Testing LED array with resistor

[galleon]

How big? 10 ohm? I just want to make sure all the connections are solid with no shorts to the heatsink before I fry a driver.

Wired in series on +24V side coming from the PSU or wired in parallel?

Thanks!!

[der_wille_zur_macht]

Can you describe your configuration? What LEDs, how many, how are they wired together, what power supply, what driver, and so on?

[galleon]

Can you describe your configuration? What LEDs, how many, how are they wired together, what power supply, what driver, and so on?

6 CREE XP-G, wired in series, 1000 mA nondimmable buckpuck, 24V 6.5A Potrans PSU

Thanks!

[der_wille_zur_macht]

Given that configuration, I would just probe the array for shorts, double-check polarities and quality of connections, and have at it.

There's really no way to accidentally overdrive/kill the LEDs given your driver, and if there are multiple shorts you don't detect you'll just get one or more LEDs not lighting up.

Testing with a resistor would mean removing the buckpuck from the circuit and replacing it with a resistor - and at that point you're not really testing the circuit "as built" since you had to modify it. And testing with a resistor isn't really going to make the most common problems any more safe.

This is all IMHO of course, other people may have different advice.

[galleon]

But say I was compelled (the buckpuck doesn't show until tomorrow), although I do understand what you're saying. Basically I'm paranoid because I've already blown one buckpuck.

[galleon]

I'm guessing it would be whatever would drop the 6.5 amps to 1 amp at 24V wired in series.

[der_wille_zur_macht]

You probably don't want to test at the full 1A current - that will require a meaty resistor. To size it correctly, you need to find the voltage drop. That is, the difference in voltage between what your power supply is providing (24v nominal, but you should check with a multimeter) and what your LEDs will need at the target current (look it up in the datasheet). Then, you use ohm's law to calculate the resistance for that voltage and current.

For instance, if you had that 24v supply and 6 LEDs that dropped 3.5v at 1A, you'd have a voltage drop of 3v: 24 - 3.5*6. Your target current would be 1A. So, your target resistor value would be 3v/1A which is 3 ohms.

Next you need to calculate the power dissipated in the resistor - to determine the power rating you'll need on it. In this example, your power is 3v * 1A = 3w. To be safe, most people would typically double the power rating to give a safety net, so you're looking at a 6w resistor. This is going to be meaty. You're not going to be able to use a common resistor for this, most are fractional wattage.

[galleon]

Ahhh, that's the step I was missing was the multiplying forward voltages.

Thanks!!!

[BeanAnimal]

Given that configuration, I would just probe the array for shorts, double-check polarities and quality of connections, and have at it.

There's really no way to accidentally overdrive/kill the LEDs given your driver, and if there are multiple shorts you don't detect you'll just get one or more LEDs not lighting up.

Testing with a resistor would mean removing the buckpuck from the circuit and replacing it with a resistor - and at that point you're not really testing the circuit "as built" since you had to modify it. And testing with a resistor isn't really going to make the most common problems any more safe.

This is all IMHO of course, other people may have different advice.

Not to mention needed current rating of the resistor!

[BeanAnimal]

Ahh I see you got to that in a subsuquent post... nevermind

I keep a drawer full of power resistors... but don't know many who do :)

[galleon]

Interesting, I have a 10 ohm 10 watt resistor. BUT, now how do you figure out how much current will be going through it? Both the forward voltage AND the current are dynamic/dependent.

[BeanAnimal]

LEDs have ZERO ability to limit current, they do however have a fixed voltage drop.

We simply size the resistor to limit the current that can travel through it. We can do this according to Ohm's law as we know that input voltage and the voltage drop over the LEDs. We simply solve for current.

Instead of yet one more Ohm's law explanation here on RC, you may be better served by visiting any of the dozens of web sites devoted to explaining Ohm's law and how it relates to driving an LED :)

[galleon]

Bean, the forward voltage depends on the current traveling through the LED, even though it doesn't vary by a lot (less than a volt over the operating range).

It's a one equation two unknown problem.

[galleon]

So lets say I pick a middle of the XP-G voltage range value, 3.25. With my 10 ohm 10 watt resistor

[24-(3.25*6)]/10 = 450 mA

450 mA at 4.5 volts = 2 watts

So this resistor would work fine for testing.

[BeanAnimal]

Bean, the forward voltage depends on the current traveling through the LED, even though it doesn't vary by a lot (less than a volt over the operating range).

It's a one equation two unknown problem.

You are thinking too hard :)

You are supplying the voltage and you know your target current, therefore you can compute the needed resistance to reach that target current at the target Vf.

[kcress]

Yes your resistor will limit around 450mA and the power is fine.

Have at it.

[galleon]

Bummer. No shorts, continuity through the circuit, no light. I hope my first buckpuck incident didn't fry them all.

[hitchikr]

Out of curiosity in all of these LED builds people are using drivers similar to the buckpucks or meanwells. If you are using a DC power supply that is providing a stable output is there any reason not to use a resistor instead of a puck to limit the voltage on the LEDs instead?

I've recently finished a 72 LED rig and most of my strands are running on pucks. I did have an unfortunate incident where I blew up a couple pucks and I now have those strands running with a sandbar resistor in place of the puck. The resistor generates a LOT more heat than the pucks but nothing that a heat sink can't easily handle. They have been running this way for about a month now without issue. At the cost of less than a $1 for a resistor that can handle a 5W load (the actual load on the resistor is 2.1 Watts) its certainly a lot cheaper to build this way. Other than a little extra math to figure out the appropriate resistance to use am I missing something? Is there something inherently wrong with this design?

[kcress]

Bummer. No shorts, continuity through the circuit, no light. I hope my first buckpuck incident didn't fry them all.

If they are blasted they usually look different. Compare them to never powered LEDs. Use a magnifier.

[kcress]

Out of curiosity in all of these LED builds people are using drivers similar to the buckpucks or meanwells. If you are using a DC power supply that is providing a stable output is there any reason not to use a resistor instead of a puck to limit the voltage on the LEDs instead?

I've recently finished a 72 LED rig and most of my strands are running on pucks. I did have an unfortunate incident where I blew up a couple pucks and I now have those strands running with a sandbar resistor in place of the puck. The resistor generates a LOT more heat than the pucks but nothing that a heat sink can't easily handle. They have been running this way for about a month now without issue. At the cost of less than a $1 for a resistor that can handle a 5W load (the actual load on the resistor is 2.1 Watts) its certainly a lot cheaper to build this way. Other than a little extra math to figure out the appropriate resistance to use am I missing something? Is there something inherently wrong with this design?


Using just a resistor and a voltage controlled supply works fine.

Down side: You can have less efficiency. You have no dimming capability. You will have no active "current limiting". Which means if one LED shorts the current will jump up dramatically with predictable results.(You can use a fuse to limit the negative chain of events in the case of shorting LEDs.)


Up side: You can have higher efficiency than any controller. Vastly less expensive. Simple.