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NTC thermistor functions are somewhat complex, and it is usually more efficient to use a lookupHi,
Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
Thank you for helping.
Huub
You don't give enough information. How precise does your conversion needHi,
Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
On Mon, 27 Jun 2005 19:19:58 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Hi,
Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
You don't give enough information. How precise does your conversion need
to be? A simple linear approximation works just fine for some ranges and
for some applications. A second order curve handles many others.
For a more precise curve fit, search for the Steinhart-Hart thermistor
equation.
Thank you for this info. Could you tell me what those a, b and c stand for?
Rich Webb wrote:
On Mon, 27 Jun 2005 19:19:58 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Hi,
Is there any formula for converting a given resistance of the
(Siemens) NTC K164/1K5 into temperature in Celsius? I'm not familiar
with components, but I need to use it in software.
You don't give enough information. How precise does your conversion need
to be? A simple linear approximation works just fine for some ranges and
for some applications. A second order curve handles many others.
For a more precise curve fit, search for the Steinhart-Hart thermistor
equation.
Thank you for this info. Could you tell me what those a, b and c stand for?
Thank you for helping.
Huub
Ok, found it..
by (1/(1/T))-273.15 I get a rediculous temperature in Celsius. Hence, IOn Mon, 27 Jun 2005 19:19:58 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Hi,
Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
You don't give enough information. How precise does your conversion need
to be? A simple linear approximation works just fine for some ranges and
for some applications. A second order curve handles many others.
For a more precise curve fit, search for the Steinhart-Hart thermistor
equation.
This formula gives 1/T with T in Kelvin. When I rework this to Celsius
Don't assume that because S-H is the "best" approximating formula thatRich Webb wrote:
On Mon, 27 Jun 2005 19:19:58 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Hi,
Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
You don't give enough information. How precise does your conversion need
to be? A simple linear approximation works just fine for some ranges and
for some applications. A second order curve handles many others.
For a more precise curve fit, search for the Steinhart-Hart thermistor
equation.
This formula gives 1/T with T in Kelvin. When I rework this to Celsius
by (1/(1/T))-273.15 I get a rediculous temperature in Celsius. Hence, I
suppose I'm doing something wrong. Can you tell me how I should do it?
On Tue, 28 Jun 2005 20:10:45 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Rich Webb wrote:
On Mon, 27 Jun 2005 19:19:58 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Hi,
Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
You don't give enough information. How precise does your conversion need
to be? A simple linear approximation works just fine for some ranges and
for some applications. A second order curve handles many others.
For a more precise curve fit, search for the Steinhart-Hart thermistor
equation.
This formula gives 1/T with T in Kelvin. When I rework this to Celsius
by (1/(1/T))-273.15 I get a rediculous temperature in Celsius. Hence, I
suppose I'm doing something wrong. Can you tell me how I should do it?
Don't assume that because S-H is the "best" approximating formula that
it's appropriate for real world use. Do you really need (and can you
afford the computational burden) of a 0.1 degree approximation? What's
the spec'd accuracy of the thermistor and associated components?
For A = 1.139456E-3, B = 2.323002E-4, C = 9.579158E-8, and R = 10000
then 1/T = 298.164 and Tc = 25.014. It's just plugging in the numbers.
In what units are A, B and C? inches or mm? I used mm.
taken that A, B and C are the physical measurements of the thermistorOn Tue, 28 Jun 2005 20:10:45 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Rich Webb wrote:
On Mon, 27 Jun 2005 19:19:58 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Hi,
Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
You don't give enough information. How precise does your conversion need
to be? A simple linear approximation works just fine for some ranges and
for some applications. A second order curve handles many others.
For a more precise curve fit, search for the Steinhart-Hart thermistor
equation.
This formula gives 1/T with T in Kelvin. When I rework this to Celsius
by (1/(1/T))-273.15 I get a rediculous temperature in Celsius. Hence, I
suppose I'm doing something wrong. Can you tell me how I should do it?
Don't assume that because S-H is the "best" approximating formula that
it's appropriate for real world use. Do you really need (and can you
afford the computational burden) of a 0.1 degree approximation? What's
the spec'd accuracy of the thermistor and associated components?
For A = 1.139456E-3, B = 2.323002E-4, C = 9.579158E-8, and R = 10000
then 1/T = 298.164 and Tc = 25.014. It's just plugging in the numbers.
Still assuming that I can use Steinhart - Hart for all thermistors:
taken that A (diameter of NTC disc), B (width of NTC disc) and COn Tue, 28 Jun 2005 20:10:45 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Rich Webb wrote:
On Mon, 27 Jun 2005 19:19:58 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
Hi,
Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
You don't give enough information. How precise does your conversion need
to be? A simple linear approximation works just fine for some ranges and
for some applications. A second order curve handles many others.
For a more precise curve fit, search for the Steinhart-Hart thermistor
equation.
This formula gives 1/T with T in Kelvin. When I rework this to Celsius
by (1/(1/T))-273.15 I get a rediculous temperature in Celsius. Hence, I
suppose I'm doing something wrong. Can you tell me how I should do it?
Don't assume that because S-H is the "best" approximating formula that
it's appropriate for real world use. Do you really need (and can you
afford the computational burden) of a 0.1 degree approximation? What's
the spec'd accuracy of the thermistor and associated components?
For A = 1.139456E-3, B = 2.323002E-4, C = 9.579158E-8, and R = 10000
then 1/T = 298.164 and Tc = 25.014. It's just plugging in the numbers.
Still assuming that I can use Steinhart - Hart for all thermistors:
They are not the physical dimensions of the thermistor. They areStill assuming that I can use Steinhart - Hart for all thermistors:
taken that A (diameter of NTC disc), B (width of NTC disc) and C
(distance between the legs) are the physical measurements of the
thermistor (Siemens NTC K164/1K5), I used metrics mm: A = 4mm, B = 1.5mm
and C = 5.5mm. When I convert these to inches: A = 0.15748031", B =
0.059055118" and C = 0.21653543".
Tried in my workroom at 19C:
Tc = (1 / (0.15748031 + (0.059055118 * ln(1686.64)) + (0.21653543 *
(ln(1686.64)^3)))) - 273.15 = -273.139 C
So, where do I go wrong?
www.ILXLightware.com in which I had to fill in those empirical values.They are not the physical dimensions of the thermistor. They are
empirical values determined by measuring at least three resistance -
temperature pairs and then solving the resulting simultaneous equations
for A, B, and C [*]. Modulo manufacturing variations, they are constant
for a given thermistor material and fabrication.
[*] or use a generalized non-linear curve fitting app, like Mathcad,
which makes the number crunching easy and also plots the residuals.
I've noticed that now. Via google I found a spreadsheet at
No html server at that address.They are not the physical dimensions of the thermistor. They are
empirical values determined by measuring at least three resistance -
temperature pairs and then solving the resulting simultaneous equations
for A, B, and C [*]. Modulo manufacturing variations, they are constant
for a given thermistor material and fabrication.
[*] or use a generalized non-linear curve fitting app, like Mathcad,
which makes the number crunching easy and also plots the residuals.
I've noticed that now. Via google I found a spreadsheet at
www.ILXLightware.com in which I had to fill in those empirical values.
After doing so, I found out I got the some crazy result: 273 C. Below
are my values.
Your resistance figures are five significant digits but your temperatureT (C) R (Ohm)
24 1518,49
29 1268,01
30 1198,8
32 1080,08
34 997,11
36 931,79
38 883,4
40 819,68
42 749,06
44 694,89
46 679,53
48 633,78
50 558,56
100 214,65
If you could show me how you resolve or calculate the a, b and c with
them, I know where I go wrong.
Datasheet for K164 series:Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
On Thu, 30 Jun 2005 17:12:55 +0200, Huub <hdotvdotniekerkathccnetdotnl
wrote:
They are not the physical dimensions of the thermistor. They are
empirical values determined by measuring at least three resistance -
temperature pairs and then solving the resulting simultaneous equations
for A, B, and C [*]. Modulo manufacturing variations, they are constant
for a given thermistor material and fabrication.
[*] or use a generalized non-linear curve fitting app, like Mathcad,
which makes the number crunching easy and also plots the residuals.
I've noticed that now. Via google I found a spreadsheet at
www.ILXLightware.com in which I had to fill in those empirical values.
After doing so, I found out I got the some crazy result: 273 C. Below
are my values.
No html server at that address.
T (C) R (Ohm)
24 1518,49
29 1268,01
30 1198,8
32 1080,08
34 997,11
36 931,79
38 883,4
40 819,68
42 749,06
44 694,89
46 679,53
48 633,78
50 558,56
100 214,65
If you could show me how you resolve or calculate the a, b and c with
them, I know where I go wrong.
Your resistance figures are five significant digits but your temperature
measurements are only two, so don't expect that the approximating
function will give results to better than two sig figs.
The curve fitting is done with Mathcad, plugging the S-H formula into
its general curve fitting engine.
A = -9.009306E-4, B = 7.678299E-4, C = -3.481274E-6
R = 1686.64 --> Tc = 23
I assume that these results are derived from my empirical values?
curve.On Mon, 27 Jun 2005 19:19:58 +0200, Huub
hdotvdotniekerkathccnetdotnl> put finger to keyboard and composed:
Is there any formula for converting a given resistance of the (Siemens)
NTC K164/1K5 into temperature in Celsius? I'm not familiar with
components, but I need to use it in software.
Datasheet for K164 series:
http://www.epcos.com/inf/50/db/ntc_02/00720073.pdf
Use R/T characteristic curve 1013:
http://www.epcos.com/inf/50/db/ntc_02/01470178.pdf
Steinhart-Hart thermistor equation:
http://www.betatherm.com/stein.htm
1/T = A + B * ln(R) + C * ln(R)^3
where T = T(Celsius) + 273.15
Use three points (endpoints and midrange) from your R/T curve (#1013)
to generate three simultaneous equations. Then solve for A, B, and C.
Note that A, B, and C are not material constants.
- Franc Zabkar
Thank you. I figured such, but it looks like I got directed to the wrong
Here's a good Excel spreadsheet:They are not the physical dimensions of the thermistor. They are
empirical values determined by measuring at least three resistance -
temperature pairs and then solving the resulting simultaneous equations
for A, B, and C [*]. Modulo manufacturing variations, they are constant
for a given thermistor material and fabrication.
[*] or use a generalized non-linear curve fitting app, like Mathcad,
which makes the number crunching easy and also plots the residuals.
I've noticed that now. Via google I found a spreadsheet at
www.ILXLightware.com in which I had to fill in those empirical values.
The Epcos tables suggest this should be 108.2 ohm (= 1500 x 0.072139).After doing so, I found out I got the some crazy result: 273 C. Below
are my values.
T (C) R (Ohm)
24 1518,49
29 1268,01
30 1198,8
32 1080,08
34 997,11
36 931,79
38 883,4
40 819,68
42 749,06
44 694,89
46 679,53
48 633,78
50 558,56
100 214,65
^^^^^^
The method and formulae are evident from the spreadsheet.If you could show me how you resolve or calculate the a, b and c with
them, I know where I go wrong.
Using the data for 25C, 50C, and 100C, I get:Thank you for helping.
spreadsheet I get a good result. However, implementing the equation inOn Thu, 30 Jun 2005 17:12:55 +0200, Huub
hdotvdotniekerkathccnetdotnl> put finger to keyboard and composed:
They are not the physical dimensions of the thermistor. They are
empirical values determined by measuring at least three resistance -
temperature pairs and then solving the resulting simultaneous equations
for A, B, and C [*]. Modulo manufacturing variations, they are constant
for a given thermistor material and fabrication.
[*] or use a generalized non-linear curve fitting app, like Mathcad,
which makes the number crunching easy and also plots the residuals.
I've noticed that now. Via google I found a spreadsheet at
www.ILXLightware.com in which I had to fill in those empirical values.
Here's a good Excel spreadsheet:
http://www.reed-electronics.com/tmworld/software/stinhart.zip
This article may be of help. It includes some VB code.
http://www.reed-electronics.com/tmworld/article/CA187515.html
After doing so, I found out I got the some crazy result: 273 C. Below
are my values.
T (C) R (Ohm)
24 1518,49
29 1268,01
30 1198,8
32 1080,08
34 997,11
36 931,79
38 883,4
40 819,68
42 749,06
44 694,89
46 679,53
48 633,78
50 558,56
100 214,65
^^^^^^
The Epcos tables suggest this should be 108.2 ohm (= 1500 x 0.072139).
If you could show me how you resolve or calculate the a, b and c with
them, I know where I go wrong.
The method and formulae are evident from the spreadsheet.
Thank you for helping.
Using the data for 25C, 50C, and 100C, I get:
A = 0.0015481
B = 0.00023793
C = 0.00000016842
Using the data for 0C, 25C, and 100C, I get:
A = 0.0015469
B = 0.00023826
C = 0.00000016547
- Franc Zabkar
Thank you for this information. Indeed, when filling in my values in the
This should be a "*".Franc Zabkar wrote:
On Thu, 30 Jun 2005 17:12:55 +0200, Huub
hdotvdotniekerkathccnetdotnl> put finger to keyboard and composed:
They are not the physical dimensions of the thermistor. They are
empirical values determined by measuring at least three resistance -
temperature pairs and then solving the resulting simultaneous equations
for A, B, and C [*]. Modulo manufacturing variations, they are constant
for a given thermistor material and fabrication.
[*] or use a generalized non-linear curve fitting app, like Mathcad,
which makes the number crunching easy and also plots the residuals.
I've noticed that now. Via google I found a spreadsheet at
www.ILXLightware.com in which I had to fill in those empirical values.
Here's a good Excel spreadsheet:
http://www.reed-electronics.com/tmworld/software/stinhart.zip
This article may be of help. It includes some VB code.
http://www.reed-electronics.com/tmworld/article/CA187515.html
After doing so, I found out I got the some crazy result: 273 C. Below
are my values.
T (C) R (Ohm)
24 1518,49
29 1268,01
30 1198,8
32 1080,08
34 997,11
36 931,79
38 883,4
40 819,68
42 749,06
44 694,89
46 679,53
48 633,78
50 558,56
100 214,65
^^^^^^
The Epcos tables suggest this should be 108.2 ohm (= 1500 x 0.072139).
If you could show me how you resolve or calculate the a, b and c with
them, I know where I go wrong.
The method and formulae are evident from the spreadsheet.
Thank you for helping.
Using the data for 25C, 50C, and 100C, I get:
A = 0.0015481
B = 0.00023793
C = 0.00000016842
Using the data for 0C, 25C, and 100C, I get:
A = 0.0015469
B = 0.00023826
C = 0.00000016547
- Franc Zabkar
Thank you for this information. Indeed, when filling in my values in the
spreadsheet I get a good result. However, implementing the equation in
my program on Linux in C++, this happens:
for 1/K:
K1 = a + (b + (log(R))) + (c * pow(log(R), 3))
^^^
Which value are you using for R? Is it 1686.64?to get C I invert this, i.e. 1 / (1/K), and subtract 273,15
for directly into C:
tc = (1 / (a + b * (log(R)) + c * power(log(R),3))) - 273,15
Using your a, b and c I get -221,165 C with both equations. As far as I
can see, both equations are correct, but obviously something is wrong.
Which value are you using for R? Is it 1686.64?
I use the actual value read from the interface.
Can you copy and paste your actual C++ code into your next post?
See below.
I suspect your error is in the last term. Perhaps you are evaluating
the following ...
[c * log(R)] ^ 3
... instead of ...
c * {[log(R)] ^ 3}
- Franc Zabkar
This is the calculating part of my code: