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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.
You are correct, of course. I didn't notice the obvious path for the cap to the transistor when it turns on and shorts the remaining charge on the cap to ground.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.
The Cree datasheet does so specify the maximum allowed pulse width (at or less than 0.1ms) at a duty cycle of 1/10th or less for it to survive 100mA pulses.
Sorry, I missed that. According to that, I should be OK, since the >100 mA pulses are a lot less than 0.1 mS wide.
Thats not exactly right: the absolute Ipeak at 25C is 100mA and the pulse width has to be <= 100uSec, so its not really OK.
In a Joule Thief the battery energy is accumulated in the inductor inductance of the circuit and this stored inductive energy is what provides the current pulse, not the battery.A battery cell that is nearly dead cannot supply 100mA pulses unless it has as pretty big capacitor in parallel.
Al--post the circuit? I'm interested in any low-tech voltage converters like these.