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Circuits, electronics and noise
Average output of a full-wave rectifier
Average (DC) value of a full-wave rectified sine with an ideal diode bridge or centre-tapped supply, and a resistive load without smoothing.
- Ref
- ELN-13
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- Vdc [V]
Full-wave rectification flips each negative half-cycle, so the average over the period is 2Vp/π ≈ 0.637 Vp, twice the half-wave value. Vp must be the peak actually seen across each conducting path: half the secondary voltage for a centre-tapped transformer, and the full secondary for a bridge, less the two diode drops.
Average (DC) value of a full-wave rectified sine with an ideal diode bridge or centre-tapped supply, and a resistive load without smoothing. It is twice the half-wave value.
Calculator
- Average of a half-wave rectified sine, Vp/π3.82 V
- RMS of the rectified wave, Vp/√28.485 V
- Ratio of the average to the peak, 2/π0.6366
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 12 V peak gives 7.639 V average after full-wave rectification, twice the 3.82 V of the half-wave case.
- The full-wave RMS equals the RMS of the source, 8.485 V, since squaring removes the sign.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Vp Sine peak voltage at each conducting path | 0.5 to 500 V | 0.3183 | 318.3 | 318 | +10% | +10 % |
Sweep of Sine peak voltage at each conducting path
| Sine peak voltage at each conducting path (V) | Average output voltage (V) |
|---|---|
| 0.5 | 0.3183 |
| 1.186 | 0.7548 |
| 2.812 | 1.79 |
| 6.668 | 4.245 |
| 15.81 | 10.07 |
| 37.49 | 23.87 |
| 88.91 | 56.6 |
| 210.8 | 134.2 |
| 500 | 318.3 |
Sweep one variable
Variables and how each one contributes
Average output voltage
What the equation gives you. Everything else feeds it.
Sine peak voltage at each conducting path
The peak voltage presented to each conducting path, in V. Scales the average in direct proportion. Raising it by 10 % (to 13.2) moves the result by +10 %, from 7.639 to 8.403 V.
Worked example with the values above
- The peak is Vp = 12 V.
- Double it and divide by π: 2 × 12 / 3.1416.
- Vdc = 7.6394 V.
Assumptions and limits
- Ideal diodes with zero forward drop for the rectifier formulas, and a resistive load.
- A sinusoidal supply and, for the ripple formula, a small ripple compared with the peak.
Common mistakes
- Using the peak value where an RMS or average value is needed.
- Forgetting that a full-wave rectifier recharges twice per supply cycle, so the ripple frequency is 2f.
- Ignoring the minimum Zener current, or its power rating.
Statistics and uncertainty
Rectifier formulas assume an ideal diode, so a real diode drop of about 0.7 V lowers the results for small peaks. Zener results depend on the supply, the Zener voltage tolerance and the load current, so use worst cases.
Where it is used
- Power supplies, battery chargers and meter circuits.
- Simple voltage references and regulators for sensors and microcontroller inputs.
Maths you need
- Average and RMS values: the mean and root-mean-square of a waveform
- Diode: conducts in one direction
- Capacitor discharge: ΔV = I Δt / C