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

# Zener regulator output voltage and minimum RL

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

- Asked in: 2081 Bhadra, 3+3 marks
- 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-load-regulator-rz-6v8
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Given Vz = 6.8 V at Iz = 5mA, rz= 20 ohms and Vs = 10 ± 1 V, Rs = 500 ohms and Izk = 0.2mA. Determine the output Voltage of Zener voltage regulator without load. Find the minimum value of load resistance such that the Zener diode operates in Zener breakdown region.

## Short answer

Using Zener model Vz0=6.7V, rz=20Ω, the no-load output voltage is 6.83V. The minimum load resistance to stay in breakdown is 1.53kΩ when the supply drops to 9V.

## Answer

**Given:** Vz = 6.8 V at Iz = 5 mA, rz = 20 Ω, Izk = 0.2 mA, Vs = 10 ± 1 V, Rs = 500 Ω.

#### Zener model
Vz0 = Vz − rz Iz = 6.8 − 0.1 = 6.7 V (Zener = 6.7 V source in series with 20 Ω)

#### (a) Output voltage without load (Vs = 10 V)
I = (10 − 6.7)/(500 + 20) = 6.346 mA

**Vo = 6.7 + 20 × 6.346 mA = 6.83 V**

For Vs = 9 V: Vo = 6.7 + 20 × 4.423 mA = 6.79 V; for Vs = 11 V: Vo = 6.87 V. So a ±1 V supply change changes Vo by only ±40 mV (±0.04 V).

#### (b) Minimum load resistance to stay in breakdown
At the limit, Iz = Izk = 0.2 mA, and Vo = 6.7 + 20 × 0.0002 = 6.704 V.

Worst case is the lowest supply, Vs = 9 V:

Is = (9 − 6.704)/500 = 4.592 mA

IL = Is − Izk = 4.592 − 0.2 = 4.392 mA

**RLmin = 6.704 V / 4.392 mA = 1.53 kΩ**

(At the nominal 10 V supply, Is = 6.592 mA, IL = 6.392 mA, and RLmin = 1.05 kΩ.)

**Answer: Vo (no load) = 6.83 V; RLmin ≈ 1.5 kΩ (worst case, Vs = 9 V).**

## Check yourself

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

1. What is Vz0 for Vz = 6.8 V at 5 mA with rz = 20 Ω?
   A. 6.9 V
   B. 6.7 V
   C. 6.8 V
   D. 5.0 V
   Answer: B. Vz0 = 6.8 - 20 x 5 mA = 6.7 V. (Key point: Zener model)

2. What is Vo with no load at Vs = 10 V?
   A. 6.83 V
   B. 6.7 V
   C. 10 V
   D. 6.0 V
   Answer: A. I = 3.3/520 = 6.346 mA; Vo = 6.7 + 0.127 = 6.83 V. (Key point: No-load output)

3. Why use Vs = 9 V to find RLmin?
   A. The highest supply is worst for the load
   B. The nominal 10 V is always worst
   C. Supply does not affect RLmin
   D. The lowest supply leaves the least current for the load while keeping Iz ≥ Izk
   Answer: D. Lower Vs gives smaller Is, so RL must be larger. (Key point: Worst case supply)

4. What is RLmin at Vs = 9 V?
   A. About 1.05 kΩ
   B. About 500 Ω
   C. About 1.53 kΩ (6.704 V / 4.392 mA)
   D. About 6.8 kΩ
   Answer: C. Is = (9 - 6.704)/500 = 4.592 mA; IL = 4.392 mA. (Key point: RLmin)

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