The circuit shown below is working as a 8 V dc regulated voltage source. When 12 V is used as input, the power dissipated (in mW) in each diode is; (considering both zener diodes are identical) _________.

Solution

Physics sir · 3 steps

Both 200 Ω resistors carry the same current, and the 8 V splits equally across the two identical zeners.

1. Find the current. No load is connected at Vₒ, so one current flows through both resistors and both zeners in series. The resistors share the leftover 12 - 8 = 4 V across 200 + 200 = 400 Ω.

   I = (12 - 8)/400 = 0.01 A = 10 mA

2. Voltage per diode. The two zeners are identical and together hold 8 V. So each one has 4 V across it.
3. So. Power in each diode is P = VI = 4 × 0.01 = 0.04 W, which is 40 mW.

The common slip: 80. Using the full 8 V across a single diode gives 8 × 10 = 80 mW, which is the power of both diodes together, not each.

The idea to keep. For a zener regulator, first get the series current from the voltage dropped across the resistors. Then use the voltage across one diode, not the whole stack, for that diode's power.

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