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I working on DC LED street light driver using this LED street light look like picture bwelow. enter image description here

I don't know much about this LED but it use about 30V to drive and I design CC drive for it. The white LED usually got Vf about 3V. My lamp got 90 LEDs on it, so I assumed the schematic look like this (10 LED in series x9 in parallel). Why we can use LEDs in parallel in this situation? I remembered we should not parallel the LEDs.

  1. Vf of LEDs can be difference.
  2. Temperature make VI curve changed.

Note: This schematic is just my assumption. If you has real schematic please share.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Voltage Spike Nov 24 '20 at 19:02
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    \$\begingroup\$ They are matched by the supplier. I have designed electronics for for industrial lasers for cutting steel, which use same arrangement of large parallel banks of series diodes. They were very high power and the need for matched power output per LED is far more critical there. Using matched LEDs in parallel is standard technology. \$\endgroup\$ – TonyM Nov 24 '20 at 19:38
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Here's an excerpt from some Yuji LED datasheet:

enter image description here

So yes, if you order in quantity you can get LEDs with good Vf match. Wiring them in series strings means the Vf mismatches add and substract which increases the probability that total Vf will be closer to (number of LEDs x average Vf).

If they are mounted on an aluminium PCB which keeps them all at the same temperature, Vf under load will track pretty well. This means they can be arranged in series/parallel strings without resistors or other current sharing/limiting devices. This is very common in LED lights.

If the LEDs are not binned by Vf, and especially if they are at different temperatures, then it wouldn't be a good idea to parallel them.

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    \$\begingroup\$ Just for completion, can you link to the LED data sheet. I'm not disputing what you say BTW. \$\endgroup\$ – Andy aka Nov 23 '20 at 16:09
  • \$\begingroup\$ With passives the assumption of normal distribution of values only applies at quantity, any given batch or even lot, which you get from a distributor is not necessarily going to be randomly distributed about the mean. Is that also the case with LEDs? \$\endgroup\$ – crasic Nov 23 '20 at 17:11
  • \$\begingroup\$ @Andyaka I've added the link. \$\endgroup\$ – bobflux Nov 23 '20 at 18:29
  • \$\begingroup\$ @crasic I have no idea what the actual distribution is, but considering the tight binning I'd bet it's uniform in the interval. So the more you wire in series, the closer the Vf sum will be to a gaussian... \$\endgroup\$ – bobflux Nov 23 '20 at 18:31
  • \$\begingroup\$ Cheers @bobflux... Hold on - it must be a different LED to the one in your answer because it doesn't have the same range of bins. \$\endgroup\$ – Andy aka Nov 23 '20 at 18:46
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You should not parallel leds, when you do not take the time to ensure they match in forward voltage drop, current draw and temperature curves, or more. The advice to not parallel leds is an often regurgitated statement without critical thinking. Like anything else, if you look into the reasons behind it and the math and practical application you know its just more complicated.

Parallel leds are used in multiple applications and its not a problem when you match the leds. Commercial products like the street lamp are one of those applications.

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An incoming or source-inspection process could match the LEDs reasonably well. Still, there's probably a low-ish resistor on each series string to compensate for any residual string-to-string Vf difference that may come up.

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  • \$\begingroup\$ Hey, marker-downer, look at actual lamps. They either use separate current drivers for each string, or there is a resistor for each string that shares a current source. \$\endgroup\$ – hacktastical Nov 23 '20 at 19:03
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    \$\begingroup\$ in a long chain, the PCB trace or thin wire resistance is enough to stabilize. \$\endgroup\$ – dandavis Nov 23 '20 at 21:01
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    \$\begingroup\$ Not the down voter and I haven't seen alot of these boards but the handful I have seen of street lamps and tvs use a single driver output with no discreet resistors. Do you have any examples? \$\endgroup\$ – Passerby Nov 24 '20 at 2:08
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    \$\begingroup\$ I didn't downvote you either, but using series resistors is not common in well-designed high power lighting. Instead they're putting binned parts in series to get the variation in total forward voltage down low enough that it doesn't matter. If the manufacturer needed a series resistor, it probably means they were too cheap to buy binned diodes and are hoping you don't notice that they're selling you crap. \$\endgroup\$ – user1850479 Nov 24 '20 at 2:19
  • \$\begingroup\$ @dandavis No it isn't, not ever. Anyone making that assumption has a PCB trace waiting to turn into a fuse, \$\endgroup\$ – Graham Nov 24 '20 at 11:59
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If you add the LEDs into long serial chains connected in parallel, you add a lot of very small resistors to the chains. The short and wide connection between two chips has a very small resistance, much less than a milliohm. But a chain of 10 or more has small total resistance stabilizing the current through parallel chains.

A street lamp for 110 V AC may use chains of up to 50 LEDs in series. Using a small number of long chains there will be no problem.

90 LEDs may be connected as 15 chains of serial 6 LEDs each, but also in 5 chains of serial 18 or 3 chains of serial 30 each. I would avoid 2 chains of 45 in series, 5 or 6 chains should be a good compromise.

If longer chains of LEDs are used, sorting of the LEDs to get very similar chains would be easier and more efficient.

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    \$\begingroup\$ Reads as supposition, or is it your own product design/manufacturing experience? \$\endgroup\$ – TonyM Nov 24 '20 at 10:37
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Mismatch and thermal runaway destabilize parallel LEDs. Matching, parasitic resistance, and isothermal design stabilize parallel LEDs. It requires excellent knowledge of your parts and careful analysis to determine which effects dominate. If you don't have the expertise, time, and resources to acquire that knowledge and perform that analysis, it is safer to avoid paralleling LEDs without additional stabilization. But for a product produced in huge quantity, the extra work of carefully calculating stability may pay for itself by saving a few pennies per unit.

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  • \$\begingroup\$ So you're saying that a buyer would get a load of LEDs and try and match them themselves? Rather than the manufacturer selling matched parts. \$\endgroup\$ – TonyM Nov 24 '20 at 14:49
  • \$\begingroup\$ Don't know what combination of means is actually used, nor am I a LED expert. The details are probably trade secrets. In silicon, parts from the same wafer are usually well-balanced. If this is also true of LEDs it might make sense for the lamp manufacturer to procure them by the wafer. \$\endgroup\$ – John Doty Nov 24 '20 at 16:26
  • \$\begingroup\$ So this answer is just a guess? \$\endgroup\$ – TonyM Nov 24 '20 at 17:16
  • \$\begingroup\$ @TonyM No, I am listing the factors involved. The details are presumably trade secrets, but the approach to making this work is as I said. Since the details are application-specific, it isn't terribly important to know them unless you're trying reverse-engineer the product. Otherwise, you have to fill in the details for your application yourself. \$\endgroup\$ – John Doty Nov 24 '20 at 18:21
  • \$\begingroup\$ I'm making sure that my first comment was a fair summary of the answer. You're listing what you imagine you might do but from no experience so it's an unqualified guess. Sure, downvoting. \$\endgroup\$ – TonyM Nov 24 '20 at 18:44

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