David Bowling's Continuous Charging Device

[ramset]

Yes Miles Your acting like a doting Den Mother , give the fellows some air...

Go make us some cookies ...

[MileHigh]

I have already explained as to why battery A discharges  more than battery B in the two series batteries.
If you take the time to look at the circuit,you will also see why.

Okay so you are going to redo the current draw measurements by the inverter sometime in the future and get it right.

I just want to try to understand how you explain that battery A always discharges before battery B.  I attached your diagram and I added the battery designations.  I think that I may have an explanation but I want to try to understand your explanation first.

Here is what Carroll said:

<<< The fact is the battery that is in series (battery A) and connected to the load ALWAYS goes down faster than the other series battery (battery B).  As far as I know none of us have been able to come up with an explanation for why that happens. >>>

Here is your explanation:

<<< I can answer that question for you Carroll.
Batter A(the series battery connected to the load),is only in a series configuration.
Battery C(the charging battery) is in a parallel configuration.
But battery B,(the series battery in the middle of the other two batteries)is in both a series and parallel configuration. It is in series with battery A,but in parallel with battery C.
Depending on the load type(E.G inverter,DC motor) will depend on how well that parallel path is between battery B and C,but in almost all cases,there will be some sort of partial parallel connection between battery B and C. How well this parallel path is,will depend on how much battery B and C equalize,and so battery B will receive some partial charging from battery C,but battery A will always be in series,and will be the one that gives up most of it's charge.
Battery A is the one you would want to swap out for battery C,and battery C would take the position of battery A. Battery B might only have to be changed with battery C every 3 of 4 cycles. >>>

Okay, when you say, "Battery C(the charging battery) is in a parallel configuration," it's because battery A's positive is "pointing upwards" in the diagram and battery C's positive is also "pointing upwards" which sort of looks like "positive meeting positive" and so you use the term "parallel configuration."  If I am making a mistake let me know.

So for battery B, it is obviously in series with battery A, but battery B and battery C have positives "pointing upwards" which is sort of like "positive meeting positive" and that is also called a "parallel configuration."  So battery B is in series with battery A  and also in "parallel" with battery C.

Here is your explanation for battery A discharging faster than battery B:

<<< Depending on the load type(E.G inverter,DC motor) will depend on how well that parallel path is between battery B and C,but in almost all cases,there will be some sort of partial parallel connection between battery B and C. How well this parallel path is,will depend on how much battery B and C equalize,and so battery B will receive some partial charging from battery C,but battery A will always be in series,and will be the one that gives up most of it's charge. >>>

You make reference to "some sort of partial parallel connection between battery B and C."  Here is where you have lost me.  Battery B's output goes into battery A and then through the inverter before it gets to battery C.  So how can you call that a "parallel" connection?

You say that "battery B will receive some partial charging from battery C" but when you look at the circuit the current always flows clockwise through the loop, and therefore it is impossible for battery B to receive some partial charging from battery C.

So from what I can see, I can't understand how you are claiming that battery A will discharge faster than battery B because I do not see any partial charging of battery B from battery C at all.

Am I missing something?

MileHigh

[tinman]

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Am I missing something?

MileHigh
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Okay so you are going to redo the current draw measurements by the inverter sometime in the future and get it right

Take another look at the scope shot i provided MH,and look at the RMS and Mean value.
The CVR in the DMM is 200 milliohms,and the DMM reads a current of 3 amps-->how close is them apples

I just want to try to understand how you explain that battery A always discharges before battery B.  I attached your diagram and I added the battery designations.  I think that I may have an explanation but I want to try to understand your explanation first.
So from what I can see, I can't understand how you are claiming that battery A will discharge faster than battery B because I do not see any partial charging of battery B from battery C at all.

Perhaps if you use cap's instead of batteries ,it will become clear.
Take into account the series internal resistance total of battery A and B,and then the parallel resistance total of batteries B and C. You will see that battery A will always have a higher circuit resistance than that of batteries B and C dose in the circuit.

Below is a picture i just took of a test i not long ago carried out--just for your question.
It was my belief that the same should apply for capacitors as it dose for the batteries. The 3 caps(which i am sure you have seen me use many times over the years) are identical caps--10,000uf,63 volt high current caps. Caps A and B were charged to 12 volt's,and cap C i charged to 9 volts. I then completed the circuit using a 12 volt LED.
I tried the same test 3 times,and each time i rotated the position of the caps,and the results were always the same. As you can see,cap A always discharges the most,while cap B ends with a higher voltage. Cap C of course charges up much higher than it's starting voltage.
So the same applies for caps as it dose for the batteries.

Only battery Bs negative is hooked to battery Cs negative,and the parallel connection of the two positive terminals is via the circuit it self--and that includes battery A. The internal series resistance of batteries A and B is added together,while the internal parallel resistance of batteries B and C are halved--just the same as series connecting or parallel connecting normal resistors.
So battery B has a lower resistance value toward battery C,while battery A has a higher value resistance toward battery C.

I am finding it hard to word in a way you may understand,even though i know exactly what i mean.
More power is delivered from battery C to battery B,because of the higher resistance between battery C and A,and so some of that power is dissipated by this resistance.
Say we have our 3 batteries.
1 battery is at 13.5 volt's,and the other two are at 11 volts. We will charge the two batteries with 11 volts,with the battery that has 13 volts-only we will insert two resistors in the jumper leads.
the battery with 13 volts we will call battery C,and the two with 11 volts ,we will call A and B.
The 3 negatives are all linked together by jumper leads. From battery Cs positive to battery Bs positive,we will put a jumper lead that has 1 ohm resistance. From battery Cs positive to battery As positive,we will use a jumper lead that has 2 ohms resistance. When we do this,we see that battery B will charge faster than battery A.
I hope this all makes sense.


Brad

[MileHigh]

Brad:

I can't explain your capacitor example but it is interesting and could merit some more study.  Is that really a 12-volt LED in the picture?  I thought that little panel LED lights that are white might be about 5 volts, not 12 volts.

As far as your theory goes with the series and parallel stuff, you are really just making up stuff on your own and inventing a belief system that has no real merit.  And I hand out a big FAIL to all of the other people reading and contributing to the thread that say nothing.  You are supposed to challenge each other, and not always be compliant mush with respect to each other.  Sure, be open to new ideas and new ways of looking at things, but be just as open to debating them and rejecting them if you think they don't make sense.  I am just saying this in general terms, don't read anything extra into it.

The simple view of the setup is a current loop with battery A and B as the source, and the inverter and battery C as the load.  KCL and KVL will always apply.  Current flows clockwise all the time and going around the loop:  From the bottom you go up in potential from battery B and up in potential from battery A and then you drop down in potential through the inverter and then you drop down in potential from battery C and then you are back at home base.

In your 'invention' you say that battery C is closer to battery B and so there is less resistance between them or it's a "parallel" resistance.  Battery A is further away so it has less affect on battery C or something like that.  These things all sound interesting but the bottom line is that this is a single current loop with voltages that go up and down as you travel through the loop.  You are trying to reinvent the wheel to come up with an explanation but it will not work.  You can clearly see that battery A and battery B export power and the inverter and battery C absorb the power exported by battery A and battery B.  The only thing that all three batteries "see" is the voltage across them, and the same current flowing through them.

However, within that "reality constraint" there still has to be an explanation for why battery A discharges faster than battery B.

I am not going to debate your viewpoint anymore.  If you want to stick with it that is fine with me.  However, what I am going to do is make a new post where I offer up a possible theory for why battery A discharges faster than battery B.  I am not here for the long haul but I did put some thought into this mystery for fun and I will share my thoughts about it.  If some of you find what I have to say interesting then feel free to explore it further.  I don't plan on defending my idea or pursuing anything further with this.

MileHigh

[MileHigh]

Here is the basic premise for my explanation for why battery A discharges more quickly:  We know that the lighter the load on a battery, the more energy that you can extract from that battery.

The reason for this is that when a battery drives a load power is dissipated in the internal battery resistance and in the load.  The larger the load resistance is compared to the internal battery resistance, the more efficient the power transfer is, and therefore you can extract more energy from the battery.

The key to this is that the inverter draws current from the set of batteries as a very short spike of current.  The spike may be so short that it is easily affected by other circuit elements.  My theory is that the spike of current is not identical in battery B.  There is some stray or inherent inductance and capacitance in the setup such that battery A outputs a relatively sharp spike of current, but battery B's output is low-pass filtered and as a result the spike is spread out over time.  That means there is a lower current flow over a longer time in battery B and that translates into less losses to internal resistance in battery B and/or a more efficient exporting of energy from battery B.

Below you will see a simplified example done just to get a handle on things and the numbers do add up.  I also make an assumption to give me better numbers.  I make a "battery B-prime-prime" where I assume that the internal resistance of the battery is non-linear with respect to current draw, and the lighter the current draw, the lower the internal resistance.  Of course you can easily measure battery internal resistance vs. current draw and find out for yourself.

Here is what I come up with in a nutshell in a very simple model:  Battery A outputs the current pulse that goes into the inverter.  Battery A gets it's energy from the current pulse from itself, and from a capacitor that is between the two batteries.  After the current pulse is done, then battery B charges up the capacitor much more slowly and sluggishly.  That slow charging of the capacitor is a more efficient process. (see the numbers below)

Note:  In my crunching notes below in ny conclusion I reverse the order and say that battery B charges up the capacitor first.  It really makes no difference and you can look at it either way.

The net result is this:  For every current pulse, battery A loses more energy to internal resistance than battery B.  There are millions of current pulses so over time battery A discharges faster than battery B.

Number crunching:

Battery A-prime:  12 volts, with one ohm internal impedance
Battery B-prime:  12 volts, with one ohm internal impedance
Battery B-prime-prime:  12 volts, with 0.5 ohm internal impedance

Battery A-prime:  Apply 5 ohm load for one second gives 2 amps for one second, 10 watts dissipated in load, 20 Joules of energy put into load.
2 watts dissipated internally, 2 Joules.total internal dissipation.
Total energy expended:  22 Joules, efficiency 90.9%

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What if on Battery B-prime the current is 1/2 amp for 4 seconds?

Load now looks like (11.5V/0.5A) = 23 ohms.  5.75 watts dissipated in load for four seconds, 23 Joules of energy put into load.
0.5 watts dissipated internally for 4 seconds, 2 Joules total internal dissipation.
Total energy expended:  25 Joules, efficiency 92%

Battery B-prime is more efficient in transferring energy into load than battery A-prime.

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What if on Battery B-prime-prime the current is 1/2 amp for 4 seconds?

Load now looks like (11.75V/0.5A) = 23.5 ohms.  5.875 watts dissipated in load for four seconds, 23.5 Joules of energy put into load.
0.25 watts dissipated internally for 4 seconds, 1 Joule total internal dissipation.
Total energy expended:  24.5 Joules, efficiency 95.9%

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Simplified model:  Battery B-prime-prime fills up a capacitor with 95.9% efficiency, and then Battery A-prime coupled with the capacitor discharges into the load with 90.9% efficiency.

Therefore over time Battery A will discharge more quickly than Battery B.