# How do you know when a 160 AH Lion Battery is Charged?



## Karter2 (Nov 17, 2011)

> .... how would you sense it is 90% charged?


 Monitor the voltage ??
What chemistry are the cells ? ..LiFePo4 ?
find the makers spec sheet and charge to the recommended voltage.


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## bob635 (Dec 19, 2012)

The original Thundersky spec. said to charge their batteries to 0.05 C. 

WE have been doing that for over 3 years and have had great battery performance. 

For 160 AH cells, this would be 8 amps. Does this mean you could hook up a 5 amp charger Indefinitely?

It seems like for the larger cells, you may have to regulate charge based on voltage.


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## palmer_md (Jul 22, 2011)

it is just a formula for getting the correct number of amp-hours into the cell.

formula is :
charge at constant current (maximum the charger can output) until the voltage rises to a preset voltage. Hold this voltage (constant voltage) and keep decreasing the current until you reach some minimal current. Stop charging and you should be "full".

If you have a 5 amp maximum current charger then you would follow the same formula, but when you reach the constant voltage point, you are "full" instantly because you are already below the minimal current setting.


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## Elithion (Oct 6, 2009)

palmer_md said:


> Hold this voltage (constant voltage) and keep decreasing the current.


You seem to imply that the charger actively reduces the current in the CV phase.

I am not sure that a clarification is needed, but just in case it is:

In the CV phase, a CCCV charger just holds the voltage constant, and has no control on the current. 

The current in the CV phase of charging decreases because of simple physics.
As the battery gets charged, its internal voltage (Open Circuit Voltage) increases, approaching the charger's CV.
As the difference between the charger's CV and the OCV decreases, so does the current.
Also, as the battery gets charged, the cell resistance increases, which too contributes to the reduction in current.

So, in the CV phase, the charger has nothing to do with the current decrease; it's all due to the physics of the cells.

True, there are charger that include a "profile" and may reduce the CV after some time. That does affect the current. But the primary mechanism of current reduction is the cells being fully charged.


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## Karter2 (Nov 17, 2011)

> ..But what if you charge the battery with only a 5 amp charger.


Is this a genuine question ?
charging a 160Ahr cell with a 5A charger seems highly impractical !


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## Siwastaja (Aug 1, 2012)

To answer the question, indeed the CV phase "stopping" current of 0.05C is selected so that no overcharging takes place at the specified voltage (3.65V), and the cell becomes exactly fully charged. So, at a lower "stopping" current, overcharging may take place, because some current _will_ flow as there is voltage differential between open cell voltage and charger voltage. In fact, it is possible that the charger never reaches the CV phase at all.

You can compensate lower "stopping" current by selecting a lower CV voltage, too. As an extreme example, you can use the safe "floating" voltage which can be applied practically forever. For LiFePO4, this safe floating voltage would be somewhere around 3.4V. You can terminate the charging manually when you are at this voltage, and with current this low, you should have a cell that is probably at least 99% charged but surely not overcharged.

The reason why you don't always charge just to 3.4V and just hold it there for long enough is that getting the last bit of charge into the cell would take about forever. But you are taking forever anyway with that charger so this is what you should do in this case. (Higher than open-circuit voltage helps current flow more quickly, but then you have to terminate with correct timing.)


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## bob635 (Dec 19, 2012)

The charger are rated at 8 amps and have been great with the 80 AMP cells. 

The current sensor controlling the charger senses 0-6 amps so this was sensitive enough for the cut off current of 4 amps. 

However, with the bigger 160 ah cells the cut off current should be 8 amps. 

It just seems that cutting off at 8 amps is a level at which you are still putting significant energy into the battery. 

Or is this just an impedence and the current energy is actually being converted to heat.?


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## palmer_md (Jul 22, 2011)

Elithion said:


> You seem to imply that the charger actively reduces the current in the CV phase.
> 
> I am not sure that a clarification is needed, but just in case it is:
> 
> In the CV phase, a CCCV charger just holds the voltage constant, and has no control on the current.


yes, poor wording on my part. I need to avoid posting from iPhone so I can easily read what I wrote and proof it. Your explanation is much clearer. Thanks for clarifying.


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## Ziggythewiz (May 16, 2010)

bob635 said:


> It just seems that cutting off at 8 amps is a level at which you are still putting significant energy into the battery.


It's all relative. To a 160AH cell 8A is just the same as 4A to the 80AH cell. It's still C/20, it just seems like a lot because our minds don't scale things as well.

I would stop the charge earlier. If you want them just 90% full that's nowhere near 3.65 VPC anyway.


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## PStechPaul (May 1, 2012)

In my testing of the "3600 mAh" Li-Ion cells I purchased recently, I charged a single cell at 0.5A until it reached the target of 4.20V, and then I reduced the charge to 250 mA at which the voltage dropped to 4.14V. Then I continued to charge and the voltage stayed at that level. The charge I applied was 959 mAh. When I removed the charge current the voltage dropped to 4.08V open circuit, and drifted down to about 4.06V. I interpret that as the cell having an internal resistance of about 0.06/0.25 = 0.24 ohms. 

Do those figures indicate that it was fully charged, or could I have kept charging at 250 mA until it reached 4.20V and thus add more mAh? I kept the 250 mA on the cell for about 8 minutes, which might have added about 35 mAh. The measured capacity on a subsequent discharge was 964 mAh at about 1.75A, starting at 4.04V open circuit and dropping to 3.67V at 1.85 amps, and ending at 2.69V at 1.35A, recovering to 3.05V after removing the load. This seems to indicate an internal resistance of (3.05-2.69)/1.35 = 0.27 ohms.

How accurate might it be to determine the SOC by measuring the voltage under load or charge? It seems like you could at least determine if the cell is at full charge or fully depleted. Is this correct? I realize that most DIY EVs use LiFePO4, which may have different characteristics, but the OP apparently has Li-Ion so my figures may coincide with his 160 Ah cell if the chemistry is identical.


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## bob635 (Dec 19, 2012)

Is there a way to monitor the KW put into a battery during charging?


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## Siwastaja (Aug 1, 2012)

bob635 said:


> Is there a way to monitor the KW put into a battery during charging?


Start here:
http://www.diyelectriccar.com/forums/showthread.php?t=6535
http://www.diyelectriccar.com/forums/showthread.php?t=11708

You can monitor charging power (unit: Watt) by measuring both voltage and current.

You can monitor charge, or energy put into the batteries (unit: Joule or Watt-hour) by integrating charging power over time, this is, taking measurements frequently and summing them up.


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