Calculating Capacitor Value for 30 LEDs in Series on 240V AC with Full-Wave Rectifier

The discussion addresses the design challenge of powering 30 white LEDs (20 mA) in series from a 240V AC source using a full-wave rectifier. The original approach uses a series resistor to drop excess voltage, which generates excessive heat. To reduce power dissipation, it is suggested to replace or supplement the resistor with a capacitor providing capacitive reactance equivalent to the resistor's voltage drop, placed on the AC side before the bridge rectifier. However, capacitive current limiting can cause high inrush currents at AC peak voltage, potentially damaging LEDs, so a small series resistor (about one-quarter the capacitive reactance) is recommended for current limiting and surge protection. The capacitor should be rated for at least 630 VAC and designed to withstand line voltage spikes up to 2 kV, with a bleeder resistor for safe discharge. Calculations involve determining the capacitive reactance (Xc = 1/(2πfC)) to match the desired current limit. For 240V AC (338V peak), the voltage drop across LEDs (approx. 3.4V forward drop each) leaves a significant voltage to be dropped by the resistor/capacitor. Alternative solutions include using constant current regulators like the Supertex CL2 series (CL2N3, CL2K4) to maintain LED current without large power dissipation. The CL2 devices require a minimum voltage overhead (~5V) and have maximum voltage ratings (up to 90V), so multiple LED strings in series may be needed. Combining a resistor and capacitor for voltage/current regulation is also advised to optimize efficiency and reliability.
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