# Voltage Divider DC 1000W for LED

I´ve done some research but not clear solutions are given to this problem, so I will post my case here in case somebody can help. Thanks in advance.

I want to use a DC power supply to power up 121 LEDs. Each of these LEDs is taking 460 mA, so, mounted in parallel, I need an outcome of 55A for the complete LED series.

The problem comes now, the DC power supply that I have found most suitable (RSP 2000-24) can be configured at 21V minimum, and these LEDS need something in between 19V and 20V, so I need to drop the voltage from 21V to 19.5V with the 55A current flow. (They could probable work with the 21V but these LEDs are expensive and I dont want to screw it up)

I´ve found some resistors which can support up to 250W, so I could use several of them I guess... I suspect is not a good solution. I have also found some voltage regulators but none of them seem to stand the 1000W that I need..

Any ideas would be really appreciated!

EDIT: Thanks for your replies first of all. I will answer here the questions mentioned by Russell McMahon so we can tackle the problem better.

Why are you using those particular LEDs?

The set of LEDs working togheter will be used as a flash light for camera shooting. There will be 6 camera shots delayed 500ms among them. So the LEDs will be turned on 6 times each 500ms for a duration of just 1ms when the flash signal closes the circuit formed by the PSU and the LEDs. That is why I´m not regarding heat sinking, because they will only be shining for 1 ms.

Since the duration of the pulse of just 1ms, I think we may have problems with the Rise up Time & Setup time of the PSU, I´ve thought about using a capacitor so the current can flow immediately after closing the circuit´s switch and the LED´s can light up sincronized with the camera shot. Any suggestions on this issue are appreciated as well.

Do you need a high CRI?

Yes. Otherwise object colours will appear distorted in relation to reality.

How long do you want these LEDs to last?

As much as possible. They will only be flashing 6 times for 1ms each time the camera shoots.

For the specifications I need, this manufacturer resulted to be the cheapest and fastest in delivery. Cree was more expensive and takes longer time.

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Can't you choose a PSU that supplies the correct voltage, and use as many of those as needed? If you really need to use resistors to drop excess voltage, I would use one per LED. – Wouter van Ooijen May 9 '14 at 14:29
Parallel LEDs can be a bad idea - the voltage needed for one can be just right but for another can be too much and it fries - these are highly non-linear devices and you can't use simplistic ohm's law predictions - are all 6 inch nails the same length - ask the guy who sleeps on them - he makes sure they are all the same length or he'll be stabbed by the tallest. LEDs are the same IMHO – Andy aka May 9 '14 at 15:01
As much as I appreciate your apparent temerity, I would recommend some timidity when you're talking about current in the tens of amps and powers in the hundreds of watts. At the very least, start with maybe just 10-12 LEDs in a string and see if you can get that working. – Nick T May 9 '14 at 22:31
this is Daniel, the manager of the webstore where you purchased the LEDs, I am very glad that you discussed the questions about our products. Meanwhile I apologize for the problem caused by us, since we are not very professional on power source, but we can provide any details about the COBs, maybe there will be some info can help you. So please don't hesitate to contact me. FYI, we suggest the voltage not over 20.5V, otherwise the voltage curve will be extremely changed and bring a high risk of burning. – user42925 May 16 '14 at 13:17

Edits to this answer have subtly but significantly changed things I was trying to say. I've changed some back, but it gets hard and time consuming and it's easy to make a mess unawares. I could have just done a "revert" but that would be a sad thing to do as there were numerous worthwhile changes.
If editing please try to retain the sense of each point being edited. Fixing typos and spill chucking is fine. Adding poncy formula renderings is fine. If you feel you wish to change phraseology or grammar rather than technical content I'll hazard that you'll find that there are many more worthy candidates which warrant prior attention.

A major problem is that you have not told us what you are trying to achieve but rather how you are hoping to achieve whatever it is that you are doing. A good overall description is liable to result in a superior solution.

Things to consider:

• Why are you using those particular LEDs?

• Do you need a high CRI?

• How long do you want these LEDs to last?

• Do you care or know anything about this manufacturer?

These LEDs MAY be extremely good and have a long lifetime.
The webpage and site look good and the published specifications are good...
However, experience has shown that for other than well known manufacturers, it is highly likely that the LEDs will be be sub-standard, having a short lifetime and low efficiency.

LEDs of this power level and this cost essentially must be driven with constant current(CC). While it is possible to get CC power supplies which control a resistive pass element these dissipate excessive energy in high power applications. What you want is a CC power supply that adjusts adjusts its voltage as required to control current so that energy dissipation in the driver is minimised. What the LED manufacturer terms the "largest value" in his spec sheet (which is why I used the term here but it was edited out) = 20 V is the maximum value = Vf, max that may appear across the LED at rated current, but usually actual values will usually be lower and will vary between devices.
If you parallel LEDs the differences in Vf will at best result in different light output from each LED, and at worst may cause a cascade of failures over time. (Some LEDs having low Vf hog the current and starve high Vf LEDs. The low Vf LEDs dissipate excessive power and die early leaving fewer LEDs to carry the same current and the process accelerates. )

It is not obvious that you have considered LED heat sinking or other thermal management. [My reference to LED heatsinking was edited out but I was specifically meaning that as other heat dissipatioon had been mentioned, making the LED aspect more noticeable by its ommission. /. Most of the almost 2000 W input will be dissipated in the LEDs as heat and will require substantial heat sinks.

The LEDs are specified as having 100-120 lumens per watt (l/W) or thereabouts. If they achieve this then about 25% of the input power is converted to light and 75% as heat you must deal with. Cree is selling 200 l/W LEDs and they have achieved 350 l/W the lab. However, most LEDs are far lower. You can buy 120-150 l/W LEDs from top manufacturers, but many lesser-known manufacturers claim much but deliver less. High CRI will likely reduce l/W. Looking at their graphs the methods they use to (claim to) achieve include bandwidth notching (where part of the spectrum is suppressed) which reduces l/W if it is done by adsorbing emitted light. Even Cree appear to do this rather than solely using phosphor spectrum tailoring.

Anyways, if you budget for about 1500 W of heat dissipation and say 80 °C their Topr,max) LED at the manufacturer's external measurement point, and keep an ambient temperature of 30 °C (optimistic) you need a 0.03 degree C per Watt heatsink [just try and buy one!]. This requires an extremely good (or large) heat sink.

Liquid cooling or well-designed fan cooling in conjunction with large heat sinks is required for such power levels. If you were to pack the 13.5 mm square LEDs at, say, 20 mm centres (6.5 mm = 1/4" between each LED) they would occupy about 220 mm square (~9" square). Evacuating 1.5 kW of heat with a lower than 50 °C rise (given no hot spots) will be "interesting".

Simple test:

• Take a 1 kW heater element.
• Enclose in a say 9"x 9" metal box - say filled with oil. (Do not short electrically).
• Attach cooling methods of your choice.
• Operate element at 1 kW.
• Monitor oil temperature. MUST stay <= 80C.

Success?
OK. Try 1.5 kW.

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I can't overemphasize the heat sinking aspect. Even using a wildly optimistic estimate of 0.5 W/W efficiency, you're looking at about 1 kW of heat dissipation. That's comparable to a medium-sized space heater, and will most likely require active cooling. – duskwuff May 9 '14 at 18:45
I think the average LED dissipates about 70% of the electrical power you put in. Unfortunately, unlike every other lighting technology, none of that power is rejected for "free" (radiatively) and instead stays right there unless you conduct it away in a very deliberate manner. – Nick T May 9 '14 at 22:38
@NickT: Right, I'm being very, very, very optimistic (and making my numbers simpler) by assuming that 50% of the energy will be emitted as light. (20% to 30% is more realistic.) Either way, you're looking at an incredible amount of heat — you should keep a fire extinguisher handy. – duskwuff May 9 '14 at 23:58
The LEDs are rated at a CLAIMED 100-120 l/W or thereabouts. If they achieve something like this then about 25% goes out the front as light and ~ 75% as heat. Overall if you budget ABOUT 1500W dissipation and say 80 C LED outer (their Tmax -) and say 30C ambient )(dangerously optimistic) you have 80-30 = 50C so 50 C/1500 W = 1/30th Watt/degee_C. Liquid cooling or competently designed fan cooling of large heatsinks is required. Pack these as say 20mm centres and they occupy about 220mm square (~~ 9" square). – Russell McMahon May 10 '14 at 3:09

Driving LEDs is best achieved by a Constant Current power supply. A Constant Current power supply adjusts the output voltage to deliver the current you have set.

You will need several power supplies, each one driving a number of LEDs in series. The number of LEDs will depend on the PSU drive capability and the LED requirements.

Each PSU will adjust its voltage to supply the chosen current. You need to make sure the expected forward Voltage of the LED is in the output Voltage range of the PSU.

Here's an example. Put the N LEDs in series and drive with a PSU at N*Vforward. You could use LXC35-0450SW to drive 3 LEDs in Series (450mA at 60V).

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Now that you're talking about pulsed operation you make the power supply problem "easy": just use a big enough capacitor bank to supply the current briefly. Say you want the power supply to sag no more than 0.1 V due to a pulsed load of 55 A for 1 ms.

$Q = I t$

So you're talking about a charge transfer of 0.055 C.

Now for your capacitor bank,

$Q = C V$, or $C = Q/V$

So you need about 0.55 Farads of capacitance. That is really a pretty large capacitor, but not unachievable. You will probably need to be pretty careful about how you connect it to the load to ensure it can be discharged as quickly as you want through the interconnecting parasitic resistances and inductances.

You do not need a high-power PSU. After each discharge, the capcitors can be recharged "slowly" to prepare for the next flash. Since you talk about 500 ms between flashes of 1 ms, you have a duty cycle of 0.2%, and your PSU will only need to provide about 110 mA at 20 V (about 220 mW).

On the other hand, pulsed operation opens up a whole new can of worms: Can these devices even be switched that quickly? Will the color rendering be the same when pulsed as under continuous operation? Do you need to operate the LEDs on lower power when setting up your shot to be able to check color balance? etc., etc. ...

Also it's highly likely the lifetime of these LEDs will be shortened by thermal cycles, possibly even worse than consistent operation at high temperatures. So you really should still pay attention to heat-sinking to maximize the life of your flash.

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If you're not concerned about the power loss, I'd say the easiest, cheapest, and safest way to go is just get 121 ~3.26ohm 1 Watt resistors and put them in series with your LEDs.

I came up with 3.26 ohms by (21-19.5)/0.46 = 3.26
1.5V * 0.46A = 0.69 Watts.

Any of these would probably work

If you want to halve the number of resistors you use, you could put them in series of two and use a resistor the same way while using the 42 V powersupply that's on the same link that you had. Just double the resistor value and wattage so that you can drop double the voltage/power over them.

A constant current supply is going to burn off the excess power as well unless you get a SMPS for each of them separately.

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