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

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How can I replace the hot series resistor with a capacitor to drive 30 white LEDs in series from 240 V AC with a full-wave rectifier?

Use a capacitive dropper on the AC side of the bridge rectifier, sizing the capacitor so its reactance provides the same current-limiting effect as the resistor [#21662693] Add a small series resistor to limit the high inrush current at turn-on, and fit a bleeder resistor across the capacitor so it discharges when power is removed [#21662693][#21662700] The capacitor must be a mains-rated, high-voltage part because line spikes can be severe; one reply suggested designing for about 2 kV surge withstand on 240 VAC, and another suggested at least 630 VAC for the capacitor [#21662693][#21662700] For 30 white 20 mA LEDs, the string itself only drops roughly 96–105 V, so the remaining voltage is what the dropper must handle [#21662695] The thread also recommends a constant-current LED driver such as the CL2 as the best way to drive LEDs, with a resistor and/or capacitor used only to drop the excess voltage if needed [#21662701][#21662705]

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#1 21662691 21 Jun 2012 03:44

Elba R Dalet Elba R Dalet

Anonymous

Post #1
21662691 21 Jun 2012 03:44

I am making blocks of 30 LED to be conected to 240V using a Full-wave rectification, and a resistor to drop voltage.
This resistor produce much heat.
How can calculate and use a capacitor insted of the resistor?

Regards

Elba
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#2 21662692 21 Jun 2012 03:52

Aniruddh Kumar Sharma Aniruddh Kumar Sharma

Anonymous

Post #2
21662692 21 Jun 2012 03:52

Dear Elba
Pls Share the Schematic .
1:-what is the forward curent rating of LED's.
2:-What is the emiting color of LED.
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#3 21662693 21 Jun 2012 08:37

DAVID CUTHBERT DAVID CUTHBERT

Anonymous

Post #3
21662693 21 Jun 2012 08:37

Elba,

you can use a capacitor having a capacitive reactance equal to the dropping resistor value. The capacitor is placed on the AC side of the bridge rectifier.

One issue is when the AC is switched ON at the peak of the AC waveform. With the resistor the current is limited but with the capacitor the peak current will be quite high and may damage the LEDs. To limit the current during this event a series resistor can be use with the capacitor. A resistor of perhaps 1/4 the reactance of the capacitor would work.

To make this circuit reliable with commercial power it must be designed for the expected line-to-line spike voltage. I don't remember what this is for 240 VAC but for 120 VAC it is 1 kV. So I would design the circuit to withstand 2 kV spikes. The diode voltage is limited by the LEDs but the capacitor voltage is not. It must be rated for 2 kV. And the series resistor must be rated for a peak of 2 kV. The series capacitor should have a bleeder resistor to discharge it when power is removed. And there is the question of can the LEDs survive the spike current? With the resistor-only circuit the current is limited to some value but with the capacitor and 1/4 value resistor circuit the current is four times higher during the spike.

Using a capacitor introduces several new issues but does reduce the power dissipated. It's all an engineering tradeoff.
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#4 21662694 21 Jun 2012 11:11

Steve Lawson Steve Lawson

Anonymous

Post #4
21662694 21 Jun 2012 11:11

Here's a capacitive reactance calculator (and formula):

http://www.electronics2000.co.uk/calc/reactance-calculator.php
#5 21662695 21 Jun 2012 11:22

Steve Lawson Steve Lawson

Anonymous

Post #5
21662695 21 Jun 2012 11:22

Does that mean you have 30 LEDs in series with 240V (which is more like 240 * 1.41 = 338V peak (less the forward drops of the rectifiers), If so, for LEDs with a forward drop of 3.4V (assuming standard 20ma white LEDs [you didn't say], that's 338V - 1.4V - 30 * 3.4 = 234.6V that must be dropped by that series resistor. No wonder it's getting hot!

I.e. a series string of standard white 20-30ma LEDs is going to drop, typically, only 96V to 105V, depending on the forward drop of the LEDs (which can vary a bit from part to part, and over time).
#6 21662696 21 Jun 2012 15:20

Elba R Dalet Elba R Dalet

Anonymous

Post #6
21662696 21 Jun 2012 15:20

The 30 LED are connected in serie, they are White, 20 mA.
But if you have a better idea, I will appreciate.
Thanks in advance
#7 21662697 21 Jun 2012 15:23

Elba R Dalet Elba R Dalet

Anonymous

Post #7
21662697 21 Jun 2012 15:23

Thanks so much
#8 21662698 21 Jun 2012 15:24

Elba R Dalet Elba R Dalet

Anonymous

Post #8
21662698 21 Jun 2012 15:24

Thanks Steve.
Regards
#9 21662699 21 Jun 2012 15:27

Elba R Dalet Elba R Dalet

Anonymous

Post #9
21662699 21 Jun 2012 15:27

Thanks Steve, It is very helpful
Regards

Elba
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#10 21662700 21 Jun 2012 23:56

Aniruddh Kumar Sharma Aniruddh Kumar Sharma

Anonymous

Post #10
21662700 21 Jun 2012 23:56

With my experiments i have found that a capecitor of value 1mfd can provide a forward current of 60 mA. Select the capecitor value accordingly.
The capacitor should be rated to atleast 630 V AC and it will be good to provide a bleeder resistance accross that.
this combination should be put infront i.e. before the bridge rectifier.
#11 21662701 22 Jun 2012 08:58

Steve Lawson Steve Lawson

Anonymous

Post #11
21662701 22 Jun 2012 08:58

You mentioned that there will be more than one string of 30 LEDs. There is enough [Voltage] "head room" to drive two of those strings in series (and almost three, but there needs to be enough voltage left over to provide a "current regulation" effect and you should follow the manufacturers spec which usually sets the MAX LED forward voltage at 4V (though that is usually when driven at 30ma so you may be able to get away with 3.6V max but check the spec [which doesn’t always give such details].

Another possibility is a little two pin IC called the "CL2" :
http://www.mouser.com/Search/Refine.aspx?Keyword=cl2n3 [Digi-key doesn't have it]
http://www.supertex.com/pdf/datasheets/CL2.pdf

The CL2 is a constant current source/sink [doesn't matter where it is put in the series string]. This is the BEST way to drive LEDs [with a constant current]. But you will need to make sure that the voltage across the CL2 never exceeds 90V -- so there needs to be enough LEDs in series with it to drop the voltage down to that max (i.e. not even 60 will be enough, so add one or more zeners to drop the rest (use more than one to distribute the power)).

AND MAKE SURE THESE LEDS ARE ISOLATED/INSULATED SO THERE IS NO SHOCK HAZARD!!!
#12 21662702 22 Jun 2012 09:12

Steve Lawson Steve Lawson

Anonymous

Post #12
21662702 22 Jun 2012 09:12

Above I provided a link to the Mouser catalog page for CL2N3. A better choice is the CL2K4. It has a TO-252 (D-PAK) package and can take 2W (the CL2N3 is a TO92 and can handle only 600mW -- not enough for your application, unless you take up the slack with zeners).

http://www.mouser.com/ProductDetail/Supertex/...=Wgksyu2ekN8hKHaZfvdGQw4JQgWWFy1KikRrMe7IubE=

But, they only have two left (with 3Week lead time) so:

http://www.alibaba.com/trade/search?SearchText=cl2k4-g&IndexArea=product_en&fsb=y

Happy hunting
#13 21662703 22 Jun 2012 09:17

Steve Lawson Steve Lawson

Anonymous

Post #13
21662703 22 Jun 2012 09:17

Also, try calling SuperTex (the manufacturer). See if they can locate a distributor for you. Their contact info is on this document:

http://www.supertex.com/pdf/misc/CL2PSS.pdf
#14 21662704 22 Jun 2012 13:36

Elba R Dalet Elba R Dalet

Anonymous

Post #14
21662704 22 Jun 2012 13:36

Hi Steve, first, thanks a lot.
I never heard about that CL2, and really looks great.
But I mention, I will connect them to 240V AC, Do you have any suggestion how drop the remain voltage?

And again thanks

Regards
#15 21662705 22 Jun 2012 13:49

Steve Lawson Steve Lawson

Anonymous

Post #15
21662705 22 Jun 2012 13:49

Actually, just use a resistor (I suggested one or more Zeners, but because of the constant current, a resistor will behave like a constant voltage device, once the CL2 begins to regulate).

Use the formula: R = (V-5)/20ma where V is the excess voltage you want to drop. Its V-5 because the CL2 needs a minimum of 5 volts to operate properly (i.e. it current regulates over a voltage range of 5 to 90 volts).

If that will result in a higher wattage resistor than you desire, then also include a series capacitor on the AC side (as described in other replies). Divide the resistance R between the resistor and the reactance of the capacitor.
#16 21662706 22 Jun 2012 14:54

Elba R Dalet Elba R Dalet

Anonymous

Post #16
21662706 22 Jun 2012 14:54

Thanks. It sound great
I will let you know the result
Regards

Elba
#17 21662707 22 Jun 2012 14:55

Elba R Dalet Elba R Dalet

Anonymous

Post #17
21662707 22 Jun 2012 14:55

Thanks David, it will help me
Regards
#18 21662708 22 Jun 2012 14:56

Elba R Dalet Elba R Dalet

Anonymous

Post #18
21662708 22 Jun 2012 14:56

I´m taking note

Thanls
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Topic summary

✨ 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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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]