Why are my LEDs burning out?

[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

 

[marconi]

For Brouhaha: The LEDs have arrived. I have just connected 4 LEDs with a 500 Ohm resistor ( made from two 1kOhm 1/4 W resistors in parallel) across a 24V dc regulated power supply to do a 24 hour soak test. Current 20 to 30mA. Turned on at 1115am.

 

[DPG]

For Brouhaha: The LEDs have arrived. I have just connected 4 LEDs with a 500 Ohm resistor ( made from two 1kOhm 1/4 W resistors in parallel) across a 24V dc regulated power supply to do a 24 hour soak test. Current 20 to 30mA. Turned on at 1115am.

I would put a fiver on that being absolutely fine.

(And I'm tight!)

 

[ColinACJL]

I would put a fiver on that being absolutely fine.

(And I'm tight!)

A typical forward voltage for a red LED is 2 volts, so for 4 LEDs would be about 8 volts. So that would be say 16 volt across the 500 ohms of the ballast resistor, that means 32 mA forward current.

 

[marconi]

For Brouhaha:


TruOhm CR-50 560R 0.5W Carbon Film Resistor - Pack of 100 - https://www.rapidonline.com/truohm-cr-50-560r-0-5w-carbon-film-resistor-pack-of-100-62-0550

The forward voltage drop Vf I measured for each white LED were: 2.85V. 2.99V, 2.89V and 2.87V (or about 3V). These were measured after 22 hours being on. See attachment.

At link are the details of the required series voltage dropper resistor R which one calculates as follows:

PSU voltage - (number of series LEDs x Vf) =

24- (4 x 3) = 12V

R = 12/Led current

R = 12/0.02 = 600 Ohms.

Nearest value is 560 Ohms.

Power dissipation is (24 - 12) squared/560 = 144/560 = 0.26W Next higher power rating is 0.5W.

LED current is 12/560 = 0.021mA

You could use a much cheaper single 560 Ohms 0.5 Watt resistor instead of the Dynaohm to run 4 white LEDs using a 24V power supply. Other colour LEDs have different nominal forward voltages Vf.

See:
How to calculate the series resistor for an LED – Stompville - https://stompville.co.uk/?p=37


I will still do the test on the dynaohm + LED circuit when it arrives from the USA.

 

[DPG]

For Brouhaha:


TruOhm CR-50 560R 0.5W Carbon Film Resistor - Pack of 100 - https://www.rapidonline.com/truohm-cr-50-560r-0-5w-carbon-film-resistor-pack-of-100-62-0550

The forward voltage drop Vf I measured for each white LED were: 2.85V. 2.99V, 2.89V and 2.87V (or about 3V). These were measured after 22 hours being on. See attachment.

At link are the details of the required series voltage dropper resistor R which one calculates as follows:

PSU voltage - (number of series LEDs x Vf) =

24- (4 x 3) = 12V

R = 12/Led current

R = 12/0.02 = 600 Ohms.

Nearest value is 560 Ohms.

Power dissipation is (24 - 12) squared/560 = 144/560 = 0.26W Next higher power rating is 0.5W.

LED current is 12/560 = 0.021mA

You could use a much cheaper single 560 Ohms 0.5 Watt resistor instead of the Dynaohm to run 4 white LEDs using a 24V power supply. Other colour LEDs have different nominal forward voltages Vf.

See:
How to calculate the series resistor for an LED – Stompville - https://stompville.co.uk/?p=37


I will still do the test on the dynaohm + LED circuit when it arrives from the USA.

The traditional methods are often the best.