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Read a Pt100

M

Marco Trapanese

Jan 1, 1970
0
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
 
U

Uwe Hercksen

Jan 1, 1970
0
Marco said:
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.
Hello,

if the needed temperature range is small and the resolution is large, a
small lookup table will do. You may use a stepwise linear aproximation
of the formula for a larger temperature range and smaller resolution.

Bye
 
T

Tauno Voipio

Jan 1, 1970
0
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


You may be better off by dropping most of the electronics.

Get a high-resolution A/D converter (e.g. LTC 2420) and one
precision resistor from the refefence voltage to the A/D
input. Connect the Pt from the A/D input to ground.

This will lose most of the converter range, but what is
left will be good enough for 0.1 C resolution from -70 C
to +250 C, and this is better than the precision of most
of the sensor elements.

For calculation, a third-degree polynomial computed with
32 bits will be accurate enough and computable with a
small AVR. I computed the coefficients by putting the
circuit equations into an Excel spreadsheet at 1 degree
increments and doing a regression fit for the polynomials.
Inverting the equations is for masochists only.

Been there, done that.
 
N

Nico Coesel

Jan 1, 1970
0
Marco Trapanese said:
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?

Thats quite complicated. What accuracy do you need? Better use a
pt1000 element. Feed from 3.3V through a 2.2k resistor and read with
an ADC.
 
S

Spehro Pefhany

Jan 1, 1970
0
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

Assuming you're in no hurry for the results, you can do a binary
search using the Callendar-Van Dusen equation and get EXTREMELY good
results. Just loop the evaluations for a fixed number of iterations--
it really won't take much code at all.



Best regards,
Spehro Pefhany
 
T

Tauno Voipio

Jan 1, 1970
0
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

You may be better off by dropping most of the electronics.

Get a high-resolution A/D converter (e.g. LTC 2420) and one
precision resistor from the refefence voltage to the A/D
input. Connect the Pt from the A/D input to ground.

This will lose most of the converter range, but what is
left will be good enough for 0.1 C resolution from -70 C
to +250 C, and this is better than the precision of most
of the sensor elements.

For calculation, a third-degree polynomial computed with
32 bits will be accurate enough and computable with a
small AVR. I computed the coefficients by putting the
circuit equations into an Excel spreadsheet at 1 degree
increments and doing a regression fit for the polynomials.
Inverting the equations is for masochists only.

Been there, done that.

--

Tauno Voipio- Skjul tekst i anførselstegn -

- Vis tekst i anførselstegn -

You really don't need high resolution ADC for a PT100 temperature
measurement. A good reference and a 10bit ADC is enough for 0,5 degree
accuracy

Regards

Klaus

Then you do need more analog circuitry over the only reference resistor.
The added precision resistors cost more than the A/D.
 
S

Spehro Pefhany

Jan 1, 1970
0
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

You may be better off by dropping most of the electronics.

Get a high-resolution A/D converter (e.g. LTC 2420) and one
precision resistor from the refefence voltage to the A/D
input. Connect the Pt from the A/D input to ground.

This will lose most of the converter range, but what is
left will be good enough for 0.1 C resolution from -70 C
to +250 C, and this is better than the precision of most
of the sensor elements.

For calculation, a third-degree polynomial computed with
32 bits will be accurate enough and computable with a
small AVR. I computed the coefficients by putting the
circuit equations into an Excel spreadsheet at 1 degree
increments and doing a regression fit for the polynomials.
Inverting the equations is for masochists only.

Been there, done that.

--

Tauno Voipio- Skjul tekst i anførselstegn -

- Vis tekst i anførselstegn -

You really don't need high resolution ADC for a PT100 temperature
measurement. A good reference and a 10bit ADC is enough for 0,5 degree
accuracy

Regards

Klaus

I've used Pt100 sensors that are good to a millidegree or better (they
cost as much as a low end car). Even the cheap ones are as stable as
that under benign conditions.

In North America anyway, we tend to use thermocouples for the sloppy
applictions (like plastics and packaging machines) and RTDs are
reserved for the more demanding applications.
 
J

John Devereux

Jan 1, 1970
0
Fred Bloggs said:
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?
 
S

Spehro Pefhany

Jan 1, 1970
0
Fred Bloggs said:
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.

BTW, I don't think that ADC sample code could be written in a more
inefficient way. Horrible, even considering it's free.


Best regards,
Spehro Pefhany
 
T

Tauno Voipio

Jan 1, 1970
0
Fred Bloggs said:
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 physics of the sensor does not have any discontinuities
at 0 C. The different constants are used to keep the error
funnel at its narrowest around zero C.
 
J

John Devereux

Jan 1, 1970
0
Spehro Pefhany said:
Fred Bloggs said:
On Nov 15, 8:08 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...
 
Spehro Pefhany said:
On Nov 15, 8:08 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!

Would you rather have the change made at 12.3456C? 0C seems to be a pretty
good place to make a "fit" change, to me (as long as there isn't a
discontinuity there). Easy on the algorithm, anyway.
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...

That just shows how "fishy" the Kelvin scale is. ;-) It would look pretty
messy on the Rankin scale, too.
 
M

Martin Brown

Jan 1, 1970
0
Spehro Pefhany said:
On Nov 15, 8:08 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!

To be fair they do show the way the positive model function goes
pearshaped at around -20C on their graph Fig4 of errors at negative T.
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...

Their algebra is a bit ropey too. There is no need at all for Z4 it can
be divided out of the numerator terms as a constant and the problem can
with care be recast in fixed point arithmetic.

BTW if your hardware has a fast integer divide then it is worth
considering rational approximations to the inverse function which have
smaller errors than polynomials using the same number of coefficients.
 
S

Spehro Pefhany

Jan 1, 1970
0
Spehro Pefhany said:
On Nov 15, 8:08 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..
 
J

John Devereux

Jan 1, 1970
0
Tauno Voipio said:
Fred Bloggs said:
On Nov 15, 8:08 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 physics of the sensor does not have any discontinuities
at 0 C. The different constants are used to keep the error
funnel at its narrowest around zero C.

Thanks yes that makes more sense. It was just one of those things that
could not be exactly physically correct. Like taking the logarithm of a
distance. Or travelling at the "square of the speed of light" like I
remember in some dodgy scifi.
 
J

John Devereux

Jan 1, 1970
0
Spehro Pefhany said:
Spehro Pefhany said:
On Thu, 17 Nov 2011 10:12:33 +0000, the renowned John Devereux


On Nov 15, 8:08 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..

Great explanation, thanks.
 
U

Uwe Hercksen

Jan 1, 1970
0
John said:
Thanks yes that makes more sense. It was just one of those things that
could not be exactly physically correct. Like taking the logarithm of a
distance. Or travelling at the "square of the speed of light" like I
remember in some dodgy scifi.

Hello,

lets have a look a both equations and compare them.

RRTD(t) = R0[1 + At + Bt^2 + C(t – 100°C)t^3] (for t <= 0°C)
RRTD(t) = R0[1 + At + Bt^2] (for t >= 0°C)

A = 3.9083 * 10^–3 °C^–1
B = –5.775 * 10^–7 °C^–2
C = –4.183 * 10^–12 °C^–4

The coefficents A, B and C are the same for both equations, C is used
only in the first equation and C is very small.
Both equations have the same result for 0 °C, R0.
Because C is very small compared to A and B, C has a (small) effect only
for very low temperatures.

You may use an excel sheet to calculate values of both equations for
some values between 0 and -273 °C, when you compare both results, you
will see the effect of C.

Bye
 
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