Forum Discussion
Freep
Apr 14, 2021Explorer
From https://diysolarforum.com/threads/coulomb-counter-not-strictly-compatible-with-lifepo4.21238/
well, only an old retired engineer here, so I will take a shot at making it "simple" ;) (said absolutely every engineer as they start their presentation)
An amp is just a coulomb/sec
or
1coulomb = 1amp * 1sec.
the rate of flow of the electron does not give a hoot about the voltage so it literaly is the number of electrons that have flowed.
energy is the relationship between voltage and current.
if you know how many amp-seconds worth of electrons you put into something, well, when you have started to pull that many out you are running on empty. Sounds like people are trying to compare "current" and "energy"...they are related but different.
the "energy" is a function of voltage * coulombs used; and therefore the potential energy in the cell is dropping as the voltage drops even with a constant coulomb draw... the flat discharge nature of the lifepo4 means that the window of constant energy draw (consistant voltage and current) is also mostly flat until the end.
Once you have hit the point of rapid voltage drop you ARE getting less "energy" for the same coulomb draw; but the end is near anyway due to the sharp dropoff. Your coulomb counter is still valid, if you want to maintain a specific power output, you will draw more coulombs as the voltage drops...shunts measure coulombs and you will not get out more than you put in(if you do, quick, get a patent!!!).
you can tell how long you have left before dead just by coulomb counting...
If I have 200coulombs left and I am drawing them thru the shunt at some rate, then its easy to solve for 0.
You could probably take the standard discharge curve of the end of the lifepo4 and create a simple two point equation from it once you hit the drop-off to actually estimate the amount of energy left. The voltage will keep changing and coulombs will increase to sustain the energy demand.
Or, you can just count coulombs, when you hit zero, game over hehe
You are using the energy but its really the flow of electrons you are controlling.
That said, its way simpler to just accept that once you drop under 3volts, you are done. If you have been counting coulombs all along, you know how many amp-seconds you have left if you want to predict exactly how many watt-seconds remain.
well, only an old retired engineer here, so I will take a shot at making it "simple" ;) (said absolutely every engineer as they start their presentation)
An amp is just a coulomb/sec
or
1coulomb = 1amp * 1sec.
the rate of flow of the electron does not give a hoot about the voltage so it literaly is the number of electrons that have flowed.
energy is the relationship between voltage and current.
if you know how many amp-seconds worth of electrons you put into something, well, when you have started to pull that many out you are running on empty. Sounds like people are trying to compare "current" and "energy"...they are related but different.
the "energy" is a function of voltage * coulombs used; and therefore the potential energy in the cell is dropping as the voltage drops even with a constant coulomb draw... the flat discharge nature of the lifepo4 means that the window of constant energy draw (consistant voltage and current) is also mostly flat until the end.
Once you have hit the point of rapid voltage drop you ARE getting less "energy" for the same coulomb draw; but the end is near anyway due to the sharp dropoff. Your coulomb counter is still valid, if you want to maintain a specific power output, you will draw more coulombs as the voltage drops...shunts measure coulombs and you will not get out more than you put in(if you do, quick, get a patent!!!).
you can tell how long you have left before dead just by coulomb counting...
If I have 200coulombs left and I am drawing them thru the shunt at some rate, then its easy to solve for 0.
You could probably take the standard discharge curve of the end of the lifepo4 and create a simple two point equation from it once you hit the drop-off to actually estimate the amount of energy left. The voltage will keep changing and coulombs will increase to sustain the energy demand.
Or, you can just count coulombs, when you hit zero, game over hehe
You are using the energy but its really the flow of electrons you are controlling.
That said, its way simpler to just accept that once you drop under 3volts, you are done. If you have been counting coulombs all along, you know how many amp-seconds you have left if you want to predict exactly how many watt-seconds remain.
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