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Topic: Can't get correct value from thermistor (Read 1 time) previous topic - next topic

erbedo

Hi
    I have a 50K thermistor (this one), which I cannot get to work correctly.

My set up is the following:
5V -- T -- A -- 10K -- GND
Where T is the thermistor, A is the analog PIN.

The problem is that I got weird temperatures like 100 celsius degrees, even if I am sure that the SH equation is correct, since I read the same values as in the datasheet table.

Any idea about that?

Thanks

Arrch

Have you tried manually measuring the resistance of the thermistor and verifying the temperature from the datasheet?

robtillaart


Can you post your code?
and/or the SH equation?

Such a table is work for - http://arduino.cc/playground/Main/MultiMap -

Think 11 - 15 entries should be enough to get an acceptable approximation although I do not know your requirements
Rob Tillaart

Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -
(Please do not PM for private consultancy)

oric_dan

Since you have a 50K@room_temp thermistor, is it possible your curve is expecting to see
R = 50K rather than 10K?

erbedo

My code is:
Code: [Select]

#include <math.h>
#define READING_PIN 5
#define SERIESRESISTOR 10000

double R1 = 10000.0; //resistance put in parallel
double V_IN = 5.0;

double A = 9.6564E-04;
double B = 2.1069E-04;
double C = 8.5826E-8;
double K = 9.5; // mW/dec C - dissipation factor

double SteinhartHart(double R) {
// calculate temperature
  double logR  = log(R);
  double logR3 = logR * logR * logR;

  return 1.0 / (A + B * logR + C * logR3);
}

void setup() {
  Serial.begin(9600);
}

void loop() {
  double adc_raw = analogRead(READING_PIN);

  double V =  (adc_raw / 1024) * V_IN;
  Serial.print("V =");
  Serial.print(V);

  //calculate resistance
  double R_th = (R1 * V ) / (V_IN - V);
  Serial.print("V   R_th = ");
  Serial.print(R_th);

  double kelvin = SteinhartHart(R_th) -  V * V / (K * R_th);
  double celsius = kelvin - 273.15;
  Serial.print(" Ohm  -  T = ");
  Serial.print(celsius);
  Serial.println("C");
  delay(1000);
}


For the map function, I think that is not the problem. I can correctly map my resistance value to temperature values, but is my resistance value which is wrong, not its mapping.

I also tried with 5 10KOhm resistors in series, but this hasn't work either.

michael_x

#5
Nov 02, 2012, 09:32 am Last Edit: Nov 02, 2012, 10:02 am by michael_x Reason: 1
According to your datasheet the Thermistor has 50k @25°C , which should result in V =0.83, if I understand your ascii wiring scheme correctly.

At lower temperatures you get a higher resistance,  e.g. 100k @10°C, which would result in V = 0.45.
Higher temperatures provide a higher voltage, e.g. 10k @62.5°C yields V =2.50

Is that in line with your first debug output, [edit:]and what you read with a voltage meter between A5 and GND ?
If no, check your wiring (or correct my understanding)
If yes, check the maths.

Papa G

Your equation for Rth assumes a set up like so:

Code: [Select]
5V -- 10K -- A -- T -- GND

michael_x

That deserves a karma brownie, PapaG.

And it provides a much bigger voltage swing that way.

Papa G


dhenry

Quote
I got weird temperatures like 100 celsius degrees


Did you get the right V?

Did you get the right Rth?

Did you get the right T?

erbedo

Using PapaG help, I got it working! Thanks!

Docedison

@ dhenry... did you read the whole post... you know, from the first posting??

Bob
--> WA7EMS <--
"The solution of every problem is another problem." -Johann Wolfgang von Goethe
I do answer technical questions PM'd to me with whatever is in my clipboard

dhenry

You can also rewrite your Rth calculation to fit your setup:

Rth = (Vin - V) * R1 / V.

Papa G


Using PapaG help, I got it working! Thanks!


You're welcome, erbedo.

You can try an experiment if you're curious. If you select 50K for your series resistor instead of 10K, you'll notice that the voltage vs. temperature curve is far more linear over the range 0-100 Celsius. So much so that, if your interest is a thermometer for measuring the temperature of your environment, you might not even need the Steinhart-Hart function.

If you wire your circuit the way you had it originally, and use the right equation, you get a response curve that is fairly linear in the temperature range above 40 degrees Celsius but as michael_x commented, you don't get much of a voltage variation so your resolution is less. If you substitute the 50K resistor again, you get a fairly linear curve at room temperatures but again, not much resolution.

So, you see, there are a few trade-offs you can consider in your design.

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