David Bowling's Continuous Charging Device

[tinman]

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%

---------

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.

---------

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%

--------

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.

Looks to me that you have taken my explanation about the differences in combined internal resistances,and added your own spaghetti  to it. Of course the batteries have capacitance,and that is why i did the same test with the capacitors,and asked you to use capacitors instead of batteries to work it out. But once again, your automatic fail button toward me was pushed,and only because you have no idea as to what is going on--this is clearly seen in your last post.

Once again we see you ducking,and not delivering any type of theory before some one else has put there's  forward. Then the auto 'fail' button is hit. You then take there theory,word it different,add your own spagetti,and claim some sort of distorted victory.
We see you do this all the time MH,but some of what you posted above shows how i was correct in a mater on another subject--i will get to that tonight when i get home from work,but you should be carful what you post,as you are going against your self this time,and you have just proven me correct on another subject you accused me being wrong in.
What go's  around comes around.

The LED is 12volts,and it is also not polarity conscious.


Brad

[TinselKoala]

The LED is 12volts,and it is also not polarity conscious.

Can you give us some specifics about this special LED? Part number, datasheet, where you bought it, etc.

If it is "not polarity conscious" and runs on 12 volts, then maybe it has some circuitry inside, contains two actual diodes in anti-parallel, or something else of that nature. LEDs are _diodes_, hence are inherently polarity conscious, and for a white LED to run on 12 volts it will usually have at least a current-limiting resistor or a (JT-like) pulsed or buck-type power supply or regulated current-sink type circuitry along with it, or perhaps 4 diode junctions in series.

[minnie]

   I can't ever say that I understand what Johan is on about.
   One thing he came up with, the quote about a fool and a
   fanatic certainly hit the nail on the head!
         John.

[MileHigh]

Looks to me that you have taken my explanation about the differences in combined internal resistances,and added your own spaghetti  to it. Of course the batteries have capacitance,and that is why i did the same test with the capacitors,and asked you to use capacitors instead of batteries to work it out. But once again, your automatic fail button toward me was pushed,and only because you have no idea as to what is going on--this is clearly seen in your last post.

Once again we see you ducking,and not delivering any type of theory before some one else has put there's  forward. Then the auto 'fail' button is hit. You then take there theory,word it different,add your own spagetti,and claim some sort of distorted victory.
We see you do this all the time MH,but some of what you posted above shows how i was correct in a mater on another subject--i will get to that tonight when i get home from work,but you should be carful what you post,as you are going against your self this time,and you have just proven me correct on another subject you accused me being wrong in.
What go's  around comes around.

Brad

The fail was you characterizing the draw from the inverter in the form of a train of current spikes as being "inductive" as well as not properly measuring them.  The only thing people have right now is a next-to-useless aliasing pattern that "proves that the spikes exist."

There is a lot of baggage in that reply.  I simply offered up my own unique and independently thought out possible explanation for why battery A discharges before battery B.  I rejected your explanation which doesn't even make sense and put forth a possible explanation myself that had no relation to your "explanation."  And I crunched some numbers to see if at least there is some merit to my possible explanation and there is.  People are free to run with that and do tests and experiments to see if it is true or not.  I can think of all sorts of measurements that could be done with a DSO to investigate this theory as well as some simple bench tests to see if it has merit.

I would think that anybody reading your posting would be just as surprised as me about your crazy allegations.

MileHigh

[tinman]

Can you give us some specifics about this special LED? Part number, datasheet, where you bought it, etc.

If it is "not polarity conscious" and runs on 12 volts, then maybe it has some circuitry inside, contains two actual diodes in anti-parallel, or something else of that nature. LEDs are _diodes_, hence are inherently polarity conscious, and for a white LED to run on 12 volts it will usually have at least a current-limiting resistor or a (JT-like) pulsed or buck-type power supply or regulated current-sink type circuitry along with it, or perhaps 4 diode junctions in series.

Sorry,looking at the LED closely,it is actually 24 volt rated--no other information about it on the can. I would say that there is Some sort of circuitry in the can,but it would have to be small.
They are close to the ones in the link below,but have the small bayonet type base. Sizes are about the same.

http://www.mouser.com/ds/2/109/MDEL586Q001-31687.pdf


Brad