---
title: "Full bridge rectifier operation"
subject: "Basic Electrical and Electronics Engineering"
chapter: 3
key: "semester_1/basic_electrical_and_electronics_engineering:full-bridge-rectifier"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/full-bridge-rectifier
site: BCT Engineering
---

# Full bridge rectifier operation

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

- Asked in: 2081 Baishakh (new course), 6 marks
- Video: [Full Wave Bridge Rectifier](https://www.youtube.com/watch?v=Kl8IOESVWlM) — Neso Academy
- 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/full-bridge-rectifier
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-3

## Question

Draw the full bridge rectifier circuit and explain its operation. Express its equivalent average dc output voltage.

## Short answer

A full-bridge rectifier uses four diodes in a bridge to conduct on both half cycles, producing a pulsating DC output with frequency twice the input. The average DC output voltage is Vdc = 2Vm/pi ≈ 0.637 Vm, where Vm is the peak AC voltage.

## Answer

A full-bridge rectifier converts AC into pulsating DC using four diodes connected as a bridge, so both half cycles of the input are used.

```text
        A (AC in)
       /  \
     D1    D4
     /      \
   (+)-RL-(-)
     \      /
     D3    D2
       \  /
        B (AC in)
```

D1 and D2 conduct together on one half cycle; D3 and D4 conduct on the other. Each pair connects the load to the supply.

**Operation:**
- **Positive half cycle** (A positive, B negative): D1 and D2 are forward biased, and D3 and D4 are reverse biased. Current flows A → D1 → load (from + to −) → D2 → B.
- **Negative half cycle** (B positive, A negative): D3 and D4 conduct. Current flows B → D3 → load (again from + to −, the same direction) → D4 → A.

The current through $R_L$ has the same direction in both half cycles, so the output is a pulsating unidirectional waveform with two pulses per cycle (frequency $2f$).

**Average (DC) output voltage.** If the secondary peak voltage is $V_m$ (ignoring diode drops):

$$
V_{dc}=\frac{1}{\pi}\int_0^{\pi}V_m\sin\theta\,d\theta=\frac{2V_m}{\pi}=0.637\,V_m
$$

$$
I_{dc}=\frac{2I_m}{\pi}, \qquad V_{rms}=\frac{V_m}{\sqrt2}, \qquad \text{ripple factor}=0.48, \quad \eta_{max}=81.2\%
$$

With silicon diodes, two diodes conduct at a time, so $V_{dc}\approx\frac{2(V_m-1.4)}{\pi}$. The peak inverse voltage of each diode is $V_m$. No centre-tapped transformer is needed.

## Check yourself

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

1. In the positive half cycle (A positive), which diodes conduct?
   A. D1 and D2
   B. D3 and D4
   C. All four
   D. D1 and D3
   Answer: A. D1 and D2 are forward biased in one half cycle; D3 and D4 in the other. (Key point: Conducting pairs)

2. How does the load current behave over the two half cycles?
   A. It reverses each half cycle
   B. It flows in the same direction, giving two pulses per cycle (frequency 2f)
   C. It is zero in the negative half
   D. It has frequency f
   Answer: B. Both pairs pass current through RL in the same direction, giving full-wave pulsating DC. (Key point: Load current direction)

3. What is the average DC output voltage (ignoring diode drops)?
   A. Vm/pi
   B. 0.318 Vm
   C. 2 Vm/pi = 0.637 Vm
   D. Vm/sqrt(2)
   Answer: C. Vdc = (1/pi) integral of Vm sin(theta) from 0 to pi = 2Vm/pi. (Key point: Average output)

4. With silicon diodes, which statement is correct?
   A. Vdc = 2(Vm - 0.7)/pi, one diode conducts
   B. Vdc = 2Vm/pi unchanged; PIV = 2Vm
   C. Vdc = Vm - 1.4; centre-tap needed
   D. Two diodes conduct, so Vdc is about 2(Vm - 1.4)/pi, PIV = Vm, no centre-tapped transformer
   Answer: D. Two diodes are in the path, each dropping 0.7 V; PIV of each is Vm. (Key point: Diode drops and PIV)

## Related in Basic Electrical and Electronics Engineering

- [Zener circuit output voltage and current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Azener-circuit-120v-5k) — Basic Electrical and Electronics Engineering, chapter 3

## Related in other subjects

- [Full wave bridge rectifier operation](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Afull-wave-bridge-rectifier) — Basic Electronics Engineering, chapter 2, asked 4×
- [Full bridge inverter; dc-dc output power control](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Afull-bridge-inverter-dc-dc-control) — Advanced Electronics, chapter 7
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