Mixing Battery types in 1 Bank

[colemj]

(In response to TeddyDiver)

I mentioned what I thought the flaw was - a static application of Ohm's law across a dynamic situation (charge/discharge). Applying V/R=I is good for instantaneous (in other words, static) measurement. Also, the treatment of two batteries as independent instead of a combined equivalent circuit.

My reasoning (actually, this is a fact) is that individual batteries cannot have differing voltages when combined together while charging or discharging. So during a charge cycle, even if the one battery is taking more current than the other at any given point in time, as its voltage and internal resistance increases, its current acceptance decreases (Ohm's law). Since the two batteries are tied together, the "hungry" battery voltage cannot get above the other battery voltage. So it sits there with the same V, higher R and lower I until the other battery achieves the same SOC and its voltage allows the equivalent bank voltage to increase.

My reasoning is not flawed because I assume a charger is allowed to complete a charge cycle. If so, both batteries will be at the same absorption voltage for enough time that both batteries internal resistances stop accepting current at that voltage. In other words, fully charged.

When discharging, the same applies - the batteries cannot have different voltages when combined into a single bank. In SC's example, R cannot decrease on one battery while I and V stays the same. V cannot decrease on that battery while it is tied to another battery until that other battery's V decreases. So the only option left to it is to decrease I until the other battery's condition causes the bank to drop the combined V and allow it to drop its R and produce more I.

If you would like to change the parameters of this discussion so that a large current is put into a combined bank for a short period of time and then taken off, the batteries separated from the bank and individual SOC's measured - then, yes they can have different states of charge. So would two identical batteries, along with the individual cells within a single battery.

The flaw in reasoning on the other side of this discussion is that the individual AGM and FLA characteristics are being applied without consideration of the two combined into a single circuit and charged/discharged as a bank.

So everyone has to ask themselves this one question: "Can two batteries combined together have different voltages?". The answer to this question is the key to understanding the dynamics of charging and discharging a multi-cell bank.

Again, we have been discussing particular AGM's and FLA's - ones whose manufacturers recommend an absorption voltage of 14.4V for both of them. This is not an unusual recommendation for both types. If one was to combine large traction batteries, which have an absorption recommendation of 15.2V, with those FirstStart AGM's mentioned earlier, which have a recommendation of 14.2V, then this would be a problem with either over-charging or under-charging one or the other.

Mark

[colemj]

Originally Posted by sailorchic34

With parallel batteries, one FLA, one AGM, At a given voltage drop, the AGM will flow more amps then the FLA, due to the lower internal resistance. There is a 600% difference between in the internal resistance between AGM and FLA. AGM will flow more amps per unit of voltage drop. Once the load stops, current flows from the FLA to the AGM till the voltage is balanced again. OK That's over simplified, but the loads are not at all balanced between the batteries.

No, sorry but this is very wrong. How many amps can a 100Ahr AGM battery deliver until it reaches 10.5V? How many can a 100Ahr FLA battery deliver until it reaches 10.5V? What is the amps/unit of voltage drop from 12.7-10.5V for both?

If AGM's delivered 600% more than FLA, nobody would be using FLA for anything and the AGM marketing would be seriously stressing this point.

Again, hooked together, FLA is not parasitic to AGM in either charge or discharge. Yes, you can argue at the microscopic level they are unbalanced, but that is also true for the same types of batteries, as well as for individual cells in a single battery. You can also argue that you can design an experiment where a high load is quickly applied, removed, the batteries separated and terminal voltage measured - and find an instantaneous difference. But that experiment would not be representative at all of anything in real-life usage.

Mark

[colemj]

Originally Posted by FlyingCloud1937

Extreme example
Firstly, if you take one discharged cell and one fully charged cell and put them in parallel, the voltages will be equal immediately. It won't take any time (well, a microsecond maybe...). This is because the point where the the positive (or negative) ends are connected is a single point. A single point has a single potential, by definition.

The way this is explained is that current will start flowing from the charged cell to the discharged cell, and this current will produce a potential difference through the internal resistances of the cells. The charged cell will experience a voltage decrease due to the current it is delivering, and the discharged cell will experience a voltage increase due to the charging current it is receiving. The current will adjust itself until these exactly balance each other.

Lloyd, I understand this point, but it is hardly applicable to a real-life example of a combined bank. Nobody is talking about taking a fully discharged battery and combining it with a fully charged battery and counting that as a single cycle. We are talking about batteries in equilibrium going through charge/discharge cycles together over time.

What you describe does happen on the microscopic level in a dynamic way, but those dynamics are very fast - we are not talking about one battery supplying 75% of a load while the other just sits there until the load is removed - at which time that battery sucks down the other.

This is the way people are thinking about it, which is their sticking point here, but it is not the way it works in reality.

Mark

[colemj]

Originally Posted by sailorchic34

Oh agreed there are some pretty neat chemical dynamics evolved more then just V/R=I.

Just to be clear, I am only talking about Ohm's Law here - not any chemical or other dynamics.

Mark

[sailorchic34]

Agree that the voltage will be the same across multiple batteries in parallel. I think we all agree on that. But the current accepted and discharged at the AGM will always be higher then the FLA due to the differences in the internal resistance. They are not in balance and do not charge or discharge at the same rate.

So what happens is the AGM's will recharge faster then the FLA. Both at the same voltage. But as the AGM has 1/6 the internal resistance it will flow or accept more current then the FLA. Then they (AGM) will sit in bulk until the FLA rise enough for the charger to switch to absorption. Both can accept the same charge voltage, but current flow is very different. So under charge conditions the AGM will tend to overcharge waiting for the FLA to catch up.

Just because the battery voltage is the same does not at all mean that individual batteries, AGM and FLA, have the same SOC. The AGM's will always accept more amps faster at a given charge voltage then the FLA. Conversely they will flow more amps at a given voltage sump then VLA.

In other words the AGM's will work harder and fail much sooner.

Think of two 1000 gallon tanks mounted 10 feet above grade. Both hold the same amount. But one tank has a 2" pipe the other 1/2". The size difference represent resistance to flow. Both pipes are tied together and continue on with a 2" pipe. Assuming the same water level in both tanks the tank with the 2" pipe will flow more and drain faster then the tank with the lower flow rate. When the flow stops the water level (capacity) balances, flowing from the tank with the smaller pipe (higher resistance to flow) to the tank with the large pipe. It's a simplified generalization, but applies to how AGM and FLA would work in the same circuit.

[donradcliffe]

Originally Posted by sailorchic34

In other words the AGM's will work harder and fail much sooner.

Think of two 1000 gallon tanks mounted 10 feet above grade. Both hold the same amount. But one tank has a 2" pipe the other 1/2". The size difference represent resistance to flow. Both pipes are tied together and continue on with a 2" pipe. Assuming the same water level in both tanks the tank with the 2" pipe will flow more and drain faster then the tank with the lower flow rate. When the flow stops the water level (capacity) balances, flowing from the tank with the smaller pipe (higher resistance to flow) to the tank with the large pipe. It's a simplified generalization, but applies to how AGM and FLA would work in the same circuit.

Pretty good analogy, which points out the flaw in your analysis. What happens to the flow from the tank with the 2" pipe when its level starts to fall below the level of the tank with the 1/2" pipe?? At some point that tank is going to be nearly empty, and most of the flow will be coming from the other tank. In battery speak, as soon as the AGM goes to a lower SOC, the FLA will pick up more and more of the load, and vice versa during charging.

[sailorchic34]

Lets do the math together.

AGM resistance = .0033 ohms
FLA resistance = .0200 ohms

From Ohms law I=V/R

Assuming a .2V sump from 12.6V

AGM = .2V/.0033 = 60 amps

FLA = .2V/.0200 = 10 amps

Total amps delivered 70 amps


Voltage drop is the same in both cases. At no time are both batteries putting out the same amount of amps. In reality its not quite this drastic as there are other chemical reactions in play, but no way are both battery's putting out the same amount of amps or accepting the same amount of amps under charge. Voltage is the same though.

The chemistry and resistance is different. Of course I'm ignoring wiring losses and as the AGM discharge the voltage will sump more and FLA will flow more amps. At some point the amps will balance. But by then the AGM is mostly depleted.

[sailorchic34]

Originally Posted by donradcliffe

Pretty good analogy, which points out the flaw in your analysis. What happens to the flow from the tank with the 2" pipe when its level starts to fall below the level of the tank with the 1/2" pipe?? At some point that tank is going to be nearly empty, and most of the flow will be coming from the other tank. In battery speak, as soon as the AGM goes to a lower SOC, the FLA will pick up more and more of the load, and vice versa during charging.

As the head falls the smaller piped tank will flow more, but never the same, until the first tank is mostly empty. While the system pressure stays the same the tank with the higher water level will actually have a higher static head for most of the time. The dynamic system head will be lower due to energy used to place the water in motion.

Yes the FLA will flow more as the voltage drops, but the AGM will discharged far faster then the FLA will.

[colemj]

The tank analogy is only correct, with relation to batteries, if the pipe connecting the tanks is very large with respect to the "load" pipes. In this case, the tanks will empty at the same rate.

How you describe ohm's law working is true for an instantaneous point in time, but not true dynamically. The AGM will not "fill up" and sit in bulk mode waiting for the FLA.

It cannot because they have the same voltage. So at any given instantaneous voltage change during charging (let's pick 13.1-13.2V), the AGM will accept whatever current it can at 13.1V and its internal resistance increases accordingly - causing that current to decrease or halt until the FLA also accepts enough current to move the combined bank to a voltage higher than 13.1V. Ohm's Law.

So as the combined bank moves from 13.1V to 13.2V, the amperage taken by each battery is the charge needed to move each battery to the 13.2V SOC. There is no way the AGM can suck up huge amounts of current boiling away waiting for the FLA to catch up. Any transient "catching up" by the FLA is done while the AGM is accepting little to no current. Again, Ohm's Law - you cannot simultaneously increase current and resistance while holding a constant voltage

This goes on until, at the end of a charge cycle, both batteries are fully charged and neither has either sat around sucking down huge amount of over-current nor parasitically dragged the other down.

And I am really stuck on your claim that the currents accepted and discharged by an equal-sized AGM will always be higher than that of the FLA. How does this happen? There cannot magically be more "stuff" packed into the AGM. And if the two batteries are connected together, how can they not be in balance?

OK. Maybe someone else who really knows this can chime in? I will certainly be happy to be corrected in my understanding and reasoning on this, but right now I see errors in the opposite argument.

Mark

[colemj]

Originally Posted by sailorchic34

Lets do the math together.

AGM resistance = .0033 ohms
FLA resistance = .0200 ohms

From Ohms law I=V/R

Assuming a .2V sump from 12.6V

AGM = .2V/.0033 = 60 amps

FLA = .2V/.0200 = 10 amps

Total amps delivered 70 amps


Voltage drop is the same in both cases. At no time are both batteries putting out the same amount of amps. In reality its not quite this drastic as there are other chemical reactions in play, but no way are both battery's putting out the same amount of amps or accepting the same amount of amps under charge. Voltage is the same though.

The chemistry and resistance is different. Of course I'm ignoring wiring losses and as the AGM discharge the voltage will sump more and FLA will flow more amps. At some point the amps will balance. But by then the AGM is mostly depleted.

Your mistake is in treating the two batteries separately. I don't argue that. But combined in a bank, your reasoning is not correct.

Using your example, your AGM will deliver 630 amps as it goes to 10.5V (well, not really, but let's ignore the dynamics). Your equal sized FLA will deliver 105 amps. How do you reconcile this?

Do you really think in a combined bank the AGM will be a depleted battery many hours early while the FLA is happily pumping current into a load?

Alternately (if this is your meaning), do you think it matters if the FLA delivers its current to the AGM vs. the load?

Mark

[sailorchic34]

Originally Posted by colemj

Your mistake is in treating the two batteries separately. I don't argue that. But combined in a bank, your reasoning is not correct.

Using your example, your AGM will deliver 630 amps as it goes to 10.5V (well, not really, but let's ignore the dynamics). Your equal sized FLA will deliver 105 amps. How do you reconcile this?

Do you really think in a combined bank the AGM will be a depleted battery many hours early while the FLA is happily pumping current into a load?

Alternately (if this is your meaning), do you think it matters if the FLA delivers its current to the AGM vs. the load?

Mark

Oh where to begin. No the two batteries are in the same circuit not separate. Same voltage drop, different resistance and different amp rates

When charging the internal resistance between AGM and FLA is not linear. So the AGM does not wait for the FLA to catch up. At a 0.01V difference a AGM will flow/charge 3 amps while the FLA connected in the same circuit flows .33 amps. The internal resistance for the AGM is 600% lower then for FLA. Even though the resistance may rise as voltage increase, it never ever comes close to FLA internal resistance. This is why if you mix batteries, the AGM will be overworked.

A 100 AH AGM can accept 500 amps charge load without overheating. It would be fully charged in less then 30 minutes. A FLA would barf, er overheat at a 500 amp charge rate.

So yes its quite possible to overcharge a AGM battery when mixed with FLA.

Battery chargers don't know the internal resistance of the battery, they only switch from bulk to acceptance based on a set point voltage.

As to the question about 10.5V and amps, actually the AGM assuming fully charged and discharging at 10.5V would have a sump of 2.1V so yes it could discharge at 700 amps for a short period while the FLA at the same conditions would only discharge 105 amps. Quite correct. That's Ohm's Law. Surprisingly, that's what I've been trying to say. The chemistry and charge rates are not the same...

The main point I'm making is that AGM both under load and charging will flow faster then the FLA can. It's physic's. Yes sometimes the FLA will recharge the AGM but with a 50 amp/hr load the AGM will be dead long before the FLA gets to 50% SOC. You can't use one SOC for different batteries that have way different charge/ discharge rates. At rest they will balance out. But under dynamic load or charge the AGM will be working harder.

[colemj]

Originally Posted by sailorchic34

A 100 AH AGM can accept 500 amps charge load without overheating. It would be fully charged in less then 30 minutes. A FLA would barf, er overheat at a 500 amp charge rate.

So yes its quite possible to overcharge a AGM battery when mixed with FLA.

Battery chargers don't know the internal resistance of the battery, they only switch from bulk to acceptance based on a set point voltage.

No, a charger could not put 500 amps into a combined bank even though the AGM could take it. It can only put in current proportional to the combined acceptance rate at a given voltage. The AGM will not be scarfing up current and the FLA will not be boiling. You cannot overcharge the AGM this way.

Battery chargers ONLY know the internal resistance of the battery bank (well, the total resistance is more accurate). Their only function is to provide as much current as the bank will take until a set voltage is reached. At that point, it provides only as much current that will maintain that voltage. That voltage will be the same for each battery by definition. If left at 14.4V and supplied current, a battery's internal resistance increases and the current decreases (ohm's law). So in your example, the AGM current acceptance will drop faster than the FLA's but at the end of the charge cycle, both will be charged and none will be damaged.

Mark

[goboatingnow]

Ahem, a little Kirchoffs here would go along way.

In discharge Two batteries with differing series resistance ( i.e. equivalent output impedance) may actually have currents flowing into OR out of the combining node. Ie at any time, one battery could be charging , while the other is discharging and vice versa or they could be both discharging, the exact current distribution can only be computed knowing the series resistance at the given SOC.

Hence in discharge both batteries will contribute current, but as Sailorchic says the one with the lower resistance will "work " harder , i.e. it will see deeper discharges BUT only if the current demand is such that the smaller battery cannot actually deliver its proportion of the current. In normal use with fractional C discharges, the batteries will lower their SOC percentage evenly.

On charging both batteries will ultimately charge up according to their acceptance rates ,thats true ( up to a point) , (sorry edit here bad english) both will reach the absorption point at the same time ( i.e. the same voltage) but the time in absorption mode can mean that the FLA or the AGM depending on relative capacities will be too long possibly resulting in gassing.


Depending on the settings for the absorption mode, this extended period could ( i stress could ) cause gassing and venting in sealed batteries.

While it is correct to say that the FLA and the AGM terminal voltage will rise together during charging, that does not means the % SOC is rising together, Again using Kirchoffs , you can see that in charging there can be current flowing from one battery to another, Thats not an issue. ultimately the system will reach full charge

Where the problems arise is in typical real life we do not recharge to full, hence you will find that one battery , typically the FLA, will be charged to a lower SOC then the AGM, resulting in it having poor discharge recharge cycle, resulting in premature failure.

As an aside remember that batteries given a big enough charger, do not in fact self regulate their charge. LA is greedy , it will allow its terminal voltage to be held up and current delivered enough to destroy itself. It only self regulates within an agreed voltage range

The analogy of combining start batteries and house batteries is misleading, typically the starter battery is virtually at full charge and hence receives little current from the combining process and most LA can withstand absorption level voltage unto about 14.8 virtually indefinitely. However run a higher long absorption phase and that battery will indeed suffer.

In practice, combining batteries of different configurations with fixed charging regimes can and does lead to lazy battery syndrome, poor cycling and premature failure.


dave

[sailorchic34]

Ah, Kirchoffs. I could not remember the fellows name. Thanks Dave. I pretty much agree with what you said.

[sailinglegend]

Originally Posted by sailorchic34

As the head falls the smaller piped tank will flow more, but never the same, until the first tank is mostly empty......

Many thanks for introducing the Water Tank analogy - it is often the only way to explain volts amps and resistance to people who just can't understand it.

BUT, it still appears that it doesn't work for some people. but I would persist in using this analogy, but you need to make the tank analogy more realistic.

May I suggest that each tank really has a series of many tanks inside it and each internal tank discharges into the next and when that is full it discharges into the next, and so on. The pipe connecting these internal tanks together is 2" in the AGM tanks and 0.5" in the FLA tanks.

Now your two tanks can be joined together by a very large pipe fed by a constant flow pump. This will fill up the first tank within both the FLA and AGM tanks - the water pressures will be the same so this is the end of the absorption phase. A water level detector senses this and switches the pump to a constant pressure pump and the AGM tank with its bigger internal pipe will be filling up more slowly because it has to force water through a 2" pipe to the next tank, but it will still be filling much faster than the FLA tank. At this point the "adaptive charging algorithm" in the water pump sets a timer for say three hours, by the end of which the last of the internal tanks in AGM tank will be nearly full and the pump will switch to its lowest pressure which is just their to maintains leaks and evaporation. If the pump was switched off then some of the FLA tanks would still be empty and the AGM would start to allow its water to slowly help to fill the FLA tanks. In doing so some of the AGM's water would be lost to the FLA tanks. If the pump was left on the FLA tank would eventually fill also.

So in this scenario it is not the AGM battery that overcharges, but the FLA that under charges.

Now apply this to tank discharging. The AGM will allow more water to flow out more quickly than the FLA tank, and when the flow stops the FLA will have to help top up the AGM until the water levels equalize. So an already partly empty FLA tank has now lost more of it's water to the AGM tank.

I don't know if this helps??????