Are Joule thieves hard on LEDs?

[audioguru]

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.

 

[MrAl]

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.

 

[crutschow]

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.

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.

 

[ronv]

Another app. note on pulsing.

https://www.avagotech.com/docs/AV02-0871EN

 

[MrAl]

Another app. note on pulsing.

https://www.avagotech.com/docs/AV02-0871EN

Hi there ronv,

That app note is interesting, but i think it's only for red LEDs, and wont be much help for say a white LED. I'll have to read it more carefully soon, but from the graphs it looks like mainly they are dealing with red LEDs and didnt consider the white or else something else is wrong. For example, for their graph of relative intensity vs foward current they show almost a perfect linear relationship, while for the white LED this is not true at all and is very different. Strange i think.

 

[carbonzit]

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.

 

[ericgibbs]

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.

 

[carbonzit]

Thats not exactly right: the absolute Ipeak at 25C is 100mA and the pulse width has to be <= 100uSec, so its not really OK.

Dang, you're right. I went back and rechecked the plot (I for the LED), and the pulse width is about 300 µS @ 100 mA. So I guess we can conclude that yes, Joule thieves are at least a little hard on LEDs.

Plus, according to that Avago app note, ideally, pulsed LEDs should be driven by a square wave. (But at least a triangular wave, which is approximately what the JT outputs, is better than a sine wave.)

So I guess I won't invest all my money in JT stocks just yet ... however, I'm going to continue to build and experiment with them.

 

[ericgibbs]

The point I consider that you are not taking into your calculations, is that a Joule Thief is designed for semi exhausted 1.5V batteries, [thats why its got that name] try the test at 0.8V or 0.9V

 

[audioguru]

A battery cell that is nearly dead cannot supply 100mA pulses unless it has as pretty big capacitor in parallel.
LED pulse widths that are less than 30ms appear to be dimmed.
So although I have never seen a Joule Thief (but I have many solar garden lights with a similar circuit) it will probably not produce much light.

 

[MrAl]

Hello again,

I used the same circuit for a 12v to 9v converter, except i put two secondary windings (one isolated) on the transformer instead of just one because i needed the 9v isolated. So yeah, it makes a cheap and simple low power voltage source isolator too. It was used to drive a volt meter that needed an isolated DC supply to run.

 

[carbonzit]

Al--post the circuit? I'm interested in any low-tech voltage converters like these.

 

[crutschow]

A battery cell that is nearly dead cannot supply 100mA pulses unless it has as pretty big capacitor in parallel.

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.

 

[MrAl]

Al--post the circuit? I'm interested in any low-tech voltage converters like these.

Hi,

The circuit is the same as the Joule Thief, but with a second secondary winding. The second secondary winding then has a Schottky diode to rectify the output, and a filter capacitor to filter the DC, and also a zener to limit the output voltage. The zener is across the capacitor. In other words, a half wave rectifier and filter on the output of the second winding and that's all there is too it really. It can power small circuits or LEDs etc.