Explain half-wave rectification with a diode
Read a half-wave rectified graph and explain how one diode blocks one half-cycle of alternating current.
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 single series diode do to an alternating current in a half-wave rectifier?
What you will learn
- Distinguish half-wave rectification from an alternating waveform
- Explain how one diode produces half-wave rectification
- Relate diode direction to the conducting half-cycle
A diode conducts in one direction
A diode allows conventional current through when it is forward-biased and blocks current when reverse-biased. Connect one diode in series with a load on an AC source: during one half-cycle the diode conducts, while during the opposite half-cycle it blocks.
With an ideal diode and a chosen orientation, the load voltage follows the positive half-cycles and is zero during negative half-cycles. Reversing the diode orientation selects the opposite half-cycles. The output is pulsating unidirectional voltage, not steady direct current.
The output graph keeps alternate half-cycles
On a time graph, half-wave rectification has one series of positive (or negative, if reversed) lobes separated by intervals at zero. The input sinusoid crosses the axis each half-period; the half-wave output contains a pulse once per input cycle.
The ideal graph ignores a diode's forward voltage drop and other component losses. A real diode reduces the output magnitude slightly. Do not draw the blocked half-cycles as inverted output: they are removed at the load in this ideal rectification model.
Pause and explain
How many output pulses appear per input cycle in an ideal half-wave rectifier?
Worked example
An ideal diode is oriented to pass positive input half-cycles to a resistor. Describe the load-voltage graph and what changes if the diode is reversed.
Show the worked solution
- Identify which half-cycles forward-bias the diode.
- During reverse bias, the load current is blocked.
- Reverse the diode and the selected polarity swaps.
Answer The output has positive half-sine pulses separated by zero intervals. Reversing the diode gives negative pulses from the opposite half-cycles.
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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.
Sketch in words the ideal load voltage for an input that begins with a positive half-cycle, using one diode oriented to pass positive current.
Give me a hint
Keep the conducting lobes and identify the blocked intervals.
Compare my reasoning
- The first positive half-sine reaches the load.
- The following negative half-cycle is blocked, so load voltage is zero.
- This pattern repeats: one positive pulse for each input cycle.
Positive half-sine pulses are separated by zero-voltage intervals.
Look for these in your work
- Pass half-cycle identified
- Blocked half-cycle shown as zero
- Cycle pattern repeated
A half-wave rectifier's diode is reversed. What happens to the ideal load waveform?
Give me a hint
Determine which input polarity now forward-biases the diode.
Compare my reasoning
- Reversing the diode reverses its conducting direction.
- The former blocked half-cycles now pass; the former passing half-cycles are blocked.
- The output polarity reverses while retaining one pulse per input cycle.
Negative half-wave pulses reach the load, with zero during the opposite half-cycles.
Look for these in your work
- Conducting half-cycle swapped
- Polarity stated
- Pulse rate identified
A learner draws a half-wave rectifier output as positive pulses during both input half-cycles. Identify the circuit change needed for that output and explain why one diode is insufficient.
Give me a hint
Compare the one-diode output with a bridge rectifier.
Compare my reasoning
- One diode blocks one of the input half-cycles.
- A four-diode bridge routes both input half-cycles through the load in the same direction.
- The learner's two-pulse output is full-wave, not half-wave, rectification.
A four-diode bridge rectifier is needed; a single diode passes only one half-cycle.
Look for these in your work
- Single-diode limitation stated
- Bridge function explained
- Waveform named correctly
Common mistakes
- Calling the pulsating output steady d.c.
- Drawing blocked half-cycles inverted at the load
- Forgetting diode orientation selects polarity
- Ignoring the stated ideal-diode assumption in a graph calculation
What can you explain now?
What does a single series diode do to an alternating current in a half-wave rectifier?
Compare with the explanation
Passes one half-cycle and blocks the other
The diode conducts in one direction only.
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Make it stick
How does a diode's orientation determine which half-cycles appear across a load?
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