Rectifiers
Turning AC into one-way current.
Lesson at a glanceShow the key points
- One diode passes only positive half-cycles.
- Two diodes with a centre tap conduct alternately: full-wave.
- Average output V₀/π (half) or 2V₀/π (full).
Wall sockets supply alternating voltage, but phone chargers and radios need direct current. A diode, which only lets current through one way, is the first step in converting one into the other.
Half-wave rectifier
A rectifier turns an alternating voltage into a one-way (direct) voltage. The simplest has an AC source, one diode and a load resistor R in series. Let the source voltage be V = V₀ sin ωt.
- Positive half-cycle: the diode is forward-biased, so its resistance Rj is small. The current is i = V₀ sin ωt/(R + Rj) ≈ (V₀/R) sin ωt, and nearly the whole source voltage appears across R.
- Negative half-cycle: the diode is reverse-biased and its resistance is huge, so the current is practically zero.
The load current always flows the same way, but only for half of each cycle: half-wave rectification.
Full-wave rectifier
To use both halves, take the AC from a transformer secondary with a centre tap C. Each end, A and B, goes through its own diode (D₁ and D₂) to a common point E, and the load sits between E and C.
- When A is positive relative to C (and B negative), D₁ conducts and D₂ blocks: current flows A → D₁ → E → load → C.
- Half a cycle later B is positive: D₂ conducts and D₁ blocks: current flows B → D₂ → E → load → C.
Either way the current goes through the load from E to C. The output is a train of positive humps, twice as many as before.
Average output
Averaged over a cycle, a half-wave output gives V₀/π, and a full-wave output 2V₀/π. Real silicon diodes need about 0.7 V before they conduct well, which trims a little off every hump. In practice a large capacitor across the load smooths the humps into a nearly steady voltage.
A 12 V (peak) AC supply feeds a 1.0 kΩ load through an ideal diode in half-wave mode. Find the peak load current and the average load voltage. What would the average be with a full-wave circuit giving the same peak?
- 1A diode passes current in the positive half-cycles only: half-wave rectification.
- 2Two diodes with a centre-tapped transformer conduct alternately: full-wave rectification.
- 3Output is pulsating DC; average V₀/π (half) or 2V₀/π (full) for ideal diodes.
Check yourself
Keep exploring
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