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Using varistor instead of pot. to autom. adjust res. in joule-thief for bat.life
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Distelzombie:
Hi

Sry for the complicated title. XD
I made a joule-thief a while ago that uses a potentiometer to be able to adjust input current. As you can imagine, the higher the resistance the longer the battery lifetime and the lower the light output. If the resistance is high (125kOhm) the lifetime is really great (1550h) and the LED is sensibly bright. But the LED stops lighting up at about 700mV battery voltage. (Output voltage too low for LED, therefore circuit breaks) The battery regenerates to about 770mV, but doesn't light up again.
If I then lower the resistance it can light up bright again until battery is less than 400mV, improving the lifetime by many hours. To fully use the battery's capacity I have to dynamically adjust the resistance of the pot.

So I got the idea to use a varistor instead.
Would that work? Is there even a varistor that goes from 125kOhm to 400Ohm over a voltage range of 1.55V to 0.40V? Can I bodge one together and will it still be as efficient as it is now?

Thank you for your help and answers. :D
JS:
I don't think you will find such a part, one option would be to use a low voltage micro, sleeping till you get down to 0.8V where it still clocks, at that point you wake it up, add a resistor in parallel with a locking switch circuit and turn it off completly.

How many extra hours you get with the lower resistor?

The micro will take something while in sleep mode, none after the switch, so your battery life might improve very little for the added complexity. Instead of the micro it could be a smartly designed comparator.

Also, the switch isn't trivial, bjt would chew too much voltage from load, mosfet won't even turn on, jfet might be the only option, but I don't know...

I'm asking how much battery life you get as there isn't much energy after 0.7V to drain from the battery.

JS

Distelzombie:
I`m currently testing. But I made a mistake: 400Ohm is too low and it drained too fast. (One hour) However I think I would only get one day extra out of it, or a few hours. (Educated guess xD)
For reference: I used an Energizer MAX AAA battery for the test. It has about 1150mAh at 25mA discharge. I use max 1.71mA with fresh AAA battery at 125kOhm. The datasheet of the battery only goes to 25mA so I can't tell how many mAh it has with 1.71mA and less discharge.
Am I an idiot and just have to calculate something with 1550h to get the true mAh? The average current over 1550h is ~0.77mA. But the current is changing over time as well as the voltage. Does it fall logarithmically? Can someone give me an example calculation, please? xD
The circuit was designed for AA batteries.

Edit: forgot to add 1.71mA
JS:
Now I know, it doesn't worth it. You just depleated your battery at 0.7V. You have a two month battery life and you want an extra day? Tolerance in the used resistors not to talk about difference in LEDs will chew that extra day

JS

Distelzombie:
I forgot to add something above.

Well, it's not the battery life in the first place. The LED does get pretty dim at the end. (it's a good one so you could still walk around no problem in darkness)
I want to more or less flatten the light curve, actually. :D
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