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

[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.

[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.

[JS]

Change the resistor with a CC source. Again now, jfet might be your bet here, a single one and a single resistor will give flat enough current for most of the range.

Ohhh, use a batterizer! Bateroo time again!

JS

[Distelzombie]

I don't want to use a micro for that.

Hm... It doesn't have to be analog. What if I use a couple diodes with different breakdown voltages that are in series with a resistor? Eeeeh... That's not quite working. (it's inverse)
Voltage comparator, you said? If I use a couple of these I can maybe have 2 or 3 different resistor values that switch on at different voltages? Just like what I imagined with a couple diodes, but inverted. The less voltage the more resistors get put in parallel.
Would that work? At these voltages? (1.55V, ~1.25V, ~1.00V and 0.80V)

---- after your new comment - - - -

A Batterizer xD
It already is one... Wait would that actually work? Is that a CC source? Hm... I'd rather do that myself, but if that thing is the most efficient CC source...

Also could you help me with the calculation above? I never used log

[wraper]

At 700mV there is only a tiny percentage of the energy left in the battery. By discharging to such voltage you are just asking it to leak electrolyte into the device.

[Distelzombie]

You're right about energy. But they are high quality batteries that very rarely leak.

Anyway: I found a component called current regulating diode or current limiting diode. They are essentially JFETs with source already tied to gate.
They sound like the perfect thing and they are available at these currents. But until now I can only find some with voltages at about 100 to 50V.
And at these low voltages they are working almost identical to a regular resistor. (As I read in datasheet)  I therefore assume a usual JFET would act identical.
Well I go search for a fitting one. Thank you JS

Edit: Oh no... I can't read the datasheets properly! What do I have to look out for?
Wikipedia says Idss, OK. But I never find a chart wich puts voltage in relation to Idss! How do I find out which one works? 
And do I need N or P channel?
Dies this one work? https://www.digikey.de/product-detail/de/toshiba-semiconductor-and-storage/2SK879-Y-TE85LF/2SK879-Y-TE85LF-TR-ND/4503064
It looks like it does 1.6mA over the whole voltage range I need. (Id to Vds) On the other hand none of it makes sense. Why negative current? Does it go over 1.6mA at 1.5v? 

[Distelzombie]

Edit: Oh no... I can't read the datasheets properly! What do I have to look out for?
Wikipedia says Idss, OK. But I never find a chart wich puts voltage in relation to Idss! How do I find out which one works? 
And do I need N or P channel?
Dies this one work? https://www.digikey.de/product-detail/de/toshiba-semiconductor-and-storage/2SK879-Y-TE85LF/2SK879-Y-TE85LF-TR-ND/4503064
It looks like it does 1.6mA over the whole voltage range I need. (Id to Vds) On the other hand none of it makes sense. Why negative current? Does it go over 1.6mA at 1.5v? 

Ups, sorry. I always forgot people read my posts before I edit them, and then never come back because there is seemingly nothing new going on. 
XD

[JS]

I don't know the parameters of your circuit, but at such low voltage is quite hard to work, even worse without knowing what is actually happening.

The bateroo will keep a constant voltage at the output, you place a resistor and your working conditions are constant till your battery is depleted.

Now, I think you are over engineering this, if it works just fine with a resistor you might be facing a headache that doesn't worth for some extra battery life that doesn't worth it.

Now, I guess the jfet and single resistor won't cut it, not enough voltage to play around with. Another option, jfet, gate tied to ground (battery -), other two terminals to the battery + and anode of the LED. Find a suitable LED which makes a current that make sense at the voltage ranges of the battery. Next improvement, Si diode from V+ (or LED anode) to gate, resistor from there to ground. Now you have a more constant current with the varing V+. Finding the right jfet is the tricky part here.

JS

[Distelzombie]

I don't know the parameters of your circuit, but at such low voltage is quite hard to work, even worse without knowing what is actually happening.

The bateroo will keep a constant voltage at the output, you place a resistor and your working conditions are constant till your battery is depleted.

Now, I think you are over engineering this, if it works just fine with a resistor you might be facing a headache that doesn't worth for some extra battery life that doesn't worth it.

Now, I guess the jfet and single resistor won't cut it, not enough voltage to play around with. Another option, jfet, gate tied to ground (battery -), other two terminals to the battery + and anode of the LED. Find a suitable LED which makes a current that make sense at the voltage ranges of the battery. Next improvement, Si diode from V+ (or LED anode) to gate, resistor from there to ground. Now you have a more constant current with the varing V+. Finding the right jfet is the tricky part here.

JS

hi thanks for you answer
Find a LED that makes a suitable current? What do you mean? A second led?

What do I search for in case of the JFET? I mean, what property do I have to look out for? And HOW do I look for it?
Also, P or N channel? xD

XD sorry it's like all the possible questions. XD

[JS]

it seems like you already have an LED, I'm talking about the jfet.

You need a jfet that with ~-0.5V at the gate you get the desired current. Look at the Id/Vgs plots.

JS

[mmagin]

If you're willing to use a proprietary ICs, there are a number of constant current output boost converters intended for running off a single cell.  e.g. https://www.diodes.com/products/power-management/led-drivers/low-voltage-dc-dc-led-drivers/part/ZXSC300

[Distelzombie]

I finally found the picture or the schematic I drew. (And roughly simulated) It's attached.
The resistor in the picture is where the pot is. Also there is a second transistor in parallel with the one in the picture. This one is a different type and it's glued with thermal paste to the one in the schematic. I'm not sure if the paste was necessary, but I figured, since they sit flat-face to flat-face (ones pins are mirrored since diff. type) on the board, I might as well couple them thermally.

I can't connect anything directly to the LED. It would probably stop oscillation. So it has to be in place of the resistor.
I assume I also can't use the Batteroo or Batterizer. I guess they don't put out flat DC and would interfere, or they just ruin the efficiency. But I'll see where I can get one.

@JS does it even matter if it is P or N channel? I take a look around, but the Ig Vgs  or Vds plots are with negative current. Does that matter? Why is it negative?

@mmagin Hm. Oh: "The circuit generates constant current pulses" nope, not gonna work, I think. (How is that constant xD)

I'll try to find a JFET and one of these batterizeroos if they don't cost too much.
Thank you very much so far
If you have more ideas now that you have the schematic, please tell me.

[Distelzombie]

The Batterizer wouldn't work and it would be stupid to use one.
So I need a JFET. But I still can't search for it. I have no idea what kind I need.

Would this one work? https://m.reichelt.de/JFETs/2SK-246/3/index.html?ACTION=3&LA=446&ARTICLE=2178&GROUPID=8417&artnr=2SK+246&SEARCH=Jfet&trstct=pos_7

Is that even in the right direction? What about the Voltage? 50V? Would it still do just 1mA at 1V? I can not read the sheet.

[JS]

The curves on that jfet looks about right, I don't know your LED, the current you want at it and what voltage it takes for that current.

This is what you want to look
You saud your LED worked down to 0.7V with 200Ω, I don't know the current but I'll take 1mA

You can do 1mA at about 200mV with this. Which is like a 200Ω resistor. That's with about 0V at Vgs. I'd say you need to swap the diode and the resistor in the circuit I told you, as you want close to 0V when battery is empty and more like 1V when battery is full. A diode from gate to V- and a resistor from gate to V+ would make something like this happen, keeping about 1mA.

You could also try attaching the gate to V- directly but the current would be lower when the battery is full, maybe posw enogh, if not you could use a resistor in parallel with the jfet source and drain.

All this is V+ to drain, source to LED anode, LED cathode to V-.

JS

[Distelzombie]

The curves on that jfet looks about right, I don't know your LED, the current you want at it and what voltage it takes for that current.

hi
I assume you mean my last link. The current to the LED is provided by the joule thief, as well as the voltage. Hooking anything up to the LED itself would probably break the oscillation.  (Also the picture you made showd Vds instead of what you said earlier: Vgs)
The current I want at the place of the resistor in my circuit is 1.7mA down to whatever. This would be the best, because the transistors get quite hot with a fresh battery and a bigger current, also the brightness is really great.
The voltage at the place of the resistor should not change from what it does with just the resistor. It should just be battery voltage.

in the circuit I told you

Shouldn't I short source to gate in order to make the JFET a constant current source? Current limiting diode? (Like, what wikipedia said - although I can't find that anymore  )
Also it is a normal LED. Its just a good one that is very efficient. (Nichia nspw500gs-k1)
I have a LTspice file of the circuit ready to simulate (except the double-transistor) if you want it.

Ok... so I buy me some of these JFETs, attach battery positive to drain, short gate and source and connect these to the transformer. Sounds rather easy.

EDIT: Oh, could you help me with that calculation, please?
As much as I never actually used JFETs, I also never actually used that log key on my calculator. (I assume it is needed here)

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?  xD

[JS]

hi
I assume you mean my last link. The current to the LED is provided by the joule thief, as well as the voltage. Hooking anything up to the LED itself would probably break the oscillation. 

Sorry, I forgot about the joule-thief on the middle of all this, I was thinking for the LED current limit. I don't know how the joule thieve will interact with the jfet. Post a pict of the circuit please!

(Also the picture you made showd Vds instead of what you said earlier: Vgs)

Look again, the curves are Id vs Vds for various Vgs (the different curves)

The current I want at the place of the resistor in my circuit is 1.7mA down to whatever. This would be the best, because the transistors get quite hot with a fresh battery and a bigger current, also the brightness is really great.
The voltage at the place of the resistor should not change from what it does with just the resistor. It should just be battery voltage.

For 1.7mA you need about 0.6V to 0.7V between gate and source, so the diode from source to gate and the resistor from gate to ground persists. The question is if this circuit does work for you.

Shouldn't I short source to gate in order to make the JFET a constant current source? Current limiting diode? (Like, what wikipedia said - although I can't find that anymore  )
Also it is a normal LED. Its just a good one that is very efficient. (Nichia nspw500gs-k1)
I have a LTspice file of the circuit ready to simulate (except the double-transistor) if you want it.

Ok... so I buy me some of these JFETs, attach battery positive to drain, short gate and source and connect these to the transformer. Sounds rather easy.

Jfets work as CC in many configurations, not just with gate and source shorted. With gate and source shorted, this fet works as a 5mA CC source. One way of making it is using a resistor between gate and source and taking the current out of that resistor and gate node. This way as the current at the resistor is the programmed constant current, and the voltage at the gate is that current times that resistor it all stays where you want. I'm proposing a diode to fix that voltage and not having to deal with the losses in that resistor.

EDIT: Oh, could you help me with that calculation, please?

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?  xD

This will change with your new circuit, also this calculations are only ball park to estimate battery life, then it will depend on the actual battery and as you said not much info to low mA discharge rates.

JS

[Distelzombie]

Duuuude... First you forget what that circuit even is and then that there is already a schematic.
The schematic is 5 posts up from this one. https://www.eevblog.com/forum/beginners/using-varistor-instead-of-pot-to-autom-adjust-res-in-joule-thief-for-bat-life/msg1655183/#msg1655183
No problem. I didn't include it directly in the post after all.

This will change with your new circuit, also this calculations are only ball park to estimate battery life, then it will depend on the actual battery and as you said not much info to low mA discharge rates.

Yea, but still this will help me doing such calculation later. But if you don't want, I can't force you. No problem

Thanks for your help. I think I finally understand what to do. (Burn everything!) xD
No. I get me one or two of these JFETs and try around.
"The question is if this circuit does work for you." Yea

[JS]

Sorry, I missed everything here I guess...

I'll take a closer look tomorrow, but that schematic is asking for a redrawn, badly drawn schematics brings a lot of confusion and just redrawing them makes things much more clear!

JS

[Distelzombie]

Well. Not much to change there. I don't know why you think it is badly drawn. 

I just removed the unnecessary switch that was there for testing purposes. If you have a simulation program I also uploaded the LTspice file, including the transformer I made for it.

[Distelzombie]

Sry. I realized I had too many custom parts. I changed the SPICE file to work in stock LTspice. Schematic is still the same.

[JS]

I will check it later, my LTS is a bit rusty as I usually use other tools, I'll see what I can do.

The placement of the parts inbschematics is quite a big deal, at the univ just by redrawing simple circuits as teachers gave them to us I got my bodys to understand it quite better, even if I got it as it was drawn. I haven't put much attention to yours yet, just a quick look to understand what's going on but not enoght to modify it to what you want.

JS

[mmagin]

@mmagin Hm. Oh: "The circuit generates constant current pulses" nope, not gonna work, I think. (How is that constant xD)

Possibly I was unclear, the datasheet circuit replaces the whole joule thief circuit.  I believe the implication is that the average led current is roughly constant and the frequency fast enough to not be flickery.