# What dictates C charge rate?



## lnpurnell (Sep 27, 2012)

Hi all,

I have been looking through the interweb on the definition of C charge rates and how they are calculated by manufacturers for each battery.

Can someone tell me how manufacturers come up with their 'C' rating for a specific type of cell?

Is it the internal resistance of the battery the factor?

For example the rating for the Calb CA series cells is 3C but is this to 100% or can it achieve higher C rating before hitting the CC/CV level?

Thanks!

Leigh


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## Roy Von Rogers (Mar 21, 2009)

Calb CA cells.

Max Continuous Discharge: 3C 

10 second Discharge: 10C 


Roy


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## IamIan (Mar 29, 2009)

OEM C rates tend to be a combination of several battery effects all summed up... to a number that makes the engineer's / user's job easier.

The goal of the C rate limit recommendations is to prevent or greatly reduce the amount of permanent cell damage.... without making it so complex that everyone involved has to have a degree about electrochemical reactions.

- - - - - - - -

So yes internal resistance plays a part ... so does cell initial temperature ... ambient temperature ... the SoC ... the diffusion rate of that particular battery ... etc.

And there is also a difference between the continuous C rate and the Pule C rate 10s or 1s ... because as the usage context changes , you get different results.

- - - - - - - - 

Some people push the limits of the cells ... and the cells might take it for a while ... but it can come at the cost of shortened operational life time ... others baby the cells but getting longer operational life times.


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## Karter2 (Nov 17, 2011)

I thought the OP was asking about CHARGE rates ?


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## Roy Von Rogers (Mar 21, 2009)

Karter2 said:


> I thought the OP was asking about CHARGE rates ?


 
Doesn't matter, its a two way street.

Roy


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## Karter2 (Nov 17, 2011)

Roy Von Rogers said:


> Doesn't matter, its a two way street.
> 
> Roy


 So , why then do manufacturers quote a different max charge rate vs the max discharge rate ??
( which was the OP question)


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## Roy Von Rogers (Mar 21, 2009)

Karter2 said:


> So , why then do manufacturers quote a different max charge rate vs the max discharge rate ??
> ( which was the OP question)


 
Its called CYA, its a fact that those CA cells can handle 3c charge rate or even more, whether or not it shortens its life time, hard to tell, all the tests I've seen show no appreciable heating or loss of capacity at 3C, which is generally a good indicator.

But its actually irrelevant, cause where you going to get 3C from, unless its a real small cell.

Roy


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## IamIan (Mar 29, 2009)

IamIan said:


> without making it so complex that everyone involved has to have a degree about electrochemical reactions.


But to be fair ... if you want to get more into the details of how / why / etc ... Perhaps one good place is the attached 150page pdf... beware, it is a little bit dry and technical.


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## Karter2 (Nov 17, 2011)

Roy Von Rogers said:


> But its actually irrelevant, cause where you going to get 3C from, unless its a real small cell.
> 
> Roy


 Agreed, generally not relevant but there are situations..
..I bet the F1 guys would like to have much higher charge rates..
..They have about 5 secs to push 60kW back into their KERS packs. !


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## lnpurnell (Sep 27, 2012)

Karter2 said:


> I thought the OP was asking about CHARGE rates ?


Yes Karter, 

I was talking about charge rates, there is a distinct difference between the the charge and discharge rates in most cells. I just wanted to know why they are so different.

Leigh


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## lnpurnell (Sep 27, 2012)

Roy Von Rogers said:


> Its called CYA, its a fact that those CA cells can handle 3c charge rate or even more, whether or not it shortens its life time, hard to tell, all the tests I've seen show no appreciable heating or loss of capacity at 3C, which is generally a good indicator.
> 
> But its actually irrelevant, cause where you going to get 3C from, unless its a real small cell.
> 
> Roy



Hi Roy,

So why would a Calb CA cell for example have a 3C charge rate and has been rated for 12C discharge, if your theory is correct wouldn't these numbers be interchangeable? 

I am working on a project that has a 200KW DC source so I do have enough power.

Leigh


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## lnpurnell (Sep 27, 2012)

IamIan said:


> But to be fair ... if you want to get more into the details of how / why / etc ... Perhaps one good place is the attached 150page pdf... beware, it is a little bit dry and technical.


Thanks!

I will take a long, hard look at the 150 pages. :-D

Leigh


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## IamIan (Mar 29, 2009)

lnpurnell said:


> I was talking about charge rates, there is a distinct difference between the the charge and discharge rates in most cells. I just wanted to know why they are so different.


That pdf is a good place to start with the chemical reasons why there is a difference ... but it would also be worth noting the marketing effects on the OEM claims ... Higher Discharge Rates are more marketable than higher Charge rates are ... both are nice ... but 10C vs 5C discharge will sell better than 10C vs 5C Charge rates.

Part of that is infrastructure ... how common is a 100kw line available to charge a 10kwh battery.

Part is usage patterns ... most people's operational conditions require the battery to be discharged much faster than charged ... they often have more time to allow it to charge... ~8 hours over night to charge a BEV battery is not as much of a inconvenience as ~8 Hours to do the commute trip.


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## John Metric (Feb 26, 2009)

I have this testing done on my cells before we quote a C rate. The C-rate for discharging gives different power levels. You can check these out at my facebook page "Lonestar EV Racing"


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## lnpurnell (Sep 27, 2012)

John Metric said:


> I have this testing done on my cells before we quote a C rate. The C-rate for discharging gives different power levels. You can check these out at my facebook page "Lonestar EV Racing"


Hi John,

These seem to all be from LiPo cells, I am assuming that the charge cycle of a LiPo cell would be significantly different compared to a LiFePo4 cell. 

Can anyone else shed some light on this?

I have been reading into the PDF and understand that the charging regime for the first couple of cycles is important because the SEI and cell properties have not been fully formed but after that what stop the cell from being charged as fast as it can be discharged?

Thanks

Leigh


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## kennybobby (Aug 10, 2012)

@ John Metric 

What was your charge rate in these tests--same as the discharge?


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## pm_dawn (Sep 14, 2009)

Hi !

In the data sheet for the A123 pouch you can see the relasionship between charge and discharge rates pretty well.

http://www.raceyard.de/tl_files/Newsletter/Dateien/A123-AMP20-M1HD-A-1-Data-Sheet.pdf

it shows best discharge power rate in 10s pulse to 45C @ 1,6v
and best charge power rate in 10s pulse to 14,5C @ 3,8v 

regards
/Per


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## Roy Von Rogers (Mar 21, 2009)

lnpurnell said:


> Hi Roy,
> 
> So why would a Calb CA cell for example have a 3C charge rate and has been rated for 12C discharge, if your theory is correct wouldn't these numbers be interchangeable?
> 
> ...


The question is why would you want to do that in the first place unless you were racing a vehicle, and at that point ask some who have done so, there are a few in here. I know for a fact you can put a lot more in those CA cells than stated by CALB. But you need to understand that they will always be on the conservative side, for reason I stated before.

Just because you have such a source available doesn't mean you have to use it, why stress the cell, unless you have a good reason to do so.

If you have a need to know more, go to EVTV and look at some of the tests done there. Don't take my word or anyone else's here, watch the tests and judge for yourself.


Roy


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## major (Apr 4, 2008)

pm_dawn said:


> Hi !
> 
> In the data sheet for the A123 pouch you can see the relasionship between charge and discharge rates pretty well.
> 
> ...


Hi Per,

Your C-rate calculations are best case for discharge and worst case for charge. If you use the nominal power of 65 Watts (@ 1 hour) for your base, then the maximum discharge E-rate = 23 and charge E-rate = 17. They also list a nominal discharge power of 1200W. From the 10s pulse graph, that would be at 50% SOC. The charge power for 10s at 50% SOC would be 950W. From both of these comparisons, it looks like charge power can be 70 to 80% of discharge power.

Different ways of looking at it 

major


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## Karter2 (Nov 17, 2011)

Roy Von Rogers said:


> The question is why would you want to do that in the first place unless you were racing a vehicle......,


 With respect Roy, .... it doesnt matter why he might want to... he was just asking why the difference.
...and he didnt say he was going to do it.!



Roy Von Rogers said:


> ....Just because you have such a source available doesn't mean you have to use it, *why stress the cell*,....


 ..and there you seem to be giving a reason , but one that still needs a technical explanation[/QUOTE]


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## Coulomb (Apr 22, 2009)

> Why the difference?

From the paper, it seems that lithium plating happens with high charge current. Lithium plating is not completely reversible. It's something to do with the electropotentials as to why that doesn't happen on discharge.


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## John Metric (Feb 26, 2009)

kennybobby said:


> @ John Metric
> What was your charge rate in these tests--same as the discharge?


I was just responding to the person posting about how to test.
I don't have a good reason why the recharge rate is so low compared to the charge rate
You can see at the tail end of my discharge charts, the cell rebounds on its own then we begin to recharge. We watch cell temperature.
I think recharge was about 1C here.


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## Karter2 (Nov 17, 2011)

John Metric said:


> I don't have a good reason why the recharge rate is so low compared to the charge rate.


 John,
what do you mean by "recharge rate " ?


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## John Metric (Feb 26, 2009)

The specified c rate for recharging 
It is usually much less than the discharge rate. 
On the lipo cells its 5C charge and 100C discharge. 
I think the original topic was about why is the charge rate so significantly different that the discharge rate. 

I posted about how we test discharge rates. They are a family of curves with decreasing capacity and temperature.


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## Karter2 (Nov 17, 2011)

sorry John, i uderstand the Charge and Discharge rate discussion ...
...but you were comparing "charge" rate to "recharge" rate ??
What is the difference between charge and recharge ?
Or did you mean to say "discharge rate " ?
Quote:
Originally Posted by *John Metric*  
_I don't have a good reason why the recharge rate is so low compared to the charge rate._


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## tomofreno (Mar 3, 2009)

Karter2 said:


> sorry John, i uderstand the Charge and Discharge rate discussion ...
> ...but you were comparing "charge" rate to "recharge" rate ??
> What is the difference between charge and recharge ?
> Or did you mean to say "discharge rate " ?
> ...


 He means CHARGE. You charge the battery, or you can say you REcharge it after it is discharged.


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## Karter2 (Nov 17, 2011)

> He means CHARGE. You charge the battery, or you can say you REcharge it after it is discharged.


 I get that 100%,...
so,.. are we then to understand from this following comment, that RECHARGING after a discharge is much lower "C" rate than a CHARGE "C" rate at any other time. ??


John Metric said:


> I don't have a good reason why the recharge rate is so low compared to the charge rate.


 If so, i would like to understand more , as i never heard of this before.

OR.. did John mean to say..
.._I don't have a good reason why the recharge rate is so low compared to the DISCHARGE rate_...
???


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## lnpurnell (Sep 27, 2012)

I think I have my head around this now.

The lithium plating is the main issue, but I am surprised that manufacturers have not really pushed this issue now that fast charging is becoming more prolific. 

Having the infrastructure and batteries to push 200KW into a pack would remove the issue of supposed 'Range Anxiety' and keep the packs lighter than trying to cram 80Kwh of batteries into a chassis.

If a Leaf could run for 100 miles but then charge to 90% in less than 10 minutes, there would be a whole different picture in the electric car market.

For example the Enerdel performance cells can easily do 5C charge rate, at 100% DOD they last over 3,000 cycles at 95% original capacity (at 0 - 5 degrees c).

These cells have been out for a couple of years, if a cheaper version of these cells could be made then packs would only have to stay at the 20Kw mark but appeal to a wider variety of people.

You would be pushing the boundaries if you required a larger pack (say Tesla's 80KwH pack and were wishing to charge it to 90% in 10 minutes) then the power input requirements would be approaching the Megawatt range for each charging station.

The main approach for Tesla is 180 mile charge in 20 minutes which means that a significant portion of their pack is left idle (which is good for life-cycle, but means that a large weight is carried around on a constant basis)

It is unlikely that Tesla will increase their supercharge stations power output; but their main restriction is what current rate their battery's will take . The newer Gen 3 vehicle will likely have a smaller battery (as costs in 3 years won't have reduced significantly to offer the 80KwH option for alot the cars price range). Which means that the Gen 3 car will likely have a 30-40KwH pack (to give higher range than other competitors) but the battery chemistry will allow the use of the 120 Kw chargers on a smaller pack (higher C rate). 

The main contention for the adoption of electric cars has never been "The vehicle can only do 100 miles". The issue has been "I have to wait 6 hours after doing 100 miles"

Leigh


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## Roy Von Rogers (Mar 21, 2009)

This is somewhat the same problem when gasoline vehicles first came on the market, very few gas stations, many took along extra gas so they can travel long distances.

Matter of fact in those days many farm vehicles would run on fuel oil, for it was cheaper and easier to get.

I have an old Farmall tractor that has two tanks, a large one was used for fuel oil, a smaller aux tank was gasoline. You would start it with gasoline and get it hot by closing the louvers on radiator, once it got hot, you switched over to fuel oil.

As times go by, these problems with EV's will also be solved, just have to give it some time. As we get lighter vehicles, the range will be increased to 300 miles, and once you drive that far, your ready for a break anyway, so 30 to 45 minutes to recharge, wont be a big deal.

And for local driving it will be charging at home.

Roy


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## pm_dawn (Sep 14, 2009)

major said:


> Hi Per,
> 
> Your C-rate calculations are best case for discharge and worst case for charge. If you use the nominal power of 65 Watts (@ 1 hour) for your base, then the maximum discharge E-rate = 23 and charge E-rate = 17. They also list a nominal discharge power of 1200W. From the 10s pulse graph, that would be at 50% SOC. The charge power for 10s at 50% SOC would be 950W. From both of these comparisons, it looks like charge power can be 70 to 80% of discharge power.
> 
> ...


Hi Major !

I know your a smart guy. So I'm a little confused about your statements here.
This thread is about C rates. I tried to find the C-rates of the A123 pouches by using the datasheet. In the Power graph it is stated Max voltage and min voltage. I assumed that Max voltage occurs when charging and min voltage at discharge. If the Resr of the battery is constant the max power out of a battery occurs at half the Vnom, I guess that is why the graph states Vmin to 1,6V (3,2/2). If you take that into account then it would give the figures that I stated. 
The discharge current for max power would be: 
1450w = 1,6V x 906A which is about 45C pulse

At charge on the other hand this would give:
1100w = 3,8v x 290A which is about 14,5C pulse

So there is a huge difference in charge/discharge current capability.
But the power is not that different.

The core difference i think is the upper voltage for charging that limits the power. And the damages a high charge voltage can do.

But I'm no expert in how theses cells work internally.

Regards
/Per


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## major (Apr 4, 2008)

pm_dawn said:


> major said:
> 
> 
> > Hi Per,
> ...


Hi Per,

That was my point.


pm_dawn said:


> I tried to find the C-rates of the A123 pouches


 I do not think you find valid C-rates from back calculating from nominal power specifications. The power graph shows an average value of power for a particular SOC yet you treated it as an instantaneous value at a specific ending voltage. Hence my comment about the best case for discharge and worst case for charge.

Regards,

major


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## pm_dawn (Sep 14, 2009)

Found this datasheet today.

seems they have reworked it a little bit.

http://assets.buya123batteries.com/images/a123/AMP20_data_sheet.pdf

Little different power graph there, pretty flat over the SOC. Odd but any way.

I agree with you on some parts Major.

But lets look at these numbers....

I made a model including a ESR of the cell to try to find out where this is heading current wise.

I found some numbers online from people that has been testing real genuine A123 pouches, They state about 1,7mOhm in ESR at discharge fairly high SOC and warm cells. And about 1,5mOhm at charge of cells.

OCV of 3,35v and discharge current to 500A gives voltage at 2,5v and total 1250w of discharge power.

Same OCV but charge current 320A gives a voltage of 3,83v and a charging power of 1225w.

That is 25C charging for 1250w and 16C charging for 1225w.

That is 64% C-Rate in charge mode with regards to discharge current if the cell can take the same Power in as out.

If we look at the power graph from the above datasheet we can see that the best case for charging power is almost 1000w and for discharging about 1400w.
In my data model that gives:
Charging: 1000W at 3,75v and about 267A that is a C-rate of 13,3C
Discharging: 1400w at 2,33V and 600A that is a C-rate of 30C

So from the datasheet that indicates best case charging power to about 1000w and best case discharging power to almost 1400w. 

Give a C rate difference in these cases to 13,3C for charge and 30C for discharge........

Tell me what I have misunderstood about these figures.......

Regards
/Per


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## major (Apr 4, 2008)

pm_dawn said:


> Tell me what I have misunderstood about these figures.......


You are drawing specific case current conclusions from nominal or typical or general power information. I don't see that as valid reasoning. That is all I am saying. You are arbitrarily setting your charge C-rate too low in my opinion. Why limit to 3.83V/c during charge? That is likely an open circuit limit.

If you want C-rate limits, get them from the manufacturer, don't back calculate them from some power graph. It's not the same thing.


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## pm_dawn (Sep 14, 2009)

Well i guess that means that i'm pulling useless numbers out of my a.s !

But fact still remains: for the same given power level, in and out of a cell the discharge current will always be higher than the charge current. 
Pure simple Facts 

So even if the cell CAN do the same charge Power as discharge Power it will not be even close in the respect of C-Rates in and out. Ohms law for that weird ESR of the cell will dictate that. 

Or maybe I just pulled Ohms law out of there to 



Back to the core question: Why the difference ?

My theory is that it has to do with the breakdown of electrolyte in conjunction with lithium plating. 
If we try to cram enough power into a cell the cell voltage will rise above the breakdown level of the electrolyte. Maybe that is the biggest problem.

Could be a reason to look even more into using the lower voltage chemistries like LiFepo4 since that would give a little more headroom up to the level where breakdown occurs if the electrolyte is fairly same mixture. 

Regards
/Per


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## major (Apr 4, 2008)

pm_dawn said:


> Well i guess that means that i'm pulling useless numbers out of my a.s !
> 
> But fact still remains: for the same given power level, in and out of a cell the discharge current will always be higher than the charge current.
> Pure simple Facts
> ...


Yes per,

Maybe there are differences in the physics between charging and discharging. But I believe that for short durations in useable SOC range, the allowable C-rates are pretty close to equal for charge and discharge. I don't think I was at the limits of the EESS I used for work done with hybrids, but I did use the same current limits for both charge and discharge, so I would see higher power under braking (charge) than motoring (discharge).

Here is an example of a Lithium based EESS with a limit of 130 kW for discharge and 170 kW for charge. http://www.hybridrive.com/lithium-ion-energy-storage-system.asp Again, power; not C-rate.

If you have a particular Lithium battery and plan to hit it hard with charge, I'd suggest you ask the engineering department at that cell maker to get the right answer.

Regards,

major


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## pm_dawn (Sep 14, 2009)

Yes I think you're right Major !

It all comes down to the battery !
But that is also the big problem, it is almost impossible to get any hard info about any batteries unless you sign 45 pages of sellyoursoulnondisclosureyadayada, and order a test order of about 1000 pieces. If you even get to talk to any of the real producers of batteries.

REgards
/Per


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## lnpurnell (Sep 27, 2012)

One of my original thoughts when opening this thread was how the manufacturer calculated the 'C' rate in the first place. 

Is the 'c' rate calculates over a 100% charge of the battery then the current level is averaged out?

Would this mean that if we chose a charge level of 80% (and kept out of the CV curve) then the charge rate could be higher. 

Does anyone have any evidence of when 'lithium plating' occurs? Is it during the whole charge cycle or is it at specific points in the cycle? (such as the last 10% where most heating issues are observed)

Leigh


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## jddcircuit (Mar 18, 2010)

lnpurnell said:


> One of my original thoughts when opening this thread was how the manufacturer calculated the 'C' rate in the first place.
> 
> Is the 'c' rate calculates over a 100% charge of the battery then the current level is averaged out?
> 
> ...


Leigh
I have the same question.

I am getting real close to running an experiment to test out 3C charge/discharge for some CALB CA40 cells. I am avoiding the CV phase altogether. I seem to be able to detect the point where the internal resistance starts to increase at both ends of the charge profile with the constant current being applied. I want to stay out of this high resistance region. I am also noticing other plateaus in the slope of the voltage profiles that may indicate other SOC phase changes that may be significant.

I plan to apply the same IR slope thresholding to terminate charge and discharge for a least two stacks of cells. One stack will use a 1C rate cycle and the other 3C. I have individual cell voltage monitoring designed for negligible parasitic loads so this test is also trying to determine if there is any SOC drift during the life cycle testing.

I am considering a control group that will use the CC - CV charge profile to squeeze every Ah into the cells but the main limitation for this test group scenario is it will take a long time.

As far as Lithium plating goes. I am doing some dry runs of my test setup on some old CALB SE series cells and just had a very irregular voltage profile in one of the cells within the six series stack. I think this may reflect some type of breakdown or formation within the cell during that cycle. I don't want to jump to conclusions but I am eager to see what the data logs have to show.

Regards
Jeff


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## IamIan (Mar 29, 2009)

lnpurnell said:


> Does anyone have any evidence of when 'lithium plating' occurs?


JensGroot 150page pdf posted on post #8 previously points out some of that evidence.

The short version.

It can happen at any point in the curve ... it is less likely to occur in some parts of it than other parts ... and it is less likely to happen in some conditions than others.

Low temperatures it becomes more likely.
High Charge rates it becomes more likely.
High SoC it becomes more likely.

Perspective is also important ... If it does not happen enough or fast enough ... it just isn't significant ... for example ... the loss of 10 individual molecules per cycle from the whole battery pack , is still an insignificant amount ... even if technically it is happening to those 10 molecules every cycle... we just wouldn't care that it was still happening.

- - - - - - -

'Li Plating' is only one mechanism ... there are other degradation mechanism as well... those lumped under ... 'corrosion' ... 'gassing' ... etc.

The short version ... of all of it ... the closer toward its limits ... the less well the battery will do in the long run ... be it temperature , Charge Rate, discharge rate , SoC , DoD , etc.... and the more of those combined the effects can be compounding... ie high charge rate combined with low temperatures is worse than just the high charge rate by itself at less extreme low temperatures.


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## major (Apr 4, 2008)

Here's an interesting find:










This is for an EnerDel cell. From: http://www.enerdel.com/wp-content/u...-Moxie-Prismatic-Cell_ENERGYSTORAGE_Final.pdf It is a bit different than the chart Per showed earlier. It also uses power, not C-rate. But if, just for kicks, you used the nominal cell voltage (3.65V/c) and back calculated, you'd get relatively equal* C-rates for charge and discharge at opposing DOD, if you follow what I mean. Notice the logarithmic scale for power. That translates into some very high C-rates (for 10 second pulse).


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## IamIan (Mar 29, 2009)

major said:


> But if, just for kicks, you used the nominal cell voltage (3.65V/c) and back calculated, you'd get the same C-rates for charge and discharge at opposing DOD, if you follow what I mean.


'very similar' but not the same ... and there are C rate differences shown there ... favoring higher C rates for discharge.

There is a clear ( although small ) difference between discharge and charge for power levels ... If you flipped them both to track from peak power to minimum power , as you suggested ... that graph shows a higher discharge power than a charge power from peak all the way down.

And that has bigger implications for C rate ... on discharge the voltage dips ... on charge the voltage rises ... if discharge and charge both showed the same max watts of power , that means that the discharge is a higher C rate ... and the charge is a lower C Rate... but with that graph showing slightly higher power for discharge ... that means an even bigger C-Rate gap.

- - - - - 

I feel like I did too much hand waving and not enough detail / explanation.

If it helps to visualize the power graphs ... I flipped that last one for charge and discharge ... see attached.

- - - - - 

Another example of the concept on a different battery ... The A123 - 20Ah pouch cell power graph with a much larger gap between charge and discharge power levels ... still has much higher discharge C rates than charge C Rates... see attached.

Minimum Discharge 10% SoC ~800 W @ 1.6v =~500 Amps ... ~25C
Minimum Charge at 90% SoC ~875 W @ 3.8v = ~230 Amps ... ~11C

At that point that cell , that OEM lists about ~2x more discharge C-Rate than charge C-Rate.

I like the A123 power graph better because they tell you the end point max and min cell volts... the other points might be more fuzzy ... but at least it defines those points... better than just a nominal Voltage... but because of the dV=IR I suspect we would need more than ~2x the power difference in charge vs discharge curves , just for the C-Rates to get very close to each other.

Although ... along the lines that Major suggested ... the difference might very well be mute ... and for all 'practical' purposes they may be functionally the same in the application it is used.


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## major (Apr 4, 2008)

IamIan said:


> Another example of the concept on a different battery ... The A123 - 20Ah pouch cell power graph with a much larger gap between charge and discharge power levels ... still has much higher discharge C rates than charge C Rates... see attached.
> 
> Minimum Discharge 10% SoC ~800 W @ 1.6v =~500 Amps ... ~25C
> Minimum Charge at 90% SoC ~875 W @ 3.8v = ~230 Amps ... ~11C
> ...





> that OEM lists about ~2x more discharge C-Rate than charge C-Rate


The OEM is not listing C-rate information. He is showing some typical or nominal power curves. IMO, you are drawing similar invalid conclusions from the same data as pm_dawn was in post #17.


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## IamIan (Mar 29, 2009)

major said:


> The OEM is not listing C-rate information. He is showing some typical or nominal power curves. IMO, you are drawing similar invalid conclusions from the same data as pm_dawn was in post #17.


Not quiet.

They give you all the information needed ... the rest is just math.

As long as V=IR and P=VI ... the only way the charge C-Rate could be equal to the discharge C-Rate ... is if it shows a higher Charge power rate ... and none of the graphs to date show that ... to date all the power graphs show greater discharge power rates ... which thanks to V=IR and P=VI ... requires those Charge C-Rates to be less than the Discharge C-Rates.

The Min Volts and Max Volt Data listed on the A123 sheet I referenced ... is why it is NOT a invalid conclusion.

If you know power and volts ... then you have defined the amps
If you know the Ah of the cell and the amps ... then you have defined the C-Rate.

~800 Watts listed on power graph for discharge power ... 1.6v listed on power graph as the minimum voltage for that point ... in order to get ~800 watts from 1.6v that discharge MUST be at ~500Amps ... and it is not invalid... it also tells you it is a 20Ah cell ... 500Amps from a 20Ah cell ... is a 25C rate... that is 100% valid conclusion from the data on that graph.

It is not the same as the previous power graph conclusion that was based on nominal voltage... there is a significant difference between using nominal voltage over the whole power curve for a specific point on the curve ... instead of using the actual voltage for a given point on the curve.

And As I wrote previously ... sense they only gave the voltage for two points ... It only defines the Amps ... and thus the C-Rate for those two points.... ~25C discharge @ 10% SoC and ~11C charge @ 90% SoC... All the rest of the points on the power curve we don't have a defined C-Rate for ... but for those two points that we have all those pieces ... the C-Rate is defined.

Although we may not have the other points defined ... we do still know V=IR and P=VI ... thus we know that in order to have the same C-Rate the charge power MUST be GREATER than the discharge power ... Any power graph that shows Charge power equal or less than discharge power , is showing you the charge is a smaller C-Rate than the Discharge.... even if they don't give you all the other data ( like voltage points ) in order to define by how much.


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## major (Apr 4, 2008)

IamIan said:


> ~800 Watts listed on power graph for discharge power ... 1.6v listed on power graph as the minimum voltage for that point ... in order to get ~800 watts from 1.6v that discharge MUST be at ~500Amps ... and it is not invalid... it also tells you it is a 20Ah cell ... 500Amps from a 20Ah cell ... is a 25C rate... that is 100% valid conclusion from the data on that graph.


Why not use 3.0 Volts? It said 1.6V was a minimum, not actual. And I listed why I thought your method was invalid when replying to Per. I don't want to repeat myself. I am not saying all charge C-rates are higher than discharge C-rates for similar powers. I am saying you are drawing invalid conclusions. I think it is similar in nature to previous misunderstanding regarding SOC and energy. Power is not charge rate.


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## IamIan (Mar 29, 2009)

major said:


> Why not use 3.0 Volts?


Because 3.0v is not what was listed for a given reference point ... we were given a min volt and a max volt... neither of those was 3.0v.



major said:


> It said 1.6V was a minimum, not actual. And I listed why I thought your method was invalid when replying to Per. I don't want to repeat myself.


Yes I did see your explanation for why you thought it was invalid ... which is why I then tried to explain why that is not the case.

You compared it to the conclusions in post #17 ... I explained what the significant differences are ... and why that comparison is incorrect.

As for 1.6v minimum voltage:
Dude , seriously ...  ... it tells you the minimum voltage for the discharge ... and it shows you the power at the minimum point ... the C-Rate for that point is defined ... voltage goes down as you discharge ... especially as you get to 10% SoC ... especially at the power rates listed... the 10% SoC point on discharge is the only reasonable point I see for the minimum voltage to be on such a discharge like that.

And I explained that none of your reasons removes the requirement of V=IR and P=VI on a power graph to show charge and discharge to be the same C-Rate.



major said:


> I am not saying all charge C-rates are higher than discharge C-rates for similar powers.


ok ... and that is the exact opposite of what I am claiming ... I am the one pointing out to you ... that V=IR and P=VI ... REQUIRES ... at the same power the Charge C-Rate MUST be lower ( not higher ) than the discharge C-Rate at that same power... even if it is only a graph of the power rates , we still know this.

And because it does require this to be true ... any power graph that does not show the charge power rate higher than the discharge power rate IS showing a smaller charge C-Rate... even if it is a power graph , that does not change this requirement.

Once the charge power rate is higher than the discharge power rate ... we would then need other data in order to know which one charge or discharge is at a higher C-Rate.

for example:
3.45v rest voltage ... if it will produce a dV of 0.1v from dI of 20A from a 20Ah cell .. then a discharge of 20Amps pulls down that voltage to 3.35v times 20 amps = 67Watts of power from a 1 C discharge ... if charged at 20 amps the voltage goes up to 3.55v times 20 Amps = 71Watts of power for the same 1C rate.... The ONLY way to get the same C-Rate is if the charge power rate is HIGHER than the discharge power rate.



major said:


> I am saying you are drawing invalid conclusions. I think it is similar in nature to previous misunderstanding regarding SOC and energy. Power is not charge rate.


And I'm trying to explain why my conclusions are valid... and some of the conclusions you've made on this issue are not.

I never said Power was charge rate ... I pointed out they are related ... and they are ... by the following.

Power = Volts x Amps
( Change in Voltage ) = ( change in Amps ) x Resistance
C-Rate = Amps / ( Cell Ah Capacity )

These three are what make the power graph requirements I listed:

The Charge power rate must be HIGHER than the discharge power rate in order to have the same C-Rate ... even if it is a graph of power this still holds true.

You claimed:


> But if, just for kicks, you used the nominal cell voltage (3.65V/c) and back calculated, you'd get the same C-rates for charge and discharge at opposing DOD, if you follow what I mean.


And I showed this conclusion of yours is incorrect... from just about every way to look at it.

The power graph you posted does not show what you claim ... it shows the opposite ... flipped as you asked ... the charge power is slightly less than the discharge power ... even with the voltage you listed it would result in a slightly smaller charge C-Rate ... and that even incorrectly ignores V=IR and P=VI... and if I did this same flip method with other power graphs the gap from charge to discharge is even larger.

When you include the requirement that happens from V=IR and P=VI ... that tells you that flipped like that the max to min C-Rates for charge are smaller than the max to min C-Rates for discharge... and it shows you that the average C-Rate for charge is less than the average C-Rate for discharge.

If you don't flip it , as you asked for ... V=IR and P=VI requirement would still disagree with you ... that requires the charge C-Rate must be less than the Discharge C-Rate for about ~75% of that power graph ... the last ~25% of that power graph is undefined for that V=IR and P=VI requirement... but we still know from the flip we did the max , the min, and the average are all less.


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## major (Apr 4, 2008)

IamIan said:


> Because 3.0v is not what was listed for a given reference point ... we were given a min volt and a max volt... neither of those was 3.0v.


3.0V is between the minimum and maximum. So it fits the criteria. 



> What dictates C charge rate?


 The title of the thread. I am saying that these power curves do not dictate the charge C-rate, which you are saying near as I can tell. I think using such data to establish charge current limits will artificially keep the regeneration and fast charging levels lower than the product's capability. 

If this chart was intended to be used to relate (or dictate) maximum charge and discharge currents, or C-rates, why didn't the manufacturer use C-rate as the vertical scale instead of power? 

Notice I used


> just for kicks


in my example calculation because I was using Per's (and apparently your) logic. 

You can use these graphs to draw whatever conclusions you want. I am entitled to my opinion, which is that your conclusions are invalid. It is also my opinion that generally speaking, as long as other limits are observed (like temperature, SOC, etc), the charge and discharge current (or C-rate) limitations are nearly equal. In other words, you can charge a cell as fast as you can discharge it, over a broad range of SOC, but not to 100%.


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## major (Apr 4, 2008)

IamIan said:


> major said:
> 
> 
> > IamIan said:
> ...


O.K. Let's look at what you said there and compare to my calculation. 

800 Watt discharge is about 15% SOC as indicated by the chart. The open circuit voltage for the cell would be about 3.3V. The internal resistance is 0.0011Ω. The cell voltage at 267A would then equal approximately 3.0V. The cell power would equal approximately 800W. The C-rate is 13.35.



IamIan said:


> ... is a 25C rate... that is 100% valid conclusion from the data on that graph.


I just showed that 100% valid conclusion of yours is incorrect, seriously, Dude


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## major (Apr 4, 2008)

IamIan said:


> major said:
> 
> 
> > I am *not* saying all charge C-rates are *higher* than discharge C-rates for similar powers.
> ...


I said not higher; you say not higher. Yet you say it is the exact opposite  Sometimes it appears you just like to disagree with me.


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## IamIan (Mar 29, 2009)

major said:


> O.K. Let's look at what you said there and compare to my calculation.
> 
> 800 Watt discharge is about 15% SOC as indicated by the chart. The open circuit voltage for the cell would be about 3.3V. The internal resistance is 0.0011Ω. The cell voltage at 267A would then equal approximately 3.0V. The cell power would equal approximately 800W. The C-rate is 13.35.
> 
> I just showed that 100% valid conclusion of yours is incorrect, seriously, Dude


Not even close. 

Your conclusion is incorrect.... because of several errors you made along the way.

#1> The requirement about that comes from V=IR and P=VI you did not address at all... and remains 100% intact and still disagrees with the conclusions you've posted about Power rate graphs and how that reflects on the C-Rate difference from charge to discharge.

#2> I answered 100% honestly .. about , why I did not use 3.0v ... that is 100% correct and accurate as to why I did not use 3.0v ... and my reasons remain 100% intact ... only 2 voltage points are listed on the graph ... max 3.8v and min 1.6v ... your 3.0v is not listed... that is why I did not use it.

#3> If you wish to try and continue to insist that the 800W discharge point was not the 1.6v minimum ... by all means please explain what higher SoC point you thick had that lower minimum voltage point during the discharge?? ... like it or not ... the graph lists 1.6v as the minimum ... and that 800W 10% SoC Discharge point is the only reasonable point on that graph the cell would be at that minimum voltage point... Higher SoC points should be at a higher than minimum voltage.

#4> You did not read the graph correctly... Look at the graph ... 800W discharge is at 10% SoC ... your claim of 15% is incorrect...see attached... the 800W discharge point is circled... it is 10% SoC.

#5> Your guess of what the rested cell voltage would be is only a guess ... and 3.3v you guessed at ... you came up with based on faulty SoC data.

#6> Your guess of what the cell's internal resistance would be is not part of the data on the graph ... it is an assumption about that cell that you are making ... even if you picked a resistance value that seems to help your position.

#7> If you had used a correct SoC% ... you would have had a lower rested cell voltage ... which in order to get the same 800W out of a lower voltage ... would have increased the Amps of Current... and thus also increased the C-Rate.

#8> If we applied the assumed 0.0011Ohms of resistance you assumed to the other parts of the graph ... it still will result in the same ... for the vast majority of this cell Charge C-Rate being less than the Discharge C-Rate.

#9> Also on the Attached I included a Black Line ... due to the V=IR and P=VI requirements explained previously ... we know that for every part of this power graph to the right of that black line the Charge C-Rate MUST be less than the discharge C-Rate... to the left of the black line we need other data to know ... which we happen to have for one point thanks to the graph telling us 1.6v is the minimum voltage... so we also know the C-Rate at that minimum voltage point.

#10> The % of the graph from #9 that needs additional data is tiny ... the vast majority of the cell curve clearly shows it is a smaller C-Rate for charge than discharge ... due to the requirements of V=IR and P=VI.


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## IamIan (Mar 29, 2009)

major said:


> I said not higher; you say not higher. Yet you say it is the exact opposite  Sometimes it appears you just like to disagree with me.


Sorry if it gave that impression ... I was not disagreeing just to disagree.

I interrupted that quote of yours to mean ... you were not saying x ... where x was that it was higher ... and I agreed ... ok that's what you are not saying ... but I was not claiming x , where it was higher ... I was claiming it was not higher.


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## major (Apr 4, 2008)

IamIan said:


> Sorry if it gave that impression ... I was not disagreeing just to disagree.
> 
> I interrupted that quote of yours to mean ... you were not saying x ... where x was that it was higher ... and I agreed ... ok that's what you are not saying ... but I was not claiming x , where it was higher ... I was claiming it was not higher.


I am glad that is clear


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## IamIan (Mar 29, 2009)

major said:


> I am glad that is clear


In hindsight ... perhaps I should have phrased it like ... it seems no one seems to being claiming that x that you were also not claiming.


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## major (Apr 4, 2008)

IamIan said:


> major said:
> 
> 
> > 800 Watt discharge is about 15% SOC as indicated by the chart. The open circuit voltage for the cell would be about 3.3V. The internal resistance is 0.0011Ω. The cell voltage at 267A would then equal approximately 3.0V. The cell power would equal approximately 800W. The C-rate is 13.35.
> ...


My values are reasonable and fairly accurate. They conform to Ohm's Law. 


IamIan said:


> as to why I did not use 3.0v ... and my reasons remain 100% intact ... only 2 voltage points are listed on the graph ... max 3.8v and min 1.6v ... your 3.0v is not listed... that is why I did not use it.


When you travel on the interstate highway and see road signs which say maximum 70mph and minimum 40mph, do you drive at only 70mph or 40mph, or at any reasonable speed in between those limits?


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## major (Apr 4, 2008)

IamIan said:


> #4> You did not read the graph correctly... Look at the graph ... 800W discharge is at 10% SoC ... your claim of 15% is incorrect...see attached... the 800W discharge point is circled... it is 10% SoC.












O.K. Even though I think this is picky, at 10% I read 782W and at 15% I read 804W. My number is reasonable. And a percent or two or rounding error isn't going to make a big difference.


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## IamIan (Mar 29, 2009)

major said:


> My values are reasonable and fairly accurate. They conform to Ohm's Law.





major said:


> O.K. Even though I think this is picky, at 10% I read 782W and at 15% I read 804W. My number is reasonable. And a percent or two or rounding error isn't going to make a big difference.


I agree ... they are fairly reasonable ... and conform to Ohms Law.

But ... I think I am also being fairly reasonable ... and I am also conforming to Ohms Law.

We have differences in what we are including and or not including ... And I see some issues with some of the differences.

#1> Perhaps the biggest about that 10%SoC point ... I'm using the minimum voltage that they list ... you are not ... I don't see any other more reasonable point on the graph to have that minimum voltage they list.

#2> You used a reasonable rest voltage ... But even if we want to ignore the min volts they listed ... I don't think the voltage you used was correctly applied to this context ... the unloaded ( but not rested ) cell voltage at 10% SoC during that discharge curve ... will not be that 3.3V you assumed ... Ohms law should be applied to the unloaded , but current voltage ... not the rested voltage... it seems you used a much higher rested voltage.

#3> I also agree with you ... that low point for discharge itself also seems to me like nit picking ... sure I have my view about it ... and you have yours ... but the exact C-Rate doesn't change the results ... either way.

Look at what the C-Rate would be for the top of the charge curve ... feel free to make your reasonable assumptions for a ~95% SoC voltage and a ~875W charge C-Rate ... using your same assumed ~0.0011Ohms of resistance to determine the dV ... that Charge C-Rate is less than the discharge C-rate your reasonable assumptions get from the discharge.

The V=IR and P=VI requirement that was explained shows the entire right side of that black line is a lower C-Rate for charging ... Even if we assume the graph is wrong about 1.6v minimum and instead use your 3.0v minimum ... and we assume your resistance Ohms ... none of that changes the results ... the vast majority of the curve would still be a lower C-Rate for charging than it is for discharging ... the C-Rates are not the same ... and yes there is enough data on this power curve to show us that ... even easier to do so , if we start making the reasonable assumptions you included.


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## major (Apr 4, 2008)

IamIan said:


> ... the vast majority of the curve would still be a lower C-Rate for charging than it is for discharging ... the C-Rates are not the same ... and yes there is enough data on this power curve to show us that ... even easier to do so , if we start making the reasonable assumptions you included.


All this Ohm's Law exercise is academic. It still boils down to the fact you are using a manufacturer's power curve to figure C-rate which I think is invalid.


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## IamIan (Mar 29, 2009)

major said:


> All this Ohm's Law exercise is academic. It still boils down to the fact you are using a manufacturer's power curve to figure C-rate which I think is invalid.


I don't see why? ... By all means please tell me what part of this logic I present bellow is invalid? ... I know you write that you think it is invalid ... but why ? ... Please Explain ??? 

I'll walk through it again.

- - - - - - - - - - - - 
#1> V=IR ( Ohms Law )

When you apply charging amps to a cell ... the voltage goes up.

When you apply discharging amp load to a cell ... the voltage goes down.

- - - - - - - - - - - -

#2> (Power) Watts = Volts x Amps

At the same power , when the volts go up the Amps go down.
At the same power , when the volts go down the Amps go up.

- - - - - - - - - - - 

#3> C-Rate = Amps / Ah

More Amps with same Ah cell = Higher C-Rate
Less Amps with same Ah cell = Lower C-Rate.

- - - - - - 

The above 3 points which are all completely well established and accepted points as far as I know.

From that above ... a requirement comes out of it ... that is not opinion ... but follows directly from those points.

Only when the Charge power Watts is higher than the discharge power watts is it even possible for the Charge C-Rate be equal to the discharge C-Rate... it might not be ... but at least it would be possible then.

There is NEVER any time when the Charge power is equal or less than the discharge power ( on the same cell ) and the C-Rate for charge is equal to the C-Rate for discharge.

- - - - - - - 

That requirement that comes out of V=IR and P=VI ... means that on the entire right side of the Black line I put on the previous A123 Power Curve ... that the Charge C-Rate Must be less than the Discharge C-Rate ... on the left side we would need additional data to compare the charge and discharge C-Rates ... but on the Right side of that line ... those 3 points above leave no other option.

As far As I am aware ... This concept holds 100% completely true and firm ... on ANY Ohm's law obeying battery power curve ... even if it is just a power Curve ... it still tells us this much about the Charge and Discharge C-Rates.


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## major (Apr 4, 2008)

IamIan said:


> I don't see why? ... By all means please tell me what part of this logic I present bellow is invalid? ... I know you write that you think it is invalid ... but why ? ... Please Explain ???


Please reread my posts in this thread. I have explained it several times. And it is not about Ohm's Law. I am proficient with that. It is about deducing charge from energy and specifics from generalities. You're drawing conclusions from assumptions which I feel is an invalid method. I do not see where that power curve dictates a charge C-rate, which is what the OP was after. In my opinion, it should not be used to set current limits for charging.


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## PStechPaul (May 1, 2012)

Here is how I understand this. A lithium cell has a fairly consistent characteristic voltage, about 3.4 volts, and there is also an equivalent series resistance, perhaps 10 milliohms. At this point, to add charge at 10 amperes, the applied voltage needs to be 100 mV higher, or 3.5 volts. To discharge at 10 amps, the output voltage will be 100 mV lower, or 3.3 volts. Thus it takes 3.5 watts to charge at 1 ampere while the battery can supply only 3.3 watts at the same current.

However, this ignores the possibility that the ESR may not be the same for charge and discharge, as some of it may be due to chemical action and not simple resistance. This is really just charge and discharge efficiency, and is probably much better than the example I show. The same amount of power is lost as heat in either case, and the overall efficiency must take into account the losses under both conditions. So if this is a 10 Ah battery charging and discharging as 1C, the total losses are 2 watts, with an input of 35 watt-hours and output of 33 watt-hours.

This might be interpreted as a charging and discharging efficiency of 1/34 or 2.9% and overall 5.8%. As for determining C rate for charging and discharging, I think this is due to the chemistry and construction of the battery. The maximum discharge rate might be determined by the point of maximum power which will occur at half nominal output voltage. It is always possible to pump higher power into a cell than can be obtained at discharge, because higher voltage can always be applied. But the actual recommended C rates will be determined by the manufacturer based on reasonable expectations of lifetime and cycles.

That's my understanding, anyway. My assumptions may not be accurate.


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## major (Apr 4, 2008)

PStechPaul said:


> So if this is a 10 Ah battery charging and discharging as 1C, the total losses are 2 watts, with an input of 35 watt-hours and output of 33 watt-hours.


 That sentence is a strange combination of charge, current, power and energy. But it is again off topic, IMO. The OP question is what dictates charge C-rates. Applying Ohm's Law to power levels or power charts isn't the way to answer that question. Tell him about the physics of the SEI layer or something real. The charge/discharge equation is reversible and I haven't heard a valid argument that indicates to me that the charge C-rate needs to be lower than the discharge C-rate limit.


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## IamIan (Mar 29, 2009)

major said:


> Please reread my posts in this thread. I have explained it several times. And it is not about Ohm's Law. I am proficient with that. It is about deducing charge from energy and specifics from generalities. You're drawing conclusions from assumptions which I feel is an invalid method.


I've re-read your posts on this thread ... a few times now ... I don't see how any of it proves or explains the relationship I've been pointing out being invalid ... recently in Post #57 ... and yes the relationship in question is partially about Ohms Law... that's the first needed point #1 of the 3.

If you think some of it does ... please by all means point it out ( what post# ) or explain how it does apply ( or prove invalid ) the relationship I showed recently in my post #57?

Please point out what conclusions and assumptions you referrer to me making in the relationship in question ... post #57... I see me using 100% valid proven points ... in a 100% valid connection ... to point out a 100% valid relationship... if you think not ... please explain?

I am not getting specifics from generalities ... X>Y does not tell you what X is ... nor does it tell you what Y is ... just that X>Y... nor is that relationship deducing charge from energy ... that is not what it does.

I need more than for you to tell me you think it's invalid ... I've shown my hand ... I showed the basis for the relationship ... I showed where it comes from ... I need more than 'because I say so' ... Please Explain , how / why this relationship is invalid ... I am only seeing that relationship (post#57) being valid , any way I slice it.

And I do honestly want to know what about that relationship (post#57) you see an error with , or something that would make the relationship invalid.

 It looks 100% valid to me.


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## major (Apr 4, 2008)

IamIan said:


> I've re-read your posts on this thread ... a few times now ... I don't see how any of it proves or explains the relationship I've been pointing out being invalid ... recently in Post #57 ... and yes the relationship in question is partially about Ohms Law... that's the first needed point #1 of the 3.
> 
> If you think some of it does ... please by all means point it out ( what post# ) or explain how it does apply ( or prove invalid ) the relationship I showed recently in my post #57?
> 
> ...


You don't get it. I am not arguing that your math is wrong. I am not saying that power is higher or lower on charge or discharge. What I am saying is that all that is not addressing the OP question. Calculating the current for a given case does not tell him (or me) what dictates the allowable C-rate for charging a cell. Why can't I charge at the same rate as I discharge? What dictates that?


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## IamIan (Mar 29, 2009)

major said:


> You don't get it. I am not arguing that your math is wrong. I am not saying that power is higher or lower on charge or discharge. What I am saying is that all that is not addressing the OP question. Calculating the current for a given case does not tell him (or me) what dictates the allowable C-rate for charging a cell. Why can't I charge at the same rate as I discharge? What dictates that?


AHHHH ... I think , I get it now.

The Post #57 about the relationship on a battery power graph between charge/discharge power and charge/discharge C-Rate , IS , itself valid ... the invalid part is ( that relationship ) does not itself directly explain why the OEM didn't just force any arbitrarily larger Watts of power on charge than they did on the posted power graphs ... to thus get any arbitrary charge C-Rate they may want.

Yes ... I did not get that at all ... I thought you were claiming the relationship I was talking about , itself as invalid... and I just didn't see how.

Thanks , for clarification.

- - - - - - - - 

I suspect part of the 'why they don't' ... is because of how the different chemicals at the anode and cathode are effected by the differences that happen during Oxidation vs Reduction... See Attached Buttler-Volmer Kinetics overview.


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## major (Apr 4, 2008)

IamIan said:


> AHHHH ... I think , I get it now.


Finally 


IamIan said:


> I suspect part of the 'why they don't' ... is because of how the different chemicals at the anode and cathode are effected by the differences that happen during Oxidation vs Reduction... See Attached Buttler-Volmer Kinetics overview.


That looks symmetrical, charge to discharge, but like the note says, they were not at the limits. They mention the mass diffusion transfer rate limit for current but give no indication of difference between charge and discharge.


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## pm_dawn (Sep 14, 2009)

Hi !

I'm just gonna throw in two specsheets here that show higher charge power capabilities than discharge in much of the DOD region.

http://www.eigbattery.com/eng/product/F014.pdf
and
http://www.eigbattery.com/eng/product/F007.pdf

They look pretty nice......

I tried to look through the Groot study.
And found this.....
"For the PHEV cycles the charge rate appears to be a very important factor for the ageing; a reduction from 3.75 C-rate charge to 2 C-rate charge increased the cycle life with almost 100%."
Thats with the A123 cells.

I guess that is a comparison between the Cycle C and Cycle E

On the otherhand if you compare the Cycle E and the Cycle D which have both pretty much the same charge/discharge rates but different SOC-range the Cycle D with only 20% SOC range is the winner with a 300% longer cycle life.

It would have been really interesting to have a test that had the same currents, the same SOC-range but at different SOC -avg.


REgards
/Per


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## IamIan (Mar 29, 2009)

major said:


> Finally


Sometimes I'm a snail 



major said:


> That looks symmetrical, charge to discharge, but like the note says, they were not at the limits. They mention the mass diffusion transfer rate limit for current but give no indication of difference between charge and discharge.


Correct ... It is very symmetric ... That Butle-Volmer Kinetics graph does not itself go into the effect on the cathode and anode it's just pointing out one aspect of what each would experience at different total cell rates ... ie what the Cathode sees is not exactly what the Anode sees.

We also know that the Cathode is not exactly the same chemicals nor chemical reactions as the Anode.

That graph itself doesn't go into everything else ... it's just that one piece.

- - - - - - 

As for the diffusion rates ... that piece isn't shown there ... but the work of others ... does show a bit about the diffusion rates change at different current rates ... attached is a graph about some work about that , that came out of the University of Paris in February 2013.

Of course even with both those pieces ... one would still not have a complete picture ... there is still more to it... And Although I know a bit ... I would not presume to infer I know everything about it ... because I don't... But I am left to wonder ... because I suspect the OEM had good reason for doing the different C-Rates they did ... unfortunately they didn't list what they were. .. so we're left trying to piece it together.


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## major (Apr 4, 2008)

IamIan said:


> ... because I suspect the OEM had good reason for doing the different C-Rates they did ... unfortunately they didn't list what they were. .. so we're left trying to piece it together.


Well, from Per's last post I followed a link and suspect that these charts you've been using were calculated in accordance with http://avt.inl.gov/battery/pdf/PLUG_IN_HYBRID_Manual Rev 0.pdf 

If that is the case, then it supports my contention that conclusions about a cell's maximum charge C-rate should not be based on those charts. Such conclusions would only be valid for the battery pack configuration submitted for the specific PHEV application and hence have certain limitations which would not otherwise apply.


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## Siwastaja (Aug 1, 2012)

Hi major,

As I see it, you are implying that there is no difference is mechanism between discharge and charge and hence, the same current ratings should apply.

It's commonly stated that li-ion may not be charged at all below zero degrees C or at least it must be charged at very slow rates. Otherwise lithium plating occurs. However, there is no problem in discharging below zero degC. This clearly shows that there is a difference.

Further, the temperature limit is soft, arbitrary limit. The probability of lithium plating more likely gets higher when the temperature is lower AND the current is higher. So, we could deduce that there is some current limit that applies even at 20 deg C to avoid lithium plating.

Do you think that this is incorrect and do you have data to support it?

What I mean, it's a groundbreaking claim that an additional limit for charging current wouldn't apply even though it's a well known concept. Extraordinary claims requires extraordinary evidence.

You can find information on things that happen during high-current charging that do not happen during high-current discharging, such as lithium plating and gassing. But are these relevant in reality? It may be that the heating by internal resistance becomes the shared limiting factor in both charge and discharge that masks any additional charge limitations. This may be heavily temperature dependent and also product dependent.

If you are right, this would be great news.


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## IamIan (Mar 29, 2009)

major said:


> Well, from Per's last post I followed a link and suspect that these charts you've been using were calculated in accordance with http://avt.inl.gov/battery/pdf/PLUG_IN_HYBRID_Manual Rev 0.pdf
> 
> If that is the case, then it supports my contention that conclusions about a cell's maximum charge C-rate should not be based on those charts. Such conclusions would only be valid for the battery pack configuration submitted for the specific PHEV application and hence have certain limitations which would not otherwise apply.


Which charts are you referring to? ... please be more specific.

I've posted a variety of charts ... some I know have nothing to do with the DoE link you reference.

- - - - - - - - 

While I understand and agree with the caution about keeping the context of a test in mind , when looking at the results of that test... especially if the results might be used for a different context than the initial test.

I am not sold on the idea that ... data from context A not being valid at all for context B ... I don't see it quiet that black and white ... it depends on what conclusion one taking from it ... and how different the other context will be.

- - - - - - - 

For Example:


JensGroot said:


> For the PHEV cycles the charge rate appears to be a very important factor for the ageing; a reduction from 3.75 C-rate charge to 2 C-rate charge increased the cycle life with almost 100%.


For a different ... Non-PHEV context ... maybe one would not see the same effect ... maybe it is only 90% or 80% instead of 100% ... who knows ... maybe it's 120% or 150%?? ... and outside of that context ... one definitely should keep the context of that data was obtained in mind ... but without specific evidence showing otherwise ... I don't think one would be justified to assume 0% or 2,000% ...just because they are a BEV charging at a 3.75 C-Rate instead of a PHEV... there was something chemically happening and being shown there in the PHEV test... differences to a different context is reasonable ... but that doesn't mean the other test tells you absolutely nothing... If a BEV used it's battery ~99% identically to the PHEV , they should expect to see a very similar results to the PHEV , even if they are not a PHEV.


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## major (Apr 4, 2008)

Siwastaja said:


> Hi major,
> 
> As I see it, you are implying that there is no difference is mechanism between discharge and charge and hence, the same current ratings should apply.


Hi Si,

Don't get me wrong. I am not implying anything. I don't know. But I look at the cathode and anode reaction equations and they both are indicated to be reversible. I am just interested in the answer to the thread title. What is it that dictates this? And how do we know? And how do we use that to our advantage? 

Much of this thread has been about drawing conclusions based on charts which turn out to calculated from tests done on battery packs designed for specific applications and have had constraints placed on them which influence the data and may not be representative of real limits. Certainly no reasons for those charge rates have been shown. I'd like to find out the real deal.

major


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## major (Apr 4, 2008)

IamIan said:


> Which charts are you referring to?


From this post:


IamIan said:


> I like the A123 power graph


Which was the same one from Per in post #17. Which is the one you used for about a hundred calculations. Which is similar to the I posted in post #40. And finally the ones in the links from Per's post #65. All graphs of pulse power characteristics.


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## IamIan (Mar 29, 2009)

major said:


> From this post:
> 
> Which was the same one from Per in post #17. Which is the one you used for about a hundred calculations. Which is similar to the I posted in post #40. And finally the ones in the links from Per's post #65. All graphs of pulse power characteristics.


Thanks for clarification.

Just a thought.

The relationship between C-Rate and Power Rate for Charge/Discharge I described earlier ... has another implication for Chemical Energy.

As long as the chemical energy of the reaction is a set amount of energy under some set conditions ... which it is ... the C-Rate for charge vs Discharge can not be 100% equal for any Ohmic system ... maybe very close , but not 100% equal.



major said:


> I am just interested in the answer to the thread title. What is it that dictates this?


One on point answer would be ... the criteria of the test ... be it the ones you reference or some other one test criteria.

Different criteria give different C-rates ... is it a 10s pulse C-Rate ... or a continuous C-Rate ... is it the average of continuous charge or discharge from 100%DoD to 100%SoC ... or from 20-80 ... what temperature is it at ... What's your minimum cut off Voltage ... max charge voltage ... etc ... If you start changing the criteria of the test you will get different C-Rate Results.


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## lnpurnell (Sep 27, 2012)

Wow, a lot of replies in such a short time 

OK, so what I can gather from a quick read is that the 'C' charge rate has nothing to do with the charge curve and people should rely on the manufacturers specifications for this charge profile.

What I am asking now is how the manufacturers calculate the 'C' charge rate in the first place?

I understand that lithium plating can occur at all stages of the charge cycle and the risk of dendrite formation is increased with higher charge rates; but is this the limiting variable that manufacturers calculate it from? 

I am going to throw some theories out and please shoot down the incorrect ones:

- C charge rate is calculated by the internal resistance of the cell.
- C charge rate is calculated by the reduction in charge/discharge cycles of the cell.
- C charge rate is calculated by the shorting of a cell over a certain current input.
- C charge rate is calculated by the increased risk of a cell shorting (how is this calculated?)


Is the 'c' rate calculated over a 100% charge cycle of the battery? and what are the variables that manufacturers use in the calculations?

There must be a standard that is used for the industry..

Leigh


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## IamIan (Mar 29, 2009)

lnpurnell said:


> I am going to throw some theories out and please shoot down the incorrect ones:
> 
> - C charge rate is calculated by the internal resistance of the cell.
> - C charge rate is calculated by the reduction in charge/discharge cycles of the cell.
> ...


I suspect a combination of all of the above... plus others.



lnpurnell said:


> Is the 'c' rate calculated over a 100% charge cycle of the battery? and what are the variables that manufacturers use in the calculations?
> 
> There must be a standard that is used for the industry..
> 
> Leigh


The Most widely used that I am aware of ... is the one major linked to:



major said:


> http://avt.inl.gov/battery/pdf/PLUG_IN_HYBRID_Manual Rev 0.pdf


There is no law that I know of ... or that sort of thing ... only 'false advertising'

So if they know higher C-rates will give fewer cycle life ... they use lower c-rates for advertised graphs of cycle life with.

If they know Peukert Effects at higher C-Rates will show less usable capacity ... they use lower c-rates for graphs of capacity.

If you know higher Peak 10Second Power Pulses are valuable to customers ... they show higher C-Rates for that condition.

The C-Rate for Power Pulses is almost never used for the cycle life graphs ... etc.

That is one of the reasons a standard of some kind is nice ... it helps to compare products tested in a similar method.

But ... as major also already pointed out ... there should be some caution about trying to apply the data from one test method to a condition that you know will have different conditions than what the test had.



major said:


> If that is the case, then it supports my contention that conclusions about a cell's maximum charge C-rate should not be based on those charts. Such conclusions would only be valid for the battery pack configuration submitted for the specific PHEV application and hence have certain limitations which would not otherwise apply.


 But ... no matter what 'standard' you picked ... it would never be possible for any standard to be an equal fit for every possible other types and combinations of conditions.

If you test at low C-Rates it isn't 100% valid for high C-Rate usage ... if you test at high C-Rates it isn't 100% valid for low C-Rate Usage ... if you test at low temperatures it isn't 100% Valid for high temperatures .. etc.... etc

The testing gives you an idea ... but it is always useful to try and see what the testing method was for the results they got.

For power graph of the 20Ah puch cell from A123 ... they list this method:


> 10s Pulse Power Capability vs State of Charge at 23°C, Using FreedomCAR HPPC
> Vmax = 3.8 V, Vmin = 1.6 V


How closely or how far away that test method is to your specific application ... will completely vary from person to person's different application conditions.


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