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The LED emitters in my outdoor lamps blew out and I need to replace them but I have to rebuild the emitter from scratch because replacement parts are not available. The whole lamp was made in China. They are fed by AC.

I have the following: - 1 Luxeon K High Power LED Array. Forward voltage requirement of 20V with 730 lumens at 350mA. Perfect sized emitter and I have a good heat sink to mount it to. The Part number is LUXEON-K-COBWW08, an 8 emitter array.

  • 1 AC to DC power converter. It is a MagTech AC Dimmable - 18-Watt, 700mA Constant Current LED Driver. I don't need it to be dimmable at the switch, but I don't mind adding a dimmer to the inner workings of the lamp. The Part number is M18-U24-0700

  • I can buy any resistor needed. That's not the issue.

Curiously, the DC output from the AC converters I measured to be 30V. Both of them. They are rated to be dimmable between 14-24 VDC. This is at a constant 700mA.

When I bought the drivers, I planned a 24V power to a 20V requirement and got planned for a resistor of 12 ohms. (24-20)/.35mA = 11.4 ohms. The lights are rated at 350mA. Any more current and they'll burn up.

Since the output is 30V, should I swap the 12 ohm resistor for a 30-35 ohm resistor? (30-20)/.35=28.5ohms

Is there a simple way of hardwiring a variable drop in voltage? I can set the output with my multimeter and leave it in the housing of the light.

I'm worried that, within this outdoor light, wiring a 35 ohm resistor will require me to radiate off too much heat. I'll need a 20 watt resistor.

This is getting out of my league and I think ledsupply.com sold me incorrect equipment.

Thanks.

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  • \$\begingroup\$ What load was attached when you measured 30 V output from the supply? The point of a constant current driver is that it will change output voltage so that it always (within limits) produces the same current (in this case, 700 mA). Trying to use it as a 30 V voltage supply is asking for trouble. \$\endgroup\$ – The Photon Aug 24 '15 at 2:11
  • \$\begingroup\$ I tested the voltage output with a Fluke multimeter in DC mode to see what it was outputting. I wanted to test it before hooking it up to the LED and resistor. I had called Ledsupply.com and he selected these to drive the 20V drop Luxeon LEDs. He calculated that at a max of 24V (which was what was stated on the specs), it would adequately power the LED. Now I'm scratching my head here.... \$\endgroup\$ – Daniel Woodward Aug 24 '15 at 2:39
  • \$\begingroup\$ The question was "What load was attached when you measured 30 V output from the supply?". If you would please answer the question The Photon asked, he may be able to help you. Assuming you had no load when you measured your supply, this is entirely consistent with his comment. \$\endgroup\$ – WhatRoughBeast Aug 24 '15 at 3:31
  • \$\begingroup\$ Sorry. I don't understand the question. I don't know what the load is/was. I attached the AC/DC, 700mA Magtech driver to an AC 120V outlet. I tested the + and - DC terminals of the driver with my multimeter. \$\endgroup\$ – Daniel Woodward Aug 24 '15 at 3:38
  • \$\begingroup\$ Measuring it without a load, ie the leds on, will not tell you what it really should be voltage wise. Its like trying to test a speaker with the volume on mute. \$\endgroup\$ – Passerby Aug 24 '15 at 7:44
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Let us calculate the resistance of the LED at the operating voltage:

R = V/I

V= 21 v (typical), I = 350 mA, so R = 60 Ohm.

The LED driver operates at maximum 24 volt with 700 mA, so it works with maximum 34 Ohm load. Any load with higher resistance will give an unpredictable current result that may damage the load. So you can not simply connect 35 ohm resistor with the 60 Ohm LED in series.

To divide the current, you can connect the load in parallel (not in series) with 60 Ohm resistor. If it is difficult to find exactly 60 Ohm resistor, you can use 56 Ohm/10 Watt resistor that is available in digikey. It gives 338 mA and 20.276 Volt in the LED.

enter image description here

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  • \$\begingroup\$ Whoa. Wait a minute... So the LED doesn't need a resistor in it's part of the parallel circuit? Won't it burn up? Oh wait, of course, the driver won't deliver more than 700mA. I get it. I'm gonna stop by Frys tomorrow. If I had to choose between higher or lower resistance in the resistor, which should I choose? \$\endgroup\$ – Daniel Woodward Aug 24 '15 at 3:12
  • \$\begingroup\$ I prefer the lower resistance that give the LED lower current, safer. In fact, the LED resistance will be lower as increasing temperature. \$\endgroup\$ – Oka Aug 24 '15 at 3:18
  • \$\begingroup\$ I now understand what I have and what I need to do. I have one final question. It seems like my LED driver is overkill for the diode I purchased. The circuit you've planned requires a 56 ohm resistor to split the current. Would I be better off just buying a 350mA 22V constant current driver? It wouldn't need a resistor but would be maxed out when in use. Doing this, I wouldn't have to dissipate the extra heat from the resistor. The lamp is enclosed, outside in Texas. Heat is obviously an issue. The model is: ledsupply.com/led-drivers/… \$\endgroup\$ – Daniel Woodward Aug 25 '15 at 2:28
  • \$\begingroup\$ Yes, certainly it is better if you use a 350mA constant current driver for a 350mA LED. Alternatively, you can purchase additional 350mA LED :) and use a 700mA driver to drive 2x350mA LEDs in parallel (must have same type & manufacturer) \$\endgroup\$ – Oka Aug 25 '15 at 2:58
  • \$\begingroup\$ You guys are awesome... Now my sister, who is a professional sculptor/artist, wants me to design an array of 5mm LEDs for an art display box. I may have opened a huge fan of worms now that she thinks I'm some kind of wiring wizard (I am not). Still.... Basic electrical wiring is a fun/hard/interesting departure from my usual work. \$\endgroup\$ – Daniel Woodward Aug 27 '15 at 15:23
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It would be best if you had a LED power supply with the proper output current. Given the low cost of these supplies this would be my preferred solution.

However if you have a dimmer I think you could try driving it at 700 mA with 50% dimming, since the average heat dissipation will be the same and the datasheet shows that this LED (LXK8-PW30-0008) has been tested up to 1A. But there is a risk of burning it if you accidentally change the dimmer setting.

As for the discrepancy in voltage that's quite normal for a LED power supply with no load, since they regulate the current and not the voltage. 30 V is probably the value at which the open circuit protection kicks in and caps the voltage (without it, the power supply would increase the voltage without bounds trying to get 700mA through the non-existent load).

When selecting a LED power supply you need to match the output current, and to make sure that the output voltage range includes the LED forward voltage. It's no problem if the power supply is able to go above that.

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