The mental picture
A reservoir bridges gaps in inflow. A capacitor can supply charge between replenishment pulses. More load current makes its voltage fall faster. A regulator is another stage with its own headroom and limits.
The principle
Here we consider the low-voltage side of a 50 Hz transformer and bridge rectifier. Both half-cycles recharge the reservoir, ideally one hundred times per second. Reservoir DC is closer to the sine peak than its RMS value, minus diode losses.
Try it yourself
Power supplies & reservoirs
Double load current and observe ripple. Then double C. This shows why the same capacitor gives less reserve under heavier load.
Model limits: Simplified 100 Hz reservoir model for 50 Hz full-wave rectification, fixed 1.4 V bridge drop and constant load current. Not a mains construction guide.
A worked example
A 100 mA load and 2 200 µF give about 0.455 V peak-to-peak ripple with 100 Hz replenishment. 12 V RMS corresponds to about 17 V peak before diode losses.
A closer look · equations and assumptions
The tool replaces real charging pulses with instantaneous recharge and constant-current discharge. Real transformer and wiring resistance limit charging current. Regulators need headroom; electrolytics face ripple-current and temperature stress. Switching regulators add switching frequency, control-loop and layout considerations.
A common pitfall. A larger capacitor does not cure every power-supply weakness. It can increase inrush and charging-pulse currents, stressing other parts.
Workshop takeaway
Measure voltage under load, not only at idle. For first experiments, use a suitable finished low-voltage supply.
Sources & transparency
Original illustrations and educational calculations. No measurements of a specific product. Edited 11 October 2026.
Work it out · answers unfold
Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.
12 · An IC needs an extra 20 mA for 10 µs. What ideal capacitance limits voltage droop to 0.1 V?
Show worked answer
- Extra charge: I · Δt = 0.2 µC.
- C = ΔQ/ΔU.
- At least 2 µF in the ideal model.
13 · A decoupling capacitor has 0.1 Ω ESR. A current step is 0.5 A. Find the immediate ESR voltage step.
Show worked answer
- Apply Ohm’s law to ESR.
- ΔU = ΔI · ESR.
- ΔU = 50 mV; further capacitive droop follows.
14 · Why place a small decoupling capacitor next to the IC rather than 30 cm away?
Show worked answer
- Wires have inductance.
- Fast current changes create voltage drops across it.
- Short supply and return paths reduce loop area.
16 · Distinguish signal coupling from supply decoupling by the current path.
Show worked answer
- Coupling: capacitor in series with signal, blocks steady DC.
- Decoupling: capacitor across supply, supports local current peaks.
- Same component type, different purpose.