LiFePo4 -- Parallel or Not?

[T1 Terry]

Originally Posted by evm1024

So basically you are saying that I cannot parallel 3, 100AH batteries and pull out 300 amps

Is that what you are saying - That the current draw is limited to the 1 C rate of the individual batteries?

And that this is a function of the cells or the BMS in the dropin?

The BMS in the drop in will have a maximum current limit, anywhere from 0.5CA to 1CA in most quality drop in batteries. They might have a high surge current, but continuous output is generally between these two rates. It is a function of the battery disconnect within the battery that limits the maximum continuous current. It will either over load an electronic device or melt the contacts on a mechanical device.
It might be able to deliver 300 amps from any one of the 100Ah batteries, but it won't deliver that 300 amps for 20 mins. 60 mins @ 100amps = 100Ah, therefore 20 mins @ 300 amps should also equal 100Ah, but it won't because at high current draws like that, Peukert's law starts to come into play, some electrical energy will be converted to heat energy and lost.
This heating of the electrolyte also shortens the cycle life, just how many cycles before serious capacity is lost really depends on how often the high load is drawn and for how long.

T1 Terry

[hzcruiser]

Originally Posted by T1 Terry

The BMS in the drop in will have a maximum current limit, anywhere from 0.5CA to 1CA in most quality drop in batteries. [...]

I believe we're on the same page here. It's just the details that are being presented differently.


We all agree that the BMS will limit the current in a way that it disconnects the batt altogether if that threshold has been exceeded, right? It won't just limit the current to, say 1C, and not let more current through. For that you would need a very beefy MOSFET switching setup like the ones in motor controllers.



However your earlier statement in post #41 says that "putting several LFPs in parallel only increases the capacity, not the maximum current". IMHO this is what caused some friction in the discussion.


Now it just dawned on me that maybe your assumption might be that there is only one BMS between all the LFPs in parallel? In that case, you're right, the maximum current is still capped by that one BMS and to whatever its threshold is. Going over that limit will cut off all the batts!



Yet that's something most people won't like: the whole boat goes dark from one second to the other.


Hence my assumption was always to have one BMS per LFP battery and then they could deliver the sum of all batts, albeit for a very limited time, as you mentioned.



It comes down to sizing the batts correctly for the maximum expected load, and if there is a short in a consumer, the breaker for that device should disconnect it, leaving the rest of the boat online.

[evm1024]

Originally Posted by T1 Terry

The BMS in the drop in will have a maximum current limit, anywhere from 0.5CA to 1CA in most quality drop in batteries. They might have a high surge current, but continuous output is generally between these two rates. It is a function of the battery disconnect within the battery that limits the maximum continuous current. It will either over load an electronic device or melt the contacts on a mechanical device.
It might be able to deliver 300 amps from any one of the 100Ah batteries, but it won't deliver that 300 amps for 20 mins. 60 mins @ 100amps = 100Ah, therefore 20 mins @ 300 amps should also equal 100Ah, but it won't because at high current draws like that, Peukert's law starts to come into play, some electrical energy will be converted to heat energy and lost.
This heating of the electrolyte also shortens the cycle life, just how many cycles before serious capacity is lost really depends on how often the high load is drawn and for how long.

T1 Terry

You did not answer my question.

If I put 3 batteries of 100 ah each in parallel will I be able to draw 300 amps from them?

Let's eliminate any BMS, and stipulate that the batteries are rated for 1C discharge rates, and that the wiring is of the gauge to support 100 amps from each battery and 300 amps to the load.

[s/v Jedi]

This is silly. If the resistance/impedance of multiple parallel circuits is equal, then current *will* be shared equally. Not a matter of opinion.

This is why I add the “if wired correctly” to my earlier comment. Many still do this incorrectly, claiming it doesn’t matter.

The hzcruiser setup is different, with batteries distributed over the boat, bringing them close to the loads they power while being connected in parallel. That works great as well but the load is not shared as evenly, so the battery close to the windlass needs to be able to supply 100% of the power used by the windlass. If a single battery would not be able to, then he can put two of them in that location, connected to the system with equal resistance cabling to that they share the load 50/50.

Then the charging part: this is just fine as well. The batteries furthest away from the charge source will take longer to charge and probably some creativity with absorption time is needed to get enough charge into them but with lfp you don’t need to fully charge so it’ll be okay. He could even make standardized packs and rotate them like done with car tires.

[evm1024]

Originally Posted by s/v Jedi

This is silly. If the resistance/impedance of multiple parallel circuits is equal, then current *will* be shared equally. Not a matter of opinion.

This is why I add the “if wired correctly” to my earlier comment. Many still do this incorrectly, claiming it doesn’t matter.

The hzcruiser setup is different, with batteries distributed over the boat, bringing them close to the loads they power while being connected in parallel. That works great as well but the load is not shared as evenly, so the battery close to the windlass needs to be able to supply 100% of the power used by the windlass. If a single battery would not be able to, then he can put two of them in that location, connected to the system with equal resistance cabling to that they share the load 50/50.

Then the charging part: this is just fine as well. The batteries furthest away from the charge source will take longer to charge and probably some creativity with absorption time is needed to get enough charge into them but with lfp you don’t need to fully charge so it’ll be okay. He could even make standardized packs and rotate them like done with car tires.

Exactly! Very well said.

[T1 Terry]

Originally Posted by evm1024

You did not answer my question.

If I put 3 batteries of 100 ah each in parallel will I be able to draw 300 amps from them?

Let's eliminate any BMS, and stipulate that the batteries are rated for 1C discharge rates, and that the wiring is of the gauge to support 100 amps from each battery and 300 amps to the load.

OK, the BMS inside the battery case along with the cabling inside the case is rated to the continuous load the manufacturer specifies. There are no current measuring devices or current limiting devices in these drop in batteries, if you put a 300 amp load on any of the 100Ah batteries, eventually something will let the smoke out because it is not rated to supply 300 amps. a 4 x 100Ah quality cell battery built with the cells is series and adequate links used between the cells will allow the 300 amp load to be drawn from it. there will be some voltage drop, but not substantial until the battery is drained, then it will fall like a stone.
Here is a graph from testing I performed towards the latter part of 2013.

The blue line is combined battery voltage and can be read on the scale to the left, the other 4 lines are cell voltage and read on the scale on the right.
as you can see, at rest with only the inverter idle load connected, the cell voltages are close to the same, yet under a 94 amp load, the diverge, this is just an example of the difference between the cells in any 4 cell group.

So, what does the in case drop in BMS use to trigger a battery disconnect? Cell voltage, not likely but maybe, terminal voltage, most likely.

So, this was the voltage drop with 94 amps on a 90Ah cell LFP battery, try to picture what the voltage drop through under rated cables inside a drop in 100Ah battery with 300 amps applied .... enough to drop the terminal voltage to 10.5v? This is the average voltage used for drop in battery low voltage disconnect. So the full 300 amps isn't going to be drawn from the first battery with the lowest resistance between the load and the battery terminal, but it will have to handle the lions share until the voltage drops to 10.5v, then the next in the string copes the lions share and so on down the line.
So, short answer yes, long answer, for how long will you be able to pull that 300 amp load ...... a 300Ah battery built from 100Ah cells in parallel x 4 sets in series will supply that load for the full 1 hr plus a bit, how long will the 3 x 100Ah drop ins supply the 300 amp load, a graph of the actual test would be nice

T1 Terry

[T1 Terry]

Originally Posted by s/v Jedi

This is silly. If the resistance/impedance of multiple parallel circuits is equal, then current *will* be shared equally. Not a matter of opinion.

This is why I add the “if wired correctly” to my earlier comment. Many still do this incorrectly, claiming it doesn’t matter.

The hzcruiser setup is different, with batteries distributed over the boat, bringing them close to the loads they power while being connected in parallel. That works great as well but the load is not shared as evenly, so the battery close to the windlass needs to be able to supply 100% of the power used by the windlass. If a single battery would not be able to, then he can put two of them in that location, connected to the system with equal resistance cabling to that they share the load 50/50.

Then the charging part: this is just fine as well. The batteries furthest away from the charge source will take longer to charge and probably some creativity with absorption time is needed to get enough charge into them but with lfp you don’t need to fully charge so it’ll be okay. He could even make standardized packs and rotate them like done with car tires.

If you connect each battery with identical resistance cables to a common point where the load or charging will be connected, it will go close to a 50/50 share or 33.3% share across 3 batteries etc .... as long as the internal battery resistance is identical. The one most able to supply by having the highest terminal voltage under load will supply the majority of the current, that is just the way it is, electricity will always flow via the path of least resistance.

T1 Terry

[s/v Jedi]

Originally Posted by T1 Terry

If you connect each battery with identical resistance cables to a common point where the load or charging will be connected, it will go close to a 50/50 share or 33.3% share across 3 batteries etc .... as long as the internal battery resistance is identical. The one most able to supply by having the highest terminal voltage under load will supply the majority of the current, that is just the way it is, electricity will always flow via the path of least resistance.

T1 Terry

Now you agree with me but before you went against this same position from me. I know, I know, it’s just that everything must be argued against, right? Your last sentence is just to top it off, you have no idea who you are dealing with and it’s showing

[evm1024]

Originally Posted by T1 Terry

If you connect each battery with identical resistance cables to a common point where the load or charging will be connected, it will go close to a 50/50 share or 33.3% share across 3 batteries etc .... as long as the internal battery resistance is identical. The one most able to supply by having the highest terminal voltage under load will supply the majority of the current, that is just the way it is, electricity will always flow via the path of least resistance.

T1 Terry

Perhaps this the the cause of the confusion.

Electricity does not flow via the path of least resistance.

That is a common misunderstanding.

Electricity will flow via every path. The currents will divide in inverse proportion to the paths resistance.

[T1 Terry]

Hmmm.... electrical flow is the direction of current flow isn't it? The presence of a voltage reading at zero load does not mean the current will flow at any given value and that voltage maintained .... isn't that how a resistor functions in the real world?


I'm ready to surrender,there is little value in me continuing when theoretical starts to overshadow just how things will function in the real world and I'm left trying to defend my position just using real world results, they will never match theoretical results ....

T1 Terry

[s/v Jedi]

Originally Posted by T1 Terry

Hmmm.... electrical flow is the direction of current flow isn't it? The presence of a voltage reading at zero load does not mean the current will flow at any given value and that voltage maintained .... isn't that how a resistor functions in the real world?


I'm ready to surrender,there is little value in me continuing when theoretical starts to overshadow just how things will function in the real world and I'm left trying to defend my position just using real world results, they will never match theoretical results ....

T1 Terry

No, the laws of nature are 100% represented in real world applications. EVM is absolutely correct but instead of acknowledging that, you counter with something that isn’t incorrect by itself, but irrelevant to that what you are arguing about.

So to make this simple, lets take two identical drop-in batteries. When connected into a parallel bank with correct wiring and both fully charged, what we get is double the capacity and double the current rating. As long as balance is maintained, each does 50%.
Don’t divert from this hypothetical setup or to other scenarios because this is the core of the issue where you previously stated that even though the capacity doubles, the current rating does not.

[hzcruiser]

Originally Posted by T1 Terry

Here is a graph from testing I performed towards the latter part of 2013.

The graph shows nicely how the voltage sags under load, from >3.3V to <3V per cell. An almost 10% drop which allows the other LFPs that might have had a lower starting voltage, to add to the overall current.

And here I thought someone stated voltage sag was not a thing with LFPs?

[Afrinus]

Jedi, I think what Terry is trying to say is that you are right as long as all resistance (cables AND battery AND cell internal) is EQUAL.

Reality is that is not going to be the case, and therefore the difference

At least that's what I get from this, and also how I understand this

Originally Posted by s/v Jedi

Now you agree with me but before you went against this same position from me. I know, I know, it’s just that everything must be argued against, right? Your last sentence is just to top it off, you have no idea who you are dealing with and it’s showing

[s/v Jedi]

Originally Posted by Afrinus

Jedi, I think what Terry is trying to say is that you are right as long as all resistance (cables AND battery AND cell internal) is EQUAL.

Reality is that is not going to be the case, and therefore the difference

At least that's what I get from this, and also how I understand this

That’s what hzcruizer mentioned iirc. But in that case, current is shared in ratio to the differences and not a “winner takes all”. Even when one parallel path has double the resistance then it still takes 1/3 of total current. I can only assume Terry knows that which is why I am confused as to what he means...

[evm1024]

That is what puzzles me too. Terry has been around a long time and has decades of experience with LiFePO4. Surely he knows circuit laws that have been around since 1845.

I can only speculate that he has observed something that he assumed violated those same circuit laws. And came up with an incorrect understanding as to what was happening.

The trick is to figure out what exactly he is saying and find the actual cause. This is hard to do because of the tangents that keep popping up. Tangents that appear to support his assumptions but do not really have anything to do with it.

The notion that current comes from one battery then the next then the next is just wrong. We will get nowhere until that is understood.

Drop-in batteries are not simply strings of cells. They have microprocessor controlled BMS's which have their "own agenda". And drop-ins are not created equally. Even still drop-in do not communicate between each other and say "Hey, you supply all the current, I'm going to sleep till you are depleted".

Many drop-ins are cheap kit and state that they cannot be wired in parallel. I imagine that the BMS in each battery ends up interacting poorly with the BMS in other batteries.

Plus as Will Prouse has shown us many of those drop-ins have internal wiring that will not support 1C discharges.

On the other hand top shelf drop-ins are designed for parallel operation and have internal wiring that supports their continuous discharge rates. Makers like Victron come to mind. And if anyone were to look in the solar world many installations are using drop-ins in parallel to increase current capacity.

Well designed drop-in batteries will increase current capacity when in parallel.

Poorly designed drop-in batteries will behave poorly (but still increase current) when in parallel. This is a design fault of the battery not a failure of Kirchoff's laws. And to extrapolate poor design as a failure of laws that have 175 years of proof behind them is missing the mark.