So I had an idea for an LED lamp. I've got some old flashlights that have older model CREE LEDs.

My idea is that I'd put the LEDs in parallel, alternating polarity so that for each phase of AC, half of the LEDs will be conducting.

The array of LEDs would be in series with a resistor and a fuse to drop voltage from 110 VRMS from the outlet to 3VRMS over the LEDs and prevent the LEDs from breaking if something catastrophic happens (and to compensate for my own failings.)

The LEDs have a 5V reverse voltage and a 3.3V forward voltage at 3A at 80 deg C, so I figure if I limit to 3 VRMS I should't hit the reverse voltage and make badness happen.

With this in mind ... I am only passingly familiar with anything more advanced than kirchov's and ohm's laws. I can't apply ohm's law to the diodes, so what do I need to know to figure out the resistor value for n parallel diode pairs (Or maybe n sets of 5 in series and put those in parallel)? or is it just a bad idea?

  • \$\begingroup\$ If you are only passingly familiar, throw a transformer in the mix. Transformerless designs are inherently dangerous. \$\endgroup\$
    – jippie
    Oct 18, 2013 at 19:16
  • \$\begingroup\$ I assume you mean to use a transformer to step the voltage down to the 3.3VRMS, but I'd still need a resistor to limit current wouldn't I? Why is it dangerous? \$\endgroup\$
    – Daniel B.
    Oct 18, 2013 at 19:18
  • \$\begingroup\$ Not necessarily step the voltage down to 3.3V, yes you'd still need a resistor. But you want the transformer for galvanic separation from the mains. Mains power can potentially kill people. An added safety is that you have to deal with a lower voltage, which is safer too. You want a voltage below 40V, that is commonly regarded as a safe voltage. \$\endgroup\$
    – jippie
    Oct 18, 2013 at 19:22
  • \$\begingroup\$ Your LED's are really rated for 3A? \$\endgroup\$
    – jippie
    Oct 18, 2013 at 19:25
  • 1
    \$\begingroup\$ @jippie: The XM-L's, yes. \$\endgroup\$ Oct 18, 2013 at 19:28

1 Answer 1


First, using a resistor to drop from 120V to 3V is very, VERY inefficient. You will be burning off about 90% of your energy in heat. You need a transformer here, at the very least. Just doing the math: If you need 3A/3.3V for light, then 3A/120V is 360 Watts of power, of which you're extracting 9.9W for the actual diodes, and thus burn about 350W in heat in the resistors.

Second, LEDs in parallel is unreliable, because the actual voltage drop of the LEDs will vary a little bit because of process variance, and thus some of the LEDs will draw a lot more current than others. This will result in the lower-voltage LEDs burning out, and the higher-voltage LEDs being dim. To fix this, you need one current limiting resistor per LED, rather than a single one for the gang.

Finally, what's the point of wiring half of them each way? Why not use a bridge rectifier?

  • \$\begingroup\$ "Why not use a bridge rectifier?" - Indeed! And a switching mode current regulator and all diodes in series? Because the asker never heard of such things... \$\endgroup\$
    – johnfound
    Oct 18, 2013 at 21:03
  • \$\begingroup\$ Id' actually looked at current drivers beforehand (I'm assuming that's a switching mode current regulator?). My original plan was indeed to have the LEDs in series but then I decided I wanted to try running it from AC and that's where I went to this conclusion. I was thinking I could do this for less monies, but it's looking like I was wrong, once I buy transformers and rectifiers and things, I might as well just buy a manufactured current supply. \$\endgroup\$
    – Daniel B.
    Oct 19, 2013 at 1:29
  • \$\begingroup\$ As I said in the question, I wasn't sure it was a good idea. My fears were founded, so here's your accept and thank you for your help :) \$\endgroup\$
    – Daniel B.
    Oct 19, 2013 at 1:30

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