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Zener current and power dissipation

Short answer

Series current Is = (18−10)/250 = 32 mA, load current IL = 10 mA. Zener current Iz = 32−10 = 22 mA. Power dissipated Pz = 10 V × 22 mA = 220 mW.

A load resistance of 1 kΩ is connected across a Zener voltage regulator with Zener diode of 10 V. Find the Zener current and power dissipated across the Zener diode. [Figure: supply VS = 18 V in series with RS = 250 Ω feeding a 10 V Zener diode (VZ = 10 V, reverse biased) connected in parallel with load RL = 1 kΩ.]

The Zener diode holds the load voltage at Vz = 10 V (it is in breakdown because the open-circuit load voltage 18 × 1000/1250 = 14.4 V is greater than 10 V).

 Vs=18V o--[ Rs=250Ω ]--+--------+
                        |        |
                      Zener    RL=1kΩ
                      Vz=10V     |
        o---------------+--------+

Current through Rs

Is = (Vs − Vz)/Rs = (18 − 10)/250 = 0.032 A = 32 mA

Load current

IL = Vz/RL = 10/1000 = 0.010 A = 10 mA

Zener current (KCL)

Iz = Is − IL = 32 − 10 = 22 mA

Power dissipated in the Zener

Pz = Vz × Iz = 10 × 0.022 = 0.22 W = 220 mW

Answer: Iz = 22 mA, Pz = 220 mW.

(Power in Rs = 8 × 0.032 = 0.256 W; power in the load = 0.1 W.) The Zener should be rated above 220 mW, e.g. a 10 V, 0.5 W Zener.

✓ Answer checked by Claude Opus 5.5 (high effort)

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