2015-06-30 RF Interference and LED Lighting

I talked to a club member about LED lights and how much money they saved in electricity.  But he said he stocked up on incandescent lights before they were banned so he could have them for his radio room.  He has had problems with LED lights causing RF interference with his radios, so he went back to using old incandescent lights.

I think it’s a waste of electricity to use incandescent lights. But regular LED lights have this interference problem. There is an alternative: use a power supply that is linear, not a switching power supply. It’s easy to do.

All that is needed is a transformer with a 12 VAC, 1/2 amp or more secondary, a 1 amp bridge rectifier and a 1000 uF or more, 25 VDC capacitor. An easier way to do this is to use an old 12 VDC AC adapter or wall wart, rated at a half amp or more. This has to be the old unregulated kind with a heavy transformer in it. The advantage of the separate parts is you can put .01 uF capacitors across the bridge rectifier if there is still a small amount of RF interference.

The output of a 12 VAC transformer/rectifier/filter capacitor combination is around 15 to 18 volts DC, depending on load. Say for instance you want a 3 watt light. You fasten four 1Watt warm white LEDs to a strip of aluminum which acts as a heat sink. Some LEDs use a ‘star’ base with screw holes, but I bought LEDs without a star, so they have to be fastened to the aluminum with thermal glue. I blogged this recently (Electronic Goldmine, SKU G18580A, a package of five 1 watt warm white LEDs which periodically goes on sale). I bent the two contacts up so they wouldn’t get stuck in the glue. The next step is to wire all four of them in series with some wire. The negative terminal will have a very small “-” sign next to the LED body.

Each 1 watt warm white LED will drop about 3.2 volts at the current they are rated at, which is about 330 mA. For 4 LEDs that’s about 12.8 volts DC. If the voltage across the output of the wall wart or capacitor is about 16 VDC at 250 mA, that would be about 13 ohms for the resistor. This can be adjusted to get the right current. For 13 ohms at .25 amp, that’s 3.25 volts DC. The power dissipated by the resistor is .25 multiplied by 3.25 or .8125 watt, so a 1 watt or more resistor is needed.

Or two 27 ohm, 1/2 watt resistors in parallel would also work. You could put a switch so that 1 or 2 resistors would be connected, to allow for dimming to half brightness. I think this would be a good 1 evening project. The aluminum heat sink might be a piece of 1 inch or 25 mm angle stock, which can be screwed to a vertical surface.

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