KendallP
Grants Pass, OR
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LScamper wrote:BFL13 wrote:
"battery voltages--both go up after the initial hit"
Did you let the 6s recover after running them down to 50% before connecting them. Both battery voltages going up makes me think that the 6s were just recovering after discharge. The important thing is the difference voltage between the two batteries, that is what drives the current.
BFL is easily one of the most obsessive battery testers on these boards. I would be shocked if he didn't take that into account... and so would numerous other members.
Cheers,
Kendall
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LScamper
Los Alamos NM
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KendallP wrote:
"BFL is easily one of the most obsessive battery testers on these boards. I would be shocked if he didn't take that into account... and so would numerous other members."
I agree, however he said he ran the battery down to 50% and that would have a voltage of about 12.23V. Yet his test started at 11.95V?
He is the only one that can answer that, we will have to wait.
Lou
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BFL13
Victoria, BC
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I did wait a half hour to let the voltage "bounce back," but it was still rising very slowly, as it would for the rest of the day. I may have overshot the 50% a little, where I was trying to use SG for that but the voltages I posted are what happened.
Perhaps it has something to do with that initial rush of current that knocks down the voltage of the AH pail so much. Then it too would do some "bounce back?"
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Salvo
California
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It's a lot harder to model a charging battery. That's because charging battery resistance is all over the map. Battery discharge resistance is fairly constant. The slope of the curves shown below is representative of internal battery resistance. The smaller the slope, the lower the resistance.
During charging, resistance increases drastically as SOC reaches 75%.
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dclark1946
Richardson,TX,USA
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Salvo wrote:It's a lot harder to model a charging battery. That's because charging battery resistance is all over the map. Battery discharge resistance is fairly constant. The slope of the curves shown below is representative of internal battery resistance. The smaller the slope, the lower the resistance.
During charging, resistance increases drastically as SOC reaches 75%.
Salvo,
I would expect the battery impedance to be the slope of voltage vs. current (R = delta V/delta I) rather than slope of voltage vs SOC. Does it work out to be the same?
Dick
Dick & Karen
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LScamper
Los Alamos NM
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Salvo now I have worked myself into a corner. I think you have said there is nothing showing that the resistance of a 12V battery and a pair of 6V batteries is any different. Now after looking much to close at your plots (I have seen them may times but did not pay much attention to them) and thinking about putting some of it into the simulation I see a problem. Must be how I'm interpreting the plots.
The plot shows that the voltage at C/3 at 100% SOC is about 11.75V. Now the problem.
232 AH battery current at C/3 = 77.3A
115 AH battery current at C/3 = 38.3A
The voltage drop is the same but the current is only about half. This implies the the resistance of the 115 AH battery is 2 time the resistance of the 232 AH battery.
A pair of 6s are about 232 AH, A 12V battery is about 115 AH. So now what am I doing wrong or do the 12V batteries have higher resistance? Or do you have to have a separate plot for every battery AH rating and not a general plot for all lead acid batteries?
I should never have gotten started on this!!
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LScamper
Los Alamos NM
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Salvo thanks. I understand that the dynamic resistance goes up with lower SOC and it is nearly constant to around 50%. That is why I was going to ignore the change and only look at the response in a limited range.
I don't understand your resistor capacitor equivalent circuit. If you have a constant current source the voltage across the capacitor will climb at a constant rate to infinity no matter what size the resistor is.
I don't have a problem with using different lookup tables for different batteries. The question on my mind is that it seem people that use high power draw items say 12V batteries provide more current than 6Vs. This seems to be wrong according to this plot. It seems that 12V batteries have 2 time the internal resistance than 6s. Not sure what resistance to use for 12V, 20 mOhms?
This will fit with dclark's answer and seems to make sense.
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dclark1946
Richardson,TX,USA
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LScamper wrote:Salvo now I have worked myself into a corner. I think you have said there is nothing showing that the resistance of a 12V battery and a pair of 6V batteries is any different. Now after looking much to close at your plots (I have seen them may times but did not pay much attention to them) and thinking about putting some of it into the simulation I see a problem. Must be how I'm interpreting the plots.
The plot shows that the voltage at C/3 at 100% SOC is about 11.75V. Now the problem.
232 AH battery current at C/3 = 77.3A
115 AH battery current at C/3 = 38.3A
The voltage drop is the same but the current is only about half. This implies the the resistance of the 115 AH battery is 2 time the resistance of the 232 AH battery.
A pair of 6s are about 232 AH, A 12V battery is about 115 AH. So now what am I doing wrong or do the 12V batteries have higher resistance? Or do you have to have a separate plot for every battery AH rating and not a general plot for all lead acid batteries?
I should never have gotten started on this!!
I will take a shot at your question to Salvo as I am also waiting for a response from him in the post just before your last one. It would be reasonable for the 12V battery with half the AH capacity of the 6V to have twice as much impedance since the cells are smaller. However when you put the two six volt batteries in series, the battery impedance adds. For the two 12V batteries in parallel the total inpedance will half the single battery imprdance so you would end up with the same total battery impedancefor two six volt batteries in series or two 12V in parallel.
Dick
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Salvo
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That's a good question. As you state resistance is calculated as delta V over delta I. From these graphs we can do that calculation. However I wasn't interested is a specific resistance. My purpose was to show how resistance changes with SOC.
As long as the C3, C5, C10 plots are parallel to each other (which they are in discharge), resistance does not change. For example the C5 & C3 plots are about 0.3V apart over 50 to 100% SOC. We can now do the calculation delta V / delta I. Since delta V remains unchanged at 0.3V and delta I doesn't change resistance stays constant.
The charge curve is a totally different story. At 50% SOC, the voltage difference between C/5 & C/10 is 0.2V. At 90% SOC, the difference is 1.2V. That means the resistance at 90% is 1.2/0.2 = 6 times greater than at 50%.
Getting back to your question. We can replace the battery with an equivalent circuit of a capacitor and series resistor. If you apply a constant current source to this circuit the voltage will ramp up. The ramp has a fixed slope dependent on current and capacitance. Up to 75% SOC the charge plots are fairly linear. Above 75% the ramp is not linear. Voltage climbs extremely fast. I don't think the capacitance all of a sudden got smaller, and charging current didn't change, that means resistance is getting bigger.
At about 110% SOC the curves level out. The C/5 plot is at 16.3V and probably starting to gas pretty good.
dclark1946 wrote:
Salvo,
I would expect the battery impedance to be the slope of voltage vs. current (R = delta V/delta I) rather than slope of voltage vs SOC. Does it work out to be the same?
Dick
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Salvo
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The simple equivalent circuit I'm talking about is a capacitor and resistor in series. The voltage will ramp at a constant rate if you charge this circuit with a constant current. The resistor provides a voltage offset, it has no effect on the slope of the rising voltage rising during charging. At about 75% SOC something abnormal happens. The resistor begins to increase in resistance. At 90% SOC the resistance is 6 times greater than at 50%. As resistance increases, the charging voltage ramp in no longer constant. The greater the resistance the greater the voltage rise. You can see the ramp and then non-linear voltage rise in this volts per cell plot.
Regarding 12V vs. 6V resistance. Yes, 6Vs have half the resistance as 12Vs. But you need two 6Vs in series to get 12V. Series resistance adds, so now both have the same resistance.
Two 12V batteries in parallel will have half the resistance of a single 12V battery. I would use 20 mOhm for the 12V's and 5 mOhm for the 6Vs. Each battery bank (consisting of two batteries) will now have 10 mOhm resistance.
Edit, my measured 10 mOhm resistance is with a 220 AH battery bank. If you had a 440 AH bank then I would assume resistance is 5 mOhm.
LScamper wrote:
I don't understand your resistor capacitor equivalent circuit. If you have a constant current source the voltage across the capacitor will climb at a constant rate to infinity no matter what size the resistor is.
I don't have a problem with using different lookup tables for different batteries. The question on my mind is that it seem people that use high power draw items say 12V batteries provide more current than 6Vs. This seems to be wrong according to this plot. It seems that 12V batteries have 2 time the internal resistance than 6s. Not sure what resistance to use for 12V, 20 mOhms?
This will fit with dclark's answer and seems to make sense.
* This post was
edited 12/29/11 10:09am by Salvo *
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