Spehro Pefhany said:
On Thu, 17 Nov 2011 10:12:33 +0000, the renowned John Devereux
On Nov 15, 8:08Â am, Marco Trapanese <
[email protected]>
wrote:
Hello,
to acquire a Pt100 I set up a board with INA326. Full bridge: the upper
resistors are 100k (say R1 and R2), the lower 100R (R(t) and R3).
The INA326 is programmed to gain 10 (4k and 200k). Supply voltage is 3.3V.
The start equation is:
R(t) = R0(1 + At + Bt^2)
My output is:
Vo(t) = G(V+ - V-)
where G is the INA326 gain, V+ and V- both arms of the bridge.
Thus:
Vo(t) = G[Vs*R(t)/(R1+R(t)) - k]
where k is the reference voltage from the fixed network divider.
Well, but I need t(Vo)!
Inverting the formula leads to a quite complex expression, and I need to
fit it into a small AVR.
This is the right way or there is something else that I'm missing?
Thanks!
Marco
The mathematical expedients here apply to any method of measuring
R(t)
http://www.analog.com/static/imported-files/application_notes/AN709_0.pdf
There is something weird that bugs me about those RTD equations.... How
is it possible that a different equation applies to "positive" and
"negative" temperatures?
The same equation is used in the Callendar-Van Dusen equations- but
one of the coefficients is different. That's a bit odd, but it's not
unusual to have different equations valid over different regions.
But the 0 degrees C point is totally artifical, and seems to bear no
relation to anything physical - unless we are talking about water which
we are not!
On the Kelvin scale one equation describing the resistance of platinum
suddenly takes over from another when we get to 273.15K?
Seems a bit fishy...
Perhaps it does seem so. It's a useful division point because many
applications (eg. comfort heating) only need to work above 0°C, so the
simpler version can be used. The reasons are historical and practical.
The RTD resistance is specified at 0°C and 100°C. Callendar came up
with an equation that used measurements at an additional (higher)
point (eg. freezing point of pure Zn). That agrees very well with
actual resistance over the range of 0..600°C at least (and doesn't go
all squirrely outside the optimal range like typical higher order
least squares polynomial fits- it's still quite good at more than
800°C).
Subsequently, Van Deusen added the higher order terms to reduce the
error at very low temperatures, the determination of which involves
using another convenient low temperature point such as the boiling
point of pure O2.
The equations are written in terms of degrees C, so the effect of all
three coefficients is zero at T = 0°C-- so it's a natural place to
make a transition. Kind of arbitrary, but it fits really well and is
convenient (and it matches the IEC standard exactly).
BTW, if T is constrained to be >= 0° it's possible to exactly solve
for T (R) using the CvD equations (Actually just C's contribution--
it's just a quadratic in T)-- so multiplications, divisions and a
square root. It's also pretty close at temperatures a bit below zero,
if a fudge works for you. I never get that lucky, but others might..