[NSRCA-discussion] Internal battery impedance

[Lance Van Nostrand]

Thanks for all the posts. I understand the piece of the puzzle I was missing.  Actually pieces.  bob's point is important but the basic approach is to start with a resistor of a known resistance that will give you the current where you want internal impedance measured.  Let's say I have a fresh 2S pack measuring 8v unloaded.  I buy a heavy duty resistor of 1 ohm that actually measures exactly 1 ohm (for example only).  Under load it measures 7.2v across the resistor and is therefore pulling 7.2A (I = V/R)
The voltage drop is due to internal resistance so 7.2a = 8v / (1 + internal impedance) 
internal impedance = .11 ohm

Is the idea of measuring at 1 ohm and 10 ohm a way to find an average number?  If so, I think just finding a resistor that is in the neighborhood of your worst case situation may actually be more representative.

--Lance

  ----- Original Message ----- 
  From: Bob Richards 
  To: NSRCA Mailing List 
  Sent: Sunday, March 16, 2008 6:54 PM
  Subject: Re: [NSRCA-discussion] Internal battery impedance


  The resistor does not really need to be a precision resistor, you just need to know precisely what the resistance is. :-)  And, if you are comparing or matching cells, the measurement only needs to be relative and the accuracy may not important.

  I do agree that the measurement will probably not be linear, IOW the cell impedance may vary with large changes in current. If I normally run 30 amps, I would want to make a measurement at 30 amps and probably 25 and 35 amps as well, comparing the differences in voltage at the three currents. This would be more meaningful to me than comparing the unloaded voltage with the voltage at 1 amp.

  Just M2CW.

  Bob R.


   


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