Internal resistance of a battery

Discussion in 'General Electronics Chat' started by Jas9, Nov 14, 2010.

  1. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    hello friends,
    i am calculating internal resistance of a car battery during the charging phase. When the car is moving the alternator keeps on charging the battery if the output voltage of the alternator is greater than the battery voltage. Basically to proceed with this i have input of continuous voltage and current which is measured at the battery terminals. These signals are in phase and that means its a pure resistive circuit. So the voltage divide by the current has to give the internal resistance value. If we consider an open circuit in which the internal resistance would be calculated as,
    Ri=Voc-Vterminal/Ibatt
    but when i have to calculate the battery internal resistance during its charging phase then how i amply this formula because in this case i have voltage and current signal which are measured at its terminal and its going to be a closed circuit. So should i directly divide Vbatt/Ibatt to get internal resistance or something else?
    The other thing is the voltage and current has ac component that is some ripples so how to consider these ripples to get internal resistance?
    thanks
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
  3. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    thanks but i didnt find related answer to my above question
     
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    The internal impedance of a typical lead acid battery is around 50 milliOhms, new and can reach 1 ohm by the end of the life of the battery.

    A typical car alternator is a polyphase generator, with low ripple so you do not need to take this into account (luckily as the maths gets quite hairy).
     
  5. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    ok thanks but now as i said its a purely resistive circuit as their is no phase shift between the voltage and signal then the internal resistance of the battery would be the voltage measured at the battery terminal divided by the current measured at the battery terminal right?
     
  6. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    No, in many aspects of battery operation, the internal resistance of the battery is not the voltage measured at the battery terminal divided by the current measured at the battery terminal.
     
  7. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    ok but as i said its a pure resistive circuit then how to proceed?
    if it would have been an open circuit then
    Ri=Voc-Vterminal/Ibatt
    where
    Ri=Internal resistance,
    Voc=Open circuit voltage,
    Vterminal=Terminal voltage,
    Ibatt=Battery current

    so in case of such battery Ri can be easily calculated but while charging how it will be?
     
  8. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    No, that is the load resistance.

    You cannot measure the internal impedance directly, although Google reports several patents to do this.

    The internal resistance forms a potential divider with the external load resistance. A first line assumption is that the EMF and internal resistance of the battery remains constant so as the current increases the external load decreases.

    You should be able to work out the implications of this, bearing in mind that you are using your battery as a sink not a source.
     
  9. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    oh i c....to be honest i am banging my head now as i am very much confused in getting it work...so that means if i divide the terminal voltage by the current then it will give me the total impedance of the battery
     
  10. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    There was a very similar question in another forum recently.

    Look at post #11 which analyses the situation where there are two unequal voltages supplies in parallel with a load.

    If you take the battery as one and the generator output as the other and your automotive wiring resistance as the load, you have the same situation.

    http://www.physicsforums.com/showthread.php?t=446780
     
  11. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    ok so that means if i am correct that
    Ri=Vgen-Vterminal/Ibatt
     
  12. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Try drawing a circuit. Both the battery and the generator have their own internal impedances.
     
  13. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    i already did the circuit drawing where a alternator is parallel to battery. So their are two emf and if i take a subtraction of these two then the corresponding value will be the voltage across the resistor.
     
  14. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    What about the load resistance presented by the auto-wiring?

    Your circuit should include three resistances in a T and two EMFs.
     
  15. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    two emf is ok but how come three resistance? one would be the series resistance after then battery emf, but if i consider the alternator resistance as negligible then? if not the how this would be calculated?
     
  16. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    All sources, without exception in this universe, have an internal resistance in series with them.

    You either put in a typical value or you measure this resistance. In order to measure this resistance you need two points on the terminal voltage/current curve.

    This was explained, with example calculations, in the link I gave. Did you look at it?

    One point is usually the open circuit voltage. A second one is found by measuring the terminal voltage at known current. The series resistance is the slope of the line joining the two points on a terminal voltage/current graph.

    Many (auto) manufacturers supply open circuit (no load) and rated current terminal voltages for their alternators. You should find such information in the service handbook.
     
  17. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    thanks a lot friend i will follow this tomorrow and will let you know about my work thanks again
     
  18. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,795
    950
    The best way to determine it is a low resistance load.


    determine the resistance and short the battery across the load for about 1 second. Take the voltage reading on the battery terminals...This is how the old battery checkers worked.

    If the load was .25 ohms and open circuit voltage was 12.7 V
    then 51 amps would flow(50.8) You hook up the meter and find 11.8 volts on the battery when the load is placed across it. So there is 47 amps flowing, not 51(47.2)

    There is a .9 volt drop across the battery internal resistance when 47.2 amps flows. This gives us an internal resistance of .02 ohms (.01906)

    That's a little high for a real automotive starting battery resistance. My guess of 11.8V at that load was too low. Figure it for 12 or 12.1 volts and it would be about right.

    I believe some of them used wraps of nichrome wire for loads and they had to cool down between uses
     
  19. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    hi again,well i have attached my circuit for the battery during charging that is it is getting charged by the alternator when the vehicle is moving. The Ubatt and Ibatt are measured at the battery terminals and are assumed to be same as the alternator output excluding the resistance of the alternator. So now in the circuit Ubatt=Uc0+Ibatt*Ri+Voltage across the capacitor/rct. But i am finding it difficult to resolve this circuit maybe i have wrong voltage directions. Can you please check it out.
    thanks
     
  20. Jas9

    Thread Starter Member

    Sep 30, 2010
    36
    0
    well after going through lots of literature i conclude that during charging of a battery the terminal voltage is bigger than cell emf so the internal resistance of a cell causes the terminal voltage to change because of voltage drop across it so the internal resistance of a cell during charging would be
    Ri=Vterminal-Emf/I
    and the circuit would consist of only one resistance and a capacitor(electrochemical double layer capacitor) in series
    i hope i succeed with this assumption
    if you have any views or suggestions about this then please let me know
     
Loading...