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Need help with powering LEDs, ASAP

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You will not accurately measure the voltage across the LED because it is pulsing at a high frequency. Your meter will measure the average voltage which might be half or less of the actual voltage.

It depends. On what kind of meter you use, and what the frequency is, for instance. What do you suggest to solve this?

Another point. If I'm inaccurately measuring the LED current but all I need is a comparison between two circuits, then assuming the inaccuracies are the same in the compared circuits, they shouldn't be a problem. The comparisons are what I'm after, not an accurate measurement. One is brighter and I measure higher LED current than the other circuit. I don't have a decent accurate light meter, so this seems to be an easy and cheap solution to making a comparison of actual light output.

I'm on the fence about how using the resistor in series with the LED works or if it's inaccurate. What I did was try measuring with a bypass capacitor of various sizes across the resistor, and it didn't seem to change the measurement. Then I decided that putting a 100 uF or even larger across a 1 ohm resistor was not enough, since the reactance of the capacitor was still more than 1 ohm, even at 100kHz. So I put a 100k and 0.1 uF or larger RC low pass filter between the resistor and the meter. Well, it didn't change the measurement either. But what led me to do this was that I was calculating (and other experimenters were too) an efficiency of 85 to 95 percent, and I thought that was out of line and it was due to some inaccuracy of the voltage measurement across the resistor.

Oh, one other point. The amount of LED output is not related just to the pulse height. If the pulse width varies, so does the LED light output. And another thing is that the LED light output is not linear in relation to the energy contained in the pulse. So if I get an accurate measurement it may not reflect the exact relationship to LED light output. Some complex issues, so I hope I am making myself clear?
 
It depends. On what kind of meter you use, and what the frequency is, for instance. What do you suggest to solve this?
Use a DC coupled oscilloscope.

If I'm inaccurately measuring the LED current but all I need is a comparison between two circuits, then assuming the inaccuracies are the same in the compared circuits, they shouldn't be a problem. The comparisons are what I'm after, not an accurate measurement. One is brighter and I measure higher LED current than the other circuit. I don't have a decent accurate light meter, so this seems to be an easy and cheap solution to making a comparison of actual light output.
Then you might burn out the LEDs.
Most half-decent (not cheap Chinese LEDs on e-Bay) LEDs have a detailed datasheet so you can design a circuit that makes then bright but does not burn then out.

I'm on the fence about how using the resistor in series with the LED works or if it's inaccurate.
use simple Ohm's Law.

What I did was try measuring with a bypass capacitor of various sizes across the resistor, and it didn't seem to change the measurement.
A capacitor has nothing to do with an LED unless it is fed a pulsing voltage.

Then I decided that putting a 100 uF or even larger across a 1 ohm resistor was not enough, since the reactance of the capacitor was still more than 1 ohm, even at 100kHz.
A capacitor has reactance, not resistance. Its reactance limits the current at low frequencies, not at high frequencies. Your capacitor is parallel with the LED so it steals power.

So I put a 100k and 0.1 uF or larger RC low pass filter between the resistor and the meter. Well, it didn't change the measurement either.
If you want the LED to slowly fade and slowly brighten then the RC must be at the input of the LED driver.

The amount of LED output is not related just to the pulse height. If the pulse width varies, so does the LED light output. And another thing is that the LED light output is not linear in relation to the energy contained in the pulse. So if I get an accurate measurement it may not reflect the exact relationship to LED light output. Some complex issues, so I hope I am making myself clear?
Brightness is not linear, it is logarithmic. 1/10th the power is half as bright. Then you can see in moonlight and in sunlight (disregarding your attenuating iris).
 
Use a DC coupled oscilloscope.


Then you might burn out the LEDs.
Most half-decent (not cheap Chinese LEDs on e-Bay) LEDs have a detailed datasheet so you can design a circuit that makes then bright but does not burn then out.


use simple Ohm's Law.


A capacitor has nothing to do with an LED unless it is fed a pulsing voltage.


A capacitor has reactance, not resistance. Its reactance limits the current at low frequencies, not at high frequencies. Your capacitor is parallel with the LED so it steals power.


If you want the LED to slowly fade and slowly brighten then the RC must be at the input of the LED driver.


Brightness is not linear, it is logarithmic. 1/10th the power is half as bright. Then you can see in moonlight and in sunlight (disregarding your attenuating iris).


FIRST OFF, I think I need to make the circuit clearer. The 1 ohm resistor is in series with the cathode lead of the LED to monitor the LED current by measuring the IR drop. Since it's a 1 ohm resistor, each millivolt equals 1 milliamp current.

I put the capacitor across the 1 ohm resistor to filter out the spikes and give the voltage measurement not dependent on frequency, assuming the meter was not made for high frequencies. I DID NOT put the capacitor across the LED.

I changed the big cap across the 1 ohm to a 100k from the resistor hot lead to the meter lead, and a .1 uF cap across the meter leads. This RC low pass filter is to do a better job of filtering out the spikes going to the voltmeter.

In my reply, I said that the capacitor has reactance. One thing about reactance is that it is at a particular frequency, But in this JT circuit, the waveform is a pulse which is a series of frequencies, so the capacitor has to be viewed in the time domain, as it charges and discharges with the pulses.

The issue is not apparent brightness to the eye, the issue is the LED output in lumens or candelas, for illumination purposes. My ancient Weston light meter is what I have to measure the LED's light output.

Measuring the voltage with the DC coupled 'scope doesn't tell you what's the amount of power to the LED. It's a complex waveform and you cannot divide the peak-to-peak voltage of the pulse by 2.8 to get the RMS voltage.

The LED has the same pulses that are across the transistor, collector to emitter. Nothing is between the two, the LED is connected directly to the transistor.

Even with the inaccuracy, perhaps the method I'm using is the best given its ease and low cost.
 
Use a DC coupled oscilloscope.


Then you might burn out the LEDs.
Most half-decent (not cheap Chinese LEDs on e-Bay) LEDs have a detailed datasheet so you can design a circuit that makes then bright but does not burn then out.


use simple Ohm's Law.


A capacitor has nothing to do with an LED unless it is fed a pulsing voltage.


A capacitor has reactance, not resistance. Its reactance limits the current at low frequencies, not at high frequencies. Your capacitor is parallel with the LED so it steals power.


If you want the LED to slowly fade and slowly brighten then the RC must be at the input of the LED driver.


Brightness is not linear, it is logarithmic. 1/10th the power is half as bright. Then you can see in moonlight and in sunlight (disregarding your attenuating iris).

I blogged a copy of my reply and added a schematic to clarify the current measuring circuit. See **broken link removed**
 
@acmefixer
Have had one of these running for a little over a week now. The light was very bright, but the battery is down to .7V already and the light is very dim. Beginning to think this joule thief is a gimmicky litle toy and has no real practical application.
When I hooked th light up to my project, it was much brighter than I needed. Is there a way to cut this back which will also extend battery life? In my application, battery life is much more important than blinding brightness, bu the brightness at .7V is too little. I need it to last longer before getting to .7V.
 
I really need to keep these simple and cheap, because if they work OK, I would like to start selling the clocks with the lights in them. They look cool, but can't make them profitable with expensive parts, and won't have happy customers who constantly have to change batteries or are unsatisfied with the quality/performance of the lights.

Thanks for the help so far, everyone.

the point that everyone is missing here is that you can not run an LED for a reasonable amount of time to run a clock off of AAs. that is why you will not find a battery operated clock that leaves the lights on.

you could use an LCD and only turn the back light on for a while when a button is pushed...

the thing to remember is WHrs .... an AA delivers 1.5WHrs if you are lucky and a 20mA 3V LED takes 0.06W. divide one into the other and you get 25Hrs per AA battery, not counting conversion efficiency.

dan
 
@acmefixer
Have had one of these running for a little over a week now. The light was very bright, but the battery is down to .7V already and the light is very dim. Beginning to think this joule thief is a gimmicky litle toy and has no real practical application.
When I hooked th light up to my project, it was much brighter than I needed. Is there a way to cut this back which will also extend battery life? In my application, battery life is much more important than blinding brightness, bu the brightness at .7V is too little. I need it to last longer before getting to .7V.

Yes, I explained in an earlier reply that you can increase the resistor which will reduce the brightness and also reduce the battery current, thus increasing the battery life. If you are using a 1k then increase it to 10k, and the LED will be dimmer and the battery will last longer. I will not say that the battery will last ten times as long, but it will last longer. An even higher value resistor will give even less current drain and longer battery life.

Remember that there is no free lunch. If you have the LED running 24/7, then the battery will run down soon. Going to a D cell might give you five times the battery life. It depends on how much light you need and how long or short your battery can handle the current drain.

Reminds me of the xmas trees some of my co-workers purchased before the holidays. They have LEDs and 3 or 4 AA cells in the base. They turned the tree on for a few days and in a matter of a week the batteries were dead. They complained about it to me, and I said that it would have been best to buy the one that has a wall wart adapter and plugs into the wall, with no batteries.

Let us know how it works out.
 
The Joule Thief circuit is extremely simple so it has nothing to regulate its output power as the battery runs down so of course the LED gets dimmer.
 
A battery stores much more power than a super capacitor.

Yes, I know.

I was thinking of a way to quickly charge the supercap (with a switching circuit via a battery) and let the LED work off the slow discharge from the supercap before the switching circuit needs to switch on again - periodically using the battery source. Seems that the battery would last much longer. Maybe my theory is wrong about this.
 
If the battery charges the super capacitor in 1/10th the time that the capacitor powers the load then the battery must supply 10 times more current than the load uses and doing it, the battery will heat and waste power. You lose, you gain nothing.
 
If the battery charges the super capacitor in 1/10th the time that the capacitor powers the load then the battery must supply 10 times more current than the load uses and doing it, the battery will heat and waste power. You lose, you gain nothing.

Battery needs to supply a burst of current to charge the cap in short amount of time.

I see now. It's that "Law of Conservation of Energy" - you just can't break it!
Power in equals power out.
 
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