Bridge Rectifier Calculator
AC → DC bridge rectifier design: compute Vpeak, Vdc, ripple, PIV, diode current, and filter capacitor needs from transformer and load specs. Supports half-wave mode, 50/60 Hz, Si/Schottky diodes.
⚡ Design Calculator
📐 Worked Calculations (verified)
Vpeak = 12 × √2 = 16.97 V · Unloaded Vdc = 16.97 − 2×0.7 = 15.57 V · ripple f = 2×50 = 100 Hz · Ripple = 1/(100 × 0.0047) = 2.13 Vpp · Loaded Vdc ≈ 15.57 − 2.13/2 = 14.51 V · PIV per diode = 16.97 V · Average current per diode = 0.50 A.
Vpeak = 33.94 V · Vdc = 32.54 V · Ripple (f = 120 Hz) = 1.67 Vpp · Current/diode = 1.00 A.
Vdc = 16.97 − 0.7 = 16.27 V · ripple f = 50 Hz → Ripple = 4.26 Vpp (2× bridge).
🔗 Formulas Used
| Quantity | Formula | Notes |
|---|---|---|
| Vpeak | Vrms × √2 | Transformer secondary peak voltage |
| Vdc (unloaded) | Vpeak − n·Vd | n = 2 bridge, n = 1 half-wave |
| Ripple frequency | 2 × mains (bridge) / 1 × mains (half) | 100 Hz / 50 Hz at 50 Hz mains |
| Ripple Vpp | I / (f × C) | With filter capacitor |
| Vdc loaded (est.) | Vdc − ripple/2 | Mid-ripple point estimate |
| PIV per diode | Vpeak | Choose rating ≥ 2× for margin |
| Average current / diode | I/2 (bridge), I (half) | For diode current rating sizing |
| C for target ripple | I / (f × Vr_target) | In farads; ×10⁶ for µF |
| Vdc without capacitor | (2/π)·Vpeak − 2Vd (bridge) | Average of rectified |sine| |
❓ Frequently Asked Questions
What is a bridge rectifier?
A bridge rectifier is a 4-diode circuit that converts AC voltage into full-wave pulsing DC — both AC half-cycles are used, unlike half-wave which discards half. This calculator computes Vpeak, Vdc, ripple, and PIV from your transformer and load specs.
How do I calculate Vpeak from Vrms?
Vpeak = Vrms × √2. Example: a 12 Vrms transformer produces a Vpeak of 16.97 V. This is the maximum voltage the filter capacitor and diodes see.
Why is unloaded Vdc = Vpeak − 2×Vd?
In a bridge, current always passes through two series diodes each half-cycle, so two diode forward drops are lost. For 12 Vrms with silicon diodes (Vd 0.7 V): 16.97 − 1.4 = 15.57 V. In half-wave only one diode is in series (subtract 1×Vd → 16.27 V).
What is the ripple formula with a filter capacitor?
Vr ≈ I / (f × C), with f = 2 × mains frequency for bridge (100 Hz at 50 Hz) or 1 × for half-wave. Example: 1 A load with C 4700 µF → 1/(100 × 0.0047) = 2.13 V peak-to-peak. Loaded Vdc ≈ unloaded Vdc − ripple/2 ≈ 14.51 V.
What PIV must each diode withstand?
PIV (peak inverse voltage) = Vpeak = 16.97 V for the 12 Vrms example. Choose diodes with a reverse voltage rating (V_RRM) of at least 2× Vpeak as a safety margin — for this example a 1N4007 (1000 V) or equivalent is more than enough.
What is the difference between bridge and half-wave?
Bridge uses both half-cycles: ripple frequency is 2× mains, so it needs half the capacitance for the same ripple. Half-wave on the same example gives 4.26 Vpp ripple (2× larger) and loads the transformer with DC current. Bridge is almost always the right choice.
What if there is no filter capacitor?
Without a capacitor, Vdc = (2/π) × Vpeak − 2Vd for bridge, i.e., 9.40 V at 12 Vrms, with a ripple of 16.97 Vpp. Only really suitable for simple resistive loads like heating elements.
How does this differ from the inverting buck-boost converter?
This tool designs an AC→DC rectifier (from transformer/mains to DC). Inverting Buck-Boost Converter Calculator (#432) is a switching DC-DC converter that turns one DC voltage into another DC voltage of opposite polarity — totally different physics and formulas.
How accurate are this calculator's results?
The results are ideal values: transformer winding resistance (ESR), ±10% mains voltage tolerance, and Vd variation with current and temperature are not modeled. Treat results as a design starting point, then verify with measurements on the real circuit.
How do I choose the filter capacitor value?
C = I / (f × Vr_target). For a 0.5 V ripple target at 1 A / 50 Hz bridge: C = 1/(100 × 0.5) = 20,000 µF. Choose the capacitor voltage rating above Vpeak with margin — e.g., Vpeak 16.97 V → pick a 25 V capacitor.
Why is there a start-up surge current warning?
When first switched on, the empty capacitor behaves like a short circuit: the peak charging current can be dozens of times the normal load current. Use an NTC inrush limiter or soft-start circuit, and make sure the diodes can handle the surge current (I_FSM) per the datasheet.
When should I use Schottky diodes?
Schottky (Vd ≈ 0.3 V) saves 0.8 V vs silicon: the 12 Vrms example's Vdc becomes 16.37 V vs 15.57 V. Good for low-voltage rails (5 V / 3.3 V). Downsides: lower maximum PIV ratings and larger reverse leakage current.
Free tool by NRTechStudio