Whatever else you do, there must be a fuse close to the battery that protects the cables in case of a short circuit.
If you don't have this it may help to "fill the gap" too.
Cheers,
Peter

If you wire all the loads and charging to one battery and just run the cables to the other battery in parallel, you will kill the battery carrying the loads early in its cycle life and it will then kill the other battery because it will place a constant drain on it yet absorb all the charging current.
Ideal fix, all the positive loads and charging cables to one point and the same with the negative cables, then run equal length positive cables (of a decent conductor size) to each battery positive terminal, then the same with the negative cable. All four cable don't need to be the same length, just the two positives must be the same and the two negatives must be the same.
The half *rsed way to do it, put the load/charging negatives on one battery and the positives on the other battery while linking the two batteries in parallel, it will help share the load, but not near as well as equal length cables to a common distribution point will.

T1 Terry

Hi Terry,

While I agree with your ideal parallel wiring  configuration for more than two batteries in parallel, I am not sure I would agree that your half  *rsed approach  for two parallel batteries is not as good or even better, assuming  equal length inter battery connecting cables.

The reason I think this is that if all positives and negatives  are connected to their respective common points and the connected to each battery by equal  length (resistance) cables as you suggested, then each battery will effectively have Ra plus Rb in series with the battery (where Ra = cable resistance of cable connected to positive common and Rb = resistance of cables connected to negative common)

Alternately taking the half *rsed approach, the positives are connected directly to one battery with a cable of resistance R connecting it to the second battery. Similarly the negatives are connected directly to the second battery which is connected to the "first"  battery with a similar cable of resistance R.

Comparing the two arrangements ;

Ideal approach : cable resistance  in series with each battery = Ra+Rb

Half *rsed approach:  cable resistance in series with each battery = R

If interconnecting cable resistance R < Ra+Rb then surely the half *rsed is better in this situation.

If connecting more than two batteries in parallel then the half *rsed approach is not the best.

You have suggested the half *rsed approach is not as good as the ideal appoach for connecting two batteries in parallel, but I don't see why, and I would really appreciate your advice and your mathematical analysis if my analysis is incorrect.

Ken

 

-- Edited by kgarnett on Wednesday 2nd of October 2019 05:53:36 PM

-- Edited by kgarnett on Wednesday 2nd of October 2019 05:54:33 PM

-- Edited by kgarnett on Wednesday 2nd of October 2019 06:03:55 PM

For low and moderate loads with 2 batteries, I agree.
For high loads from more than 2 batteries, I use the "star" method which involves taking all the positives and negatives to common points with equal length cables and distributing from there.
Cheers,
Peter

The part you have missed from the equation is the internal resistance of each battery. In the half *rsed approach, the battery with the least resistance will still do all the work until its voltage drops lower than the other battery, then the load will shuffle between the 2 batteries. The reverse happens when recharging but the issue gets worse because the battery to reach the end of boost voltage first, the one with the highest resistance, calls the shots and the charging voltage drop back to absorption and float voltages. The low to moderate load on batteries connect in less than the ideal configuration for multiple similar voltage batteries as even harder on the battery with the lowest resistance, basic maths will show why that is the case.

If each battery had zero internal resistance then what you say would be correct because only the cable resistance comes into the equation. No battery has zero internal resistance, not even a lithium battery, each cell has resistance and cells connected in series means this resistance is added together to create the total internal resistance of the battery.
Alternatively, if the cells are connected in parallel the cell resistance is the average of all the cells within that parallel group. Connect this group with another equal capacity group and the resistance is group 1 + group 2. Build 4 of these together in the case of a 12v lithium battery, you have resistance 1 + resistance 2 + resistance 3 + resistance 4 = total battery resistance. This resistance will alter as the state of charge alters and the load/charging current alters, but it is only one battery and the current must pass all the cells in the battery. The resistance balances out across all the cells in parallel resulting in all the cells maintaining a much closer state of charge to the cell beside it in the parallel group and all the groups in series means the same current has flowed out of or into each group.
This is the ideal way to build a battery, every step away from this ideal level adds a degree of imbalance between each cell that makes up the total battery and adds to the likelihood of a cell failure.
So what I posted as the ideal method was to get the best possible result for multiple similar voltage batteries connected in parallel, the ideal battery is built to the required capacity at cell level and the required voltage by connecting those cells in series.

T1 Terry

jegog wrote:

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As T1 Terry has said many times that you do not connect lithium batteries together, so how lithium batteries are made is not relevant.
The controller monitors voltage and current into the circuit.
The circuit loop is the controller, the cables and the batteries.
Therefore we treat the controller as a voltage source, the cables as resistances and the batteries
as voltage sources with their internal resistances in parallel.
Kirchhoff's 2nd law states:
"The directed sum of the potential differences (voltages) around any closed loop is zero."
This means that if the cable resistances are equal then the battery voltages must be equal.
If they are not equal when first connected together the battery with the highest voltage will discharge through the other battery thereby charging it until both have the same voltage.
The internal resistances only effect the rate of charge and discharge.
Any tendency for one battery to supply more than the other, drops its output voltage due to its internal resistance and then the other battery supplies more of the load.
Think of two interconnected water tanks with filling tap and a outlet tap.
As long as the interconnection capacity exceeds the supply or drain rate it is self balancing.
The same applies to batteries.

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You didn't add in the internal resistance of each cell in each battery adding together, nor the increased resistance between the first battery in the parallel string and the last battery in the parallel string. If the internal across a battery requires a voltage higher than the rested voltage to allow a current flow and the inverse when discharging, the voltage being similar is not an indication the capacity is similar. This is where the water analogy falls over.

As an example for those interested in testing out the theory, find a lead acid battery with a resting voltage of 12.7v, not fully charged but close to it. With a clamp meter to measure current flow and a volt meter on the charging source and another on the battery, dial up 12.7v at the charging source. At this point the water analogy will hold up as there should be zero current flow. Now dial up the charging source to 12.8v. The water analogy says there should be current flow between the supply and the tank with the lower capacity, a very small current flow will be seen on the clamp meter for a very brief period, then it will stop because both the battery voltage and supply voltage is the same. Now disconnect the battery, does the voltage remain at 12.8v? If the supply had topped the battery up till its internal storage was now enough to hold 12.8v now rather than the settled 12.7v.

With the water analogy, the tank would now have increased its capacity to equal the supply capacity, did the battery do the same thing?

For those that are interested in taking this further, keep increasing the supply voltage until you see a constant current flow meaning the capacity of the battery is actually increasing, not just the terminal voltage. How high was the source voltage above the 12.7v rested battery voltage before current actually started to flow constantly indicating the battery capacity was being increased?

Kirchhoff's 2nd law applies to circuits where the source voltage is established, does it allow for a leaking capacitor within with a zenor diode in the circuit?

A hint to just how much voltage differential is required to over come the internal resistance comes from the battery manufactures charging graphs and the quality charger maintenance routine voltages. If 12.8v (rested voltage of a fully charged lead acid battery) was enough to keep a 12v battery fully charged, why is the maintenance voltage set to 13.6v?   

T1 Terry

The Travelling Dillberries wrote:

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Pretty obvious Stoney123 wanted a simple answer, (as give by Jaahn) Stoney gave up in October. Turning out to be the longest battery install in history.

Amen.

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The great thing about a forum compared to a one on one question and answer, a lot of knowledge is shared across a lot of people and for a lot longer period if the search function is used.

 

T1 Terry