3
\$\begingroup\$

Short version: I'm thinking about running LEDs in a potential over-current setup, but with a PWM-based supply. Will the LEDs suffer an early death? Said otherwise, do LED die of overheating, or really because of the over-current?

Long version: My setup consists of the following parts, all coming from various low-cost Internet shops - so no datasheet or reliable spec is available for any.

  1. One LED strip. Composed of 12 modules in parallel. Each module contains 4 components in series: one 39ohm resistor and 3 5630 white LEDs. My understanding is that it's supposed to be fed by a 12V DC supply. When doing so (with a bench power supply), each LED operating point is around 3.3V/52mA, for a total of 7.6W for the whole strip (when the resistors are considered).
  2. One LED PWM driver. A small $1 module that can generate blink patterns or dims the LEDs via an 1.5kHz PWM (confirmed with an oscilloscope).
  3. One crappy wall wart, specified at 12V, 1.5A, but which outputs ~17V when there's no load, and around 13.25V when feeding the LED strip, operating each LED around 3.6V/60mA.

I've failed to locate any datasheet for a 5630 LED that has that kind of operating point. In the 12V setup, each LED dissipates ~175mW, which seems an odd value. Assuming that they're really 1/4W LEDs (even if most 5630 appear to be 1/2W), the 3.6V/60mA operation point would still be under that, albeit marginally.

Back to the question: I'm trying to speculate how long these would live if I was to operate them at, say, 80% duty cycle - which would probably end up being at a voltage/current even a bit higher than the numbers above. Can anyone provide some background of the physics of LED wear?

Obviously, adding an ~1 ohm, 1 watt resistor in series would solve everything, but where would be the fun in doing that...

\$\endgroup\$
  • 2
    \$\begingroup\$ When you "overcurrent" a LED, you also "overvoltage" it, because of LED-specific I-V load curve. As result, you will "overpower" the LED, which will cause "overheating". The LED will die sooner due to combination of all three factors. Regarding the Physics behind, you need to ask this question on Physics forum. \$\endgroup\$ – Ale..chenski Aug 19 '17 at 4:04
1
\$\begingroup\$

LEDs, as for all semiconductor devices, die through more than just one mechanism.

The most obvious is over temperature. Get it hot enough, and things diffuse around to destroy the structures that the manufacturer carefully diffused into it in the first place. With pulsing, the situation is complicated, as the (generally small) dissipating region temperature is varying with the applied power, while the bulk of the device is at a more or less constant temperature. This time-dependent behaviour is captured in dissipation versus pulse-length graphs, when you have data.

Things that are not as well heat-sunk, like bond wires, generally have a peak current specificiation.

DC connections can fail from long term overcurrent through electromigration, where at a high enough current density, atoms actually move around and further thin already thin connections, this is a runaway situation.

Insulators can degrade, and this tends to happen faster at higher temperature.

The literature seems to suggest that keeping the LEDs cool prolongs their life. Much effort goes into packages that are possible to heat-sink.

If you have the time, and components, to burn, then you could do your own accelerated ageing tests under your own conditions. You would expect to find lifetime exponential in the temperature, where a single wear-out mechanism is operating. Where several mechanisms kick in at different temperatures, the situation is more difficult to extrapolate from. Of course, the region we are most interested in, the one where 1000h becomes 100000h, is the one that takes longest to evaluate.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.