I am thinking of driving COB LEDs (rated at 10 W normal power) at 3 W.

I want to drive 32 LEDs from one 100 W constant current driver. The driver has a maximum voltage of 144 V. The LEDs are 35 V.

I can drive a string of 4 LEDs maximum in series.

Is it safe to drive 8 of these strings in parallel?

  • 1
    \$\begingroup\$ Datasheet please. \$\endgroup\$
    – pipe
    Sep 30 '16 at 8:22
  • \$\begingroup\$ How do you intend to reduce the power consumption of the led to 3W given the input voltage of 34 V. (Range 30-34V?). \$\endgroup\$
    – Decapod
    Sep 30 '16 at 8:40
  • 1
    \$\begingroup\$ It would be best if you provide us the specs of COB (chip on board) LED and the driver (for CC [constant current] LED drivers, it is quite meaningful to know about output/rated current as well). \$\endgroup\$ Sep 30 '16 at 8:45
  • \$\begingroup\$ @pipe I provided the data in a comment to the answer below \$\endgroup\$ Sep 30 '16 at 9:20
  • \$\begingroup\$ @NickCollier Like you said in a comment, if you hand-pick the LEDs to match the Vf of each string, it should be safe. \$\endgroup\$
    – rioraxe
    Sep 30 '16 at 20:41

The LEDs are designed for a constant current supply which produces whatever voltage is necessary so reducing the current would be effective in dimming the LEDS.

Rohat calculated that the supply output would be a constant 700mA and for 3W per light you need 86mA each. Putting 8 lights in parallel would give 700/8=87.5mA per light which is about right, but that assumes they will share the current equally. The problem is that LEDs have a fairly sharp turn-on so whichever bulb has the lowest forward voltage will take almost all the current and die, then the next lowest would blow and so on. Unfortunately, whichever takes the highest current will heat up most and that will probably drop the voltage futher giving thermal runaway.

What you would need to do is add something to equalise the currents, a series resistor as shown would do but I suspect the value needs to be higher than shown in the circuit, at least the maximum forward voltage variation divided by the current variation you can tolerate. As the LEDs can take up to 350mA, that is fairly easy but you probably want to keep the brightness of the bulbs fairly similar so for example 10% brightness variation implies no more than 8.6mA difference between the bulbs. Higher values of resistor will maintain more even brightness but will increase heat dissipated in the resistors. The advantage is that including 8 resistors is a lot cheaper than separate drivers.

P.S. You probably realise that the supply voltage need only be 35V plus the resistor drop so a 144V-capable supply is much more than you need, it is really intended for driving multiple bulbs in series so that all get exactly the same current and hence brightness.

  • \$\begingroup\$ Thank you. I was aware of these problems, but assumed it might be ok as I am running the LEDs at less than 1/3 of their normal power. Also I will be water cooling the heatsink. \$\endgroup\$ Sep 30 '16 at 11:00
  • \$\begingroup\$ Table 4 of the datasheet says the forward voltage of a nominal 35V unit can be between 32.4V and 37.6V. If you had one at the low end, and others typical or high, the low voltage unit would take all the current. I note the datsheet says it will survive double the nominal current so it might not blow but without some current sharing mechanism, the brightnesses will be very different. An 8mA current variation for 5.2V difference implies series resistors of 650 ohm but those would drop 57V and dissipate 5W each, more than the bulbs! You might need to consider an active constant current limiter. \$\endgroup\$
    – Fleetfoot
    Sep 30 '16 at 11:13
  • \$\begingroup\$ Ok I see what you mean, maybe I could test the voltage drop of each one, and make sure I have each string with a good mix of high ones and low ones. \$\endgroup\$ Sep 30 '16 at 13:23
  • \$\begingroup\$ 1R is a random value in my answer. Of course it should be calculated carefully. I intended to show that a series resistor "must" be placed. As indicated, OP can reduce the string current by adding more strings. For example, 3 x 10 instead of 4 x 8 will make 70mADC of string current and hence nominal LED current - this also decrease the forward voltage of LEDs to about 32.5VDC and so the total load voltage. But still a balancing resistor must be placed in series which can have a resistance of 30V / 0.07A = 428 Ohm (390 Ohm is ideal) and a power of 0.07² x 390 = 1.9W (3W or higher recommended). \$\endgroup\$ Oct 1 '16 at 5:49
  • \$\begingroup\$ Yes, that's worth doing Nick, the spec will give the long term manufacturing range but it will be pessimistically wide and if you get bulbs from a single batch, they should have a much lower spread. That will allow a lower resistor value and consequent power dissipation. Ultimately, it depends on your application as to how much brightness variation you can tolerate for aesthetic reasons. \$\endgroup\$
    – Fleetfoot
    Oct 3 '16 at 13:04

If you have a 100W CC (constant current) driver with maximum output voltage of 144V, then the output/rated current should be about 0.7A.

If your LED has a nominal power of 10W and load voltage of 35V, then the nominal drive current should be about 0.3A. If you want the LED to have 3W output power, then the drive current should be about 3W / 35V = 86mA.

As you might guess, you cannot connect all the LEDs in series.

Let's think about this:


simulate this circuit – Schematic created using CircuitLab

It's clear that LEDs will break down because of 350mA constant current per strip.

  • \$\begingroup\$ But what if I add more strings of LEDs until the current is low enough, I was planning to use 8 strings. \$\endgroup\$ Sep 30 '16 at 9:16
  • \$\begingroup\$ I was thinking of a drive like this, uk.rs-online.com/web/p/led-drivers/8234778 and these LEDs bridgelux.com/sites/default/files/resource_media/… \$\endgroup\$ Sep 30 '16 at 9:18
  • \$\begingroup\$ @Rohat: You will not be able to bring the current down. Each led need the 350 mA to function properly. The only way to use only 3W per led is by using a PCM signal with a duty cycle of 30% active. When on during the 30% the led will need the 34V and corresponding current. In other words working with a single driver is not possible this way. \$\endgroup\$
    – Decapod
    Sep 30 '16 at 9:24
  • \$\begingroup\$ @NickCollier: See my response to Rohat. What ever the arangement will be you need 32x10W = 320W peak load supply for the leds to function properly. \$\endgroup\$
    – Decapod
    Sep 30 '16 at 9:27
  • \$\begingroup\$ @Decapod Not sure about that you saw my last statement in my answer above: "It's clear that LEDs will break down because of 350mA constant current per strip." Of course we cannot bring the total current down, because there's a constant current source. \$\endgroup\$ Sep 30 '16 at 10:54

The datasheet given indicates that the current consumption of the LED system is given at a nominal value. In other words it can be more or it can be less.

Voltage driven LED systems have an internal arangement to keep the current at the desired level. If you put such systems in series the voltage distribution might not be uniform. This leads to the risk that some units have an overvoltage and become defective. So it is not a good idea to put this type of led in a series parallel combination as asked.

What you can do is give each system each own driver and supply all the drivers together with a PCM modulated signal ( maybe dmx? or 0-10V) to come to a desired result of 3W per led (30% pcm duty cycle).

It is possible to place more then one module in parallel to a driver.

You could look at the LT 8048 DMX driver. This driver can supply a max of 4A per channel.


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