Why are my LEDs dim?
April 17, 2020 8:18 AM   Subscribe

My microwave came with two LED boards for under-lighting, with 4 LEDs each. I used a hot air rework station and soldering iron to replace the horrible original 6000K LEDs with some nice 2700K LEDs Great! But one of the boards is now much dimmer than the other (they should both be very bright). All 4 LEDs on each board light up, and swapping their positions doesn’t make a difference. Any ideas on what could have gone wrong, or how to troubleshoot? Any hope of salvaging this without buying a new board and more replacement LEDs?

For reference: the board in question and the replacement LEDs. I basically followed this method to replace the LEDs.
posted by stopgap to Technology (22 answers total) 1 user marked this as a favorite
Checked the resistors, both directly at their pads as well as between contact points elsewhere in the circuit?
posted by Stoneshop at 9:13 AM on April 17, 2020

Best answer: Measure the voltage across the LEDs? (assuming they're being driven with DC rather than PWM for intensity or something...)
posted by straw at 10:13 AM on April 17, 2020

There could be a couple of things going on here:

High color temperature light can appear as brighter than lower color temperature given the same amount of illuminance.

The new LEDs could have a lower light output per watt than the ones from the previous board.

new LEDs require a different amount of series resistance than the old ones did.
posted by Dr. Twist at 10:17 AM on April 17, 2020

I mis-read your question. straw has a better suggestion
posted by Dr. Twist at 10:27 AM on April 17, 2020

Best answer: OP says that "one of the boards is much dimmer than the other one." To clarify- this brightness difference is between two boards which have both had their LEDs swapped to the new type, NOT comparing brightness between one original board and one swapped board?

I know that most microwaves with LED lighting use "electronic transformers" to drive them. I'm almost positive that the output of these are not smooth, filtered DC voltage. I'm wondering if the output might be PWM with an asymmetrical duty cycle. I'm wondering if you installed the LEDs backwards (they're polarized) if they might be illuminating during the "off" portion of the pulse train. If you have a meter with a diode check function you can compare the orientation of the LEDs between the bright and dim boards.
posted by Larry David Syndrome at 10:31 AM on April 17, 2020

Response by poster: Yes, I replaced all 8 LEDs, four per board. One board has all 4 lit up bright, one board has all 4 lit up but dim.

I knew about the polarity issue and (I think) I installed them all the correct direction. It’s hard to make contact with my DMM, but I’m reading about 4.9V across each diode on the bright board, and 2.5V, 2.5V, 10.4V, and 0.0V across the four diodes on the dim board (but again they all light so I might not be able to get contact with the probes on the last one). I believe the power source is 12VDC. Would this be consistent with having one of the diodes on backwards?

The resistors have very small contact patches and I was having trouble keeping the probes connected long enough to measure their resistance, but I could keep trying if that could be relevant. I didn’t change anything with the resistors however.
posted by stopgap at 10:42 AM on April 17, 2020

Best answer: You could have one LED which is bad and is hogging the current. There are two resistors on the board, so I'm guessing that one of them is a shared voltage dropping resistor. If one of the LEDs is leaky or something, it could pull down the voltage for the rest of them.
posted by Larry David Syndrome at 10:51 AM on April 17, 2020

Response by poster: OK, I ordered a couple extra LEDs so I’ll try replacing the two that are giving odd readings and report back.
posted by stopgap at 10:57 AM on April 17, 2020

Response by poster: I reseated the diode that had been reading the very high voltage and now everything is bright. Unfortunately it fell off my tweezers so I couldn’t tell if it had been on backwards. My spare LEDs will remain on standby in case one of the LEDs fail in the future. Thanks all!
posted by stopgap at 11:21 AM on April 17, 2020 [3 favorites]

Best answer: assuming they're being driven with DC rather than PWM for intensity or something...

You can still measure the voltage even if it's PWMed; when you have, say, a 12V DC supply then not all multimeters will show 6V DC at 50% and 3V at 25%, but they'll usually show a value that's close enough to work with.

I'm wondering if you installed the LEDs backwards (they're polarized) if they might be illuminating during the "off" portion of the pulse train.

Sorry, PWMed DC is either full forward voltage or off. It won't be backwards at any moment.

There are two resistors on the board, so I'm guessing that one of them is a shared voltage dropping resistor.

On a board like this all the LEDs would be in series because a) it saves parts, even if that's just a few resistors at millicents each and b) LEDs in series bringing you closer to the supply voltage is the right thing to do. This is consistent with the ones on the dim board being more or less equally dim. With this particular board one of the resistors is the series resistor for the string of LEDs, the other appears to be a sense resistor of some sort connected to the third pin on the power connector.
posted by Stoneshop at 12:45 PM on April 17, 2020

Could you upload closeup photos of your finished board to give us all a look at the quality of the solder joints?
posted by flabdablet at 1:10 PM on April 17, 2020

Response by poster: It’s installed now so I don’t want to take it out to take photos, but the underside is flat aluminum — the connections between parts are all sandwiched internally. The top looks very much like in the video I linked: tiny amounts of solder holding the LEDs onto flat rectangular contact points on the surface of the board. When the LEDs are in place, they just appear to sit flat on the surface. It’s been a long time since I had done any soldering, so I’m guessing one of my solder connections wasn’t very well done the first time around.

I came across a schematic at some point a year ago when I was first considering this project, but I don’t remember how to find it. In any case, the circuit must be fairly simple. Would I be correct in thinking that I could replace the series resistor with one that has less resistance and make the boards even brighter still? I have a fixed 12VDC bench power supply I was testing with (an adjustable bench power supply was three times the cost), and my new LEDs had no trouble taking 12V each), whereas the 12V supply blew out one of the original LEDs after I had removed it (which I assume are specced to require something more like the 12V / 4 ~= 3V power level in this circuit).
posted by stopgap at 1:21 PM on April 17, 2020

Response by poster: Actually, based on my readings of 4.9V each, the total voltage is probably 20 or 24 VDC. The resistors don’t have color bands, just SMD labels 089 and 121, so I guess 8000MΩ and 120Ω.

I did end up taking a couple photos of the boards inside my microwave: one and two. You can see that the lowest LED in the first picture has too much solder on one of the connection points so it sits lifted up at a slight angle, but that was the board that worked properly the first time.
posted by stopgap at 1:55 PM on April 17, 2020

Best answer: OK. Judging from those, the issue is almost certain to be bad solder joints. You're right that the bottom right LED on the bottom board in the first photo is tilted, but so is the bottom right LED on the top board in the same photo. It looks to me like the solder hasn't wet the contacts on the LEDs properly.

To fix that, you're going to want to remove those LEDs again, and this time before you put fresh solder blobs on the board contacts you're going to want to remove as much of the existing solder as you can using an iron with a flat tip and goot wick.

The solder that was on the board straight from the production line is almost certainly a lead-free type, which has a higher melting point than ordinary hobbyist 60/40 tin/lead solder and is much more difficult to get a good joint with using hand tools. To get good results, you need to wick as much of the lead-free stuff off of there as you can.

Don't add huge blobs to the board contacts. You want just enough so that the finished joint will have wetted both the board pads and the ones on the bottoms of the LED packages; you don't want so much as to make a thick pillar of solder between them, just enough to make a continuous joint without voids.

After tinning the pads, add plenty of flux before placing the LEDs, then keep the heat on from underneath the board until you see them all kind of suck themselves down onto the board and centre themselves over its pads. That's their way of telling you that all their own pads have got hot enough to allow the solder at the top of the blobs to stay molten and that it has now wet the entire surface of the package pads and sucked the packages into place via surface tension. At that point you withdraw your hot air and let the board cool without vibrating it at all, lest you disrupt the solder's freezing process and end up making dry joints.

Have a look at several of Louis Rossman's YouTube videos on Apple logic board repair to get an idea of acceptable technique.

If you've got too much solder on the pads to begin with, then instead of surface tension sucking the packages into alignment it will try to pull the solder into a convex shape and shove the packages out askew. If that happens, you can either start over with less solder, or you can do what you'll occasionally see Louis getting away with and actually push the package down with a pair of tweezers, forcing the excess solder out of the side of the joint. That can be cleaned up after the joints have set, using a micro pencil and possibly a bit more wick. But it's better to get the blob amounts sized right from the get-go.
posted by flabdablet at 2:38 PM on April 17, 2020 [1 favorite]

Best answer: By the way, your "089" resistor is almost certainly "680", i.e. 68 ohms. An 8GΩ resistor would be pretty pointless, since that's more resistance than the air gap between the pads has likely got.
posted by flabdablet at 2:43 PM on April 17, 2020

Best answer: My guess is that your 089 resistor is being read upside down and is really 680, so 68 ohms, a typical value in the E12 series of resistor values. Also much more plausible since 8 gigohms is basically an open circuit and 68 ohms is very much in line with LED currents.
posted by How much is that froggie in the window at 2:47 PM on April 17, 2020

Response by poster: I did use wick to clean up the factory solder, but I used a fine-point tip on my soldering iron instead of a flat blade and I found it difficult to control where the new solder was being applied since it seemed to climb up the tip of the iron and harden there rather than be deposited on the board. Since it was so hard to get any solder applied at all and the goal was 8 little dots side by side, I left some of the blobs large-ish rather than keep trying over and over to get the perfect tiny dot. I used lots of flux and the LEDs basically aligned themseves once the solder melted in the hot air.

Right now both LED boards are working the way I want, but I wouldn’t be surprised if I have some of the diodes fail over time, especially since I had them over heat longer than I think would be ideal. And the three LEDs on the dim board (other than the one I re-set) all seem to have a tiny black dot in the center now, so I suspect they might be one their way to burning out.

Thanks for all the advice. If I have to revisit this project it should go better next time. I’ll probably leave the resistors alone for now. (I thought 8GΩ seemed way too big!)
posted by stopgap at 3:07 PM on April 17, 2020

Best answer: my new LEDs had no trouble taking 12V each

If that's truly the case, and your bench supply doesn't have an over-current limiter set at well under 100mA, then they might not be the type you think you have. The Digikey specs you linked make sense for a Plain Old LED, with a forward voltage of 2.75V at 60mA; but any LED that can survive having a typical 12V bench supply applied straight to its terminals would pretty much have to have an inbuilt current limiting resistor.

Assuming that all four LEDs on a board are connected in a series string, and that the design puts their operating point somewhere near 2.75V at 60mA, then if this thing is being run off a 12V power supply it would need a total series resistance of (12V - 2.75V × 4 = 1V) / 0.06A = 17Ω. But that's much lower than either of the resistors on your boards, and it would also make LED current very sensitive to minor variations in supply voltage.

Pushing 60mA through a 68Ω resistor will make it drop 4.1V; pushing 60mA through 120Ω drops 7.2V. If both were in series they'd need to drop 11.3V to draw 60mA. So it's plausible that the LED boards are built to run off 20-24V, with both resistors and all the LEDs in a simple series string. About half the applied power will be going to heating up resistors rather than lighting LEDs, so it's not very efficient, but compared to the kilowatt being sucked up by the magnetron it's small change.

it seemed to climb up the tip of the iron and harden there rather than be deposited on the board

Boards with SMD LEDs mounted directly on them are not so much a printed circuit board as a printed heatsink plate. Soldering onto that with a small iron is always going to be a pain, because molten solder always wants to move toward the place where the heat is coming from, not to where it's going to.

If I were doing what you're doing, I'd be getting the initial solder balls onto the pads by heating the board with the hot air gun until the well-fluxed pads themselves got hot enough to melt a little solder straight from the end of the reel; I wouldn't use the iron at all for that step. And I'd have pulled the solder through a folded cloth between my fingers several times before doing that, to get as much oxide off it as possible. Then more flux, place all the LEDs, and hot-air the board again to set them.
posted by flabdablet at 3:24 PM on April 17, 2020 [1 favorite]

Response by poster: Good advice — thanks again!
posted by stopgap at 3:35 PM on April 17, 2020

which I assume are specced to require something more like the 12V / 4 ~= 3V power level in this circuit

LEDs are generally not specified for a particular power supply voltage, but rather for a supply current. To a pretty good first approximation, a LED can be modelled as a perfect voltage source (i.e. a fixed voltage with zero ohms effective resistance) in series with a very low resistance (ones of ohms). The 2.75V that your DigiKey LEDs are specified to drop at the 60mA specified test current will be pretty close to what they'll want to drop at any operating current. Wire one to a supply with a voltage say ten percent lower than that and it won't light at all; wire it to a higher voltage with a similarly low effective series resistance and it will draw way way way more than its rated current and burn out.

Driving LEDs needs something closer to a constant-current supply, whose effective series resistance is much higher than the internal resistance of its LED load so that any attempt by the load to draw more current causes a compensating drop in the supply voltage. The series resistors on your LED boards are there to make a crude approximation of a constant-current supply out of the pretty-much-constant-voltage supply that the board as a whole will be connected to.

So I'm not surprised that you're seeing dark dots forming in the centres of LEDs across which you're measuring 4.9V when they're running bright. For a LED that is supposed to drop 2.75V when it's happy, that's evidence of a pretty brutal overdrive.
posted by flabdablet at 4:35 PM on April 17, 2020 [1 favorite]

Should be pretty easy to figure out the circuit. First figure out which of the input wires is ground with your volt meter with the power on, probably black. Then, using that ground for your meter, measure and write down the voltage of both terminals of every resistor and led and power pin. Any two terminals with the exact same voltage are connected together. That should give you the whole circuit. Then you can follow the voltage from the positive input wire as it drops through each device to the ground input wire.

From the voltage across the resistors you measured above, you can determine the current through each resistor and in turn, the current through the leds. If the current is less than 60 mA, you could feasibly increase brightness by tweaking the resistor values.

There are three wires going to the board. Does the board have a high and low brightness level?

I kind of doubt it is shoving 60 mA through a 120 ohm resistor. That looks like a standard 1206 which is rated for only 0.25 watts.
posted by JackFlash at 6:00 PM on April 17, 2020 [1 favorite]

Well spotted.

This Amazon review describes the process of replacing the LEDs on one of these boards. It suggests the same DigiKey parts for replacement LEDs as those linked from the OP, and it also suggests searching YouTube for hot air rework instructions, so I suspect that it might be where stopgap started from.

That review also says that "The LEDs are 12v so if you have a 12v power source, you can test them before installing them back under the microwave" which strikes me as wording that might well induce a person not super familiar with LEDs to try testing them individually with a 12v bench supply. But what this more plausibly means to me is that the board is designed to run off 12V, not that the individual LEDs are.

If that's the case, then the resistors on the LED board are probably in parallel, not in series, and physically positioned far from each other and from the LEDs for best heat dissipation.

The Amazon listing has a photo of a version 1 board, whose R1 is clearly a 62Ω resistor rather than the 68Ω one from the version 2 board in stopgap's photo. Perhaps the v1 boards were burning out too many LEDs?

If the board is indeed supposed to run off 12V and the LEDs are ending up dropping 2.5V each, then that would put 2V across the current limiting resistors. If those are in parallel, then R1 would be passing 2V / 68Ω = 29mA and R2 would be passing an additional 2V / 120Ω = 17mA, for a total LED string current of 46mA which sounds about right. Resistor power dissipation would be (2V)2 / 68Ω = 59mW for R1 and (2V)2 / 120Ω = 33mW for R2, well within spec.

If you wanted a bit more brightness you could either replace the 120Ω R2 with another 68Ω part same as R1, or solder a second 120Ω part straight over the top of it. That would get you at most an extra 17mA through the LEDs though probably a smidge less because their own voltage drops would rise a bit. They should end up quite close to the 60mA their spec sheet suggests they're fine with, which is still well under their maximum 80mA rating, and the extra 30mW of heat dissipation around R2 shouldn't cause any trouble either. The only issue would be that the board might not be big enough to keep the LEDs cool enough at the higher current level, leading to earlier LED failures.
posted by flabdablet at 3:54 AM on April 18, 2020

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