Why are my LEDs burning out?

[Brouhaha]

I think this is incorrect. The Dynaohm positive should got to battery positive according to the data sheet.

OH!! That would explain a lot! I am confused however, because I have similarly been wiring the positive power wire to the negative terminal of my LEDs and they do light up when I connect them that way (and not the opposite way). So is there a different convention regarding how the terminals on a resistor are labeled versus those on a bulb?

 

[Avo Mk8]

Hi Avo, I have the positive wire coming off of my power supply going to the negative terminal of the Dynaohm--is that correct?

As DPG replied earlier, no I'm afraid not!

I did color code the wires to the bulbs.

Brilliant - that means you can check (with multimeter) if any are at least still diodes, rather than open or short circuit!

I do have a cheap, old analog multimeter, but I've wondered about it's reliability, so maybe I should get a new one...

nothing wrong with old analogue multimeters. mine's 50+ years old and still good enough to use. If yours does resistance measurements, great! (Might need a new battery?)

 

[Avo Mk8]

I have similarly been wiring the positive power wire to the negative terminal of my LEDs and they do light up when I connect them that way (and not the opposite way).

You are connecting these the correct way.

So is there a different convention regarding how the terminals on a resistor are labeled versus those on a bulb?

The Dynaohm is NOT A RESISTOR . Sorry, not your fault, it's theirs for marketing it like they do, and everyone calling it a resistor! With an ordinary resistor it doesn't matter which way round it's connected, it does the same thing either way.
Whereas the Dynaohm must have some semiconductor devices in it, to alter the voltage across it so that the current through it stays the same.

We should assume the Dynaohm could be damaged, though I think it does have some overload features - I don't know if they save it from reverse connection. If not too costly I would suggest getting another one or two, and putting yours aside for the moment. UPDATE: just looked at the data sheet which says maximum reverse voltage 1V so I suspect it's broke!

 

[DPG]

I'd take Avo's advice. At the minute you have a Dynaohm of unknown condition, and LEDs which may have been damaged. Get a new Dynaohm, and some known good LEDs. Connect them up correctly, maybe in a string of 4, and test them.

Use your multimeter to measure LED current if it will do DC amps. Monitor it at switch on and be ready to switch off if the current is too high.

 

[Brouhaha]

Thank you Avo and DPG for your very helpful replies! It will probably be a few weeks before I can get the replacement Dynaohm and have an opportunity to try testing/rebuilding, but I will come back here either with a success story--or more questions! Thanks so much!

 

[Avo Mk8]

Thank you Avo and DPG for your very helpful replies! It will probably be a few weeks before I can get the replacement Dynaohm and have an opportunity to try testing/rebuilding, but I will come back here either with a success story--or more questions! Thanks so much!

DPG gives good suggestions in his last post.

Just to reiterate, don't wire it up as your original drawing back in post #1. You need 20mA through each string of lamps, so however many you end up with in series, four or six, have a Dynaohm for each string. Perhaps they'll give you a loyalty discount

Best wishes for getting it all working!

 

[Brouhaha]

DPG gives good suggestions in his last post.

Just to reiterate, don't wire it up as your original drawing back in post #1. You need 20mA through each string of lamps, so however many you end up with in series, four or six, have a Dynaohm for each string. Perhaps they'll give you a loyalty discount

Best wishes for getting it all working!

ok, I got new 20mA dynaohms, new LEDs and started over; I made three parallel strings of 4 LEDs and tested each string using a couple of 9v batteries (and a Dynaohm, although it's possible that I might have left out the Dynaohm at some point while testing...) maybe that caused my subsequent problem?). I then wired everything up and tried powering it with my Meanwell power source and found that one of the bulbs in one of the strings is not lighting up--but the other three bulbs in that string are lighting up. I don't even understand how that is possible, I thought if one bulb was bad the whole string would go dark. Any idea how that could happen? And any ideas why the strings were working before I wired up to power source but not afterward?

 

[Avo Mk8]

ok, I got new 20mA dynaohms, new LEDs and started over; I made three parallel strings of 4 LEDs and tested each string using a couple of 9v batteries (and a Dynaohm, although it's possible that I might have left out the Dynaohm at some point while testing...) maybe that caused my subsequent problem?). I then wired everything up and tried powering it with my Meanwell power source and found that one of the bulbs in one of the strings is not lighting up--but the other three bulbs in that string are lighting up. I don't even understand how that is possible, I thought if one bulb was bad the whole string would go dark. Any idea how that could happen? And any ideas why the strings were working before I wired up to power source but not afterward?

Sorry you still have problems.
regarding the LED not lighting up, if a diode fails 'short circuit', the others will carry on working, and the Dynaohm will adjust the current to still be correct for the others.
It could be the diode legs, or the wires to that diode, are shorted together.
It could be that you've inadvertently wired it the wrong way round, and as a result its gone short circuit (but in that case it would never have lit up when you first tried it).
It could be that the diode has just failed short circuit.

Suspect you need to replace that one diode unfortunately, if that is practicable.

 

[Brouhaha]

Sorry you still have problems.
regarding the LED not lighting up, if a diode fails 'short circuit', the others will carry on working, and the Dynaohm will adjust the current to still be correct for the others.
It could be the diode legs, or the wires to that diode, are shorted together.
It could be that you've inadvertently wired it the wrong way round, and as a result its gone short circuit (but in that case it would never have lit up when you first tried it).
It could be that the diode has just failed short circuit.

Suspect you need to replace that one diode unfortunately, if that is practicable.

ah--I hadn't realized that current could still pass through a dead bulb. I will replace that one (again!) and hopefully eventually get back here to report success!

 

[Lucien Nunes]

Yes unlike a traditional filament lamp which fails open-circuit like a fuse, an LED typically fails short-circuit especially if due to overcurrent, as the diode semiconducting junction breaks down and becomes fully conductive. Unless it has been blasted with such a high current that the bond wires have ruptured.

 

[ColinACJL]

Yes unlike a traditional filament lamp which fails open-circuit like a fuse, an LED typically fails short-circuit especially if due to overcurrent, as the diode semiconducting junction breaks down and becomes fully conductive. Unless it has been blasted with such a high current that the bond wires have ruptured.

Each string must have its own constant current device.

 

[DPG]

Each string must have its own constant current device.

I think that's what the OP has done now. He refers to more than one Dynaohm in post #27.

 

[ColinACJL]

Each string of LEDs must have its own constant current device to prevent current hogging. LEDs have a negative temperature coefficient of voltage, so the hotter they get the more current they will draw with shared constant current device.

 

[Avo Mk8]

Each string must have its own constant current device.

Sort of covered that issue in post #10 onwards!

 

[DPG]

Each string of LEDs must have its own constant current device to prevent current hogging. LEDs have a negative temperature coefficient of voltage, so the hotter they get the more current they will draw with shared constant current device.

Yeah we've covered that I think.

 

[Brouhaha]

Thank you all for the advice and for the explanations!

 

[marconi]

I have ordered some Dynaohms from a US supplier and similar leds from a UK one. I suspect that the assumption that the dynaohm regulates the current through it to 20mA whatever the potential difference across it is false. I would be surprised if such a simple device was a perfect constant current regulator.

Regarding the LEDs the current/Vforward characteristic will be similar to the plot I have drawn as A in the attachment. Above the minimum space charge voltage (about 2V) the current through it rises rapidly with any change in potential difference across the LED. This is the reason the LED current must be regulated to prevent the LED burning out due to escalating Ohmic heating. A typical LED has a forward resistance once conducting of about 20 Ohms (the slope of the upward rising part of the plot dV/dI).

To make matters worse, the forward voltage for a given forward current reduces with LED junction temperature which means as it gets hotter the forward current will increase which in turn causes more heating which in turn results a lower forward voltage and thence higher forward current - thermal runaway. This is the mechanism which leads to burnout.

If the forward voltage reduces with temperature, and the circuit is supplied by a low impedance regulated voltage source such as the Meanwell PSU, the stabilisation of LED current requires a series device which has a very small increase in current when the voltage across it increases and over quite a range of dynaohm voltage drop. Ideally, the dV/dI is such that the current decreases a little each time the the voltage across it increases a little ie: it slopes downwards.

My suspicion is that the dynaohm's voltage current characteristic slopes upwards and that there is a large increase in current for each increase in voltage across it as I have attempted to show in plot C.

The combination of the dynaohm and the LED string does not perform to regulate the current to a 'safe' LED current which is leading to LEDs burning out. To make matters worse, it is not the reduction in forward voltage of one LED - it is a reduction of 3 to 4 times the voltage of on LED since they are in series. This exacerbates the increase in current through the dynaohm as the potential difference across it rises not by one LED's reduction but by the sum of all the reductions of how many LEDs are wired in series.

Anyway a theory at the moment on why your LEDs are failing. When the dynaohm and LEDs arrive I will does some tests and let you know what I find.

 

[Avo Mk8]

Thank you Marconi for your lucid narrative.
Being a commercial product claiming to do the job (!), I had naively assumed it's performance would be adequate for the job in hand, and concentrated on why the OP was experiencing problems.
There were a number of unfortunate incidents during his experiments, incluging connecting diode strings directly across the Meanwell psu, putting 3 diode strings in parallel through one Dynaohm, and having the Dynaohm the wrong way round!
I'm hoping your experiments show that the Dynaohm product does in fact work adequately to prevent a 'runaway' situation.
The data sheet does say it reduces current as it gets hotter (!), but doesn't give any graphic detail on its actual performance:

https://www.ledsupply.com/content/pdf/led-driver-luxdrive-dynaohm_documentation.pdf

Thank you for your interest

 

[Brouhaha]

I have ordered some Dynaohms from a US supplier and similar leds from a UK one. I suspect that the assumption that the dynaohm regulates the current through it to 20mA whatever the potential difference across it is false. I would be surprised if such a simple device was a perfect constant current regulator.

Regarding the LEDs the current/Vforward characteristic will be similar to the plot I have drawn as A in the attachment. Above the minimum space charge voltage (about 2V) the current through it rises rapidly with any change in potential difference across the LED. This is the reason the LED current must be regulated to prevent the LED burning out due to escalating Ohmic heating. A typical LED has a forward resistance once conducting of about 20 Ohms (the slope of the upward rising part of the plot dV/dI).

To make matters worse, the forward voltage for a given forward current reduces with LED junction temperature which means as it gets hotter the forward current will increase which in turn causes more heating which in turn results a lower forward voltage and thence higher forward current - thermal runaway. This is the mechanism which leads to burnout.

If the forward voltage reduces with temperature, and the circuit is supplied by a low impedance regulated voltage source such as the Meanwell PSU, the stabilisation of LED current requires a series device which has a very small increase in current when the voltage across it increases and over quite a range of dynaohm voltage drop. Ideally, the dV/dI is such that the current decreases a little each time the the voltage across it increases a little ie: it slopes downwards.

My suspicion is that the dynaohm's voltage current characteristic slopes upwards and that there is a large increase in current for each increase in voltage across it as I have attempted to show in plot C.

The combination of the dynaohm and the LED string does not perform to regulate the current to a 'safe' LED current which is leading to LEDs burning out. To make matters worse, it is not the reduction in forward voltage of one LED - it is a reduction of 3 to 4 times the voltage of on LED since they are in series. This exacerbates the increase in current through the dynaohm as the potential difference across it rises not by one LED's reduction but by the sum of all the reductions of how many LEDs are wired in series.

Anyway a theory at the moment on why your LEDs are failing. When the dynaohm and LEDs arrive I will does some tests and let you know what I find.

Thanks for taking an interest in my little problem! As Avo MK8 mentioned, I made a number of bone-headed errors--so it will probably come as no surprise that your post goes way beyond my comprehension of electronics! But I will be curious to see the results of your experiments. My equipment came from a US company called LEDSupply.com. I have had a good impression of them so I am hoping it was just my own errors that caused the problems...

 

[marconi]

For clarity this is how you should connect up your circuit - see attachment.

Are you insulating the bare soldered connections and leads to the components? Heat shrink tubing is a good way of doing this. You can buy it from many electronic component suppliers and on Amazon. Just gently hold the barrel of your soldering iron against the tube and slide up and down the length of the tube until it contracts. Have a few practices to develop the technique. 3mm should work well if you are using thin connecting wire. See:

315-13002 TA37 6-2-PO-X-BK | HellermannTyton Adhesive Lined Heat Shrink Tubing, Black 6mm Sleeve Dia. x 1.2m Length 3:1 Ratio, TA37 Series | RS Components - https://uk.rs-online.com/web/p/heat-shrink-tubing/8296079?cm_mmc=UK-PLA-DS3A-_-google-_-CSS_UK_EN_Cables_%26_Wires_Whoop-_-Heat+Shrink+Tubing_Whoop-_-8296079&matchtype=&pla-524231416997&gclid=Cj0KCQiAu62QBhC7ARIsALXijXSs9qVR5NI9ztEVmarqlds-0N_2t69wFRx7WIAyFduRNwxJx5kN0o4aAtl1EALw_wcB&gclsrc=aw.ds