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Circuits, electronics and noise
Reservoir capacitor ripple after full-wave rectification
Approximate peak-to-peak ripple on a reservoir capacitor that is recharged twice per supply cycle and supplies a nearly constant load current in between.
- Ref
- ELN-14
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- \Delta Vpp [V]
After a full-wave rectifier, the reservoir capacitor recharges twice per supply cycle and supplies the load current Iload for about half a cycle in between. The charge it loses is Iload/(2f), so the peak-to-peak ripple is ΔV ≈ Iload/(2fC). This assumes a short recharge time and a small ripple, so it slightly overestimates the ripple. For a half-wave rectifier use f instead of 2f.
Approximate peak-to-peak ripple on a reservoir capacitor that is recharged twice per supply cycle and supplies a nearly constant load current in between. Valid for small ripple and a short recharge time.
Calculator
- Charge lost per half-cycle, Iload/(2f)2×10−4 C
- Approximate RMS ripple, ΔV/(2√3)0.05774 V
- Ripple if the capacitance is doubled0.1 V
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 0.02 A load from 0.001 F at 50 Hz gives 0.2 V peak-to-peak ripple, about 0.05774 V RMS.
- To halve the ripple, double the capacitance or halve the current; at 60 Hz the same circuit would give 0.167 V.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Iload Load current | 0.001 to 10 A | 0.01 | 100 | 99.99 | +10% | +10 % |
| C Reservoir capacitance | 1e-5 to 0.1 F | 0.002 | 20 | 20 | +10% | −9.09 % |
| f Supply frequency | 16 to 400 Hz | 0.025 | 0.625 | 0.6 | +10% | −9.09 % |
Sweep of Reservoir capacitance
| Reservoir capacitance (F) | Peak-to-peak ripple (V) |
|---|---|
| 1e-5 | 20 |
| 3.162e-5 | 6.325 |
| 1e-4 | 2 |
| 3.162e-4 | 0.6325 |
| 0.001 | 0.2 |
| 0.003162 | 0.06325 |
| 0.01 | 0.02 |
| 0.03162 | 0.006325 |
| 0.1 | 0.002 |
Sweep one variable
Variables and how each one contributes
Peak-to-peak ripple
What the equation gives you. Everything else feeds it.
Load current
The average load current, in A. More current drains the capacitor faster, so the ripple rises in direct proportion. Raising it by 10 % (to 0.022) moves the result by +10 %, from 0.2 to 0.22 V.
Supply frequency
The supply frequency, in Hz. A higher frequency shortens the interval between recharges, so the ripple falls. Raising it by 10 % (to 55) moves the result by −9.09 %, from 0.2 to 0.1818 V.
Reservoir capacitance
The reservoir capacitance, in F. A larger capacitor drains less, so the ripple falls in inverse proportion. Raising it by 10 % (to 0.0011) moves the result by −9.09 %, from 0.2 to 0.1818 V.
Worked example with the values above
- Interval between recharges: 1/(2f) = 1/(2 × 50) = 0.01 s.
- Charge lost: Iload × that = 0.02 × 0.01 = 2×10−4 C.
- ΔV = 2×10−4 / 0.001 = 0.2 V peak-to-peak.
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