---
title: "Zener regulator Vo, load effect, minimum RL"
subject: "Electronics Devices and Circuit"
chapter: 6
key: "semester_3/electronics_devices_and_circuit:zener-regulator-6v8-line-load"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/zener-regulator-6v8-line-load
site: BCT Engineering
---

# Zener regulator Vo, load effect, minimum RL

Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators

- Asked in: 2076 Chaitra, 2+1+2 marks; 2074 Chaitra, 2+2+2 marks (asked 2 times)
- Video: [Zener Diode as Voltage Regulator | Parameters & Example of Zener Diode as Voltage Regulator](https://www.youtube.com/watch?v=F6eGVaNcavg) — Engineering Funda
- 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-regulator-6v8-line-load
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

The 6.8 V zener diode is specified to have Vz = 6.8 V at Iz = 5 mA, rz = 20Ω and Izk = 0.2 mA. The supply voltage V+ is nominally 10V but can vary by ± 1V. Find Vo with no load and with V+ at its nominal value. Find the change in Vo resulting from connecting a load resistance RL that draws a current IL = 1 mA. What is the minimum value of RL for which the diode still operates in the breakdown region?
[Figure: supply V+ in series with a 0.5 kΩ resistor, output Vo across a Zener diode to ground]

## Short answer

Vz0=6.7V. No load, Vo=Vz0+rz*(10-6.7)/(500+20)=6.827V. Load change: ΔVo≈-(rz||R)*IL=-19.2mV. Minimum RL for Iz≥0.2mA: with V+=10V, ILmax=6.392mA, RLmin=6.704V/6.392mA≈1.05kΩ; worst-case V+=9V gives 1.53kΩ.

## Answer

**Given:** Vz = 6.8 V at Iz = 5 mA, rz = 20 Ω, Izk = 0.2 mA, V+ = 10 V ± 1 V, R = 0.5 kΩ.

#### Zener model
Vz0 = Vz − rz Iz = 6.8 − 20 × 0.005 = 6.7 V. So the Zener is 6.7 V in series with 20 Ω.

#### (a) No load, V+ = 10 V
I = (V+ − Vz0)/(R + rz) = (10 − 6.7)/(500 + 20) = 6.346 mA

**Vo = Vz0 + rz I = 6.7 + 20 × 6.346 mA = 6.827 V**

#### (b) Change in Vo when a load draws IL = 1 mA
The load current is taken from the Zener (the current in R is almost unchanged), so the Zener current decreases by about 1 mA:

ΔVo = −rz (R/(R + rz)) IL = −(20 ∥ 500) × 1 mA = −19.2 Ω × 1 mA

**ΔVo = −19.2 mV** (new Vo ≈ 6.808 V). Load regulation = −19.2 mV/mA.

Exact check: Vo = (V+/R + Vz0/rz − IL)/(1/R + 1/rz) = (0.02 + 0.335 − 0.001)/0.052 = 6.808 V ✓.

#### (c) Minimum RL for the diode to stay in breakdown
The Zener stays in breakdown while Iz ≥ Izk = 0.2 mA. At this limit:

Vz = 6.7 + 20 × 0.0002 = 6.704 V

Current in R = (10 − 6.704)/500 = 6.592 mA (at nominal V+)

IL(max) = 6.592 − 0.2 = 6.392 mA

**RLmin = 6.704 / 6.392 mA ≈ 1.05 kΩ** (at V+ = 10 V)

For the worst case V+ = 9 V: IR = (9 − 6.704)/500 = 4.592 mA, IL = 4.392 mA, so RLmin = 6.704/4.392 = **1.53 kΩ**. If the supply can drop to 9 V, RL must be larger than about 1.5 kΩ.

**Answer: Vo = 6.83 V; ΔVo = −19.2 mV; RLmin ≈ 1.05 kΩ (nominal supply), 1.53 kΩ (V+ = 9 V).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the Zener modelled for calculation?
   A. A 6.8 V source with no resistance
   B. A source Vz0 = Vz - rz Iz = 6.7 V in series with rz = 20 Ω
   C. A 20 Ω resistor only
   D. A 6.8 V source in parallel with 20 Ω
   Answer: B. Vz is specified at 5 mA, so Vz0 = 6.8 - 0.1 = 6.7 V. (Key point: Zener model)

2. What is Vo with no load at V+ = 10 V?
   A. About 6.83 V
   B. 10 V
   C. 6.7 V
   D. 5.0 V
   Answer: A. I = 3.3/520 = 6.346 mA, so Vo = 6.7 + 20 x 6.346 mA = 6.827 V. (Key point: No-load output)

3. What is the change in Vo when a 1 mA load is connected?
   A. -1 V
   B. -20 mV exactly, from rz x 1 mA only
   C. +19.2 mV
   D. About -19.2 mV, from -(rz || R) x 1 mA
   Answer: D. The Zener and R share the effect, so ΔVo = -(20 || 500) x 1 mA = -19.2 mV. (Key point: Load regulation)

4. What condition gives the minimum RL for breakdown operation?
   A. Zener current equals Iz at 5 mA
   B. Load current equals zero
   C. Zener current drops to Izk = 0.2 mA, so IL = IR - Izk
   D. Vo falls to 0 V
   Answer: C. Staying in breakdown requires Iz ≥ Izk; the rest of IR (6.392 mA at 10 V) can go to the load. (Key point: Minimum RL)

## Related in Electronics Devices and Circuit

- [Zener regulator output voltage and minimum RL](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-load-regulator-rz-6v8) — Electronics Devices and Circuit, chapter 6
- [Zener with pass transistor: Vo and Iz](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-series-pass-numeric) — Electronics Devices and Circuit, chapter 6
- [Design zener shunt regulator (5.6 V, 15 V supply)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-shunt-regulator-5v6-design) — Electronics Devices and Circuit, chapter 6
- [Zener input range, 220 Ω, 20 V, 60 mA](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-range-vi-220ohm-20v) — Electronics Devices and Circuit, chapter 6, asked 2×
- [Design 6 V Zener regulator, regulation factor](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-regulator-design-6v-12v) — Electronics Devices and Circuit, chapter 6, asked 4×
- [Zener: range of load for VL = 5 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Azener-load-range-5v) — Electronics Devices and Circuit, chapter 6

## Related in other subjects

- [Zener diode as voltage regulator](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Azener-voltage-regulator) — Basic Electrical and Electronics Engineering, chapter 3
- [Zener network: VL, IL, Iz, IR](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Azener-regulator-220-ohm) — Basic Electronics Engineering, chapter 2
- [Zener regulator design: Rs and IZM](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Azener-regulator-design) — Fundamental of Electrical and Electronics Engineering, chapter 4
- [Zener current: 18 V, 100 Ω, RL 1.2 kΩ](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Azener-current-18v-100) — Basic Electronics Engineering, chapter 2, asked 4×