100ah X2 v 200ah Deep Cycle Battery

If one battery lost a cell why would the other on discharge pretty quickly and why would two fuses prevent this?

If a cell shorts out, it's because either the electrodes (plates, grids) touch each other which is a direct short and reduces the battery terminal to 10 something Volts.
Or, active material wich was shed from the plates, collects at the bottom of the container until the tip of the pile touches the underside of the cathodes/anodes - with a similar result.

The assosiated voltage drop across the battery terminals, would present a low resistive current path to the other battery in parallel (it's like short circuiting a 2V battery).
Two fuses, one in each leg of the parallel wires will certainly prevent the resulting high current from draining the good battery.

I recommend two fuses instead of one, to keep the charge/discharge currents more or less equal between the two batteries (fuses have a little more resistance than bare wire hence add more voltage drop).

Hope this explains it?

Best regards, Peter

?

no worries, I'll give you a picture (always says more than a thousand words)...





The load/charge currents are split more or less in half, because every single current path includes one fuse-voltage-drop.
The first path goes via plus charger, plus bottom batt, minus bottom batt, right hand fuse, minus charger.
The second path goes via plus charger, left hand fuse, plus top batt, minus top batt, minus charger.
Note that there is one fuse-voltage-drop in each path.

If there was only one fuse, then one battery would see less current than the other one, because the currents split up proportional to the resistances (or voltage drops) in the two paths.
If the resistances are the same, the currents split exactly in half, thus each battery experiences the same load/charge current.

If you connected two batteries in parallel, one having 12V, the other one 10V, a short circuit will result, and high current will flow, driven by the voltage difference of 2V.
This high current will blow one of the two fuses effectively saving the good battery from getting deeply discharged.

All clear now?

Best regards, Peter

You have lost me...what size fuses?

The voltage drop across the fuse would maybe .1 volts or less.

If what you are trying to do is going to work a resistor would be better

Sorry I don't understand it.

30A fuses f.e?

Ok lets look at a current example more closesly:
you suggest .1V drop across one fuse - maybe at a current of 5A?
Take this charge current of 5A through this fuse and the calculated resistance for it would be 20mOhm (100mV/5A).
If the other path would not include another fuse, the only resistance would be the battery internal resistance of say 10mOhm.

Now, you've got two current paths, the one with battery resistance plus fuse resistance, and the other path would only consist of the battery resistance.
The ratio of the current path resistance would be (20mOhm+10mOhm)/10mOhm equals 3:1.

Thus, one battery (the one without the fuse-voltage-drop in its path) would see 3 times more current than the other one.

Quite a bit of charge/discharge imbalance is the result.

In low impedance/resistance circuits, even small voltage drops can mean a world of difference in terms of current - as can be seen from this simple example.

It's easy to overlook the very low internal resistance of a lead acid battery, which is responsible for the huge effect of small voltage drop imbalances in a parallel configuration.

Best regards, Peter

Peter,

What you say about the voltage drop across a fuse and the need for both batteries to have identical voltage drops is of course completely valid and an important consideration.

I would implement it differently though. I'd connect the positive terminal of each battery via a 30A fuse to a heavy positive charge/load line. The negative terminals of both batteries would then go straight to chassis in the usual way. This would be simpler to wire up and less likely to cause confusion.

Something we haven't mentioned - a further advantage of the 2 battery setup is that, should one fail, there is still a good one there to carry on with. Using a single battery it's "one out, all out".

Cheers

John

I'm sorry you have still lost me.

In big dollar remote power set up why don't they use this principal.

@ John,

.....not to cause confusion.....

Very valid point indeed, and sorry for my unconventional suggestion of putting the two fuses between the two batteries.

I think this example is suited to highlight the charge/discharge imbalance situation in a more conventional circuit, because it is not only electrically perfectly balanced, it also appears balanced to the eye.
But this circuit also has a significant advantage if you look at it closer.

In a real world situation, where the negatives of both batteries are connected to chassis ground, and the positives to the charge/load wire via fuses, the circuit is not inherently balanced because the wire resistances are more of a concern than in the first circuit.
In the first circuit, all it takes for balancing, is an equal wire length of the two 'fused paths' between the batteries.
The lengths of the other two wires, one to chassis ground and the other one to the charger/load plus is totally irrelevant in terms of battery charge/discharge balancing.
This circuit allows for the battery ground to be anywhere on the chassis, totally independent from the load/charger ground.

To achieve the same grade of balancing in a more conventional circuit, the negatives of the batteries need to be wired to the same ground point, the wires need to be of equal length, and the same goes for the positive connections to the load/charger.
The only way to alleviate the potential for imbalance in a conventional circuit is to use cross connectors between the batteries minus terminals, and have the chassis connection half way between them (T split or Y split).
The same goes for the positives to the load/charger.

Or, another way to make a conventional circuit balanced is by connecting the minus cross connector to chassis ground on minus battery one, and don't connect the other battery minus to chassis at all. Similar goes for the plus side above the fuses, only in the opposite way.
The only prerequisite for balancing, is that the two paths (yellow/green) need to be of equal length and wire gauge.
See picture below:


The first circuit albeit unconventional is certainly more elegant in that it does not allow room for any wire induced imbalances as much as a conventional circuit does.

The other point of yours is quite important too, when making up your mind on the number of batteries:
...if one's down you always have another one left - make it redundant.

@olcoolone,

don't worry, they do.

Best regards, Peter

what a load of rubbish.
You can do that for toy installations.
When you have a serious application you cannot protect against discharge with a fuse. In general the load current is much higher than the discharge current between the batteries and it also does only apply if the cell shortens out.
For higher demand applications (inverter installations of 100+ amps) the series of 6V batteries is better because they have more C5 or C2 time if needed.

I use 2 banks of 4 220 Ah 6V batteries. Two 6V in series and then the resulting 2 12V in parallel. Makes 2 banks of 12V 440Ah. This is much healthier than only
one bank at 880Ah for both the batteries and your back.

The only automatic protection is an expensive high current (200+ amps) relay with 2 shunts in the load circuit and the between the batteries. A significant current difference between the two shunts could open the relay to disconnect the load and to sound an alarm.

have fun
gmd

@gmd,

Good point of yours.

Agreed, if your maximum load current is more than twice as high as the failure mode current between the two batteries, fuses will offer little protection and you'll need additional circuitry.

But for loads lower than say 30 amps, which can be accommodated by fuse ratings of say 15 to 20 amps (one per leg), then these will offer protection against this type of failure mode.
The maximum theoretical current in case of a dead short cell would be about 30 amps (2V divided by the combined internal battery resistances plus fuse resistances).
Because shorts are never really 'dead shorts', this current will certainly be a bit lower again yet enough to blow the 15 or 20 amps fuse on the good battery.

Note that the scenario above doesn't take into account any load currents at the time of failure.

If a load current was present at the time of failure, then the fuse on the good battery will see the sum of the failure mode current plus load current.
If the fuse rating was chosen to be close to the maximum expected load current, then the additional failure mode current is more likely to blow the fuse.

Because it'll take several hours to drain the good battery in case the fuse survives, a low voltage alarm would certainly make sense because there will still be some time left for intervention before the battery gets pulled down to 100% DOD.

I'm aware that heavy battery cables are being used in many installations, and the fuses chosen are to protect the wiring.
But in light of the above, it makes more sense to rate the fuse according to the maximum expected load/charge current, which has the advantage of being able to use lighter gauge wiring (always staying a notch or two above the fuse rating), at the cost of longer recharging times.

High voltage (ECU controlled) alternators can moot this though, so it all depends on your setup and requirements.

To come to a conclusion and wind up this thread, there are obviously some pros and cons involved when deciding if and how to split up the required Ah between a number of batteries.

Best regards, Peter

Peter,

I think we'll have to agree to disagree on this one. The simple fact is that we need the same cable/fuse losses for each battery. There is a series string from charge/load to ground, made up of the battery, the line to the charge/load, the line to chassis and a fuse. It doesn't matter what order we string these bits together, the losses will be the same, provided the wires are the same cross section and length. Your unconventional arrangement is simply a rearrangement of the conventional one, with one of the fuses moved to the earthy end of the battery rather than the positive end.

For practical purposes, this kind of finesse is going to be wasted anyway. The batteries and wires may be at different temperatures (changing the voltage characteristics when current is flowing - the resistivity of copper is highly temperature dependent) , bends in the wires will change the crystal structure (hence resistivity) of the copper, the batteries themselves will never by identical.......

Interesting to hear your thoughts, but I can't agree with them!

Cheers

John

In 25 years of auto elec I have never heard of this idea and i can not see how it would work.

There is a design very close to Peters that uses Scotchy Diodes.

Ray
your example is a perfect situation for Peters "unconventional" circuit as far as I am concerned. While I have never seen this circuit before, which is strange as I worked for a power authority in the installation and maintenance of DC systems in sub stations. The circuit does have a great deal of merit.

BarryG