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I have been working with LED lately and I encountered the following circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

  • The bridge is DB107S 1A 1000V.
  • The LED work at 60mA, 3-3,4 Vf.
  • The board is connected to mains power with 230rms @ 50Hz.

The problem is that sometimes the LED's are getting cooked.
I believe that this is caused by voltage spikes that pass in the system since it does not have a low pass filter (RC) in the input.

Any thoughts or suggestions on how to identify the problem?

EDIT: The LEDs are 54 in total and not 27 as my first schematic shown.

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  • \$\begingroup\$ Try moving R1 to the branch of the LEDs. Now you have a circuit with potentially unlimited current. \$\endgroup\$ – Gregory Kornblum Mar 15 '16 at 18:40
  • \$\begingroup\$ 60mA is pretty high- are the LEDs rated for that? Do they have sufficient cooling or all they all crammed together? (27 * 3.2 * 0.06 = 5.2W- that's a lot of heat- and a bit of light). \$\endgroup\$ – Spehro Pefhany Mar 15 '16 at 18:45
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    \$\begingroup\$ There is a low pass filter at the input. The resistor R1 is in series (through the diode bridge) with C4 and the combined capacitance of C2 and C3, forming a series RC filter of 39 ohms and 1.24 uF. \$\endgroup\$ – jms Mar 15 '16 at 18:47
  • \$\begingroup\$ I think you're applying way too much voltage to this circuit. Therefore, that is the first step--lower the voltage. Before you catch something on fire. \$\endgroup\$ – Tim Spriggs Mar 15 '16 at 19:28
  • \$\begingroup\$ If the LEDs are rated for 60mA, remove the 680nF capacitor. Approximate current will then be (cutting some corners for ease of calculation) \$(230-27×3.2)(2\pi×50×1\cdot10^{-6})\approx 45\text{mA}\$. With both capacitors the current will be about 76mA, way too much for a 60mA LED. \$\endgroup\$ – jippie Mar 15 '16 at 20:13
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1) LED lifetime is related to thermal, the higher the temperature the shorter the life. Make your LED's happy by giving them less current and/or cool them off.

2) If your worried about voltage spikes, then find a zener diode and clamp the voltage across C4. Make sure the zener is going to be able to handle the spike and not blow up.

3) If your LED's have significant voltage drop differences (some have a 3.3V drop, and some have a 3.6V drop) some of them could be dissipating much more power than others. If this is happening, you will have to match them.

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  • \$\begingroup\$ The big fat zener gets my vote for simplicity. \$\endgroup\$ – brhans Mar 15 '16 at 19:21
  • \$\begingroup\$ Thank you for the suggestions! The LEDs are spread out evenly and in open air, so overheating is not a problem. As for the Zener, that is an awesome idea, if i am not mistaken i should find a Zener with the same voltage as the voltage across C4 and with enough watt to withstand V*I, right? \$\endgroup\$ – Lamaseed Mar 15 '16 at 21:04
  • \$\begingroup\$ @Lamaseed - No, for this circuit you want two zeners with voltage slightly (5-10 volts) higher than the LED voltage. Connect them in series, but back to back, between the inputs to the bridge. \$\endgroup\$ – WhatRoughBeast Mar 15 '16 at 21:33
  • \$\begingroup\$ @WhatRoughBeast I know it will require a series resistor to limit the current and protect the zener. Should i place it between the two zener like Z - R - Z and put R1 inside the bridge so that the LED still get the necessary current? \$\endgroup\$ – Lamaseed Mar 16 '16 at 7:06
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    \$\begingroup\$ @Lamaseed - For transient suppression you don't need a resistor. Spikes can have high amplitude but, by definition, very short duration, so their total energy is small. So a modest zener should be able to soak up the spike energy. As long as they don't appear too often, of course. The zeners should be sized so that they are not turned on at all during normal operation. \$\endgroup\$ – WhatRoughBeast Mar 16 '16 at 13:28
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I've run a simulation of your circuit, and I get current peaks near 180 mA, so it's no wonder your LEDs are frying.

Put a 2k, 10W resistor on either end of your LED string.

Alternatively, replace your input caps with 0.47 uF total.

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  • \$\begingroup\$ Thank you for the simulation, the LEDs are actually 54 thought :S \$\endgroup\$ – Lamaseed Mar 15 '16 at 20:57
  • \$\begingroup\$ If you use a smaller cap, make sure it can dissipate the power it will drop. \$\endgroup\$ – MadHatter Mar 15 '16 at 21:54
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R1 is limiting the surge current .R1 sees high peak power at switch on and must be rated for this .Wire wound resistors are better than metal film when it comes to surge rating .Surge currents will blow this thing up .Your caps are not bled .This means that a rapid turn off and switch on will give you up to twice the surge current that you think .Bleed the caps with a resistor rated for mains voltage of say 330K ohm.If you have any doubts about the voltage rating of your resistors then go for a series connection .R1 is way too small when you consider the surge rating of the LEDs .R1 only stops the diodes going bang .When you increase R1 to keep surge currents below the LED rating you will waste more power .If you dont want to waste too much more power then increase C4 .If you increase C4 you will want hundreds of microfarads for any significant benefit to happen .Also check the charging current rating of C4 .I have seen a few people blow up C4 .If you dont have the specs then bigger is better .The cap supply is simple and cheap but I have seen lots of people get it wrong and end up with something unreliable.

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