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Re: cold resistors



On Sun, 2 Jan 2000, John Denker wrote:

At 08:37 PM 1/2/00 -0600, brian whatcott wrote:

Temp (K) Res (ohms)
0.3 10^6
1 10^2
3 3
10 1
30 0.7
100 0.3 (from a graph of Rose-Innes's)

All the data except the 100K point lie in the range where I would expect a
simple fit to apply. ...

When I throw the fit I get
log(R) = 0.1136/sqrt(T) + 0.327
for this resistor.

This fits all the data modestly well. But there's so much curvature in the
graph of log(R) versus 1/sqrt(T) that I wouldn't be very happy using this
fit to interpolate or extrapolate near the low-temperature end (which is
where I would be most interested).

I guess Brian and I misunderstood the fitting function; I was using
log(R) = a - b*sqrt(T)

But perhaps I am still missing something because I don't see that your fit
even begins to resemble the data. When I use your form I get quite
different results, specifically
log(R) = 8.6/sqrt(T) - 2.7

with a *lot* of uncertainty in those coefficients and pretty ugly looking
residuals. I get a far better fit to all but the 100 degree point with
log(R) = 4.36/T - 0.5

which does at least agree with your observations 1) that fiddling with the
exponent can produce better results and 2) that the 100 degreee point may
be anomalous. It's probably just a matter of units somewhere along the
line.

Ah well. Classes start tomorrow; gotta finish my syllabi.

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm