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I would like to have some advice for constructing a little device. The main purpose is to be able to connect 2 Li-Po batteries in parallel and discharge them in the same fashion throughout the circuit.

To be specific, I want the current flow through each battery to be exactly the same at any time.

First, I would like to do it with two batteries which have the same specifications.

I have thought about this problem and I imagine a small device with two current sensors and digitally controlled potentiometers connected to a microcontroller.

But I haven't been able to find digitally controlled MILLI-ohm potentiometer (0 to 100 mOhm with a step of 1 mOhm). I am also dealing with current flow up to 60A.

My second goal is to achieve this with 2 batteries of different capacities. The 2 current flows will have to be controlled so as they are proportional to the battery capacities.

Thanks for your attention!

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  • \$\begingroup\$ You can use an MOSFET as an electronic controllable resistor. There are models that can draw 60A easy and go down to the mOhm range \$\endgroup\$
    – Bruce
    Sep 1, 2015 at 16:44
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    \$\begingroup\$ No offense to you, but I think it would be a good idea for you to take a step back and explain WHY you want to do this? Several things in your question make it seem like you are a little bit naive. For example, 0 mOhms? No such thing. Also, the goal of matching currents but ignoring voltages may not be possible, or in any event introduces some problems that you don't seem to be aware of. So I think you should provide some more background. \$\endgroup\$
    – user57037
    Sep 1, 2015 at 17:56
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    \$\begingroup\$ @Bruce, modulating a MOSFET so that it has stepped resistance from "zero" to 100 mOhms in 1mOhm steps would be quite difficult, and would require adjusting somehow for temperature. Also, 60*60*0.1 is 360 Watts. That is a lot for one transistor. Not exactly "easy." \$\endgroup\$
    – user57037
    Sep 1, 2015 at 18:00
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    \$\begingroup\$ This is not likely to be easier than having individual DC-DC converters (or a multiple input converter) with current sharing on the output. It would draw the same power from each battery until one reaches cut-off voltage and then the full load from the remaining battery. Having a description of the use case would be enlightening. Assuming or hoping for matched batteries is a recipe for failure, assume they will not be matched and work around the limitation. \$\endgroup\$
    – KalleMP
    Sep 1, 2015 at 18:51

2 Answers 2

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I like that problem !

Maybe you could try something along these lines (just from the top of my head, more to get a discussion started):

schematic

simulate this circuit – Schematic created using CircuitLab

note that left MOSFET is N-channel, right mosfet is P-channel but you need to check stability, I'm afraid even with the R9 resistor you could get all manners of oscillations from the mosfets capacitances.

The idea is that if current on the left is larger than current on the right, -input will be over +input so the op amp will move its output high , making the n-mosfet more conductive and the left one less conductive (and vice versa).

You need an op-amp with low enough offset voltage for the precision you want. It should also be able to work with low supply votage and Rail to Rail input.

However I'm not sure you can get a suitable p-mosfet for the kind of current you want (you can parallel them though). Also watch out for SOA since the mosfets will be in their linear region, they might get local hot spots on the chip (see that question and references therein)

I'd keep the resistors low enough to keep most of the battery voltage for the mosfet gates. Of course if the battery voltage is higher you've got more headroom and you can drop more voltage in the resistors.

EDIT

I think I misunderstood the question. I thought the purpose was to drain (equally) the batteries, not to share current provided to a useful load. My bad...

The suggested modification by KalleMP

schematic

simulate this circuit

wouldn't work since for a working load there wouldn't be any significant voltage left for MOSFET gate operation.

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  • \$\begingroup\$ Note that this might be pure garbage; I'd love the opinion of some of the more seasoned engineers here... \$\endgroup\$
    – Nicolas D
    Sep 1, 2015 at 18:35
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    \$\begingroup\$ A small edit of your schematic to show the load connected between the battery+ common and the sense resistor common would help people wrap their heads around this and offer critique. \$\endgroup\$
    – KalleMP
    Sep 1, 2015 at 18:48
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Thank you all ! @mkeith you guessed i am a noob. I am also french, that why i am naive. O mohm is a value really tiny in regards of 1. These two batteries are lipo batteries and i want to increase autonomy of a model on using two batteries in parallel but if one of the battery discharges faster than the other i would have to stop playing with my model before she dies even if i have power in the second battery.

@Nicolas D. The OA1 will be saturated, so that the out can only take 2 values. Maybe I am wrong but in that case the current flows in the two batteries will also be two state?

Maybe i can draw this ?! :-)

schematic

simulate this circuit – Schematic created using CircuitLab

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