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I have a 20VDC input with a buck CC LED driver, driving 48 LED strings of 6 x 3V LEDs in series.

I am underdriving the LEDs at 60% of rated power. Each LED string consumes ~33mA. The LEDs are thermally coupled on the same PCB. The LEDs will be binned by Vf.

I want to add current balancing but the idea of the reference string failing and causing the other 47 strings to fail puts me off.

I have enough strings that if even 2 fail the other strings will just share the increased current which is not much, but in this approach, the LEDs are essentially fighting each other, correct?

Would a current mirror be necessary in this scenario, or would it be better to use a single current source per string? Such as 3 x LED1642GW (16 channel LED driver.)

Edit:The total Vf of each string @33ma is 17.2v. Regarding the Headroom, I've build the PCB and it does work fine with the CC source (AL8863).

Also, the resistors on the cathode side of LEDs are not really doing anything, correct? I added them to try and equalize current.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ I think it may work well enough the way it is. If you want to make a crude current source, you can do it pretty easily by adding one NPN BJT and three resistors per string. It does eat into your headroom a tiny bit but I think it will be OK. It would be nice to know what your ACTUAL operating voltage is at 33 mA to know how much headroom is really required. \$\endgroup\$
    – user57037
    Commented Jul 27, 2022 at 5:53
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    \$\begingroup\$ Commercial LED "light engines" for streetlights, etc. routinely combine series/parallel LED strings without independent drivers or current balancing/mirroring for each string - yes even the ones from reputable 'western' manufacturers. If you're building these to that sort of 'commercial' scale and know that your LED supplier really is giving you a properly binned batch of LEDs, then you're probably already doing all the right things (under-driving and thermal coupling) to make this a non-issue. \$\endgroup\$
    – brhans
    Commented Jul 27, 2022 at 13:08
  • \$\begingroup\$ On the other hand, when driving that many LEDs it's more common to increase the length of the strings and use a higher voltage CC driver (200V or more, boost if necessary) rather than putting so many strings in parallel ... \$\endgroup\$
    – brhans
    Commented Jul 27, 2022 at 13:09
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    \$\begingroup\$ The total Vf of each string @33ma is 17.2v. Good to know that commercial applications apply similar designs. \$\endgroup\$
    – Vulxcn
    Commented Jul 27, 2022 at 18:53
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    \$\begingroup\$ Look up “COB LED array”. It’s very common to have 3-10 parallel strings, each with 3-10 LEDs in series to create a 3x3 to 10x10 array. You can’t drive them as hard as a single string, but that’s fine in commercial applications. These modules are powered by a single constant-current supply with no per-string current balancing. \$\endgroup\$
    – Navin
    Commented Jul 30, 2022 at 20:27

3 Answers 3

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Led string consumes ~33ma LED's thermally coupled on the same PCB.

Thermal coupling within the string is unnecessary, the series strings already run at the same current by design. It's thermal coupling across the strings, that is from one PCB to another, that you would need for balancing between the strings.

LED'S will be binned by vf.

What happens when you come to replace a PCB? Are you only going to buy one bin of LEDs for the entire project, and spares? That will make them more expensive and possibly unobtainable. Even then, there is a spread across a bin. The difference in temperature between PCBs means that Vf binning doesn't result in equal voltages on the strings anyway.

Or would it be better to use a single current source per string ?

Exactly. This is a 100% solution, what everybody else does (or should do), if you have the space and budget. The 'extra cost' of a CC LED driver per string is offset by the simpler electronics and lower specifications you can then accept everywhere else, like not needing to bin your LED Vf.

Also, the resistors on the cathode side of LED'S are not really doing anything, correct? I added them to try and equalize current.

They are not doing much, there's not the voltage headroom for that. But they are doing something. They are a small step in the direction of a 'current source per string', or balancing. To work properly, the resistors would be a lot larger, and you would use a higher voltage supply on the strings. But it would be better to replace the resistors with a current source per string.

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  • \$\begingroup\$ I've certainly seen this sort of series-parallel setup with multiple strings on the same metal-core PCB and a CC PSU, because I replaced a failed PSU on it. That was a bathroom light from a UK source. I've no idea how well-designed it was but it's outlasted 2 of the junk PSUs (that they bought in) - the 1st failed in warranty - I managed to get a Philips one to fit and it's been fine ever since. \$\endgroup\$
    – Chris H
    Commented Jul 27, 2022 at 14:51
  • \$\begingroup\$ Yes, all LED strings are on the same PCB. \$\endgroup\$
    – Vulxcn
    Commented Jul 27, 2022 at 18:58
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You can try something like this if you want a simple current source. I have shown this before and people always criticize it thinking that the emitter resistor needs to be much bigger or that it will run out of headroom or whatever. But this circuit actually works pretty well when you don't have enough headroom for a proper current limiting resistor.

Basically, R1 and R2 are a voltage divider biasing the base somewhere around 1 V. Q1 is a voltage follower. It will hold the emitter at about 0.5 or 0.4 V. Q1 will NOT be saturated. Beta will be high. So Ic and Ie will be almost the same.

I calculated all the resistor values pretty hastily. If you build it and find that the current is not around 33 mA, you can fine tune either R2 or R3.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ R1 might get a bit warm. It wouldn't hurt to use a 250 mW or 500 mW rated resistor. \$\endgroup\$
    – user57037
    Commented Jul 27, 2022 at 6:13
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With resistors, such an arrangement is fine without matching. With matching, absolutely it's fine even without resistors.

Note that it's better to wire arrays this way (as opposed to stacking parallel arrays), as there's a better chance the strings match closer than individual diodes (statistical balancing).

Note also, 20V input isn't much headroom for the CC buck to operate in (those better be 3.0V absolute maximum!). Be sure you have adequate headroom. Including if this array is long (so that the voltage seen by strings might vary with position!).

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