---
title: "Zener circuit output voltage and current"
subject: "Basic Electrical and Electronics Engineering"
chapter: 3
key: "semester_1/basic_electrical_and_electronics_engineering:zener-circuit-120v-5k"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/zener-circuit-120v-5k
site: BCT Engineering
---

# Zener circuit output voltage and current

Basic Electrical and Electronics Engineering · Chapter 3: Introduction to Electronics Engineering

- Asked in: 2081 Chaitra (new course), 6 marks
- Video: [Solved Problems on the Zener Diode](https://www.youtube.com/watch?v=kYtebfx-yHc) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/zener-circuit-120v-5k
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-3

## Question

For the circuit shown below, find: i) the output voltage ii) the voltage drop across series resistance iii) the current through Zener diode. [Figure: 120 V source in series with $R = 5\ k\Omega$; Zener diode of 50 V in parallel with 10 kΩ load; $I$ is total current, $I_Z$ Zener current, $I_L$ load current.]

## Short answer

With Zener on, output voltage is 50V. Voltage drop across series resistance is 70V. Total current is 14mA, load current 5mA, so Zener current is 9mA.

## Answer

**Given:** $V_i=120$ V, $R=5\ \text{k}\Omega$, $V_Z=50$ V, $R_L=10\ \text{k}\Omega$.

**Check Zener state.** With the Zener removed, the load voltage is

$$
V_L=\frac{10}{5+10}\times120=80\ \text{V}>50\ \text{V}
$$

So the Zener is in breakdown (ON) and holds the load at 50 V.

**(i) Output voltage:**

$$
V_o=V_Z=50\ \text{V}
$$

**(ii) Voltage across the series resistance:**

$$
V_R=120-50=70\ \text{V}
$$

**(iii) Zener current.**

$$
I=\frac{V_R}{R}=\frac{70}{5\ \text{k}\Omega}=14\ \text{mA}
$$

$$
I_L=\frac{50}{10\ \text{k}\Omega}=5\ \text{mA}
$$

$$
I_Z=I-I_L=14-5=9\ \text{mA}
$$

**Answer: (i) 50 V, (ii) 70 V, (iii) 9 mA** (total current 14 mA, load current 5 mA).

## Check yourself

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

1. Why is the 50 V Zener assumed to be in breakdown in this circuit?
   A. With the Zener removed, the load voltage 80 V exceeds 50 V
   B. Because the source is 120 V, which is greater than the load resistance
   C. Because the series resistance is smaller than the load resistance
   D. Because the Zener current is always larger than the load current
   Answer: A. Without the Zener, the divider gives 10/(5+10) x 120 = 80 V, which is above V_Z = 50 V, so the Zener conducts in breakdown. (Key point: Checking Zener state)

2. When the Zener is in breakdown, what is the output voltage?
   A. 80 V
   B. 70 V
   C. 50 V
   D. 120 V
   Answer: C. A Zener in breakdown holds the load at its Zener voltage, so V_o = V_Z = 50 V. (Key point: Output voltage)

3. How is the voltage drop across the 5 kΩ series resistor found?
   A. V_R = V_Z = 50 V
   B. V_R = 120 - 50 = 70 V
   C. V_R = 120 x 5/15 = 40 V
   D. V_R = 50 - 120 = -70 V used as 70 V from the load
   Answer: B. The source voltage minus the output voltage appears across R: 120 - 50 = 70 V. (Key point: Voltage across series resistance)

4. How is the Zener current obtained?
   A. I_Z = I + I_L = 14 + 5 = 19 mA
   B. I_Z = V_Z / R = 50/5k = 10 mA
   C. I_Z = I - I_L = 14 - 5 = 9 mA
   D. I_Z = I_L = 5 mA
   Answer: C. Total current I = 70 V/5 kΩ = 14 mA splits into load current 50 V/10 kΩ = 5 mA and the Zener current, so I_Z = 9 mA. (Key point: Zener current)

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