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 |
Frequently Asked Questions - AC to DC Voltage Calculator:
How do I convert AC RMS to peak voltage?
Vpeak = Vrms × √2.
What is the DC output of a full-wave rectifier (no filter)?
Vdc_avg = 2×Vpeak / π (use Vpeak = Vrms√2).
How much voltage drop do diodes cause?
Silicon diodes ≈0.6–0.8 V each. A bridge uses two drops (≈1.2–1.6 V total).
How to estimate filtered DC with a reservoir capacitor?
Approx Vdc ≈ Vpeak − Vdrops − ΔVpp/2 with ΔVpp ≈ I/(f_ripple·C) and f_ripple=2f for full-wave.
Which ripple formula is correct for full-wave?
ΔVpp ≈ I / (2·f · C) (ripple frequency is 2f for full-wave rectification).
How to choose C for allowable ripple?
C ≥ I / (f_ripple · ΔVpp_allowed). For full-wave, use f_ripple = 2f.
What is Vdc for half-wave rectifier without filter?
Vdc_avg = Vpeak / π.
Can I ignore diode drop for rough estimates?
Only for very rough estimates at high voltages; for accuracy subtract diode drops from Vpeak.
How does load current affect DC level?
Higher load current increases ripple and reduces average DC when filtering is used; it doesn’t change Vpeak but increases ΔVpp.
Does transformer secondary RMS equal DC after rectification?
No — RMS is AC measure; DC after rectification depends on peak and rectifier type and filtering.
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