--- title: "Zener with pass transistor: Vo and Iz" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:zener-series-pass-numeric" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/zener-series-pass-numeric site: BCT Engineering --- # Zener with pass transistor: Vo and Iz Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2076 Ashwin, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/zener-series-pass-numeric - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question 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] ## 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. ## Answer **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 ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the output voltage of the series pass regulator with a 12 V Zener at the base of Q1? A. Vo = Vz + VBE = 12.7 V B. Vo = Vz - VBE = 12 - 0.7 = 11.3 V C. Vo = Vz = 12 V D. Vo = Vi - VBE = 19.3 V Answer: B. The emitter follows the base 0.7 V lower, so the output sits at 11.3 V. (Key point: Output voltage) 2. How is the base current of Q1 found from the load current? A. IB = IE/(β + 1) = 11.3 mA / 51 ≈ 0.22 mA B. IB = IE x β = 565 mA C. IB = IL x (β + 1) = 576 mA D. IB = IE/(β - 1) = 0.23 mA only if β is 50 and IE is unknown Answer: A. The emitter current is (β + 1) times the base current, and IE = IL = 11.3 mA, so IB = 0.2216 mA. (Key point: Base current) 3. How is the current through R = 220 Ω found? A. IR = (Vi - Vo)/R = 8.7/220 = 39.5 mA B. IR = Vz/R = 12/220 = 54.5 mA C. IR = Vi/R = 20/220 = 90.9 mA D. IR = (Vi - Vz)/R = (20 - 12)/220 = 36.36 mA Answer: D. R is connected between the 20 V supply and the base node held at Vz = 12 V, so 8 V drops across it. (Key point: Current in series resistor) 4. What is the Zener current, and how does the transistor affect it? A. Iz = IR - IL = 25 mA, since the Zener supplies the load B. Iz = IR + IB = 36.58 mA C. Iz = IR - IB = 36.36 - 0.22 = 36.14 mA; the Zener supplies only the small base current instead of the load current D. Iz = IL + IB = 11.5 mA Answer: C. The load current is supplied through Q1 from Vi, so the Zener only has to absorb IR minus the base current. 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