AC to DC Voltage Calculator
Easily convert AC RMS to DC voltage. Includes half-wave, full-wave, diode drop, and ripple capacitor formulas for accurate rectifier design.
reservoir capacitor ripple calculator
To rectify an AC input To rectify an AC input, use first half-wave average = -2V = full-wave average =2V. Include drops of diodes (single diode ≈V bridge ≈2V with a reservoir capacitor and load I, filtered DC ≈ − V 2FC)-ripple(ripple) =f half-wave, 2f full-wave.
Formula & Table Summary:
- Half-wave:
ΔVpp ≈ I / (f · C) - Full-wave:
ΔVpp ≈ I / (2 · f · C)
P = Vdc × Iload (approx for filtered supplies).Converter table — Input → Output
| Input available | Use / Formula | Output |
|---|---|---|
| Vrms & rectifier type | Vp=Vrms√2 → Vdc,avg=Vp/π (half) or 2Vp/π (full) | Unfiltered DC average (V) |
| Vrms, diode drops | Subtract Vdrops from Vp for practical DC | Vpeak avail=Vp−Vdrops |
| Vrms, I, C, f, rectifier type | ΔVpp=I/(fripple·C) → VDCfiltered≈Vp−Vdrops−ΔVpp/2 | Filtered DC (approx) and ripple |
| Desired VDC, I, f, Vd | Choose C ≈ I / (fripple · ΔVpp,allowed) | Required C for target ripple |
rectifier smoothing capacitor calc
The AC voltage to DC Voltage Calculator is used to transform the voltage between DC and AC circuits (RMS) in rectifier circuits that are filtered or not. It includes peak conversion (Vpeak = Vrms642), average DC of rectified waveforms (half-wave/full-wave), corrections in forward-drop of a single diode (or diode bridge) and single diode (or bridge) smoothed DC using a reservoir capacitor (formulas of approximate ripple). With this tool, design and troubleshooting of power supplies and battery charging circuits are estimated by comparing the unloaded DC (no filter), filtered DC under load, and peak/peak-to-peak ripple.
rectifier voltage calculator
| Scenario | Inputs | Calculation | Result |
|---|---|---|---|
| Unfiltered full-wave (ideal) | Vrms = 12 V, full-wave | Vp = 12√2 = 16.971 V; Vdc = 2Vp/π = 2×16.971/π | Vdc ≈ 10.80 V |
| Bridge + smoothing (small ripple) | Vrms = 12 V, full-wave bridge (2×Vd), I=0.5 A, f=50 Hz, C=2200 µF, Vd≈0.7 V |
Vp=16.971 V; Vp_avail=Vp − 2Vd = 16.971 − 1.4 = 15.571 V; ΔVpp ≈ I/(2fC) = 0.5/(2×50×2200e-6)=0.5/(0.22)=2.273 V; Vdc≈Vp_avail − ΔVpp/2 |
Vdc ≈ 15.571 − 1.137 = 14.43 V; ΔVpp ≈ 2.27 V |
| Half-wave unfiltered | Vrms = 9 V, half-wave | Vp = 9√2 = 12.728 V; Vdc = Vp/π = 12.728/π | Vdc ≈ 4.05 V |
| Required C for small ripple | Target ΔVpp ≤ 0.5 V, I = 1 A, full-wave, f=50 Hz | C ≥ I / (2f·ΔVpp) = 1 / (2×50×0.5) = 1 / 50 = 0.02 F = 20,000 µF | C ≈ 20,000 µF |
