Skip to main content

Zener current: 18 V, 100 Ω, RL 1.2 kΩ

Short answer

The Zener is ON because open-circuit voltage across RL (18×1200/(100+1200)=16.62 V) exceeds VZ=10 V. Series current IR = (18−10)/100 = 80 mA, load current IL = 10/1200 = 8.33 mA, so Zener current IZ = IR − IL = 71.67 mA.

Find the Zener current in the given circuit when RL = 1.2 kΩ. Assume VZ = 10 V. [Figure: V1 = 18 V source, series resistor R = 100 Ω, Zener diode in parallel with RL = 1.2 kΩ; output Vo = 10 V.]

Zener regulator analysis (VZ = 10 V).

First check whether the Zener is ON. With the Zener removed, the voltage across RL would be

V = 18 × 1200/(100 + 1200) = 16.62 V

This is greater than VZ = 10 V, so the Zener is in breakdown and holds Vo = 10 V.

  • Current through the series resistor: IR = (Vin − VZ)/R = (18 − 10)/100 = 0.08 A = 80 mA
  • Load current: IL = Vo/RL = 10/1200 = 8.33 mA
  • Zener current (KCL): IZ = IR − IL = 80 − 8.33 = 71.67 mA
 18 V o--[100 ohm]--+--------+--o Vo = 10 V
                    |        |
                    Z       RL 1.2k
                    |        |
 0 V  o-------------+--------+--

Answer: IZ = 71.7 mA (IR = 80 mA, IL = 8.33 mA). Power in the Zener = 10 × 71.67 mA = 0.717 W, so a 1 W Zener is needed.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electronics Engineering chapter 2 questions →