How to calculate the current going between batteries?

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

  1. mascip
    Joined: Apr 2015
    Posts: 15
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: London

    mascip Junior 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.
     
  2. BertKu
    Joined: May 2009
    Posts: 2,478
    Likes: 42, Points: 58, Legacy Rep: 223
    Location: South Africa Little Brak River

    BertKu Senior 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
     
  3. DCockey
    Joined: Oct 2009
    Posts: 4,562
    Likes: 258, Points: 83, Legacy Rep: 1485
    Location: Midcoast Maine

    DCockey Senior Member

    Current will only be equal if the batteries also have identical characteristics.
     
  4. mascip
    Joined: Apr 2015
    Posts: 15
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: London

    mascip Junior 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?
     
  5. gonzo
    Joined: Aug 2002
    Posts: 14,132
    Likes: 538, Points: 123, Legacy Rep: 2031
    Location: Milwaukee, WI

    gonzo Senior 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.
     
  6. BertKu
    Joined: May 2009
    Posts: 2,478
    Likes: 42, Points: 58, Legacy Rep: 223
    Location: South Africa Little Brak River

    BertKu Senior Member

    That is what I have implied. Bert
     
  7. BertKu
    Joined: May 2009
    Posts: 2,478
    Likes: 42, Points: 58, Legacy Rep: 223
    Location: South Africa Little Brak River

    BertKu Senior 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
     
  8. BertKu
    Joined: May 2009
    Posts: 2,478
    Likes: 42, Points: 58, Legacy Rep: 223
    Location: South Africa Little Brak River

    BertKu Senior Member

    Hi Mascip,

    Bert
     
  9. BertKu
    Joined: May 2009
    Posts: 2,478
    Likes: 42, Points: 58, Legacy Rep: 223
    Location: South Africa Little Brak River

    BertKu Senior 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
     
  10. Joakim
    Joined: Apr 2004
    Posts: 892
    Likes: 52, Points: 28, Legacy Rep: 422
    Location: Finland

    Joakim Senior 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.
     
  11. philSweet
    Joined: May 2008
    Posts: 2,273
    Likes: 158, Points: 63, Legacy Rep: 1082
    Location: Beaufort, SC and H'ville, NC

    philSweet Senior Member

  12. BertKu
    Joined: May 2009
    Posts: 2,478
    Likes: 42, Points: 58, Legacy Rep: 223
    Location: South Africa Little Brak River

    BertKu Senior 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
     
  13. BertKu
    Joined: May 2009
    Posts: 2,478
    Likes: 42, Points: 58, Legacy Rep: 223
    Location: South Africa Little Brak River

    BertKu Senior 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
     
  14. BertKu
    Joined: May 2009
    Posts: 2,478
    Likes: 42, Points: 58, Legacy Rep: 223
    Location: South Africa Little Brak River

    BertKu Senior 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
     

  15. mascip
    Joined: Apr 2015
    Posts: 15
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: London

    mascip Junior 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.

    I'm considering buying this clamp Ammeter: http://www.amazon.co.uk/gp/product/...rue&ref_=ox_sc_act_title_1&smid=ADFI59O5XI21B
     
Loading...
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.