---
title: "Zener input range, 91 Ω, 8 V, 400 mW"
subject: "Electronics Devices and Circuit"
chapter: 6
key: "semester_3/electronics_devices_and_circuit:zener-range-vi-91ohm"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/zener-range-vi-91ohm
site: BCT Engineering
---

# Zener input range, 91 Ω, 8 V, 400 mW

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

- Asked in: 2082 Baisakh, 6 marks
- Video: [Zener Diode Regulation - Varying load resistance, Varying input voltage and Varying input resistance](https://www.youtube.com/watch?v=CO18uJ_LxE8) — Electric Circuits
- 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-range-vi-91ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Determine the range of input voltage Vi that will maintain the Zener diode in an 'ON' state and not exceed the maximum power rating of the Zener diode.
[Figure: source Vi, series resistor Rs = 91 Ω, Zener diode Vz = 8 V with Pzmax = 400 mW in parallel with load RL = 0.22 kΩ]

## Short answer

With a 91Ω series resistor and a 220Ω load, the Zener remains ON for Vi from 11.31V (Zener just conducts) to 15.86V (Iz reaches 50mA at max power 400mW).

## Answer

**Given:** Rs = 91 Ω, Vz = 8 V, Pzmax = 400 mW, RL = 0.22 kΩ = 220 Ω.

#### Load current and maximum Zener current
IL = Vz/RL = 8 V / 220 Ω = **36.36 mA** (constant while the Zener is ON)

IZM = Pzmax/Vz = 400 mW / 8 V = **50 mA**

#### Minimum Vi (Zener just ON, Iz ≈ 0)
Rs carries only the load current:

Vi(min) = Vz + IL Rs = 8 + 0.03636 × 91 = 8 + 3.31 = **11.31 V**

(Check with the Zener removed: Vi × 220/(220 + 91) = 8 V gives Vi = 11.31 V ✓.)

#### Maximum Vi (Iz = IZM = 50 mA)
Is = IZM + IL = 50 + 36.36 = 86.36 mA

Vi(max) = Vz + Is Rs = 8 + 0.08636 × 91 = 8 + 7.86 = **15.86 V**

**Answer: 11.31 V ≤ Vi ≤ 15.86 V.** Below 11.31 V the Zener turns OFF (no regulation); above 15.86 V its power exceeds 400 mW.

## Check yourself

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

1. How is IZM found from the power rating?
   A. IZM = Pzmax x Vz = 3.2 A
   B. IZM = Pzmax / Vz = 400 mW / 8 V = 50 mA
   C. IZM = Vz / RL = 36.4 mA
   D. IZM = Pzmax / RL = 1.8 mA
   Answer: B. Power in a Zener is Vz x Iz, so the current limit is P/V. (Key point: Maximum Zener current)

2. What is IL while the Zener is ON?
   A. 8 V / 220 Ω = 36.36 mA
   B. 8 V / 91 Ω = 87.9 mA
   C. 400 mW / 220 Ω
   D. 50 mA
   Answer: A. The load sees Vz, so IL = Vz/RL. (Key point: Load current)

3. What is Vi(min)?
   A. 8 V
   B. 15.86 V
   C. 9.36 V
   D. 11.31 V, with Iz ≈ 0 and Rs carrying only IL
   Answer: D. Vi(min) = 8 + 36.36 mA x 91 = 11.31 V. (Key point: Minimum Vi)

4. What is Vi(max) and what limits it?
   A. 11.31 V; the load stops working
   B. 15.86 V; the Zener turns OFF
   C. 15.86 V; Zener power would exceed 400 mW (Is = 50 + 36.36 mA)
   D. 20 V; the series resistor overheats at 0.7 V
   Answer: C. Vi(max) = 8 + 86.36 mA x 91 = 15.86 V, where Iz = IZM. (Key point: Maximum Vi)

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