Calculate AC to DC output voltage for full-wave, half-wave, and capacitor-filtered rectifiers based on input, diode drop, load, and capacitance.

Estimate rectifier output from an RMS AC input.

Sine-wave estimate. For center-tap, enter the RMS voltage of one half-secondary. This does not model a regulated supply or inverter.

Optional ripple estimate

Leave both load and capacitance blank to omit ripple.

DC Voltage from AC Voltage Formula

For an ideal full-wave rectified sine wave without a filter, average DC voltage is about 0.900316 times RMS AC voltage. A capacitor-filtered output approaches the peak, with losses and ripple depending on the circuit.

Vdcₐvg = (2 × sqrt(2) / pi) × Vacᵣms

Variables:

  • Where Vdc is the DC Voltage from AC Voltage (volts)
  • Vac_rms is the RMS alternating voltage (volts); for center-tap rectification use one half-secondary.

Choose the rectifier and output model in the calculator. Ideal unfiltered full-wave DC is 2Vpeak/π, about 0.636620 times peak voltage or 0.900316 times RMS voltage; half-wave is half that average. With constant diode drops, the calculator averages only the conducting portion of the waveform.

AC to DC Voltage Conversion Table (ideal average DC = (2√2/π) × RMS AC)
RMS AC Voltage (V)DC Voltage (V)
10.900
21.801
3.32.971
54.502
65.402
98.103
109.003
1210.804
1513.505
1816.206
2018.006
2421.608
3027.009
3632.411
4843.215
6054.019
9081.028
120108.038
170153.054
325292.603
Ideal full-wave average DC = (2√2/π) × RMS AC; no diode loss or filter.

How Peak and RMS Factors Differ

The peak-voltage factor is 2/π ≈ 0.636620, which gives the mathematical average of an ideal full-wave rectified sine wave over one complete cycle. Multiplying by √2 converts this to the RMS-input factor 2√2/π ≈ 0.900316. When an AC sine wave passes through a full-wave rectifier, the negative half-cycles are flipped to positive. Integrating the absolute value of sin(t) from 0 to pi and dividing by the period yields exactly 2/pi. This is the average (DC) component of the rectified waveform before any filtering or smoothing is applied.

For a half-wave rectifier, which only passes one half of the AC cycle, the average DC output drops to 1/pi (0.318) times the peak voltage, exactly half of the full-wave value. This distinction matters in circuit design: choosing half-wave vs. full-wave rectification directly determines the baseline DC voltage available to the load.

Rectifier Topologies Compared

The table compares rectifier topologies. Average-voltage factors are ideal before diode loss; the loss column assumes 0.7 V per conducting diode. Vp is the peak feeding the conducting path (one half-secondary for center-tap, line-to-line for the three-phase row). The calculator supports the single-phase topologies; three-phase is a reference only.

Rectifier Type Comparison (Vd = 0.7 V)
TopologyDiodesVdc (avg)LossRipplePIV
Half-wave10.318xVp0.7V1xVp
FW center-tap20.636xVp0.7V2x2Vp
FW bridge40.636xVp1.4V2xVp
3-ph bridge60.955xVp1.4V6xVp
Vp=Vrms x 1.414. Higher ripple freq = smaller cap needed.

The bridge rectifier is the most common topology in consumer power supplies because it achieves full-wave rectification without requiring a center-tapped transformer. The trade-off is two diode drops (1.4 V total) instead of one, which becomes significant at low output voltages.

Real-World DC Output After Filtering

The 0.636 factor gives the average DC voltage of an unfiltered rectified waveform. In practice, nearly all power supplies add a smoothing capacitor, which charges to the peak voltage minus diode drops rather than the average. The actual DC output of a filtered rectifier is substantially higher than 0.636 x Vpeak.

Unfiltered vs Filtered DC (Bridge Rectifier)
RMS ACVpeakAvg DCNo-load peak DCDiff
9V12.73V6.75V11.33V+4.58V
12V16.97V9.44V15.57V+6.13V
24V33.94V20.23V32.54V+12.32V
120V169.71V106.64V168.31V+61.66V
230V325.27V205.67V323.87V+118.19V
No-load peak DC = Vpeak − 1.4 V. Avg DC integrates a clipped sine with constant 1.4 V total drop. Small-ripple estimate: Vripple ≈ I/(2fC); loaded mean ≈ peak DC − Vripple/2.

Ripple voltage depends on load, capacitance and recharge frequency. Full-wave rectification on 60 Hz has 120 Hz ripple. At 1 A and 1000 µF, the small-ripple estimate is 1/(120 × 0.001) ≈ 8.33 V peak-to-peak. Doubling capacitance halves this estimate. Large ripple reduces the accuracy of the approximation.

Common AC-to-DC Conversions

Real-World AC-to-DC Examples
UseACDCMethodEff
USB charger120/230V5V/2-3ASMPS85-92%
USB-C PD120/230V20V/3-5AGaN SMPS90-95%
LED driver120/230V30-50VCC SMPS85-92%
PC PSU120/230V12/5/3.3VLLC SMPS80-94%
Charger120/230V13.8-14.4VLin/SMPS60-88%
Arduino9-12V5V/0.5A780533-55%
VFD480V 3ph650-680V3ph bridge96-98%
Telecom120/230V-48VSMPS92-96%

Switched-Mode vs. Linear Conversion

The 0.636 formula applies to basic rectification of a sine wave, the first stage in both linear and switched-mode power supplies. In a linear supply, the rectified DC passes through a regulator (78xx series or LM317) that dissipates excess voltage as heat. A 7805 producing 5 V from 12 V wastes 58% as heat.

Switched-mode supplies (SMPS) use high-frequency switching (50 kHz to 2 MHz) to convert voltage with far less waste. Modern GaN USB-C chargers switch above 500 kHz, enabling tiny transformers while achieving 93%+ efficiency. This is why modern chargers are far smaller than the linear wall-wart adapters of the 1990s and 2000s.

The 0.636 conversion factor represents the theoretical average DC from a full-wave rectified sine with no filtering. In a filtered supply, DC output approaches Vpeak minus diode drops, as the capacitor holds voltage near peak between cycles. The unfiltered average remains relevant for resistive loads driven directly by rectified AC, such as heater circuits or basic motor drives without capacitive filtering.