Explain full-wave bridge rectification
Explain how four diodes route both alternating half-cycles through a load in the same direction.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 21. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use period, frequency, sine graphs, current, voltage, power, diodes, capacitors and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does a bridge rectifier do during the negative input half-cycle?
What you will learn
- Distinguish full-wave from half-wave output graphs
- Explain the conducting diode pairs in a four-diode bridge
- Relate output pulse frequency to input frequency
A bridge uses alternate diode pairs
A bridge rectifier contains four diodes around the AC input and load. During one input half-cycle, one diagonal pair is forward-biased; during the next half-cycle, the other diagonal pair conducts. The diode directions steer current through the load the same way in both cases.
Thus the load current remains unidirectional while the input polarity alternates. In an ideal graph, each negative input lobe is flipped upward at the load; no half-cycle is removed. Real bridge paths include two forward-biased diodes at a time, so their voltage drops reduce the output.
Full-wave output has two pulses per input cycle
Compare the graph shapes: a half-wave rectifier has one pulse per input cycle and a zero interval for the blocked half-cycle; a full-wave bridge has two positive pulses per input cycle, one from each half-cycle. Its ripple frequency is twice the input frequency.
The bridge output is still pulsating rather than constant. A smoothing capacitor can reduce the ripple, but rectification and smoothing are separate jobs: the bridge makes the polarity one-way, while the capacitor stores charge between peaks.
Pause and explain
A 60 Hz input feeds a full-wave bridge. What is the output ripple frequency before smoothing?
Worked example
A 50 Hz supply feeds an ideal bridge rectifier. State how the diode pairs conduct and the output ripple frequency.
Show the worked solution
- A different diagonal pair conducts on each half-cycle.
- Both pairs route current through the load in the same direction.
- There are two output pulses per input cycle.
Answer Alternate diagonal pairs conduct; the load current is unidirectional and the ripple frequency is 2 × 50 = 100 Hz.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Describe the current path through the load in the two half-cycles of a four-diode bridge.
Give me a hint
Name the pair that conducts alternately and keep the load direction fixed.
Compare my reasoning
- During the first half-cycle one diagonal pair is forward-biased.
- During the next half-cycle the opposite diagonal pair conducts.
- Both paths send conventional current through the load in the same direction.
Alternate diagonal pairs conduct, but load current has the same direction in both half-cycles.
Look for these in your work
- First pair stated
- Second pair stated
- Load direction compared
Draw the qualitative full-wave output for a sinusoidal input and compare it with half-wave output.
Give me a hint
Reflect the negative input half-cycle above zero rather than deleting it.
Compare my reasoning
- The full-wave bridge turns each negative input lobe into a positive load pulse.
- There are two positive pulses per input cycle.
- A half-wave rectifier instead has one pulse and one zero interval per cycle.
Full-wave output contains positive pulses for both half-cycles; half-wave output contains pulses for only one.
Look for these in your work
- Both halves mapped
- Pulse count correct
- Half-wave comparison made
A four-diode bridge is connected to a 50 Hz AC source. Explain why its output still needs smoothing and calculate the pulse frequency.
Give me a hint
Rectification fixes direction; smoothing reduces the gaps between pulses.
Compare my reasoning
- Each input cycle produces two output pulses, so pulse frequency is 2f.
- fout = 2 × 50 = 100 Hz.
- The voltage still falls between peaks, so a capacitor is needed to reduce ripple.
The pulse frequency is 100 Hz; the bridge output remains pulsating and needs smoothing for a steadier voltage.
Look for these in your work
- Frequency doubling used
- Numerical frequency correct
- Rectification/smoothing distinction stated
Common mistakes
- Saying the bridge blocks the negative half-cycle
- Claiming all four diodes conduct simultaneously in an ideal bridge
- Reversing load current on alternate half-cycles
- Calling the full-wave output smooth constant d.c.
What can you explain now?
What does a bridge rectifier do during the negative input half-cycle?
Compare with the explanation
A second diode pair routes current through the load in the same direction
The bridge steers both polarities through the load consistently.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
How do the four bridge diodes make both input half-cycles drive the load in one direction?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.