How to calculate the current going between batteries?

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

  1. BertKu
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    BertKu Senior Member

    Nice to see that other forum members also admit to make mistakes. I made again a small mistake. Those diodes are 40 x 2 x 2= 160 Ampere only in rectangular waveform, but with normal DC maximum 20 x 2 x 2 = 80 Ampere.

    If you do it that way, and you make the leisure battery as your main battery, it is good practice to have a diode between those two batteries, otherwise the two batteries may discharge themselves into each other. I assume that both batteries are not the same , equal new and of the same size and quality. If you do put a diode between your leisure battery and starter battery, make sure it is a Shottky diode and that they can handle high currents for a short period of time. Like the ones mentioned, they can handle for 10 mS each 300 Ampere = 600 Ampere. You don't have to be worried that the one battery (your starter battery), will always be filled up when the other, leisure battery get charged by the solar panel or alternator. Also you have a guarantee, that you always can start your engine with a full battery.

    Always a good investment. Bert
     
  2. CDK
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    CDK retired engineer

    Just returned from a trip across Europe with my Fiat Ducato camper van. A very popular vehicle, used for 3 out of 4 R/V's and half the fleet of delivery vans in Europe because of the low price tag.
    There is a 95 Ah starter battery sitting almost on top of the engine and a leisure battery in the back, 15ft. away. A microprocessor controlled power module separates both batteries, yet allows them to be charged from the alternator, the grid and the solar panel or any combination.

    That all sounds great, but the alternator is set at 14.4V to compensate cable loss and the voltage drop across a diode. The starter battery and vehicle wiring get the full 14.4V, the leisure battery and its circuitry gets no more than 13V because the fridge draws almost 10 amps.

    It all results in a leisure battery that is never fully charged while traveling, a starter battery slowly boiling dry and halogen bulbs having a life expectancy of just a few 100's hours. I bought the van from the factory nearly 8 years ago and already replaced the starter battery twice and the leisure battery 3 times.
    On this last trip I replaced both headlamp bulbs that died within hours of each other. Very annoying because of all the stuff you have to remove to get access to the bulbs.
     
  3. BertKu
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    Location: South Africa Little Brak River

    BertKu Senior Member

    Indeed that is a problem. However Mascip may have a different situation. Although he can learn from your unfortunate 8 years experience, his engine may run very seldom and gets the energy for the leisure battery from the solar panels or if he feels the need to run the generator. In that case the voltage drop over the Shottky diode is so low that the starter battery does get enough energy to stay reasonable full, without it gets discharged. Now he has the situation that he is emptying both batteries. Just a point of view. A lead acid battery start charging already at 12 Volt and although a voltage drop may give him only after the Shottky diode 13 Volt and his starter battery gets only slowly filled up, his battery will always be better off then now. His starter battery is maybe far from the hot engine and does not have a bonnet to boil his battery. It totally depends on how often does he starts his engine on his boat. Just a different opinon, Bert
     
  4. mascip
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    mascip Junior Member

    Again I forgot to answer, sorry for late reply. My starter battery is charged by the alternator, my leisure battery is charged by a solar panel, and I have a VSR (Voltage Sensitive Relay) between the two, so that they both get charged when either is charging. I haven't had any issue with that, it's all good =)

    A friend of mine has separating diodes between the alternator and either battery bank, instead of a VSR. It means that when the alternator gives out 14.4v, the batteries only get 13.3v. I'm wondering to which extent and how quickly this will undercharge the batteries and kill them slowly.
     
  5. W9GFO
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    W9GFO Senior Member

    Current is equal everywhere in a series equivelant circuit. Voltage is what changes.
     
  6. mascip
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    mascip Junior Member

    That's crystal clear. I'm not sure what made you think it isn't. Maybe I said something unclear or ambiguous?
     
  7. mascip
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    mascip Junior Member

    Everywhere on this cable it's the same indeed. But the question is: can it be calculated from the state of both batteries, with a formula? I was hoping some formula which would involve the state of charge of the batteries, their capacity in Amp hours, etc.

    It's interesting at the moment when you connect them.

    It wasn't a practical question about how to measure, but rather a question about theory. Just being curious about battery models and the maths behind them
     
  8. W9GFO
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    W9GFO Senior Member

    None that I am aware of. The difference in charge, temperature and capacity between the two batteries would have to be well known for the calculation to be accurate.
     
  9. Joakim
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    Joakim Senior Member

    The voltage drop shouldn't be that big. Maybe around 1 V at the maximum current, but then it should drop below 0.5 V when the battery is full and the current is only ยจ~1 A. Many alternators have a sense wire, that can be connected to leisury battery and thus the charging voltage is compansated for the voltage drop of the isolating diode and + cable.
     
  10. Joakim
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    Joakim Senior Member

    Sure there are models for that. The simplest ones use constant internal resistance and then calculate current based on voltage difference from the terminals to the open circuit voltage. You can find the open circuit voltages for different type of lead acid batteries as a function of state of charge and temperature from e.g. here: http://jgdarden.com/batteryfaq/carfaq4.htm#ocv_soc

    This model is quite problematic, since the internal resistance is not constant (it depends on SOC, current etc.) and you don't know it. Also the open circuit voltage is not that simple. E.g. after charging it can remain well above the maximum for several hours and even handle load while remaining "too high".

    You can measure the internal resistance in many ways and you will get different answers. Without measuring you can calculate one estimate by using CCA or MCA. They are defined at 1.2 V/cell for 30 seconds. Say your battery has 1000 A MCA and 12.6 V at 100% SOC and 0C (MCA is measured at 0C). Thus (12.6V-6*1.2V)/1000A = 0.0054 ohm.

    Now you have two similar batteries at 27C. One is at 100% SOC and the other at 25%, thus the voltages are 12.65 V and 12.06 V. When you connect them with extemely low resistance cables, you have only the internal resistance of both batteries. Thus I = (12.65V-12.06V)/(2*0.0054 ohm) = 54 A. In this case both batteries have terminal voltage at 12.355V.

    You probably know that is far from the truth. It is easy to test that the voltage will drop well below 12.355 when you load the 100% battery at 54 A and you will need well over 13 V to charge 25% battery at 54 A. So in reality the current between these two batteries is very low (a few A or even less).

    So you need a much more complicated model to cope with this situation. They do exists, but you need to know much about your batteries to implement it.
     
  11. mascip
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    mascip Junior Member

    Thank you Joakim for a very thorough answer. I gathered here and by searching some more online that battery models are complex and that internal resistance (and other parameters) varies a lot, so there is no easy answer. I searched some more online and many teams of researchers develop such models, which are not accurate enough yet to answer my question. So I've given up on finding a formula, but it was a good question to explore: I learned a fair bit about the dynamic behaviour of batteries.
     
  12. mascip
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    mascip Junior Member

    PS: Calculating the internal resistance by using cranking amps, smart! The result seems surprisingly low
     
  13. Joakim
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    Joakim Senior Member

    The internal resistance has to be low to enable cranking. Typically a car or a boat engine needs 100-200 A for cranking and terminal voltage is about 11 V for 50-70 Ah starting battery.

    You can also estimate the current of connecting two batteries using charts from datasheets. Here is one: http://www.yuasabatteries.com/pdfs/NP_18_12_DataSheet.pdf

    First look at the charging curves. At 25% charged volume (~25% SOC), the charging voltage is about 12.7 V and current 0.25 C. From discharging curves, you can see, that a at 100% SOC you only get 0.05 C at 12.7 V. So the current between the two batteries must be far less than 0.25 C.

    You may find charging curves at different currents to get a better estimate.
     
  14. mascip
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    mascip Junior Member

    Interesting curves, bookmarked. Thank you!

    For the 1V voltage drop of the diodes and the resulting 13.3v on the batteries, it seems that my friend's batteries won't be able to get more than 70% charged, according to the Charge curve. Is that right?

    If my friend's alternator has a voltage sensing pin, I could wire it to charge the batteries better, but will that increase the load on the alternator, hence reducing its lifespan? (and the lifespan of the belt?)
     

  15. gonzo
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    gonzo Senior Member

    Diodes give a 0.7V drop, not 1. This is for the usual silicon type. The voltage regulator can be adjusted for it so the battery gets a proper charge.
     
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