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

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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:

--- Quote ---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.
--- End quote ---
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.

--- Quote --- in the circuit I told you
--- End quote ---
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. :D

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

--- Quote ---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
--- End quote ---

JS:





--- Quote from: Distelzombie on July 07, 2018, 10:55:51 pm ---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.  :-//
--- End quote ---
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!

--- Quote ---(Also the picture you made showd Vds instead of what you said earlier: Vgs)
--- End quote ---
Look again, the curves are Id vs Vds for various Vgs (the different curves)

--- Quote ---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.

--- End quote ---
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.

--- Quote --- 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. :D

--- End quote ---
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.

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

--- Quote ---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
--- End quote ---

--- End quote ---
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... :D First you forget what that circuit even is and then that there is already a schematic. :D
The schematic is 5 posts up from this one. :D 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. :)


--- Quote ---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.
--- End quote ---
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

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