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Circuits, electronics and noise
Average output of a half-wave rectifier
Average (DC) value of an unfiltered half-wave rectified sine with an ideal diode and a resistive load, averaged over a whole input period.
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- Circuits, electronics and noise
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- Vdc [V]
For a half-wave rectified sine of peak Vp with an ideal diode and a resistive load, the average over the whole period is Vp/π ≈ 0.318 Vp. The positive half-cycle has an average of 2Vp/π, and the zero half-cycle halves it. A real diode drop of about 0.7 V reduces the result for low peaks.
Average (DC) value of an unfiltered half-wave rectified sine with an ideal diode and a resistive load, averaged over a whole input period. A real diode also drops about 0.7 V.
Calculator
- Average of a full-wave rectified sine, 2Vp/π7.639 V
- Peak reverse voltage across the diode12 V
- Ratio of the average to the peak, 1/π0.3183
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 sine gives an average of 3.82 V after half-wave rectification, about 31.8 % of the peak.
- A DC meter on the rectifier reads this average, while its heating effect corresponds to the larger RMS value of 6 V.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Vp Sine peak voltage | 0.5 to 500 V | 0.1592 | 159.2 | 159 | +10% | +10 % |
Sweep of Sine peak voltage
| Sine peak voltage (V) | Average output voltage (V) |
|---|---|
| 0.5 | 0.1592 |
| 1.186 | 0.3774 |
| 2.812 | 0.895 |
| 6.668 | 2.122 |
| 15.81 | 5.033 |
| 37.49 | 11.93 |
| 88.91 | 28.3 |
| 210.8 | 67.12 |
| 500 | 159.2 |
Sweep one variable
Variables and how each one contributes
Average output voltage
What the equation gives you. Everything else feeds it.
Sine peak voltage
The peak voltage of the sinusoidal source, in V. Scales the average in direct proportion. Raising it by 10 % (to 13.2) moves the result by +10 %, from 3.82 to 4.202 V.
Worked example with the values above
- The peak is Vp = 12 V.
- Divide by π = 3.1416.
- Vdc = 12 / 3.1416 = 3.8197 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 average is needed.
- Averaging over the conducting half-cycle only, which gives 2Vp/π.
- Forgetting the diode drop for small signals.
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