulator, first check that the open-circuit voltage exceeds V_Z. Then the series current minus the load current gives the zener current.

Answer from the NTA final answer key. Solution by PrepWiser.