The Single-diode Rectifier
The Single-diode Rectifier
- A diode allows conventional current through one way and blocks it the other way. This is enough for half-wave rectification.
Follow the two input half-cycles
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- During one half-cycle, the diode is forward biased. A complete path exists and current flows through the load.
- During the opposite half-cycle, the diode is reverse biased. It breaks the path, so load current is zero.
- The load therefore receives one positive hump followed by one flat zero section in each input cycle.
Common mistake
The diode does not turn a negative voltage into a positive one. A single diode removes one direction; a bridge reverses it.
Half-wave power includes a second half factor
Within a remaining power hump, the mean is half its peak. The blocked half-cycle contributes zero, adding another factor of one half.
Worked example
Find a half-wave r.m.s. output
A half-wave rectifier feeds a 45 Ω resistor. The sinusoidal input has r.m.s. voltage 6.0 V. Find the output r.m.s. voltage.
- Convert input r.m.s. voltage to peak voltage.
- Use the half-wave relation.
Answer
The missing-half factor and final r.m.s. value were assessed in 2025 (9702/42/M/J/25 Q8).
Your turnhalf-wave check
An ideal half-wave rectifier has a sinusoidal input with peak voltage 10 V. Calculate the r.m.s. output voltage.
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Answer