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I bought a current source to try to power some high brightness LEDs. The LEDs take 350mA so I found a current source on digikey that matches that and ordered it:

http://www.phihong.com/assets/pdf/PDA012A-S.pdf

I just hooked it up to LED, and it is pulsing - flashing the LED about 3 times a second. I put a 10 ohm resistor in series and put a scope on that. I can see the spikes/pulses. Why is it doing this? I thought this source would deliver a constant current with nothing else in the circuit. Do I need more parts?

Thank you.

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  • \$\begingroup\$ What's the forward voltage of the LEDs? Can they be needing a lot more than the voltage the current source provide? \$\endgroup\$ – Gustavo Litovsky Jan 5 '13 at 4:04
  • \$\begingroup\$ @GustavoLitovsky If the LEDs needed a lot more Vf, they would not light up, I suspect. \$\endgroup\$ – Anindo Ghosh Jan 5 '13 at 4:21
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Assumption: LED Driver used is PDA012A-350S-R, from link provided in question. Assumption: A single LED is being driven, based on "flashing the LED about 3 times ..."

The constant current driver is going into overload, due to being forced to dissipate power far over its design limits...

From the datasheet of the LED, it has a forward voltage of 3.2 Volts. The LED driver has a minimum DC output voltage of 17 Volts. When it starts powering the LED, the 350 mA current limit is reached with the LED dissipating just about 3.2 Volts. Thus, the LED driver needs to dissipate (17 - 3.2) x 0.35 = 4.83 Watts at a minimum, to supply the 350 mA design current.

The LED driver is likely to enter self-protection reset loops, either due to this dissipation causing an overheat, or due to short-circuit protection kicking in - since the output will not see the designed voltage across it when nominal current is supplied.

Solution: Use the constant current driver with a string of between 6 and 10 of the selected LEDs, in series. This will provide the required forward voltage range, between 17 and 34 Volts.

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  • \$\begingroup\$ I wasn't thinking about a minimum voltage in the driver. Now that I see it makes sense. Thanks! \$\endgroup\$ – Rob N Jan 5 '13 at 5:55

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