How a one-way switch makes dc from ac
A sine wave has an average of zero. A diode lets current pass in one direction only, and that is enough to create a dc component. A thyristor adds a gate, so you can choose when it starts to conduct. In this module you will watch current find its way through half-wave, full-wave and three-phase rectifiers, then use them to meet voltage, current, ripple and power-factor targets.
By the end you can
1. Trace the conduction path in each circuit state, and find the average (dc) output voltage of a half-wave, full-wave or three-phase rectifier.
2. Decide from R, L, the delay angle and any dc source whether the load current is continuous or falls to zero, and say how that changes the equations.
3. Use the delay angle α of a thyristor rectifier to set the output, and explain when the same circuit becomes an inverter.
4. Choose a filter inductor or capacitor to meet ripple limits, and describe what each does to the ac line current (harmonics, power factor, peak diode current).
5. Explain how source inductance causes commutation overlap and lowers the output voltage.
Three rules for any rectifier in steady state
Every result in this module comes from three statements that hold once the circuit repeats from cycle to cycle:
Switch logic and Kirchhoff’s laws. A diode conducts when it is forward-biased. A thyristor conducts when it is forward-biased and has been gated, and it stays on until its current reaches zero. Kirchhoff’s laws tell you which devices can conduct together.
Inductor volt-second balance. The average voltage across the load inductance over one period is zero, so the average load current is Io = (Vo − Vdc)/R, whatever L is. Capacitor charge balance says the average capacitor current is zero, so the average diode current equals the average load current.
Power balance. With ideal devices the average power from the ac source equals R·Irms² plus Vdc·Io. Only the resistor dissipates; ideal inductors and capacitors do not.
The six rectifier families
| Circuit | Pulses per cycle | Uncontrolled Vo (ideal) | Controlled Vo | Lowest output ripple |
|---|---|---|---|---|
| Half-wave | 1 | Vm/π | Vm2π(1 + cos α) | f |
| Full-wave (bridge or center-tapped) | 2 | 2Vm/π | Vmπ(1 + cos α) for R, 2Vmπcos α for continuous current | 2f |
| Three-phase bridge | 6 | 3Vm,LL/π = 1.35 VLL | 3Vm,LLπcos α (continuous current) | 6f |
One question decides the equations
Does the load current ever fall to zero? With a resistor it falls to zero whenever the source cannot push it. With an inductor (and no freewheeling path) it can outlast the source and then stop. With enough inductance and a freewheeling or bridge path it never stops. This is the rectifier version of continuous (CONT) versus discontinuous (DISC) conduction in a dc-dc converter, and the tutorial shows the badge in every lab.
Each lab also reports the power factor and the total harmonic distortion of the ac source current, because a rectifier loads the ac line with a nonsinusoidal current. That is the price of using it.
How to use this module
Each tab has a short explanation, a circuit explorer, a live lab and activities. Move any slider and every number and graph updates. Activities with a number to find have a Load these settings button so everyone starts from the same circuit, and a Check button that compares your answer with the model. Open-response prompts are kept for you to download on the “Check understanding” tab. Nothing is sent anywhere; your answers stay in this page.
Suggested path: how rectifiers work, then half-wave, full-wave and three-phase in the order uncontrolled then controlled, then source inductance, then the design challenges.
How rectifiers work
Before the equations, see what a one-way switch does to a sine wave, what an ac line sees when the load current is smooth, and where the current goes in each circuit. Everything on this page responds to your clicks and sliders.
What a one-way switch does to a sine wave
The average value of the output is the dc component the rectifier is meant to create. The rms value sets the heating. Their ratio, the form factor, tells you how far the waveform is from dc. A delay angle α removes the first part of each conducting interval.
Average and rms of the rectified wave
Resistive load. For the three-phase case, Vm is the peak line-to-line voltage. With α above 60° the three-phase output touches zero and the formulas for continuous current no longer apply.
What the ac line sees
The load wants dc; the source supplies whatever current the switches pass. If the load current is smooth (large L), the source current is a series of flat blocks. Their shape fixes the harmonics. The power factor is P/(VrmsIrms), which equals the distortion factor 1/√(1 + THD²) times the displacement factor cos φ1 when no dc flows in the source.
Source current for a very inductive load
The half-wave circuit with a freewheeling diode draws current only on positive half-cycles, so its source current has a dc component. The full-wave bridge reverses the current and averages to zero. Delaying the firing shifts the current later, so the displacement factor falls even though the load did not change.
Follow the current in each rectifier
The half-wave rectifier is one diode in series with the load. It is the simplest circuit and the key to the others.
How it works
ONWhile the source is positive the diode is forward-biased and the load sees the source voltage. With an inductor the diode stays on past 180° (LOAD) until the current reaches zero.
OFFWith the diode off the load voltage is zero (or the dc source voltage), and the diode blocks the source voltage.
Where the dc comes from
The source current is only ever positive, so its average is not zero. That is the half-wave rectifier’s weakness.
The full-wave rectifier passes both half-cycles. The bridge uses four diodes; the center-tapped circuit uses two diodes and a transformer with a center tap.
How it works
ONWhen the source is positive, D1 and D2 carry the load current. ON When it is negative, D3 and D4 carry it in the same direction through the load.
Where the dc comes from
The output frequency is twice the source frequency, and the average source current is zero. The bridge needs no center tap and each off diode sees only Vm; the center-tapped circuit has one diode drop in the path but each off diode sees 2Vm.
The three-phase bridge has six diodes in three legs. At every instant the diode connected to the highest phase and the diode connected to the lowest phase conduct.
How it works
The output is always the largest line-to-line voltage, so the changeover happens every 60°: six pulses per cycle. Each diode conducts for 120° and carries the full load current.
Where the dc comes from
The output only dips to 87 % of its peak, at 6f, which is why filtering a three-phase rectifier is easy.
Compare the three
| Property | Half-wave | Full-wave | Three-phase bridge |
|---|---|---|---|
| Average output | Vm/π = 0.318 Vm | 2Vm/π = 0.637 Vm | 0.955 Vm,LL |
| Lowest ripple frequency | f | 2f | 6f |
| Average source current | Not zero | Zero | Zero |
| Source current shape (smooth load) | One block per cycle | Square wave | Two 120° blocks |
| Harmonics in the source current | All | Odd (3, 5, 7, …) | 5, 7, 11, 13, … (6k ± 1) |
| Power factor (smooth load, α = 0) | about 0.45 to 0.7 | 0.90 | 0.955 |
| Typical use | Low power | Power supplies, chargers | Industrial dc, drives, HVDC |
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Half-wave rectifiers with diodes
One diode, one load. The load decides everything else: a resistor follows the source, an inductor stretches conduction past 180°, a freewheeling diode keeps the current alive, a dc source delays the start, and a capacitor turns the output into short pulses of charging current.
Resistive load
The current is a scaled copy of the voltage while the diode conducts. The dc output and the rms values follow from integrating a half sine.
The source current is the load current, so pf = P/(Vs,rmsIrms) = 0.707. Hart’s Example 3-1 (120 V, 5 Ω): Io = 10.8 A, P = 1440 W, pf = 0.707.
R-L load
The current is a forced response (the sine the circuit would settle to) plus a natural response (a decaying exponential set by τ = L/R):
The diode stays on until the current reaches zero at the extinction angle β, which has no closed form: sin(β − φ) + sin φ·e−β/ωτ = 0. Between 180° and β the load voltage is negative while the inductor releases its energy. Only R dissipates power, and the power factor is not cos φ. Hart’s Example 3-2 (R = 100 Ω, L = 0.1 H, Vm = 100 V): β = 3.50 rad (201°), Io = 0.308 A, P = 22.4 W.
R-L load with a freewheeling diode
D2 conducts when the source is negative, so the load voltage is a clean half-wave rectified sine. The average current depends only on R, and the inductor sets the ripple:
For L/R ≫ T the current is nearly constant, D1 carries it for half the cycle (average Io/2, rms Io/√2), and so does D2. Hart’s Example 3-7 (R = 2 Ω, L = 25 mH, Vm = 100 V): Io = 15.9 A, Irms = 16.3 A, P = 534 W.
R-L load with a dc source
This models a battery charger or a dc motor. The diode cannot conduct until the source exceeds the dc voltage:
The dc source absorbs Pdc = IoVdc, the resistor absorbs R·Irms², and the ac source supplies the sum. Hart’s Example 3-5 (120 V, R = 2 Ω, L = 20 mH, Vdc = 100 V): conduction from 36.1° to β = 193°, Io = 2.25 A, Pdc = 225 W, pf = 0.54.
R with a capacitor filter
The capacitor charges to Vm and then discharges into R until the source rises above it again. The diode turns off at tan⁻¹(−ωRC) + π and back on at 2π + θon, found numerically.
A larger C lowers the ripple but squeezes the same charge into a shorter pulse, so the peak diode current rises. Hart’s Example 3-9 (120 V, R = 500 Ω, C = 100 µF): ΔVo = 43 V, ID,peak ≈ 4.5 A, and C = 3333 µF would be needed for 1 % ripple.
Lab: half-wave rectifiers
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Half-wave rectifiers with a thyristor
Replace the diode with an SCR and the output is no longer fixed. The SCR conducts only when it is forward-biased and has a gate pulse. The delay angle α sets the output.
Resistive load
The SCR waits until α, then follows the source until it crosses zero:
Hart’s Example 3-10: a 120 V source and a 100 Ω load need α = 61.2° for a 40 V average. The load absorbs 57.1 W and the power factor is 0.63. A delay angle is a loss-free way to cut power, but it makes the source current more distorted.
R-L load
The current starts at zero at α and follows the same form as the uncontrolled case with a new initial condition:
The conduction angle is β − α. Hart’s Example 3-11 (120 V, R = 20 Ω, L = 40 mH, α = 45°) has β = 3.79 rad (217°).
R-L load with a dc source
Conduction cannot begin before the source exceeds Vdc: the SCR starts at the later of α and sin⁻¹(Vdc/Vm). Delaying the firing reduces the power the dc source absorbs, with no resistor loss. In Hart’s Example 3-13 (Vdc = 100 V, R = 2 Ω, L = 20 mH, 120 V source) an α of 70° gives about 148 W instead of the 225 W an uncontrolled diode delivers.
Lab: controlled half-wave rectifiers
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Full-wave rectifiers with diodes
Two kinds of full-wave circuit rectify both half-cycles: the four-diode bridge and the center-tapped transformer circuit. They give the same load waveform. They differ in the voltage the off diodes must block, the number of diode drops in the path, and whether there is a transformer.
| Bridge | Center-tapped | |
|---|---|---|
| Diodes | 4 | 2, plus a center-tapped transformer |
| Reverse voltage on an off diode | Vm | 2Vm (Vm of one half-winding) |
| Diode drops in the path | 2 | 1 (better at low voltage, high current) |
| Isolation | None required | Provided by the transformer |
| Average source current | Zero | |
Resistive load
The source current is a sine in phase with the voltage, so pf = 1. The output is the magnitude of the source; the fundamental of the output is at 2f and the Fourier series has only even harmonics. Hart’s PSpice example shows what a real bridge adds: two diode drops cut the output and power.
R-L load
The load voltage is still the rectified sine, vo = Vm[2/π − (4/π)Σ cos(nω0t)/(n² − 1)] for n = 2, 4, 6, …. Take each frequency separately:
Hart’s Example 4-1 (Vm = 100 V, R = 10 Ω, L = 10 mH): Io = 6.37 A, I2 = 3.39 A, P = 464 W. With a large L the load current is almost constant and the source current is a ±Io square wave.
R-L load with a dc source
If the current is continuous the load voltage is the full rectified sine, and the average current is Io = (2Vm/π − Vdc)/R. The ac terms are unchanged. If Vdc is large (or L small) the current becomes discontinuous and the analysis follows the half-wave circuit with a period of π. Hart’s Example 4-3 (120 V, R = 2 Ω, L = 10 mH, Vdc = 80 V): Io = 14.0 A, Irms = 15.5 A.
R with a capacitor filter
The capacitor now discharges for only half as long, so the ripple is about half that of the half-wave circuit:
Hart’s Example 4-4 (120 V, R = 500 Ω, C = 100 µF) gives 21 V compared with 43 V for the half-wave circuit. The diode current is still a narrow pulse, so the source current has a high THD and a low power factor. Voltage-doubler circuits use two capacitors to get 2Vm without a transformer.
LC filter
The inductor smooths the current and the capacitor smooths the voltage. With continuous inductor current the output is the average of the rectified sine:
Below Lmin the inductor current touches zero and the output climbs toward Vm, as a dc-dc converter does in DCM. Compared with the capacitor-only filter, the diode current is much less peaky.
Lab: full-wave rectifiers
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Full-wave rectifiers with thyristors
With SCRs in the bridge, S1 and S2 are fired at α after the source turns positive, and S3 and S4 at π + α. With an inductive load the previous pair keeps conducting until the next pair fires, so the output goes negative for a while. That is the key to understanding the cos α law, and to inverter operation.
Resistive load
Hart’s Example 4-6 (120 V, R = 20 Ω, α = 40°): Io = 4.77 A, P = 673 W, S = 696 VA, pf = 0.97.
R-L load: continuous or discontinuous?
If the current from S1, S2 reaches zero (at β) before S3, S4 are fired (at π + α), the current is discontinuous and the half-wave analysis applies with period π. If it is still positive, the next pair takes over and the current is continuous:
Hart’s Example 4-7 (R = 10 Ω, L = 20 mH, α = 60°) is discontinuous, with β = 216°. With L = 100 mH (Example 4-8) it is continuous and Vo = (2Vm/π)cos α. The ac terms are found by Fourier series, with Vn depending on α.
R-L load with a dc source
For continuous current the load voltage average depends on α only, and the current follows from the load:
For α > 90° the average bridge voltage is negative. With a dc source that delivers power (Vdc negative in this circuit) the bridge operates as an inverter: power flows from the dc side to the ac line. Hart’s Example 4-10: 1000 W into a 100 V source from 240 V needs α = 46°. Example 4-11: 1000 W from a 110 V solar array into a 120 V line needs α ≈ 167°.
Lab: controlled full-wave rectifiers
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Three-phase bridge rectifiers with diodes
For large loads the three-phase bridge is the standard. Six diodes, six output pulses per cycle, and a dc output that is already nearly smooth.
What happens in the circuit
Kirchhoff’s laws allow only one top diode (D1, D3 or D5) and one bottom diode (D2, D4 or D6) to conduct at a time. The top diode is the one connected to the highest phase and the bottom diode is the one connected to the lowest. The output is the line-to-line voltage between them, so the pairs conduct in the order (6,1), (1,2), (2,3), (3,4), (4,5), (5,6), each for 60°.
Each diode must block the peak line-to-line voltage. The output voltage has harmonics at 6k times the source frequency with amplitudes (6Vm,LL/π)/(n² − 1). The source current has harmonics at 6k ± 1 (5th, 7th, 11th, 13th, …); these are the harmonics large installations filter. Hart’s Example 4-12 (480 V line-to-line, R = 25 Ω, L = 50 mH): Vo = 648 V, Io = 25.9 A.
With a resistive load the load current is the output voltage divided by R, so it ripples with the voltage (87 % to 100 % of its peak).
With series inductance the current is nearly dc and the source current is two 120° blocks per cycle. The load can be treated as a constant current to size the diodes and the source.
A dc source in the load (a battery or a motor’s back-emf) sets the average current through Io = (Vo − Vdc)/R, and the same continuity conditions as a single-phase bridge apply when L is small.
Lab: three-phase bridge rectifiers
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Three-phase bridge rectifiers with thyristors
Gate each SCR α after it would have started as a diode and the output follows the same line-to-line segments, slid later by α. The average output falls as cos α; beyond 90° the converter can be an inverter.
What changes with α
Hart’s Example 4-13 (480 V line-to-line, R = 10 Ω, L = 50 mH) needs α = 39.5° for 50 A. The output harmonics stay at 6k but their amplitudes grow with α. The source power factor is about 0.955 cos α, so deeper control costs power factor. For a resistive load with α > 60° the output touches zero and the cos α formula no longer applies.
Inverter operation and dc transmission
With α > 90° and a dc source that supplies power, the converter returns power to the ac system. A dc transmission link pairs a rectifier and an inverter terminal: power flow is set by the two delay angles and can reverse. Real links use 12-pulse converters, two six-pulse bridges fed 30° apart (Y-Y and Y-Δ transformers). The output ripple moves up to 12f and the ac-side harmonics 5, 7, 17, 19, … are cancelled, leaving 12k ± 1.
With a resistive load the current follows the voltage. Up to α = 60° the output stays positive and Vo = (3Vm,LL/π)cos α. Past 60° the current is discontinuous and the formula changes; compare the sweep graph with the dashed curve.
With series inductance the current is continuous even at large α, and Vo = (3Vm,LL/π)cos α holds up to 90°, where the average is zero.
Add a dc source (a motor’s back-emf, or a link at the other end). Make Vdc negative and α above 90° to see the inverter: the dc source delivers power and the ac side absorbs it.
Lab: controlled three-phase bridge rectifiers
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Real sources: commutation overlap
Every ac source has inductance. The source current cannot jump, so the load current passes from one device to the next over an angle u, during which both devices conduct. Averaged over the cycle, that lowers the output voltage.
Assume a constant load current Io (a large inductor) and a source reactance Xs = ωLs. While two devices overlap, the voltage across Ls is the source voltage (or half the difference of two line voltages in the three-phase bridge), and the current in Ls ramps. Commutation ends when the incoming device carries all of Io:
| Circuit | Overlap angle | Average output |
|---|---|---|
| Half-wave with freewheeling diode | cos u = 1 − IoXs/Vm | Vmπ(1 − IoXs2Vm) |
| Single-phase bridge | cos u = 1 − 2IoXs/Vm | 2Vmπ(1 − IoXsVm) |
| Three-phase bridge | cos u = 1 − 2IoXs/Vm,LL | 3Vm,LLπ(1 − XsIoVm,LL) |
With a delay angle the overlap satisfies cos α − cos(α + u) = kIoXs/Vm (k = 1 for the half-wave circuit, 2 for the bridges) and Vo = (p/2π)Vm[cos α + cos(α + u)] with p the pulse number factor shown in the formula reference. When the converter is used as an inverter, α + u must stay safely below 180° so the outgoing thyristors can turn off.
Lab: commutation overlap
Worked examples: change the answer and see what happens
Each example opens on the correct answer. Move any slider under “Your answer” to see what would change: the checklist shows what passes, and a grey dashed line on every graph marks the correct answer. Sliders under “Change the givens” change the problem itself, and the correct answer updates to match. Use “Ask a question” on any example.
Activities
Design challenges
Choose the parts and settings. Each design is tested at the operating points that matter (light and heavy load, or nominal and low line voltage). A design passes when every check passes.
Check your understanding
Eight questions on the ideas from this module, then reflection prompts and a downloadable copy of your responses.
Reflect
Your questions for your instructor
Questions the example panels could not answer are saved here. Copy them to send to your instructor, or download them with the rest of your responses.
Save your work
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Formula reference
Ideal devices, steady state, sinusoidal source vs = Vm sin ωt. For three-phase, Vm,LL = √2VLL. Delay angle α is measured from the natural conduction point.
Average output and rms, resistive or continuous current
| Half-wave | Full-wave | Three-phase bridge | |
|---|---|---|---|
| Vo, uncontrolled | Vmπ | 2Vmπ | 3Vm,LLπ |
| Vo, controlled, resistive | Vm2π(1 + cos α) | Vmπ(1 + cos α) | 3Vm,LLπcos α for α ≤ 60° |
| Vo, controlled, continuous current | — | 2Vmπcos α | 3Vm,LLπcos α |
| Vrms, uncontrolled | Vm2 | Vm√2 | — |
| Vrms, controlled, resistive | Vm2√1 − απ + sin 2α2π | Vm√2√1 − απ + sin 2α2π | — |
| Lowest output harmonic | f | 2f | 6f |
| Source current harmonics | all | odd | 6k ± 1 |
| Peak reverse voltage | Vm | Vm (bridge), 2Vm (center-tapped) | Vm,LL |
Load relations
Filters and continuity
Three-phase device and line currents (smooth load current)
Source inductance
k = 1 and p = 1 for the half-wave circuit with a freewheeling diode; k = 2 and p = 2 for the single-phase bridge; k = 2 and p = 3 for the three-phase bridge (with Vm,LL). The loss at α = 0 is IoXs/Vm of the ideal value for the bridges.