View Full Version : Correct drop resistor for LED

18-08-2003, 11:55 AM
I managed to assemble a microprocessor controller simulator on a piece of board. The simulator contains the mother board, EEV board, potentiometers for pressure and temperature analog inputs, switches for digital inputs and LED's for the digital outputs.

This simulator will be used by our tech's to simulate different conditions and observe the responces. Familiarizing the controller can enhance their operation, service and trouble shooting skills on this particular controller which is currently used for rooftop package and aircooled-screw/recip chillers.

Unfortunately, the resistors that drops the 220vac to the LED's are overheating.

The resistors are 22k ohms and 1 watt.

Any suggestion for a correction helps a lot. I ain't electronics, just trying hard and no buddy nearby. No reference to browse too.:confused:

18-08-2003, 02:44 PM
here is some info on AC line powered led's, including a wiring scheme for an A/C line voltage LED circuit. use of a (.47uf) capacitor is indicated for the function of lowering the voltage. The resistor merely acts to control inrush.

It appears that you are applying your LED in the style of the DC circuit application. However in the AC application, you will rely on the capacitive reactance of the capacitor in the AC circuit. the smaller the value (capacitance) of the capacitor, the greater the voltage drop.



18-08-2003, 03:30 PM
Hi herefishy,

Thanks a lot.
I never think that it's quite completed. But it helps a lot to understand.

My predecessors here used to supply with 24 vac using my style of connection without any problem. But this 220 vac of mine is quite problematic.

I'll make corrections with your diag.

18-08-2003, 04:28 PM
Be sure to test , first (on a test board). The drawings are for 110V circuits, you may need to play with it (perhaps 20uf capacitor at 300V minimum voltage rating?), I'm not fimiliar enough with the application to make any specific recommendations.


29-11-2003, 06:17 PM
Is problem already solved?

24-12-2004, 10:03 AM
22kohm on 220 volt gives 220/22000 gives 10 mAmp
220*10 mamp gives 2.200 Watt so your resistors get hot.
better to use four resistors in bridge. resistors are also having problems with this high voltage as the voltage can create sparks over them.

24-12-2004, 11:09 AM
Try this link for an LED power calculator


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