How can a DC18V ~ 36V LED driver work with DC3.6V LEDs without burning them out?

I bought some slightly more powerful leds to build a hobby project and grow some herbs. The kit has 10 LEDs that have the following spec:

Power: 3W
Voltage: DC3.6V
current::700mA


It comes with a driver:

Output Voltage: DC18V ~ 36V
Output Current: 600mA


The idea is put 10 in series and drive them from AC with that driver.

My questions is how can it work without burning them out? I know about the voltage drop, but the first LED still gets a high voltage? They are only rated ~3V.

I "tested" one LED with 9V battery and it lighted up very bright and then it was dead. Was it because of 9V or was it because it drew that much current? What am I missing here, it looks like there is some very basic thing I don't understand. I have studied the OHM law and have built some simple things, but I'm out of ideas here.

• 10 times 3.6 Volts is 36 Volts. – Jeroen3 Oct 28 '17 at 13:21
• @Badr Hari No, the first LED will not get 36 volts, because the other leg of the led will not be connected to the negative pole, the other leg to the led will be connected to the next led in the series, and so on. This way, the 36 volts will be divided between the ten leds. – mguima Oct 28 '17 at 13:27
• @mguima Thank you, looks like I'm missing some basic concepts, I need to read more. – Badr Hari Oct 28 '17 at 13:28
• Herbs huh? ..sure. – Trevor_G Oct 28 '17 at 15:55
• Possible... probably, can you afford it.. maybe not. As I said in my answer though, LED drivers are not voltage supplies, they are current supplies. The voltage output is secondary. – Trevor_G Oct 28 '17 at 16:29

Your driver will output a fixed current through a string of LEDs.

Since your LEDs are rated at 700mA @ 3.6V, this driver will indeed supply adequate current for the LEDs without burning them out.

How many LEDs you can put in series on a driver depends on the drivers voltage range, in this case 18 to 36V.

So you can attach 18/3.6 to 36/3.6 or 5 to 10 LEDs to this driver.

simulate this circuit – Schematic created using CircuitLab

With 10 LEDS the output from the driver will be around 36V. That voltage is divided across each LED, so no individual diode has more than their rated voltage.

Or more accurately, when driven with 600mA each LED will generate a voltage across it, close to 3.6V, the total of those voltages will be what you measure at the output of the driver.

When you hooked up an LED directly to the 9V battery you vastly exceeded it's rated voltage and it went out with a flash of light. You basically fused it.

• Trevor is not only right, and observant of wattage per LED in OP. He is also handsome. +1 – user2497 Oct 28 '17 at 16:46
• Thank you @Trevor and all the others for discussion. It looks like I have to go back to basics because I miss too much. – Badr Hari Oct 28 '17 at 16:50
• Blushing @user2497 – Trevor_G Oct 28 '17 at 16:51

1) NEVER connect a led to any voltage source (power supply or battery) without resistors. It could ruin the led, the power source, or both.

2) It seems that this "driver" is a constant current source of 600 mA, with minimum and maximum voltages of 18 V and 36 V. Since the led's specs are 700 mA (max), you can connect the ten leds in series, and the serie of leds can be directly powered by this driver.

simulate this circuit – Schematic created using CircuitLab

There is an advantage of running the 700 mA leds with 600 mA: they will not bright as their max, but their life spanning will be higher.

• But why did the 9V kill the led but the driver wont? The driver is much higher voltage and I doubt the current of battery is higher. – Badr Hari Oct 28 '17 at 13:27
• @Badr Hari The 9v battery acted as a voltage source and so it supplied all the current the led would "ask" for; and a common led "asks" for all the current the supply can feed to it. It's like a aquarium fish that will eat all the food you give to it, even if it'll die eating so much ;-). That's the reason for adding the resistor to the circuit. The driver that you describe SEEMS to be a constant current source, that will supply A MAXIMUM OF 600mA, this current not kill those leds of 700 mA rating. Maybe you could describe this driver more precisely. – mguima Oct 28 '17 at 13:38
• Yes, never run LEDs at max. Use 80-90%. +1. – user2497 Oct 28 '17 at 16:44
• The LED wants 3.6V across it. Trying to put 9V across it was not a good idea. Trying to put 600mA through it is fine. You can understand the LED's specs as, roughly, "you want to put up to 600mA through this LED to light it up, and when you do approximately 3.6V will drop across it". For it to work with a 9V supply, you'd need a resistor for the other 5.4V to drop across, and if you do the math (V=IR), you get 7.7 ohms, absolute minimum. But you had zero (well, the battery's internal resistance, but very low), so not surprising you exceed the LED's limitations. – David Schwartz Oct 29 '17 at 0:38
• @user2497 - having disassembled and inspected a number of commercial products built for the same purpose OP has acquired his LEDs for, I've noticed that typically they tend to use about 50-70% of the rated current of their diodes. I'm guessing they leave a large margin in order to avoid problems caused by the build-up of heat you'll get running a lot of high power LEDs in close proximity in an enclosed space. – Jules Oct 29 '17 at 12:03

Others have already covered the driver situation.

If you want to test with your 9V battery, try it with a 8.2 or 9.1 ohm resistor in series with the LED. This will limit the current to something that LED can handle. The resistor should be rated for at least 4 or 5 watts though.

You probably don't want to do this for very long though--you're right at (or maybe even beyond) the current a typical 9V battery is intended to supply, so don't expect the battery to last more than an hour or so (at best) if you use it for this. The battery will also probably get pretty warm if you pull this much current from it for very long at all.

I'm pretty sure the single largest use for 9V batteries is to power smoke alarms, so they're designed primarily for long shelf life, while providing fairly minimal current (on the order of 10s of milliamps at most).