Hı. I have 100W COB Led's.(I buy them from china) I need to give power to 16 of them. The Specs. says 36V 100W 3A for one. I know there are drivers for them but i have to light too many and i don't want to buy driver for every each one seperetaly. Is there any other ways to light them with budget? There is a picture of led.Thanks. 100W COB LED


closed as unclear what you're asking by Michael Karas, Wesley Lee, ThreePhaseEel, uint128_t, Voltage Spike Feb 26 '17 at 7:05

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  • \$\begingroup\$ VTC - This question is off topic because it is about the use of electronic devices. \$\endgroup\$ – Michael Karas Feb 25 '17 at 12:28
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    \$\begingroup\$ @MichaelKaras I first voted like you, but then I realized, OP is designing a lighting system, so yeah, imho OP's question is on-topic. \$\endgroup\$ – Marcus Müller Feb 25 '17 at 12:34
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    \$\begingroup\$ @MarcusMüller -- my interpretation is that a question involving the design of a system has to disclose/show some effort and steps, so that the question is about specific issues and not merely "I need help". \$\endgroup\$ – Wesley Lee Feb 25 '17 at 12:52
  • \$\begingroup\$ Please do research before posting (required), there are scads of posts about powering LED's \$\endgroup\$ – Voltage Spike Feb 26 '17 at 7:05

I've been playing with some Luxeon COBs wit a 1.2A test current and 2.4A max.

It's all about thermal management. You cannot pump these guys with max current without extraordinary thermal management.

On your part, the most important spec in the datasheet is the thermal resistance from junction to case (thermal pad). On the Luxeon COBs I have the thermal resistance is only 0.29° C/W.

You cannot connect these in parallel. I have tried. Even with matched forward voltages, (42.0V and 42.2V) it did not work.

I also tried 5 ohm resistors, I have thousands of 5.1 ohm 5 watt resistors left over from a telecom product I used to manufacture. I had to connect 4 them in a 20 Watt series parallel network so they didn't get so hot.

Connecting the COBs in parallel what happens is due to their thermal sensitivity there becomes a mismatch in forward voltage and current. It happened over a 12 hour period and the thermal runaway go to the point where one got all the current and the other almost none.

This COB has been to hell and back as shown here. This has gotten so hot the twice solder on the power pad liquefied. That happened at about 1.5 Amp while mounted on a 0.125" thick, 1" wide, foot long copper bar. That bar is screwed down to a 0.125" x 0.5" copper bar that is soldered to a 0.5" diameter copper water pipe.


With no fan and nothing flowing through the copper pipe the max current would be less than 0.5 Amp. At 500mA the temperature of the copper bar rose to 73.4º C in half an hour. Had I allowed it top go, the temperature would have continued to rise.

enter image description here

I always use a constant current source. My favorite is the Meanwell HLG series which range in price from $30 to $200, ranging from 60 W to 540 Watt.

If you build your own the BOM for a driver is under $2 for a buck step down.

For slightly less ($1.40 qty 10) you can use an LM3466 LED string load balancer. I will be trying this myself in the next day or two. This week I bought the TI eval board for this part AN-2182 LM3466 Demonstration Board Reference Design.

With COBs there would only be one COB per string. The COB is essentially a string of 12 LEDs (3v x 12).

enter image description here

I doubt you will be able to push them over 1 Amp without them burning up or liquid cool. So a HLG-540-42B would be sufficient or 2 x HLG-240. I recommend the B model because it has a dimmer as well as voltage and current adjustment.

Or if you want to go green. You can try this:
enter image description here
enter image description here enter image description here


I buy them from China

usually doesn't mean

I buy them from China from a well-known supplier whose specs I can trust.

So as a warning: Don't expect the specs to apply. In fact, the three numbers you posted aren't even physically possible. 3 A · 36 V > 100 W.

That paints a picture of how much I'd assume these devices work consistently. And that means that yes, since they probably aren't identical operation-wise, you must use a driver for every single one, instead of being able to string them. That's the price you pay for cheap components.

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    \$\begingroup\$ LEDs are typically rated at test current which is much less than max. 36V COBs generally have a 5V range in their forward voltage. 108W > 100W, is not significant. Most 3W LEDs cannot be run at 3 Watts. There is nothing unusual about "3 A, 36 V, 100 W." \$\endgroup\$ – Misunderstood Feb 25 '17 at 21:22

If you aren't concerned about uniformity across all modules, you might be able to use a fixed voltage source and not worry about using drivers, assuming the LEDs are fairly consistent and have a wide operating range. It's impossible to say without datasheets, and even then you have to trust the parts match the datasheet.

If you look at the voltage/current graph for a diode or LED, in the forward region the current climbs rapidly above the forward voltage (Vf). If that slope for this array covers a fairly wide voltage range at the desired current, then you could use a fixed supply. If it's very steep, you might exceed the maximum current of the LEDs and damage them. And because it's steep in general, you will likely see differences in intensity.

It's not the best way to design something like this (normally there would be one driver per module), but you can possibly get by for a one-off thing. I only suggest this as something you could try because you are on a budget. So if you want to know how to 'properly' do it as in your question title, use separate drivers.

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    \$\begingroup\$ I'd strongly discourage constant voltage over parallel LED lamps. LEDs are semiconductors. The hotter they get, the less their resistance. The hotter LED will conduct more current become hotter, conduct more... Thermal runaway \$\endgroup\$ – Marcus Müller Feb 25 '17 at 15:59
  • \$\begingroup\$ You could, however, chain them serially and use a constant current source. \$\endgroup\$ – Marcus Müller Feb 25 '17 at 15:59
  • \$\begingroup\$ @MarcusMüller I agree 100%. A situation like what I proposed would have to be adjusted and monitored carefully to be sure the current didn't reach a runaway condition. I'm suggesting operating at a lower than nominal voltage/current rating to avoid that. I don't know the OP's situation - e.g. if this is more of an experiment, or a product design. Certainly for the latter this would not be advised, as I noted. Chaining them serially doesn't really work in the OPs case I think. \$\endgroup\$ – AngeloQ Feb 25 '17 at 16:18
  • \$\begingroup\$ That may work for 20mA LEDs that do not get hot. But when you are dealing with 85° C plus. It's not going to work. \$\endgroup\$ – Misunderstood Feb 25 '17 at 23:25
  • \$\begingroup\$ @Misunderstood agreed, if the heat-sinking is not handled properly, that would quickly become a problem. \$\endgroup\$ – AngeloQ Feb 25 '17 at 23:50

Sure. Get a 48 volt, 50 amp power supply. Connect each LED to the supply with a 5 ohm, 50 watt resistor. Alternatively, connect your LEDs in series strings of 2, use an 84 volt, 25 amp power supply and 5 ohm resistors to limit each string.

As you may have gathered from the above, your requirements are not particularly realistic. For instance, has it occurred to you that 16, 100 watt LEDs are going to draw at least 1600 watts, and you run a real risk of popping your circuit breaker? (I'm assuming you live in the US, where 120 VAC is the norm.) With a simple current limit mechanism, the extra power required will definitely cause you problems. Plus, of course, you'll need a forced air cooling system to make sure the resistors stay happy. I don't recommend natural convection (although it will certainly work) due to the tendency of users to get careless and block the airflow of the unit with unfortunate results. At any rate, a thermal cutout will be a really good idea.

Without a dedicated driver for each LED, or perhaps for a string of 2 or 3, simple current limiters will dissipate a great deal of extra power, and you're already marginal on the amount of power you're pulling from the wall socket.

Oh yes, and then there's the problem of budget. Big power supplies equal big bucks.

  • \$\begingroup\$ In the US we have 120 V, but also typically 15 A breakers, so 1600 W should only blow the breaker if it's worn or shared with other loads. \$\endgroup\$ – The Photon Feb 25 '17 at 15:57
  • \$\begingroup\$ @ThePhoton - Yup. And do you trust this guy to be SURE he doesn't have anything else plugged into that power circuit? I don't. \$\endgroup\$ – WhatRoughBeast Feb 25 '17 at 16:40
  • \$\begingroup\$ Not likely to run at max wattage. Even just 2 in series would be 72 volts and serious power supply money. 540W @ $200 will likely do the job. \$\endgroup\$ – Misunderstood Feb 25 '17 at 23:31
  • \$\begingroup\$ @Misunderstood - Sorry, but I don't understand. You can't directly run LED strings from constant voltage, which is why I specified 84 volts (12 volts extra for dropping by a limiting resistor). Since each string of 2 requires 200 W, a 540 W supply will only handle 4 LEDs, so 4 supplies are required for a total of 800 bucks. \$\endgroup\$ – WhatRoughBeast Feb 25 '17 at 23:48
  • \$\begingroup\$ Just because they are 3 Amp does not mean you will be able to push 32 Amp without them burning up. Probably closer to 0.5 Amp. with little thermal management. Even a constant current source must have a voltage potential rated greater then the forward voltage o the string. The 540W supply I was thinking of is a HLG-540-42B 42 Volt Constant Current supply from Meanwell. I've done the parallel 5 ohm resistor thing, it does NOT work. Thermal runaway with two strings one 42v the other 42.2. \$\endgroup\$ – Misunderstood Feb 26 '17 at 0:31

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