FAQ
TL;DR: For 240 VAC LED strings, a mains-rated series capacitor can replace a hot dropper, but design for 2 kV spikes and add a small series resistor; it "reduces the power dissipated" yet remains an engineering tradeoff. [Elektroda, DAVID CUTHBERT, post #21662693]
Why it matters: This helps DIYers calculate cooler, safer LED mains drivers and avoid inrush and surge failures.
Quick Facts
- 240 VAC mains can present ≈338 V peak after rectification (240 × 1.41). [Elektroda, Steve Lawson, post #21662695]
- Place the dropper capacitor on the AC side of the bridge, with a bleeder resistor. [Elektroda, DAVID CUTHBERT, post #21662693]
- Typical 1 µF mains capacitor yields about 60 mA LED current. [Elektroda, Aniruddh Kumar Sharma, post #21662700]
- Use a surge design target around 2 kV for capacitor and series parts. [Elektroda, DAVID CUTHBERT, post #21662693]
- CL2 constant-current driver regulates at 20 mA with 5–90 V compliance. [Elektroda, Steve Lawson, post #21662705]
How do I size a capacitor instead of a hot series resistor?
Match the capacitor’s reactance (Xc) to the resistor value you would have used. Put the capacitor on the AC side of the bridge. Add a series resistor around one-quarter of Xc to limit inrush. Include a bleeder across the capacitor for safety. This approach cuts heat but introduces surge and switch-on considerations. As one expert noted, "Using a capacitor... reduces the power dissipated" but it’s a tradeoff. [Elektroda, DAVID CUTHBERT, post #21662693]
Where exactly should the capacitor go in a bridge-rectified LED string?
Install the capacitor ahead of the bridge rectifier, in series with the AC line. This placement provides reactance-based current limiting before rectification. Add a bleeder resistor across the capacitor to discharge it when power is removed. Keep wiring distances short and insulated, and rate all parts for mains use. [Elektroda, Aniruddh Kumar Sharma, post #21662700]
How much current does a 1 µF dropper capacitor provide?
A practical rule from bench tests: about 60 mA with a 1 µF capacitor on mains. Real current depends on line frequency, LED voltage, and tolerances. Always verify with measurements and allow margin for component variation. Increase or decrease capacitance to set your target LED current. [Elektroda, Aniruddh Kumar Sharma, post #21662700]
What voltage ratings do I need for the capacitor and series resistor?
Design for line spikes. For 240 VAC applications, allow for surge levels around 2 kV. Use a capacitor specifically rated for mains and the surge level, and ensure the series resistor can withstand the same peak. Add a bleeder across the capacitor to discharge stored energy after switch-off. [Elektroda, DAVID CUTHBERT, post #21662693]
Why does a resistor-only dropper run so hot with 30 white LEDs on 240 VAC?
After rectification, 240 VAC is about 338 V peak. Thirty white LEDs may drop only about 96–105 V at 20 mA. The remaining voltage must be burned in the resistor, which creates substantial heat. That’s why a capacitive dropper is attractive for thermal reasons. [Elektroda, Steve Lawson, post #21662695]
Do I still need a resistor with a capacitive dropper?
Yes. A series resistor limits inrush if you switch on at the AC peak, protecting LEDs and the bridge. A value near one-quarter of the capacitor’s reactance offers a practical compromise. Include a bleeder across the capacitor to discharge it on power-down. [Elektroda, DAVID CUTHBERT, post #21662693]
What is the CL2 LED driver, and can it help here?
CL2 is a two-pin constant-current regulator that sources or sinks about 20 mA. It regulates with 5–90 V across it, so keep its voltage under 90 V. Use enough LEDs in series or add zeners to absorb excess. The expert warns: ensure insulation to avoid shock hazards. [Elektroda, Steve Lawson, post #21662701]
How do I drop the extra voltage when using a CL2 on mains?
Use a series resistor sized by R = (V − 5) / 20 mA, where V is the excess voltage. The CL2 needs at least 5 V to regulate. To reduce resistor heat, split the drop between that resistor and a series AC-side capacitor. This keeps current regulation while cutting dissipation. [Elektroda, Steve Lawson, post #21662705]
Is there a quick tool to calculate capacitive reactance?
Yes. Use a reactance calculator that implements Xc = 1 / (2πfC). Enter mains frequency and your target capacitance to get Xc. Then choose the nearest standard capacitor and add the protective series resistor and bleeder. Verify with measurements under load. [Elektroda, Steve Lawson, post #21662694]
Can I run more than one 30‑LED string from 240 VAC?
There is enough voltage headroom to place two 30‑LED strings in series after rectification. Three is tempting, but keep headroom for current regulation and variations in LED forward voltage. Ensure your driver method maintains consistent current across both strings. [Elektroda, Steve Lawson, post #21662701]
What are the key safety practices for mains‑powered LED strings?
Insulate and isolate all live parts to prevent shock. Maintain creepage/clearance, add a bleeder across the dropper capacitor, and observe device voltage limits. As one expert stressed, “MAKE SURE THESE LEDS ARE ISOLATED/INSULATED SO THERE IS NO SHOCK HAZARD.” [Elektroda, Steve Lawson, post #21662701]
3‑step: How do I build a reliable capacitive‑dropper LED string?
- Compute target Xc and choose a mains‑rated capacitor; add a bleeder across it.
- Add a series resistor near one‑quarter of Xc to tame inrush and spikes.
- Place the network before the bridge; confirm surge ratings near 2 kV and test. [Elektroda, DAVID CUTHBERT, post #21662693]
How do I size components to limit inrush and spike damage?
Use a small series resistor with the dropper capacitor to limit turn‑on current at AC peaks. Choose surge‑rated parts, and design for line spikes near 2 kV. This edge case often kills LEDs when only a capacitor is used without proper damping. [Elektroda, DAVID CUTHBERT, post #21662693]
What is “capacitive reactance” in this mains LED driver context?
Capacitive reactance (Xc) is the AC “resistance” of a capacitor, set by frequency and capacitance. It follows Xc = 1/(2πfC). You pick C so Xc limits LED current without wasting heat like a resistor. A calculator helps translate current targets into capacitance. [Elektroda, Steve Lawson, post #21662694]
