Rectifiers and Inverters: Converting AC to DC and DC to AC Worked Out
A mobile charger, a DC motor drive, a battery charger and the front end of every variable frequency drive all start with a rectifier that converts AC to DC. A home inverter, a solar inverter and the output stage of a VFD do the reverse, building AC from a DC source. This article works through the key formulas with numbers you can check on the bench: diode rectifiers, capacitor filter sizing, phase-controlled thyristor rectifiers, three-phase bridges, and the basics of single-phase inverters including battery current and backup time.
Single-Phase Diode Rectifiers
| Circuit | Average output Vdc | Ripple frequency at 50 Hz | Peak inverse voltage per diode | Ripple factor (resistive load) |
|---|---|---|---|---|
| Half-wave | Vm/π = 0.318 Vm | 50 Hz | Vm | 1.21 |
| Centre-tap full-wave | 2Vm/π = 0.637 Vm | 100 Hz | 2Vm | 0.482 |
| Bridge full-wave | 2Vm/π = 0.637 Vm | 100 Hz | Vm | 0.482 |
The bridge rectifier is the usual choice because it uses the full transformer secondary in both half cycles and each diode sees only Vm in reverse. Its drawback is that two diodes conduct in series, so the output is two diode drops below the peak.
Worked example: 12 V transformer with bridge and capacitor
A 230/12 V transformer feeds a bridge rectifier made of 1N5408 diodes, followed by a filter capacitor. The load draws 1 A. Design for a peak-to-peak ripple of about 2 V.
Peak secondary Vm = 12 × √2 = 16.97 V After 2 diode drops (≈0.9 V each at 1 A): Vpeak_out ≈ 16.97 − 1.8 = 15.2 V Ripple (full wave): ΔV ≈ I / (2 f C) C = I / (2 f ΔV) = 1 / (2 × 50 × 2) = 5000 µF Choose 4700 µF: ΔV = 1/(100 × 0.0047) = 2.1 V DC output ≈ Vpeak_out − ΔV/2 ≈ 15.2 − 1.05 ≈ 14.1 V Capacitor voltage rating: at least 25 V
This output can feed a 7812 or LM317 regulator, which needs a few volts of headroom above its output even at the bottom of the ripple. Note that the transformer's no-load voltage is higher than its rated 12 V, so the capacitor must be rated for that, and at light load the output rises.
Controlled Rectifiers and Three-Phase Bridges
Replacing diodes with SCRs lets you delay the start of conduction by a firing angle α. The average output voltage then falls as α increases. This is how DC motor drives and battery chargers control output, and how HVDC converters control power flow.
Single-phase full converter (continuous current): Vdc = (2Vm/π) cos α Single-phase half-controlled bridge: Vdc = (Vm/π)(1 + cos α) Three-phase diode bridge: Vdc = (3√2/π) VL = 1.35 VL Three-phase fully controlled bridge: Vdc = 1.35 VL cos α
For a 415 V three-phase supply, the diode bridge gives 1.35 × 415 = 560 V DC, which is why a VFD's DC bus on a 415 V supply sits around 560 V, a little higher with a capacitor filter and light load. A fully controlled bridge at α = 60° gives 560 × 0.5 = 280 V. With α beyond 90°, the average voltage becomes negative and, with a suitable DC source such as a motor in regeneration, the converter returns power to the AC side, which is called inverter mode.
Inverters: Building AC from DC
A single-phase H-bridge uses four switches, Q1 to Q4. Turning on Q1 and Q4 applies +Vdc across the load; turning on Q2 and Q3 applies −Vdc. Switching the pairs alternately at 50 Hz gives a square wave. Two switches in the same leg must never be on together, so a short dead time is inserted between them.
+Vdc
┌───┴───┐
Q1 Q3
├─LOAD─┤
Q2 Q4
└───┬───┘
0 V| Output type | How it is made | Quality and use |
|---|---|---|
| Square wave | Pairs switched at 50 Hz | High harmonics; fundamental RMS = 0.9 Vdc; obsolete for most loads |
| Modified (quasi) sine | Square wave with a zero-voltage gap each half cycle | Cheap UPS; fans may hum and run hot |
| Pure sine (SPWM) | Switches at kHz with duty cycle following a sine reference, then LC filter | Low distortion; standard in modern home inverters and solar inverters |
In SPWM, a sine reference at 50 Hz is compared with a high-frequency triangular carrier. The switch is on when the sine is above the carrier. The output is a train of pulses whose average follows the sine, and an LC filter removes the carrier frequency.
Worked Example: Home Inverter Battery Current and Backup
A 12 V home inverter supplies a 600 W load. Its efficiency is 85%, and it uses a 150 Ah tubular lead-acid battery. Estimate the battery current and backup time if only 50% of capacity is used to protect battery life.
- DC input power = 600/0.85 = 706 W.
- Battery current = 706/12 = 58.8 A.
- Usable energy = 12 × 150 × 0.5 = 900 Wh.
- Backup time ≈ 900/706 = 1.27 hours, about 76 minutes.
In reality backup is somewhat less, because lead-acid capacity falls at high discharge rates (the Peukert effect) and battery voltage sags under load. A 24 V system would halve the current to about 29 A, reducing cable losses, which is why inverters above roughly 1 kVA usually use 24 V or higher.
Summary
A single-phase bridge gives 0.637 Vm average, and a filter capacitor of C = I/(2fΔV) sets the ripple. SCR converters scale output by cos α, and a three-phase bridge gives 1.35 VL. Inverters chop DC with an H-bridge; SPWM with an LC filter produces a clean sine wave. Battery current equals load power divided by efficiency and battery voltage.