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Re: cold resistors



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

Taking a sample A-B 0.27ohm resistor:
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)

I don't have a curve fitting package here. Can someone else
fit this?

I did all the following analysis using nothing more than a spreadsheet
program. Nowadays they can fit to monomial power laws, polynomials, exp(),
log(). I did it this way so that typical students should be able to
replicate the result.

And then at 07:21 PM 1/2/00 -0800, John Mallinckrodt wrote:

Make that Resistance = r*exp(k*sqrt(temperature))

Assuming that your data is good and that John's statement in true, then
your data does not all lie within the "large range" that John refers to.
I can get a tolerable, but far from excellent fit to no more than three of
these data points--not an impresssive feat for a two parameter fitting
function.

All the data except the 100K point lie in the range where I would expect a
simple fit to apply. There's not much point in a fit if you can only fit
three data points with two adjustable parameters.

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).

If I cared about this data, I would fiddle the exponent. That produces a
very, very good fit:
log(R) = 0.1908 T^(-.85) + 0.1433

This exposes the 100K point as definitely anomalous, which is hardly
surprising if you know a little physics (Debye temperature and all that).

Before trusting my life to this fit I would like to have one more
calibration point at 0.5K or thereabouts.

The fact that an exponent other than -0.5 gives the best fit reinforces my
suspicion that this isn't the same physics as Richardson and Dushmann
studied. I have no idea what *IS* the physics here. If anybody figures it
out, let me know.