---
title: "Bridge rectifier output and ripple factor"
subject: "Basic Electronics Engineering"
chapter: 2
key: "semester_2/basic_electronics_engineering:bridge-rectifier-ripple-factor"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/bridge-rectifier-ripple-factor
site: BCT Engineering
---

# Bridge rectifier output and ripple factor

Basic Electronics Engineering · Chapter 2: Diodes

- Asked in: 2067 Chaitra, 3 marks
- Video: [Bridge Rectifier - Operation, Efficiency, Ripple Factor, Advantages and Applications](https://www.youtube.com/watch?v=0mVfFWkp41U) — Padmasri Naban
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/bridge-rectifier-ripple-factor
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2

## Question

Draw the bridge rectifier circuit and its output waveform with output load resistor (RL) connected. Express the ripple factor if a smoothing capacitor C is connected to the circuit.

## Short answer

The bridge rectifier uses four diodes to rectify both half cycles, producing a full-wave output across the load. With a smoothing capacitor C, the output ripple is approximately triangular; the ripple factor is r = 1/(4√3 f C RL), where f is the mains frequency. Larger C or RL reduces ripple.

## Answer

#### Bridge rectifier circuit
```text
                    A
                    o
                  /   \
       D4 (N -> A)     D1 (A -> P)
                /       \
   N (-) o-----+         +-----o P (+)
                \       /
       D3 (N -> B)     D2 (B -> P)
                  \   /
                    o
                    B

 ac input across A and B; (X -> Y means anode at X, cathode at Y)

   P (+) o----+---------+
              |         |
             [RL]     === C (smoothing)
              |         |
   N (-) o----+---------+
```
Diodes D1, D2, D3 and D4 form a bridge; ac is applied to two corners (A and B) and RL (with C in parallel) is connected across the other two (P and N). D1 and D3 conduct in the positive half cycle, and D2 and D4 in the negative half cycle.

#### Output waveform with RL (no capacitor)
```text
 vo   /\  /\  /\  /\
     /  \/  \/  \/  \
 ---------------------
```
The output is a train of positive half-sine pulses at 2f, with Vdc = 2Vm/π and ripple factor 0.482.

#### With a smoothing capacitor C
The capacitor charges to Vm during each rising pulse and discharges through RL between pulses. The ripple is approximately triangular. The time between charging peaks is T/2 = 1/(2f), so:

- Peak-to-peak ripple: Vr(pp) = Idc/(2fC) = Vdc/(2fCRL)
- RMS ripple (triangular wave): Vr(rms) = Vr(pp)/(2√3) = Vdc/(4√3 f C RL)

**Ripple factor:** r = Vr(rms)/Vdc = **1/(4√3 f C RL)**

A larger C, a larger RL or a higher f reduces the ripple. For example, f = 50 Hz, C = 1000 µF and RL = 100 Ω give r = 1/(4 × 1.732 × 50 × 10⁻³ × 100) = 0.029 (2.9%).

## Check yourself

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

1. In a bridge rectifier, which diodes conduct together?
   A. D1 and D3 in the positive half cycle, D2 and D4 in the negative
   B. All four diodes at once
   C. D1 and D2 together always
   D. D1 and D4 in the positive half cycle
   Answer: A. Opposite pairs conduct on alternate half cycles so RL always gets current in one direction. (Key point: Bridge conduction)

2. What is the ripple factor with smoothing capacitor C in a full-wave rectifier?
   A. 1/(2√3 f C RL)
   B. 1/(4√3 f C RL)
   C. 1/(4 f C RL)
   D. 4√3 f C RL
   Answer: B. Vr(rms) = Vdc/(4√3 f C RL), so r = 1/(4√3 f C RL). (Key point: Ripple factor with capacitor)

3. How can the ripple be reduced?
   A. Decrease C
   B. Decrease RL
   C. Increase C, RL or f
   D. Remove the capacitor
   Answer: C. Ripple factor is inversely proportional to f, C and RL. (Key point: Reducing ripple)

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