Anyone with the right equipment up for some experimental testing?

[Solarfun4jim]

@Will Prowse
Any of the 'Tech heads' on here, that have the right equipment, up for running some experimental testing on a single lifepo4 cell?
A range of experiments plotted on a graph could be very illuminating(with regards specifically to solar charging and low C rates).
Keeping in mind, higher voltages drives the ion mobility and the proclamation by Will, that no degradation of the electrolyte happens until above 4.2v/cell, i thought the following tests could cover a lot of bases/outcomes and may provide a lot of guidance on future charging profile set ups.
I would love to do this myself, but as a newbie, i dont have the equipment necessary at this stage or the confidence in my abilities(as yet).
Anyone up for the challenge? Will P?

Thought these tests would be good....

Start with a 'single 3.2v cell' discharged to 2.5V each time(not a battery pack).


To compare how much, higher voltage (higher ion mobility) drives charging time…

Charge 200ah cell, 4.0v @ 60A (0.3C charge rate) (240w) disconnect at 3.6V level, so no 'cv stage' charge where amps drop to zero (how long to charge & then test total capacity gained discharging to 2.5v)

Charge 200ah cell, 3.8v @ 60A (0.3C charge rate) (228w) disconnect at 3.6V level, so no 'cv stage' charge where amps drop to zero (how long to charge & then test total capacity gained discharging to 2.5v)

Charge 200ah cell, 3.6v @ 60A (0.3C charge rate) (216w) disconnect at 3.6V level, should be able to get full cc/cv charge stage to zero amps (how long to charge & then test total capacity gained discharging to 2.5v)

Charge 200ah cell, 3.4v @ 60A (0.3C charge rate) (204w) disconnect at 3.4V level, should be able to get full cc/cv charge stage to zero amps (how long to charge & then test total capacity gained discharging to 2.5v)


To compare capacity gained, versus ‘charge rate’ at higher voltage input as per example 1 above.

Charge 200ah cell, 4.0v @ 20A (0.1C charge rate) (80w) disconnect at 3.6V level (how long to charge & then test total capacity gained discharging to 2.5v) – this should take 3x as long to charge, but how much does the lower charge rate affect capacity gained?


To compare same wattage input as first example…but at lower voltage

Charge 200ah cell, 3.4v @ 70.6A (240w) disconnect at 3.4V level (how long to charge & then test total capacity)


To compare charging using higher voltage but only to 3.4V cut off (aiming for 90% SOC capacity only)…

Charge 200ah cell, 4.0v @ 60A (0.3C charge rate) (240w) disconnect at 3.4V level so no cv stage charging (how long to charge & then test total capacity gained discharging to 2.5v)

NOTE: I trust this post wont be viewed as 'cheeky' and accepted in the spirt it was intended as regards experimentation. As i said earlier, i would conduct it myself, but that could be many months away !!!!

 

[Craig]

I like your style wish I had a lifepo4 battery I would make a run at it.

I am however not sure if you could get required amps at hose voltages though. Not saying you cant it just made me say hmmm will this work or not.

From my experience to get higher amp charging rates I had to crank up incoming volts but this is just my lame experience and I have not played with it that much

 

[Solarfun4jim]

I like your style wish I had a lifepo4 battery I would make a run at it.

I am however not sure if you could get required amps at hose voltages though. Not saying you cant it just made me say hmmm will this work or not.

From my experience to get higher amp charging rates I had to crank up incoming volts but this is just my lame experience and I have not played with it that much

I wasnt aware of that Craig. I just assumed that with the bench top supplies, those parameters could be accurately applied. Thanks.

 

[BiduleOhm]

I wasnt aware of that Craig. I just assumed that with the bench top supplies, those parameters could be accurated applied. Thanks.

You can set the PSU to whatever you want but the internal resistance of the battery will limit the current sooner or later.

 

[Solarfun4jim]

You can set the PSU to whatever you want but the internal resistance of the battery will limit the current sooner or later.

I had thought that would only happen once the cell was approaching full, ie when it switches from CC to CV. Although some of the testing is proposed at 4v, the termination was set at 3.6v. It might well be, you only get the bulk charge portion and not the absob CV state before terminating at this level? Are you saying that the current will drop to zero on the CV stage, long before it hits 3.6V ?
(I'm still learning all this....so, all info helpful.)

 

[Craig]

I had thought that would only happen once the cell was approaching full, ie when it switches from CC to CV. Although some of the testing is proposed at 4v, the termination was set at 3.6v. It might well be, you only get the bulk charge portion and not the absob CV state before terminating at this level? Are you saying that the current will drop to zero on the CV stage, long before it hits 3.6V ?
(I'm still learning all this....so, all info helpful.)

No it will not drop to zero but will max out at a certain number. I will do some tests this afternoon and report back. I will be using my LTO cells though so results will be different but I think concepts will be the same.

 

[BiduleOhm]

I had thought that would only happen once the cell was approaching full, ie when it switches from CC to CV. Although some of the testing is proposed at 4v, the termination was set at 3.6v. It might well be, you only get the bulk charge portion and not the absob CV state before terminating at this level? Are you saying that the current will drop to zero on the CV stage, long before it hits 3.6V ?
(I'm still learning all this....so, all info helpful.)

Whatever the charge portion you're in the current is directly (and only) related to two things: the internal resistance of the battery, and the voltage difference between the voltage you're charging at and what the battery voltage would be at rest.

For example the battery is just sitting here, no charging, no nothing, and it's at 13.5 V. You set your charger at 14.0 V. Let's say your battery has an internal resistance of 5 mOhm. You can then charge your battery with a maximum of: (14 - 13.5) / 0.005 = 100 A

After a few hours your battery is a lot more full and its rest voltage (if you were to disconnect the charger and wait a bit) would be 13.9 V. Then the maximum charge current would be: (14 - 13.9) / 0.005 = 20 A

That's why setting a higher charge voltage lets you charge at a higher current, and that's also why you can't always charge at a high current even if you want to (well, you can always set the charger at 20 V and you'll have the high current you want but it's not a good idea for obvious reasons...).

It's directly linked to the fact a battery isn't perfect (the basic electrical model for a battery is a perfect voltage source in series with a resistor (the internal resistance)) and the fact everything obeys physics laws, including Ohm's law

 

[Solarfun4jim]

No it will not drop to zero but will max out at a certain number. I will do some tests this afternoon and report back. I will be using my LTO cells though so results will be different but I think concepts will be the same.

Thanks craig.

 

[Solarfun4jim]

Whatever the charge portion you're in the current is directly (and only) related to two things: the internal resistance of the battery, and the voltage difference between the voltage you're charging at and what the battery voltage would be at rest.

For example the battery is just sitting here, no charging, no nothing, and it's at 13.5 V. You set your charger at 14.0 V. Let's say your battery has an internal resistance of 5 mOhm. You can then charge your battery with a maximum of: (14 - 13.5) / 0.005 = 100 A

After a few hours your battery is a lot more full and its rest voltage (if you were to disconnect the charger and wait a bit) would be 13.9 V. Then the maximum charge current would be: (14 - 13.9) / 0.005 = 20 A

That's why setting a higher charge voltage lets you charge at a higher current, and that's also why you can't always charge at a high current even if you want to (well, you can always set the charger at 20 V and you'll have the high current you want but it's not a good idea for obvious reasons...).

It's directly linked to the fact a battery isn't perfect (the basic electrical model for a battery is a perfect voltage source in series with a resistor (the internal resistance)) and the fact everything obeys physics laws, including Ohm's law

Thanks a bunch, getting the point now.

 

[Solarfun4jim]

Whatever the charge portion you're in the current is directly (and only) related to two things: the internal resistance of the battery, and the voltage difference between the voltage you're charging at and what the battery voltage would be at rest.

For example the battery is just sitting here, no charging, no nothing, and it's at 13.5 V. You set your charger at 14.0 V. Let's say your battery has an internal resistance of 5 mOhm. You can then charge your battery with a maximum of: (14 - 13.5) / 0.005 = 100 A

After a few hours your battery is a lot more full and its rest voltage (if you were to disconnect the charger and wait a bit) would be 13.9 V. Then the maximum charge current would be: (14 - 13.9) / 0.005 = 20 A

That's why setting a higher charge voltage lets you charge at a higher current, and that's also why you can't always charge at a high current even if you want to (well, you can always set the charger at 20 V and you'll have the high current you want but it's not a good idea for obvious reasons...).

It's directly linked to the fact a battery isn't perfect (the basic electrical model for a battery is a perfect voltage source in series with a resistor (the internal resistance)) and the fact everything obeys physics laws, including Ohm's law

Is it possible to calculate the DC internal resistance of a cell from the AC impedance data(which weirdly, always seems to be quoted on data sheets!)

 

[BiduleOhm]

Not really but they should be close to each other.

You can do the simple method: measure rest voltage, load the battery (ideally a pretty high load, 0.5 C is nice), measure current and voltage again. R = (Vrest - Vloaded) / I

 

[Solarfun4jim]

Not really but they should be close to each other.

You can do the simple method: measure rest voltage, load the battery (ideally a pretty high load, 0.5 C is nice), measure current and voltage again. R = (Vrest - Vloaded) / I

Thanks.....just that one of the 'cells' specification sheet i was looking at had an AC impedance of only 0.25milliohms, which seemed far removed from your previous example of 5 milliohms...thought there might be a massive difference between the two.
Much appreciate all the assistance you have given.

ps... i am really envious of all you folks that have the electrical/electronics experience....it feels like i'm missing a language....lol.

 

[BiduleOhm]

5 mOhm was just an example and with 4 cells for a 12 V pack the battery will have a 1 mOhm internal resistance, and 2 mOhm for a 24 V pack so not too far off.

Well, 95 % of what I know in electronics is self-taught, you just need motivation

 

[Solarfun4jim]

5 mOhm was just an example and with 4 cells for a 12 V pack the battery will have a 1 mOhm internal resistance, and 2 mOhm for a 24 V pack so not too far off.

Well, 95 % of what I know in electronics is self-taught, you just need motivation

Magic...gottcha now.
Hats off to you on being self taught.

ps in my testing scenarios above, all were to be done on a single cell( just for reference)

 

[Solarfun4jim]

5 mOhm was just an example and with 4 cells for a 12 V pack the battery will have a 1 mOhm internal resistance, and 2 mOhm for a 24 V pack so not too far off.

Well, 95 % of what I know in electronics is self-taught, you just need motivation

So on a single cell that is sitting at 2.5v, with a charging voltage of 4v, =4-2.5 =1.5/ 0.00025 =6000A When the cell approaches cut off point at 3.6v = 4-3.6 = 0.4/0.00025=1600A in theory? Even if you allow double for the DC internal resistance, it still shouldn't influence the charging if my sums are correct. The amperage does seem totally out of whack....so have i made some error in converting the 0.25milliohms figure???

 

[BiduleOhm]

The amperage does seem totally out of whack....so have i made some error in converting the 0.25milliohms figure???

No, that's correct.

I didn't mentionned some other things about the battery to keep things simple: the internal resistance can change with SoC (typically it'll rise at the bottom and top of SoC); the basic model is, well, basic, the actual model is a lot more complex with capacitances, more resistances, ...

Also 4 V seems a bit high, I never saw anyone use a charge voltage that high.

And when we're talking about mOhm/µOhm you can't ignore the wire resistance, 4.000 V at the charger can be 3.850 V at the battery for exemple. So you need to either account for the wire or do all your measurements at the battery terminals.

 

[Solarfun4jim]

No, that's correct.

I didn't mentionned some other things about the battery to keep things simple: the internal resistance can change with SoC (typically it'll rise at the bottom and top of SoC); the basic model is, well, basic, the actual model is a lot more complex with capacitances, more resistances, ...

Also 4 V seems a bit high, I never saw anyone use a charge voltage that high.

And when we're talking about mOhm/µOhm you can't ignore the wire resistance, 4.000 V at the charger can be 3.850 V at the battery for exemple. So you need to either account for the wire or do all your measurements at the battery terminals.

Thanks...will use your guidance. Easy enough to 'set it' at the battery terminals for accuracy.(when i eventually do this)
Accept that 4v is very high, but that was part of the experiment, to see how much the higher differential could drive the rate of charge....i suspect the benefits tail off around 3.65V and any higher will give no appreciable difference, but would like to see exactly what the difference actually is in reality. For solar charging with very low C rates, i thought it would be worth exploring.
Thanks again.

oh ps...the 4v trial was based on something Will Prowse stated, that the degradation of the electrolyte doesnt start till above 4.2V. So long as charge is terminated at 3.6v, didnt think any harm would befall the cell.

 

[BiduleOhm]

Accept that 4v is very high, but that was part of the experiment, to see how much the higher differential could drive the rate of charge....i suspect the benefits tail off around 3.65V and any higher will give no appreciable difference, but would like to see exactly what the difference actually is in reality. For solar charging with very low C rates, i thought it would be worth exploring.
Thanks again.

Yeah you're deep in the knee at this point, I bet you'll only gain 0.1 % of capacity but you'll loose a lot of life on the battery.

oh ps...the 4v trial was based on something Will Prowse stated, that the degradation of the electrolyte doesnt start till above 4.2V.

Ah, yes, I forgot a bit what was the initial subject...

 

[Solarfun4jim]

Yeah you're deep in the knee at this point, I bet you'll only gain 0.1 % of capacity but you'll loose a lot of life on the battery.

The experiment wasnt to gain capacity, it was to see if charge rate could be speedied up due to the higher mobility of ions. With solar, you are often trying to grap power quickly in short bursts etc. So long as it is terminated at around 3.6v and not 4v, then i didnt think you would loose any longevity??

 

[Solarfun4jim]

I like your style wish I had a lifepo4 battery I would make a run at it.

I am however not sure if you could get required amps at hose voltages though. Not saying you cant it just made me say hmmm will this work or not.

From my experience to get higher amp charging rates I had to crank up incoming volts but this is just my lame experience and I have not played with it that much

@Craig
Craig...just realised you are talking about a four cell battery pack. My testing was only on a single cell. I have now changed the type colour in the original post to ensure that idea's are based on single cell. Thanks