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[quote]QUOTE: Originally posted by dhuff Fred, Just for reference... Not flamin' just explain'. A rectifier converts AC current to square wave DC. LEDs work on both DC and AC power so you do not need to add the rectifier. The resistor is needed to reduce the current flow through the LED and must be added in "series" with the LED not "parallel". i.e. |pwr|--|resistor|--|LED|--|gnd| and it doesn't matter which side of the LED the resistor is on, pwr or gnd it is there only to limit the current. For most LED applications a 870 - 1k (1,000) ohm is satisfactory. The higher the resistance the dimmer the light. In some of my buildings I have put 10k resistors because all I wanted is a very dim glow of light. However, I don't recommend less than 870 ohm as an LED will burn out if to much current flows through it. LEDs come in many different sizes. From X-Lrg 10mm diameter to the new Micro-Mini 1.5mm. (http://207.234.141.88/Merchant2/merchant.mvc?Screen=CTGY&Category_Code=6_1) [/quote] WELL, Just for your reference... Not flamin' just explain'. You need a full wave bridge because LEDs are polarity specific devices and when a train on DC backs up the polarity on the track reverses. If you install a full wave and hook the "ac" terminals on the full wave to the track the LED stay on all the time, not just when the polarity is correct... I also never mentioned the word "parallel" or "series" in my answer. If the poster wished to use a LED I would assume he/she would ask how. Also not all LEDs require the same voltage and current so your recomendation to resistor size is suspect. To further the point, there is a voltage drop across the full wave that also has to be factored in. I could also use a 12v LED and full wave and it would likely work without a resistor at all. You could also buy one of the new LEDs that change color on polarity reversal. You could also go to Walmart and buy a pair of shoes with the flasing LEDs in the box and lenghten the leads to one of the LEDs, they run for months on 2 AA batteries. Not flamin' just explain'. FRED
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