I take that as high praise, coming from you. Thanks.
So without (hopefully) overthinking this, when we talk about LED current, and especially LED overcurrent, what are we really talking about? What I mean by this is that there's the instantaneous current, as shown in my simulation above, which is clearly well above the rated current for these devices. But since it's a pulsed current, certainly there's a certain amount of de-rating that should be done.
So I'm wondering what the best way is to think about this current. One thing that occurs to me, since I have at least a passing familiarity with calculus, is to take the integral of this current (that is, the volume of those spikes above). Is this a valid measure to work with? or even a useful one?
And we haven't even really touched on the nature of those spikes, with their sharp rise times, and what potential effects that might have. Even if that very tip of the spike only lasts for nanoseconds, it's still up there in the stratosphere so far as instantaneous current value goes (some of them go over 400 mA).
Thanks again for your informative reply.
Hi again Carl and Carbon,
Carl:
Yes there is no stress in the transistor when it is off, but later when it turns on it has to deal with the energy in the cap trying to short it out. Do a simulation and look at the current in the transistor collector. Then, put a Schottky diode in series with the LED before the cap and do another simulation and see how much better it is.
Carbon:
Yes high praise because asking good questions eventually leads to good answers which leads to better understanding
The bottom line is that if the manufacturer says 100ma peak and we use 105ma peak and the LED blows out, we cant go running back to them claiming, "Hey your LED blew out in my circuit, i want my money back", because they will say, "Lets see the circuit", and when they find out that we were driving it at 105ma peak they will say, "Well we did not spec that LED for that high of a peak current. We suggest that you lower the peak current and try again".
In other words, unless we want to assume all of the risk, we have no wiggle room when it comes to the peak current. With the average current we have a lot of possibilities, but not with the peak current rating. Im sure there are people out there running their LEDs at higher than 100ma peak and getting away with it, but if the LED blows they dont have any recourse but to buy another one. The choice is yours of course.
The average current comes from calculating the mean value using an integral. It's the time integral over the period divided by the period. There is some wiggle room here but the consequences of going over this rating may not be as immediate as the effects are a little different, taking much more time to show up. If we have a 20ma LED it also has another rating, usually called the life of the LED down to 70 percent of rated light output (it could be 50 percent though so you'll have to check the spec for each LED). What this means is that if we have a 20ma LED rated for 10000 hours of operation then we might expect something like that for normal use. If we run it at 30ma average however we could see a very sharp decrease in life for the LED, as it could go down as low as 2000 hours. The LED will appear to work just fine, but after 2000 hours the LED will only have roughly 70 percent of it's original brightness as it did when it was brand new. There was a very detailed study on this and the result posted on the internet somewhere, we might be able to find it again and get some more accurate results as right now i am trying to remember the exact data from way back then, but i do remember the life going down sharply as the average current increased, it was not linear by any means.
So for the two extremes (over peak current and over average current) we have two different failure modes. An over peak current can cause immediate LED failure (probably due to over voltage which has to accompany over peak current) while an over average current will cause a more gradual failure where the light output diminishes much more quickly than it would at the specified average current rating for the LED.
Note about the average current:
Iavg=(1/(T2-T1))*Intg[T1 to T2] (i(t)) dt
where
T1 is the start of the period,
T2 is the end of the period,
i(t) is the current as a function of time.
"Intg[T1 to T2]" stands for "Integral from T1 to T2 of"
Some circuit simulators may have an 'avg' function which allows you to plot the average value of a variable like current, or may allow you to perform the integration while the simulation is running.