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Lipos and Ohms law - 11/9/2012 8:15 PM   
Mike in DC


 

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Status: offline 		Someone good at Ohm's law solve this for me...

Suppose I have a 3S lipo 5,000 mah battery, each cell at 4.0 volts, for a total of 12 volts. I want to charge it at 5 amps (1C). What voltage does the charger have to put out to create 5 amps of current backwards through the battery.

Or doesn't it work that way?

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RE: Lipos and Ohms law - 11/9/2012 8:50 PM   
hugger-4641



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Status: offline 		Your question is not as simple to answere as I suspect you think it is. If the internal resistance of each cell remained constant throughout the charge/discharge cycle, then you could simply find the resistance and use Ohm's law to find the current and/or voltage that would work. Say you had a resistance of 3ohms, it would be easy to just use Ohm's law to find the voltage that will give you 5amps of current at that resistance, which in that example would be 15 volts.  However, since the internal resistance of the pack changes with temperature and as the level of charge it contains changes, the charger must adjust current and/or voltage throughout the cycle as the resistance changes in each cell.  

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RE: Lipos and Ohms law - 11/9/2012 9:38 PM   
trax de max


 

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Status: offline 		60Watts for battery
the charge at 5A for the needed 60W battery needs 12V

63 watts for 12.6V battery
to charge at 5A for the 63W battery needs 12.6V

Most chargers work there magic and beef up the input voltage/ Or decrease the voltage.
More input voltage less current is needed. Less current less heat.
Less input voltage you need more current, more current more heat. Not good.
So in both these cases a 12V supply would work, but more voltage is better.

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RE: Lipos and Ohms law - 11/10/2012 12:14 AM   
guver


 

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quote:

ORIGINAL: hugger-4641

Your question is not as simple to answere as I suspect you think it is. If the internal resistance of each cell remained constant throughout the charge/discharge cycle, then you could simply find the resistance and use Ohm's law to find the current and/or voltage that would work. Say you had a resistance of 3ohms, it would be easy to just use Ohm's law to find the voltage that will give you 5amps of current at that resistance, which in that example would be 15 volts.  However, since the internal resistance of the pack changes with temperature and as the level of charge it contains changes, the charger must adjust current and/or voltage throughout the cycle as the resistance changes in each cell.  


Can we pretend that our complete circuit has 30 milliohms (at least at one point) and figure out the required voltage?


I've never figured ohms law , but I have seen the real voltage rise from 5 amps with average pack of around 10-50 milliohms.

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RE: Lipos and Ohms law - 11/10/2012 3:01 AM   
hugger-4641



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quote:

ORIGINAL: guver

quote:

ORIGINAL: hugger-4641

Your question is not as simple to answere as I suspect you think it is. If the internal resistance of each cell remained constant throughout the charge/discharge cycle, then you could simply find the resistance and use Ohm's law to find the current and/or voltage that would work. Say you had a resistance of 3ohms, it would be easy to just use Ohm's law to find the voltage that will give you 5amps of current at that resistance, which in that example would be 15 volts.  However, since the internal resistance of the pack changes with temperature and as the level of charge it contains changes, the charger must adjust current and/or voltage throughout the cycle as the resistance changes in each cell.  


Can we pretend that our complete circuit has 30 milliohms (at least at one point) and figure out the required voltage?


I've never figured ohms law , but I have seen the real voltage rise from 5 amps with average pack of around 10-50 milliohms.



I question the accuracy of 10-50 milliohms for a dead pack, even 50 miliohms is practically no resistance at all. But again, voltage is dependent on current and resistance. If the pack was already fully charged, this might be an accurate resistance if the voltage being supplied was  to 4 to 5 volts per cell and practically no current actually being applied (less than 150 ma to be exact).

Basic Ohms law is pretty simple. In the example I used , and in your question, the formula is E=IR, that is Voltage = Current times Resistance. So for your example of 10 miliohms and 5 amps, the voltage needed would only have to be 0.05 volts, 50 miliohms at 5 amps would still only be 0.25 volts. 

A dead 3 cell pack being supplied 15 volts at a current of 5amps would have an effective resistance of 3 ohms.  But a charged pack being supplied 15 volts at a current of 150ma would have an effective resistance of 100ohms. In this case the formula is R=E/I or resistance = voltage divided by current.

The problem is , you can't just hook a mulitmeter up to a battery to check the resistance like you could a heating element or some other fixed resistance object. The resistance in a battery must be calculated by applying a load, measuring the voltage drop, then using these numbers to calculate the resistance. 



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RE: Lipos and Ohms law - 11/13/2012 9:40 PM   
Mike in DC


 

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quote:

ORIGINAL: hugger-4641

Your question is not as simple to answere as I suspect you think it is. If the internal resistance of each cell remained constant throughout the charge/discharge cycle, then you could simply find the resistance and use Ohm's law to find the current and/or voltage that would work. Say you had a resistance of 3ohms, it would be easy to just use Ohm's law to find the voltage that will give you 5amps of current at that resistance, which in that example would be 15 volts.  However, since the internal resistance of the pack changes with temperature and as the level of charge it contains changes, the charger must adjust current and/or voltage throughout the cycle as the resistance changes in each cell.  

Well, I didn't think it would be simple, but I did think the 12 volts in the battery would be taken into account some how.

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RE: Lipos and Ohms law - 11/14/2012 12:40 AM   
Truckracer


 

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Status: offline 		OK, here is the simple explanation of how a Lipo charger works. First, forget all about what voltage the battery has when not charged! It doesn't matter for our example here. Second, for a 3 cell pack, you want a charged voltage of 12.6 volts or 4.2 Volts per cell. You determine what charge current you want to charge the battery at and you set that into the charger and initiate the charge. When the charger starts its charge cycle it is in whats called "constant current" mode and will apply whatever level of voltage necessary to maintain the pre-determined charge current you selected. Most of the charge cycle takes place in this mode. As the battery charges, the voltage will slowly climb until it reaches 12.6 volts (4.2 volts per cell) then the charger will automatically switch to a "constant voltage" charge mode. It will then maintain 12.6 volts on the battery. During this mode as the battery finishes charging the charge current will taper off to almost nothing then the charger will terminate the charge. Its all really quite simple but there is a lot going on in the charger to maintain this relatively simple process.

This same charge process applies to LiPo, LiFe, Lion and even lead acid batteries. Only the final charged voltage changes for the different cell chemistry types.

So you see, you can't just apply a set voltage to a LiPo to charge it. Simple OHMs law just does not apply to battery charging.

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RE: Lipos and Ohms law - 11/14/2012 12:05 PM   
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Status: offline 		If you watch a charge current graph as a LiPo is charged, things become more obvious as to what happens.
Initially, the charge current is limited to a maximum current, usually set to 1C, or 5A in the case mentioned.
As the lipo is charged, current rollback occurs, and the voltage limit, usually 4.2 to 4.22 reduces the charge current.

When a multi-cell liPo is being charged, the charge current drops to a very low value, and the charger attempts to equalize the voltage between the cells to 4.2V. In the process, an individual cell may actually be charged to very slightly more than 4.2V, and then discharged into a lower voltage cell, or a load inside the battery charger. While this is going on, the charger may be supplying current  at a very low rate, perhaps just a few Ma.

The balance portion of the charge cycle can take a significant amount of time.  When the individual cells are not well matched, it takes even longer.
From what I've seen, the usual voltage range for this situation is usually between 4.16V and 4.21V during the balance cycle.The charger may indicate that around or up to about 10% of the total battery capacity is provided by the charger during an extended balance process. 


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RE: Lipos and Ohms law - 11/14/2012 1:46 PM   
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Status: offline 		Very nice post Truckracer!

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RE: Lipos and Ohms law - 11/15/2012 1:16 AM   
Truckracer


 

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Status: offline 		Chuckk2,

I didn't want to get into balance charging. Just wanted to describe the basic LiPo charge cycle. I figured balance charging was a subject for another post which you handled nicely.

Various chargers handle balance charging in very different ways. Depending on whether they charge through the balance tap or whether they charge through the power leads and balance through the balance tap. All very interesting stuff!

< Message edited by Truckracer -- 11/15/2012 1:43 AM >

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RE: Lipos and Ohms law - 11/15/2012 1:17 AM   
Truckracer


 

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ORIGINAL: BarracudaHockey

Very nice post Truckracer!


Thanks!

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