LED Protection Diode Identification - Page 1
Forum section: electronics / repair

author: Bryan
author_rank: Frequent Contributor
timestamp: June 30, 2019, 09:13:01 am
content: Hello:Have one of those 2 channel black box LED lights for a aquarium. One channel is dead and I have narrowed it to a blown diode and a led. Problem is I have no idea what to replace the diode with. Are these protection diodes typically Zeners, Schottky or just regular diodes. is there a way of determining what type from one of the good diodes. No markings on the diode except for a blue band. Any help much appreciated.

author: sleemanj
author_rank: Super Contributor
timestamp: June 30, 2019, 09:23:27 am
content: Is the diode directly across the LED so when the LED croaks the diode will maintain the current flow for the remaining LEDs? If so, then it must be a zener of a voltage a bit higher than the forward voltage of the LED.

author: Bryan
author_rank: Frequent Contributor
timestamp: June 30, 2019, 09:49:08 am
content: All the LEDS are in series and the diode(s) are in parallel with each LED. I just swapped out the LED and used a regular 1n4001 diode and the channel worked, but is this the right approach or should I be using a zener?. I suppose the best way to test the diode would be to rig up a octopus curve tracer and check on a scope.

author: digsys
author_rank: Unknown
timestamp: June 30, 2019, 10:15:39 am
content: It has to be a zener - measure the voltage drop across the LED when it is fully on - assume it is 2.4V, then select the next standard zener voltage up - 3.3VThere are only 2 possible failure modes for a LED - 1/ Short cct - in which case, who cares - it won't break the series chain 2/Open cct - which would break the circuit.That's what the zener is for, it will conduct in place of the LED. No need for any fancy equipment :-)

author: keymaster
author_rank: Regular Contributor
timestamp: June 30, 2019, 11:31:33 am
content: if it is a zener and the led fails , then the zener will have to be similar or bigger in strength. The led seems to be 1 watt or more and the diode no more than half a watt. In case of failure of the led ,the small zener will follow pretty fast.So i guess, the diode is a normal diode for protection of reverse voltages.

author: coromonadalix
author_rank: Super Contributor
timestamp: June 30, 2019, 02:31:43 pm
content: hum dont think its signal diodes like 1n914, in4118 ??

author: Bryan
author_rank: Frequent Contributor
timestamp: June 30, 2019, 08:59:45 pm
content: Thanks everyone. I have seen some reviews of these far east black box LEDS, and the quality and safety is somewhat er.. concerning. The diodes are SMD and from the package size I doubt very much they are rated for 3 watts. I would assume the wattage rating if zener should match the LED which are 3 watts, otherwise what's the point. Perhaps explains why the diode failed as well. Couldn't handle the 3 watts when the LED failed?. The voltage drop on the leds is 3.2 volts A 3.6v zener ??Can't see the diodes being for reverse protection as the user does not have any means of reversing the voltage.They certainly look like tiny signal diodes!!

author: digsys
author_rank: Unknown
timestamp: June 30, 2019, 11:21:25 pm
content: EXXXACERRY !! :-) Using a 3.6V Zener on a 3.2V Led is about right. Each will have a say 5% tolerance, so in a worst case, the zener may just leak a tiny bit, but that's fine. What youhaveto watch out for is Supply voltageheadroom!!The LED string is either driven by a fixed Voltage + current set resistor OR constant current source. Either way, there is a Vmax at which point NO LEDs light up !!Each time a Zener kicks in, it raises the required string voltage 0.3 - 0.XV ie You can't just use a much higher zener to avoid any leakage. What you need is a zener crowbar :-) so it collapses once triggered .. but then you can get accidental latchup etc etcI used to LED signs long ago, and there's a lot to consider

author: Bryan
author_rank: Frequent Contributor
timestamp: July 01, 2019, 10:04:32 am
content: There is 55 Leds on the board with two channels. One channel has a voltage of 92 volts and the second channel is 90 volts. I assume one channel drives 28 Leds and the second 27 Leds.Would not the zener also have to have a 3 watt rating the same as the LED?. No way these tiny SMD diodes have a 3 watt rating. On the channel that blew one LED was shorted along with the diode and a number of other LEDS were dark and look like they overheated or were overdriven.

author: digsys
author_rank: Unknown
timestamp: July 01, 2019, 10:19:41 am
content: Quote from: BryanWell, if it is a single string - 92V / 55 Leds = ~1.7V per Led. That doesn't look right for a high powered Led, I'd say max 50mA = app 90mW, which should be fine.IF there are 2 strings (1/2 ea), which is common, then that's ~3.4V per Led, which is more likely a higher powered type. Then yes, it needs the same power rating.Quote from: BryanNow we have a quandary :-) What I didn't mention is - IF a Led (or diode)doesgo short cct, AND it is NOT a Constant current type drive (ie fixed voltage), then the string current will go up by ~ 1/55x for first instance, 1/54x the second etc, rising faster each time. BUT, for only 1 shorted, I can't see the current increase being enough to cause what you see !!?Something is fishy here :-)

author: Bryan
author_rank: Frequent Contributor
timestamp: July 01, 2019, 10:54:31 am
content: Hello:Sorry, it is two strings driven by two drivers so 3 watts is correct.I think the problem is two fold, part my fault and a poor design. About a year or so I replaced some of the LEDS with different spectrum LEDS, I believe they probably have different forward voltages. this probably taxed the circuit causing the LEDS to slowly degrade. I suspect one of the LEDS finally failed shorted and the zener diode which I suspect is probably 1/2 watt did not last very long and shorted as well. Would this then not increase the current and voltage on the remaining LEDS causing the driver to eventually shut down.What I don't understand is if the LED failed shorted the zener would not work?, so then how did the zener fail?

author: digsys
author_rank: Unknown
timestamp: July 01, 2019, 11:48:32 am
content: Quote from: BryanLike I said, something is fishy here :-) .. IF the Led failed O/C first, which ismorelikely, then the Zener would overload and likely go short cct. If you test it, it should show O/C. IF it did failshort circuitfirst, it CAN'T take out the Zener. Yep, odd. There's more to the story :-)Quote from: BryanAgain, it's a stretch. It only adds 3V ish to the string, not a big jump in current. Even if it was already over-driven a bit. We'll call it Plausible :-)

author: sleemanj
author_rank: Super Contributor
timestamp: July 01, 2019, 12:08:35 pm
content: I would guess that the zener didn't last long dissipating 3+ watts and failed open circuit, hence the string finally died.The way I see it there are two possibilities for that small zener1. it is not intended to last long, long enough you can see which LED has died maybe2. the driver is smart enough to see that the voltage on the string is higher than expected (because zener voltage higher than good led voltage) and drastically limits the current to the string, again so you can see which led died/provide "emergency" lighting capabilityOption 1 is more likely, that small zener simply can't dump 3W for long and live to tell the tale. I'd say it died open circuit.

author: shakalnokturn
author_rank: Super Contributor
timestamp: July 01, 2019, 12:14:21 pm
content: Admittedly I'm lacking information to understand all this.A good starting point would be to know if the source is CC or CV and measure current on the string that has had the least modifications.A crowbar protection usually has an active voltage drop of 1V this would bring it's power dissipation down to about 1/3 of the LED's.But I see no SOD-80 crowbar protections available.A zener protection seems odd, unless the LED power is close to the Zener power, relying on a zener to overheat and go short just seems like a fire risk.When tested with a 1N4001 what polarity was used compared to the original diode?Could the original diodes just be reverse polarity protection? (Odd and maybe lame source design...)If source is CV could the zeners be there to kick in close enough to LED Vf and prevent thermal runaway in the LED's?You should really remove one of the working diodes and test as a zener to find out what it is.

author: Bryan
author_rank: Frequent Contributor
timestamp: July 02, 2019, 06:23:43 pm
content: The drivers have no specs or markings on them. I suspect CC as one driver supports 27 Leds and the other 28, so the voltages outputted by the driver are different. I may be wrong, but the size of the diodes are like a grain of rice, no way they could dissipate 3 watts. Yes, just to satisfy my curiosity will pull off a good diode and see what it shows on the scope with a octopus tester.