I need to power many 100w 32-36V LEDs... and building loads of 32V PSUs is costly and time consuming.

This is a bit of a crazy idea, and I'm aware that it's probably not the safest way to do this, but this circuit has just popped into my head:

enter image description here

The tiny detail that I'm not quite sure about is the DC voltage after the rectifier:

  • Should I calculate using 240V, so 7 LEDs connected in series across 240V would give 34.2V per LED.
  • Or, should I calculate using the peak, non-RMS voltage 340v, so 10 LEDs connected in series across 340V would give 34V per LED.

N.B. I'm aware the picture shows 8 LEDs, but it's just to visualise what I'm talking about.

If anybody has any other quick, easy, safer ways to do this then please let me know!

  • \$\begingroup\$ You need to regulate current not voltage. The circuit above will not do. \$\endgroup\$ – DamienD Sep 2 '15 at 10:05
  • \$\begingroup\$ Is there any way to do that using minimal parts? \$\endgroup\$ – BG100 Sep 2 '15 at 10:11
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    \$\begingroup\$ Sorry, this kind of power is out of my league. You could have a look here as a starter: digikey.com/en/articles/techzone/2011/nov/… \$\endgroup\$ – DamienD Sep 2 '15 at 10:33
  • \$\begingroup\$ Please give some more specificaitons about the LED. If you choose your circuit layout carefully, it may be possible to put some of them in parallel. \$\endgroup\$ – Ariser Sep 2 '15 at 13:43
  • \$\begingroup\$ LED specs: Cool white, 100w, Input Voltage: 32-36v, Forward Current: 3.5A, 6000-7000 lumen \$\endgroup\$ – BG100 Sep 4 '15 at 14:40

The circuit you have drawn will not do what you want. If I were going about things this way, this is what I would do:


simulate this circuit – Schematic created using CircuitLab

How you implement the current sink is up to you. You could have a simple current sense resistor + feedback driving a linear pass transistor (if you want to dissipate lots of heat). If this is passing 3A (which is roughly what those LEDs are pulling) and has to drop 10V of extra voltage, that's 30W of heat you need to dump off into a heatsink. Not impossible, but requires thought. You can see the basic design of a linear current sink here.

You could also implement a current-controlled buck converter for the LEDs - although rating this for the full rectified mains voltage will require consideration. The link in @Damien's comment above is a good place to start looking, specifically using something like the Fairchild FL7701. This would be much more efficient, but requires a more sophisticated design.

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    \$\begingroup\$ You would also want a rather bigger cap on the bridge. 3 amps @ 100 uF gives a discharge rate of 30,000 v/sec or 30 v/msec. Granted, you're working at 120 Hz, not 60, for a cycle time of ~8 msec. \$\endgroup\$ – WhatRoughBeast Sep 2 '15 at 14:33
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    \$\begingroup\$ Considering that each of these 100 W LED is going to produce 80 W of heat, maybe it isn't such a bad idea to go for a linear current regulator indeed. Esp. considering the OP's requirement of a low cost system... he could just use the same kind of heatsink for the MOSFET as for the LEDs. \$\endgroup\$ – DamienD Sep 2 '15 at 18:49
  • \$\begingroup\$ This is great information, thanks... I don't really understand whats going on though, what is the reason that I need to limit the current? \$\endgroup\$ – BG100 Sep 4 '15 at 8:18
  • \$\begingroup\$ Because of the way LEDs work: in the absence of a current regulator, small fluctuations in input voltage or temperature would lead to large fluctuations in current which could burn the LEDs. \$\endgroup\$ – DamienD Sep 4 '15 at 8:49
  • \$\begingroup\$ This graph shows the current-voltage relationship of a typical LED. Notice that around the forward voltage (3.8V in your case) as @Damien said, small changes in voltage = large changes in current, which affects brightness and can overdrive the LEDs. You should also take a look at this SE question which explains this more in detail. \$\endgroup\$ – stefandz Sep 4 '15 at 9:20

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