LED panel repair (row of lights out)

I know that 1 led must have went out but I can't tell which one. I have verified that it is not the driver but a link in the chain so does anyone know the best route to figure out which led went out? I can't visually tell and was thinking I would have to use a voltage meter, anyone have to deal with this before? Should I just bypass the bad led if I find it because the leds are under driven 3 watters anyway so one less wouldn't do much of an increase on the remaining row.

Thought?

99% of the time when an LED burns out it still allows current through, meaning only 1 LED will be out. When a string of LEDs goes out that usually means a loose solder connection somewhere in the string. Easiest way to check is just use the continuity setting on the multimeter, the one that makes noise, and touch the + on one LED and - on the one next to it. Once you find the one that is open a quick touch up with a soldering iron/flux will fix it, or if it is trace mounted (SMD type LEDs) you can use a wire to connect them in the trace is bad on the board.

if its the other 1% you can use the diode test function of a multimeter which lits the led very dimly. I already had this occuring on the first test of my DIY fixture...

Can you take a picture of the LED mounting board, preferably the back of it?

If it's in fact just one LED that's the issue, you can simply short that LED (put a wire between the cathode and anode) - if your driver is a constant current driver, which I would think it is.

But shoot some pictures, and I'll be happy to help.

It's a Blackstar 240 (120 watts actual). The back will be a complete bitch to deal with because there is a sheet of aluminum covering the entire board screwed by 1 million screws. I peeled the reflective material from the from of the diode array and it looks like the leads are inside the silicon circuit board and they look beefy so I am assuming its a diode or connection to diode that's the culprate.

I disconnected the driver to the unlit row and may just deal with it till I can free the panel up in a few weeks. Looks like it will be a time investment but worth it not to loose 50% of my output.

Here are a few pics from someones panel, same kind.




According to another pic of whats under that heatsink I won't have anymore access to leads or diodes then I would from the front, leads and diodes are baked into the board.

If you can get a picture of the board a bit closer, it would help a lot. Also snap a picture of one of the drivers.

Oh God those screws! I HATE those little effin screws. LOL

Multimeter is your best bet here. Flip the board over and start testing. You may get lucky and see something wrong, like a black dot on the LED or some kind of heat damage, but I've been where you are and a multimeter will tell you the truth.Then you can do the soldering thing if it's a few. At some point you'll need to add a resistor or you'll burn out more. If you can figure out the diodes resistance you really should solder the right resistor in instead. But there are others way more knowledgable here.

You could call Gotham Hydro and see if there is something they can do. A 240 with half the diodes out is not a great situation to be in if you're flowering. You could add some CREE bulbs CFLs though. Good Luck Black Mesa!

Why do you need a resistor?

lax123 said:

Why do you need a resistor?

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My guess is that since I'm bypassing 1 or more led's it will up the voltage on the remaining diodes thus for making them run a little hotter.

I'm not exactly worried about the drop in 60 watts because I'm using 2 65 watt cfls & a 7 watt led grow bulb (7 - 1 watt diodes ran at 1 watt a piece) and I just ordered 2 - 21 watt led grow bulbs that are composed of 3 watt diodes $20 a piece KILLER DEAL! I will fix this panel in a few weeks and will use a resistor if more then 1 diode is bad but I don't think 1.5 watts spread over that row will cause that much of a problem.

Also should mention that I took a scope and looked at each diode and all looked good so maybe this is just a simple seating or connection problem.

My guess is that since I'm bypassing 1 or more led's it will up the voltage on the remaining diodes thus for making them run a little hotter.

Click to expand...

When Power= Voltage * Current I and I is fixed because its a constant current source, what will the constant current source be doing?

Resistors are used for parallel strings in case 1 string goes out all of the current doesn't rush to the remaining string(s) and blow the LEDs. I don't see any reason to use a resistor on a series string.

FranJan said:

Oh God those screws! I HATE those little effin screws. LOL

Multimeter is your best bet here. Flip the board over and start testing. You may get lucky and see something wrong, like a black dot on the LED or some kind of heat damage, but I've been where you are and a multimeter will tell you the truth.Then you can do the soldering thing if it's a few. At some point you'll need to add a resistor or you'll burn out more. If you can figure out the diodes resistance you really should solder the right resistor in instead. But there are others way more knowledgable here.

You could call Gotham Hydro and see if there is something they can do. A 240 with half the diodes out is not a great situation to be in if you're flowering. You could add some CREE bulbs CFLs though. Good Luck Black Mesa!

Click to expand...

lax123 said:

Why do you need a resistor?

Click to expand...

From what I found, the lamp is driven by two drivers each capable of up to 96V at 630mA constant current. Which means that you don't need a resistor, as the driver will lower the voltage itself until 630mA flows.

BlackMesa said:

My guess is that since I'm bypassing 1 or more led's it will up the voltage on the remaining diodes thus for making them run a little hotter.

I'm not exactly worried about the drop in 60 watts because I'm using 2 65 watt cfls & a 7 watt led grow bulb (7 - 1 watt diodes ran at 1 watt a piece) and I just ordered 2 - 21 watt led grow bulbs that are composed of 3 watt diodes $20 a piece KILLER DEAL! I will fix this panel in a few weeks and will use a resistor if more then 1 diode is bad but I don't think 1.5 watts spread over that row will cause that much of a problem.

Also should mention that I took a scope and looked at each diode and all looked good so maybe this is just a simple seating or connection problem.

Click to expand...

Looking at how that lamp is build, I would be scared of using it.. Horrible heat management, standard FR4 board for the LEDs and the cheapest LEDs availble on the market.

Resistors are used for parallel strings in case 1 string goes out all of the current doesn't rush to the remaining string(s) and blow the LEDs

Click to expand...

R u sure? I think using a parallel string at all without proper circuitry will result in a blowout:
Every led is a little different then another, this will result in different combined forward voltages for every string. Therefore when splitting the current into two strings one string will get more current then another, that means it gets warmer then the other string, this decreases its forward voltage ,so even more current flows through, the other string will cool (because it gets less of the current- share) and its forward voltage increases and so on...until one string.. KABOOOOM
So I think using a resistor there would not solve anything. Or does it in another way?

CaliJoe said:

Resistors are used for parallel strings in case 1 string goes out all of the current doesn't rush to the remaining string(s) and blow the LEDs. I don't see any reason to use a resistor on a series string.

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Pretty sure this is correct.

The drivers they use would not be true constant current and their control wouldn't be that great, allowing lots of excess voltage to flow before they get it back down.

CaliJoe said:

Resistors are used for parallel strings in case 1 string goes out all of the current doesn't rush to the remaining string(s) and blow the LEDs. I don't see any reason to use a resistor on a series string.

Click to expand...

Doesn't work. Let's say you run two string, each getting 700mA. If the LEDs are driven by a constant current source and one string fails - the LED that failed cannot carry any current anymore (this is the most likely scenario. Very few short circuit). The remaining string will now have 1400 mA running through it. To save it from this you would have a resistor in parallel, not series. But this would also mean that you in normal operation would have much less than 700mA in the strings.

If you split a constant current source onto two or more strings, and one fails, you cannot save the remaining with a resistor.

PICOGRAV said:

Pretty sure this is correct.

The drivers they use would not be true constant current and their control wouldn't be that great, allowing lots of excess voltage to flow before they get it back down.

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Sorry, but it's incorrect. You use resistors when you are driving two or more strings in parallel. This is due to the unlinearity of LEDs.
When the temperature of the LED increases, the forward voltage of the LED decreases (a few mV per ºC). If you supply the LED with a constant voltage, the current will increase when the temperature increase.

The resistor you placed in series with the each string, will drop more voltage as the current increase, thereby counter-acting the behavior of the LED.

If the resistor is not in place in a circuit where the LEDs are supplied with constant voltage or constant current with multiple parallel strings, the current will increase in the worst string. This will increase the current in the string, increase the temperature of the LEDs, which again increases the temperature.. and now you have a thermal runaway.


All of the above doesn't matter if you have a constant current driver and a single string. The current will insure that the voltage is lowered when the voltage of the LED drops, as it's get warmer.

JMD is 100% correct. I was thinking fuses to save the other strings, which you also use with resistors when doing parallel strings. Don't know where my brain was when I wrote that (actually I do, at work).

CaliJoe said:

JMD is 100% correct. I was thinking fuses to save the other strings, which you also use with resistors when doing parallel strings. Don't know where my brain was when I wrote that (actually I do, at work).

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All is forgiven Good idea with the fuses.

You two should go to bed together I think