Rectifier Average Voltage⚠ unverified
Electrical / Power · Compute the average DC output voltage after rectification
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| V_peak | Vpeak | V | 1.0 | Peak input voltage |
| phases | phases | — | 1.0 | Number of supply phases; 1 for single-phase full-wave or 3 for three-phase. Default is 1 |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | Vdc | V | Average DC output voltage, in volts (V). For phase counts other than 1 or 3, ``V_peak`` is returned unchanged |
The science & history
Understanding the Parameters
- $V_{peak}$ — the amplitude of the AC waveform being rectified.
- $V_{dc}$ — the mean of the rectified waveform; it is $2/\pi\approx 64\%$ of the peak for full‑wave, or only $1/\pi\approx 32\%$ for half‑wave (which wastes half the cycle).
Derivation (Approaching a Proof)
Full‑wave rectification flips the negative half‑cycles up, giving $|V_{peak}\sin\theta|$ with period $\pi$. The average is the integral over one hump divided by its width:
$$V_{dc} = \frac{1}{\pi}\int_0^{\pi} V_{peak}\sin\theta\,d\theta = \frac{V_{peak}}{\pi}\big[-\cos\theta\big]_0^{\pi} = \frac{V_{peak}}{\pi}(1-(-1)) = \frac{2V_{peak}}{\pi}.$$
(A half‑wave rectifier averages over the full $2\pi$ with the negative half zeroed, giving half as much, $V_{peak}/\pi$. Three‑phase rectifiers stay much closer to the peak because their humps overlap.)
History
Rectification — converting AC to DC — evolved from mechanical commutators and mercury‑arc valves to vacuum tubes and then semiconductor diodes. The $2V_{peak}/\pi$ average is the textbook starting point for sizing power supplies, before the smoothing filter that turns this pulsating average into steady DC.
Related Concepts: Rectifier Ripple Factor, Total Harmonic Distortion, Apparent Power
Notes: Registry calculator rectifier-average-voltage (unverified). Single‑phase full‑wave value;
neglects diode forward drop; three‑phase uses a different (higher) factor.