Parallel Luxeons at high power - question

I want to connect a couple of K2s, or a couple of IIIs, together in parallel.

Assuming I remeber to select LEDs from the same voltage bin, am I going to be able to operate them up to their maximum rated power without one LED hogging all the power and maybe melting?

I don't really want to add balancing resistors.

I assume modders do it all the time, so this should be an easy one for someone.

Thanks

Steve

The problem with running in parallel is that one LED will inevitably have a slightly lower Vf and see more current. I bought a bunch of TVOJ stars and built a tri-lux some time ago. I always intended to wire them in series, but I found that they had signifigantly different Vf's at 1000mA - 3.3V, 3.4V, & 3.5V ... and these were from the same tray!

Assuming your power source isn't too restrictive, you won't have one LED "hogging all the power," but they will see signifigantly different current levels with only slight differences in Vf.

If your power source's nominal voltage is only slightly greater than your LEDs Vf at your desired drive level, you'll find that resistors can be simple and efficient current regulators. If you build a 3-cell light and can live with some overdrive from alkalines during their short 1.5V peak, assume a nominal working voltage of 3.6V. The less voltage you drop across the resistor, the more efficient your design. Determine your LEDs' actual Vf at your desired driv level and subtract working voltage from Vf to figure out what the resistor has to drop. Beware that this is a very rough formula and you'll have some other resistance built in to the light (switch, battery contacts) that you might want to account for.

I agree with Idleprocess, consider using resistors, one per led. In addition to what Idle says, there are two other factors that make dd'ing parallel luxeons potentially hazardous for the leds:
1) Leds experience a drop of Vf during the first hours of usage. The more current is pushed through the led the faster its Vf drops. Now consider the situation Idleprocess described above: the led with lowest Vf gets the most current and its Vf drops fastest of the three leds... and its taking even more current and the Vf drops even faster etc, etc... Now we have a nasty loop here.
2) Luxeon leds have negative temperature coefficient for Vf. The hotter led runs, the lower its Vf gets and the more it takes current when direct driven. Here we have a second nasty loop.

To sum up, direct driving parallel leds is quite unstable setting. If one of the leds is taking more power than the others, some self-feeding processes try to make this led take even more and more power. The worst case scenario is that it blows.

A bit long and technical story, but I just wanted to explain why I would prefer resistors, one for each led. Good luck with modding!

- Nereus

The safest way to run LED at their rated maximum power is to use Current Regulation in series. In other words, pushing the same, constant and steady current through ALL the LED together will ensure they never exceed their power limits.

otoh, using resistors to limit current is just insurance. As mentioned, Vf varies between samples; as they age and as the the temperature changes. So the LEDs are seldom driven optimally. Then there is a problem of the battery's dropping voltage. Worse, resistor values are only available in 10% steps, and in limited accuracy - so finding the "right" ones will never be a sure thing.

For maximum life, LEDs connected in parallel using resistors should never be driven within 30% of their rated power.

Thanks folks.

As I feared.

I never used to run even 20mA leds in parallel because of uneven current and thermal run-away.
Then I saw that loads of people committed this 'sin' and had no problems. So I started doing it and have also had no problems.

So I was hopeful, against my better instincts, that someone would say 'yes, its OK with Luxeons providing you.....'

Sadly not though.

I will have to think of another scheme. I am driving these leds from a switching power supply and really do not want to add resistors as I am already throwing away efficiency in the Schottky.

And I was hoping to avoid heatsink isolation issues.

Still, all this moaning is not going to change the facts
- I have to think up another topology with leds in series.
Looks like I may have to re-investigate the glorious-but-scary Cuk buck-boost converter.

Thanks again

Steve

Parallel at low power works OK,
Get some of those T--H LuxIII's and run them in parallel from two alkaline D cells which will not overdrive them. If you goal is something very, very simple... direct drive with no resistors works fine as long as the power in does not exceed the forward voltage.
Personally, I am a fan of putting them in series and using a current regulator. Another option is to do what Elektrolumens does, NiMH cells and 8 Luxeons with a few power resistors. Luckily, if the batteries come hot off the charger, the Luxes will suck up the initial surge with no problem since there are so many of them.

Just another 2 cents from the peanut section - are you really sure you want to run the LEDs at max power ? If you goal is to obtain approximately max power from a reasonable amount of power, then you might want to tone down the current to each LED a bit.

I have seen a few graphs of current vs output which were remarkably linear, but they were taken on special bench top heat sinks with fans. My own crude testing indicated that going over 700 - 800ma was not that useful, and going to 1.5 amps can actually result in less light than 700ma due to heating, regardless of if you use a Lux III, K2, or Lux V. (using the kind of heat sinking you actually have in many flashlights) It might be brighter for 60 seconds, but after 2 - 3 minutes, it drops quickly.

The other benefit of running parallel LEDs at a bit lower current setpoint, is that if one is hogging current a bit, it is likely still well below its max, so it is largely a non issue. (at least for amateur use). While LEDs do have a negative Vf with temperature, they also have a rather strong positive Vf with current, so they are not as likely to run away as you might think.

You can reduce the Vf loss of the diode by putting one in for each LED, as they diodes tend to have increasing Vf with current, at least the ones I use. It also does not take that much resistance to approximately balance out the LEDs - nominally 2 ohms for a 2 volt drop should do it.

I like to think of the maximum ratings of the power LEDs as "engineering headroom" to make our lives easier, not necessarily that you should design the light to run there. If you need more light, use Lux Vs or add another LED.

Good luck with your project.

HarryN said:

... are you really sure you want to run the LEDs at max power ? If you goal is to obtain approximately max power from a reasonable amount of power, then you might want to tone down the current to each LED a bit.

Click to expand...

Good point - rule of thumb is that you need to triple the current to double the lumen output. A 5W Star will outperform a 3W one when they are both driven at 3W.

LEDs are tolerant of brief overcurrent, but Lumi's figures show an output drop of 10% after 1000 hours of operation with just 5% of overdriving.

just to stir things up a bit.... it occurs to me that the Lux V is built of two parallel strings of two Lux dies in series. How do they manage to parallel these two strings without having one string hog the current?

Is it just a matter of using dies from the same wafer, or maybe from the same area of the wafer, in the hopes of having dies that are very well matched?

Is there enough resistance in the bonding wires to act as ballast resistors, encouraging even sharing?

Or is this actually a problem for the Lux V, leading to short lifetimes and reliability issues?

Steve K.

Damage to LEDs around its operating point is temperature related, as Lumiled is always pointing out. Overcurrent leads to overheating, and in individual dies, this can cause run-away current conditions.

The elements in the Lumi-V share the same thermal-base, so even if one takes a bit more current, its heat will be passed on to the other elements. Of course having the identical mix and manufacturing run helps too!

Steve K said:

just to stir things up a bit.... it occurs to me that the Lux V is built of two parallel strings of two Lux dies in series. How do they manage to parallel these two strings without having one string hog the current?

Is it just a matter of using dies from the same wafer, or maybe from the same area of the wafer, in the hopes of having dies that are very well matched?

Is there enough resistance in the bonding wires to act as ballast resistors, encouraging even sharing?

Or is this actually a problem for the Lux V, leading to short lifetimes and reliability issues?

Steve K.

Click to expand...

I think I do remember discussion of differential Vf drift of the two paralleled die strings in a LuxV contributing to its reduced lifespan compared to the LuxIII. It does make sense. You might find more on a forum search.