[NSRCA-discussion] Internal battery resistance

[James Oddino]

How about using your Wattmeter?  Take readings of voltage and current  
at a number of throttle settings.  Divide the change in voltage by the  
change in current between any two settings to get the resistance.   
This is easier to do on a spread sheet if you have a lot of settings  
and makes it easy to plot a curve of resistance vs. current.  You will  
be able to see if it is linear as well as get a reading at the current  
of interest.  Do it again with your pack after it is discharged.

Save yourself $20 and 4 hours.  Send contributions to Jim O so that we  
may continue to provide you with this great entertainment.  Hope you  
are all smiling.

Jim O




On Mar 16, 2008, at 5:42 PM, Lance Van Nostrand wrote:

> 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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