At 02:37 PM 6/1/2007, peiserma@ridgid.com wrote: >piclist-bounces@mit.edu wrote: > > At the risk of being obvious, why the diode across the LED? > >probably because the LED cannot handle 660Vac reverse voltage by itself? Maybe its the summer sun getting to me. What I said was: "At the risk of being obvious, why the diode across the LED? I'd rather see a diode in SERIES with the LED. This would have the effect of reducing the power dissipated in the resistor by half. Diode has to be rated to handle the peak voltage - 1N914 doesn't cut it in this application." I understand the need for the diode in the circuit proposed by Jinx. However . . . The point I was trying to make was that power was being wasted needlessly. A simple change (diode in series with LED rather than across) means that power is being consumed on alternate half-cycles. This reduces the power dissipated in the resistor by half. Lets put some numbers on it. 660Vac average. 5 mA current. 660 * (0.005) = 3.3 Watts. Best to use a 5W resistor. Change the circuit so that the diode is now in series with the LED. Power dissipated within the resistor magically drops by half. Now you can use a 2W or 3W resistor. THAT is the point I was trying to get across. dwayne PS - memories of retrofitting LED indicators in industrial control panels come rushing back (from many, many years ago). Keeping resistor dissipation to a minimum was important, given that the LED clusters we built way back then ran at 20 mA or so. Much better LEDs available now. dwayne -- Dwayne Reid Trinity Electronics Systems Ltd Edmonton, AB, CANADA (780) 489-3199 voice (780) 487-6397 fax www.trinity-electronics.com Custom Electronics Design and Manufacturing -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist