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<DIV><FONT face=Arial size=2>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)</FONT></DIV>
<DIV><FONT face=Arial size=2>The voltage drop is due to internal resistance so
7.2a = 8v / (1 + internal impedance) </FONT></DIV>
<DIV><FONT face=Arial size=2>internal impedance = .11 ohm</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>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.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>--Lance</FONT></DIV>
<DIV> </DIV>
<BLOCKQUOTE
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><B>From:</B>
<A title=bob@toprudder.com href="mailto:bob@toprudder.com">Bob Richards</A>
</DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A
title=nsrca-discussion@lists.nsrca.org
href="mailto:nsrca-discussion@lists.nsrca.org">NSRCA Mailing List</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Sunday, March 16, 2008 6:54
PM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: [NSRCA-discussion] Internal
battery impedance</DIV>
<DIV><BR></DIV>
<DIV>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.</DIV>
<DIV> </DIV>
<DIV>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.</DIV>
<DIV> </DIV>
<DIV>Just M2CW.</DIV>
<DIV> </DIV>
<DIV>Bob R.</DIV>
<DIV><BR><BR> </DIV>
<P>
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