# How to calculate the current going between batteries?

Discussion in 'OnBoard Electronics & Controls' started by mascip, Sep 10, 2015.

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### mascipJunior Member

Hi,

I'm curious about voltage drop between batteries. How do you calculate the current between two batteries?

Say, there's a solar panel regulator (or an alternator) sending 20A at 14V to a single leisure battery. And there's a cable going from this leisure battery to a starter battery.

Is a battery simply a generator in series with an internal resistance? If so, then is the internal resistance be on the + side of the generator, or on the - side, or both? The cable iitself can be represented as a resistance, so I could then calculate the current going between batteries.

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### BertKuSenior Member

That is not easy to calculate. It depends whether you have a thick, fat cable between the 2 batteries, or one fat one and a thin one, whether the two batteries are equal new , or one new and one old. Thus theoretical, if both batteries were equal charged and new and two same size cables, the distribution to both are 1/2 current to both batteries = solar current. Otherwise it will be a matter of measuring the various resistances and work according to the law of Ohm.
Bert

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### DCockeySenior Member

Current will only be equal if the batteries also have identical characteristics.

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### mascipJunior Member

I was just thinking: whether the internal resistance is represented before or after the generator, doesn't make any difference in terms of current calculation.

If I make a diagram where each cable is a resistance (easy to calculate from its thickness and length) and where each battery is a generator in series with an internal resistance, then Ohm's law should be pretty straightforward. Is it that simple?

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### gonzoSenior Member

No, there is an electrochemical reaction that has to be taken into consideration. The density of the acid, the physical state of the plates, the degree of sulphation, etc. all will change the charge rate of the batteries.

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### BertKuSenior Member

That is what I have implied. Bert

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### BertKuSenior Member

But how are you going to calculate the internal resistance of those two batteries? If they are not the same, then the rest of the calculations does not make sense.
Bert

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### BertKuSenior Member

Hi Mascip,

Bert

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### BertKuSenior Member

Hi Mascip,

I am inquisitive, why do you want to calculate it and at what temperatures, the one battery may be 25 degrees and the other is 18 degrees, because it was located somewhere else? Like Gonzo mentioned, there are so many variables, you have little chance to make it a success of it. Just measure it with a good current meter, much better to make conclusions. bert

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### JoakimSenior Member

A lead acid battery can be simplified with a constant voltage source + internal resistance. It makes no difference where the internal resistance is, since it is internal and you can't have any connection between the resistance and the voltage source.

However both the internal resistance and the voltage of the voltage source depend on the state of charge of the battery and thus this simple model is not enough to calculate the current.

If you have two batteries always in parallel, they will be at the same state of charge (unless they have different chemistry). But you talk about a leisure and a starter battery, which can't be in parallel. There must be a diode or a relay separating them in order to avoid draining the starter battery.

Starter battery should always be very close to 100% state of charge, since it has very little use. Thus it will not take much current, probably not more than 1-2 A after a few minutes of charging. Leisure battery is used and will take almost all the current until it is up to 70-80% and its cabability to take current starts to decay. At higher current (alternator) this happens earlier, maybe at 60% state of charge.

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### philSweetSenior Member

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### BertKuSenior Member

Yes, a nice handy meter. I wonder at what micro/milli Ampere one is able to measure. Although there are some negative reports on this meter. But I assume that if one does not reset it every time before measuring, you will have a discrepancy.
Bert

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### BertKuSenior Member

If using batteries parallel, I am using normally 2 x Shottky diodes type MBR7030WT from ON semiconductors parallel, which gives you 4 x 35 Ampere = 140 Ampere continuous and 2 x 500 Ampere for 1/2 wave or short peak current.
By placing them parallel, the current gets reasonable even distributed over those 4 diodes. You just need a small heatsink.
Bert

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### BertKuSenior Member

Sorry Mascip,

I made a mistake, I am using the 2 x 40 Ampere in each casing x 2 = 160 Ampere Schottky diodes type 40CPQ100 (100 Volt 2 x 40) x 2. The others are only 2 x 35 Ampere and 30 Volt.

I looked at the wrong datasheet. Bert

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### mascipJunior Member

Oops, I hand't noticed the answers here.

I just wanted to understand the voltage drop between my starter and leisure battery. My mistake was to connect the solar panel and alternator to the starter battery. As you said Joakim, the current is generally higher to the leisure battery, so it makes sense to connect the wire sensors and sources of energy to the leisure battery, as there is then less voltage drop to the starter.