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Zener with pass transistor: Vo and Iz

Short answer

Output voltage Vo = Vz − VBE = 12 − 0.7 = 11.3 V. Load current IL = 11.3 V / 1 kΩ = 11.3 mA, base current IB ≈ IL/(β+1) = 0.22 mA. Resistor current IR = (20−12)/220 = 36.36 mA. Zener current Iz = IR − IB = 36.14 mA.

Calculate the output voltage and the zener current in the regulator circuit as shown in figure below for RL = 1 KΩ and R = 220Ω, VZ = 12V. [Figure: unregulated Vi = 20 V feeds transistor Q1 (β = 50) as a series pass element; its base is connected through R = 220 Ω to the supply and to a 12 V Zener diode to ground; output Vo across load RL = 1 kΩ]

Given: Vi = 20 V, R = 220 Ω, Zener Vz = 12 V (at the base of Q1), Q1 with β = 50 as a series pass element, RL = 1 kΩ. Assumption: VBE = 0.7 V.

Circuit

 Vi=20V o----+------------+
             |            |C
            [R]       B---|  Q1 (β = 50)
             |        |   |E
             +--------+   +-----+---o Vo
             |                  |
           (Dz 12 V)           [RL = 1k]
             |                  |
            GND                GND

Output voltage

The base of Q1 is held at Vz = 12 V, and the emitter follows it, 0.7 V lower:

Vo = VZ − VBE = 12 − 0.7 = 11.3 V

Currents

  • Load current: IL = Vo/RL = 11.3 V / 1 kΩ = 11.3 mA
  • Emitter current: IE = IL = 11.3 mA
  • Base current: IB = IE/(β + 1) = 11.3 / 51 = 0.2216 mA (IB ≈ IC/β = 0.226 mA gives practically the same result)
  • Current through R: IR = (Vi − Vz)/R = (20 − 12)/220 = 36.36 mA
  • Zener current: Iz = IR − IB = 36.36 − 0.22 = 36.14 mA

(If the load were connected directly across the Zener, the Zener current would fall by the whole load current, to about 24 mA; here the transistor reduces the Zener's burden to the small IB.)

Power in the Zener: Pz = 12 × 36.14 mA = 434 mW. Power in Q1: (Vi − Vo) IL = 8.7 × 11.3 mA = 98 mW.

Answer: Vo = 11.3 V, Iz = 36.1 mA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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