IDE and NTC thermistor

Greetings to all!

I am trying to read temperatures in C. from an NTC thermistor and am getting this result: Temperature: -273.15 C

this is the code I am using:

int ThermistorPin = 1;
int Vo;
float R1 = 10000;
float logR2, R2, T, Tc, Tf;
float c1 = 1.009249522e-03, c2 = 2.378405444e-04, c3 = 2.019202697e-07;

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

void loop() {

  Vo = analogRead(ThermistorPin);
  R2 = R1 * (1023.0 / (float)Vo - 1.0);
  logR2 = log(R2);
  T = (1.0 / (c1 + c2*logR2 + c3*logR2*logR2*logR2));
  Tc = T - 273.15;
 // Tf = (Tc * 9.0)/ 5.0 + 32.0; 

  Serial.print("Temperature: "); 
  Serial.print(Tc);
  Serial.println(" C");   

  delay(1000);
}

Can anyone help please as to what might be wrong here?

Thank you in advance .

Hi,
Have you go this line correct?

 T = (1.0 / (c1 + c2*logR2 + c3*logR2*logR2*logR2));

It is possibly very small, zero even.
Hence;

Tc = T - 273.15;

Will be -273.15.

Tom.. :slight_smile:

Thank you Tom,

No unfortunately didn't make a difference ... still reading : Temperature: -273.15 C

Hi,
What did you try that didn't make a difference?

Serial print all your values, before the main calculation.

What is your analog input A0, A1 , A3 ? ? ? ?
How have you got the thermistor wired?

What controller are you using?
Can you post link to data/spec of your thermistor please?
Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?

Do you have a DMM to measure the input volts to the Arduino from the thermistor divider network.

Thanks.. Tom... :slight_smile:

Hi Tom,

I changed  Tc = T - 273.15; to
Tc = T -273.15;

I am not sure what you mean by serial print all the values before the calculation... e.g. Serial.print("logR2: ");

Analog input is A1

I have it wired per these instructions the difference being resistor and Thermistor are 10K

https://www.circuitbasics.com/arduino-thermistor-temperature-sensor-tutorial/

Controller is FireBeetle ESP32 ( FireBeetle ESP32 IoT Microcontroller - DFR0478 | DFRobot Electronics ) and arduino nano

Thermistor's specs attached

Boards are being powered by USB cable . DMM at 5V (where the thermistor is connected) reads 4.68V

Best regards,
John

DG 103395.pdf (227 KB)

Pin A1 on an Uno is defined as '15'. You might be trying to read a digital pin with your existing ThermistorPin definition. Try:

int ThermistorPin = A1;

Hi,

Controller is FireBeetle ESP32 ( https://www.dfrobot.com/product-1590.html ) and arduino nano

Which is it, esp32 or the Nano?
Just use the Nano.
Please post a picture of your project so we can see your component layout.
If your thermistor is 10k and the resistor is 10k, at 25C the input to the analog pin should be around 1/2 of 5V, which is about 2.5V.
Check your wiring of the thermistor and 10k resistor.

Tom.. :slight_smile:

Hi,
Try this code;

int ThermistorPin = A1;
int Vo;
float R1 = 10000;
float logR2, R2, T, Tc, Tf;
float c1 = 1.009249522e-03, c2 = 2.378405444e-04, c3 = 2.019202697e-07;


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


void loop() {


  Vo = analogRead(ThermistorPin);
  Serial.print ("  Vo =");
  Serial.print(Vo);
  R2 = R1 * (1023.0 / (float)Vo - 1.0);
  Serial.print("\t R2 =");
  Serial.print(R2);
  logR2 = log(R2);
  Serial.print("\t  logR2 =");
  Serial.print(logR2);
  T = (1.0 / (c1 + c2 * logR2 + c3 * logR2 * logR2 * logR2));
  Serial.print("\t  T =");
  Serial.print(T);
  Tc = T - 273.15;
  // Tf = (Tc * 9.0)/ 5.0 + 32.0;


  Serial.print("\t  Temperature: ");
  Serial.print(Tc);
  Serial.println(" C");


  delay(1000);
}

Vo should not be near 0 or near 1023.
Measure the resistance of the thermistor, out of circuit, on its own.
Measure the resistance of the 10K resistor, out of circuit, on its own.

If you like, take the thermistor and 10K out of circuit and connect the A1 pin to 3.3V pin and see what the serial monitor reads.
Vo should be around 660.

Tom... :slight_smile:

Thank you Tom, much appreciate your help!

It works now on arduino nano (attached) but still not working on FireBeetle (attached) . I guess some compatibility issue although as stated by the manufacturer is a 100% compatible with Arduino. A1 Vin on Firebeetle is 2.21v (When connected to thermistor and 5v) as you suggested.

A1 to 3.3v on firebeetle reads : Vo =4095 R2 =-7501.83 logR2 =nan T =nan Temperature: nan C

so nothing like 660.

I guess I have to look more into the firebeetle esp32 board .

Thank you again very much for your help!

Rather than taking multiple screen-shots you can copy and paste the text from Serial Monitor.

Looks like the "1023.0" in your sketch should be "4096.0" since the ESP32 uses a 12-bit ADC instead of the Arduino typical 10-bit ADC.

Hi,
As @johnwasser has pointed out, the ESP is 12bit that is 4096 counts from ADC, so change this line;

 R2 = R1 * (1023.0 / (float)Vo - 1.0);

To;

 R2 = R1 * (4095.0 / (float)Vo - 1.0);

It looks like the ESP32 input is giving valid values with the thermistor connected, just the raw data to scaled data calculation needs changing to the different ADC.

Tom... :slight_smile:

Apologies for the screenshots!

it looks I am getting there slowly ... the problem now is that the ambient temperature is 22 (verified with a digital thermometer), after uploading the sketch it starts from 16.55C and if I heat up the thermistor, temperature never goes above 26.

  Vo =1719	 R2 =13821.99	  logR2 =9.53	  T =289.70	  Temperature: 16.55 C
  Vo =1718	 R2 =13835.86	  logR2 =9.54	  T =289.68	  Temperature: 16.53 C
  Vo =1718	 R2 =13835.86	  logR2 =9.54	  T =289.68	  Temperature: 16.53 C
  Vo =1719	 R2 =13821.99	  logR2 =9.53	  T =289.70	  Temperature: 16.55 C
  Vo =1722	 R2 =13780.49	  logR2 =9.53	  T =289.78	  Temperature: 16.63 C
  Vo =1715	 R2 =13877.55	  logR2 =9.54	  T =289.60	  Temperature: 16.45 C
  Vo =1712	 R2 =13919.39	  logR2 =9.54	  T =289.53	  Temperature: 16.38 C
  Vo =1715	 R2 =13877.55	  logR2 =9.54	  T =289.60	  Temperature: 16.45 C
  Vo =1742	 R2 =13507.46	  logR2 =9.51	  T =290.27	  Temperature: 17.12 C
  Vo =1925	 R2 =11272.73	  logR2 =9.33	  T =294.78	  Temperature: 21.63 C
  Vo =1984	 R2 =10640.12	  logR2 =9.27	  T =296.25	  Temperature: 23.10 C
  Vo =2025	 R2 =10222.22	  logR2 =9.23	  T =297.27	  Temperature: 24.12 C
  Vo =2057	 R2 =9907.63	  logR2 =9.20	  T =298.07	  Temperature: 24.92 C
  Vo =2065	 R2 =9830.51	  logR2 =9.19	  T =298.27	  Temperature: 25.12 C
  Vo =2073	 R2 =9753.98	  logR2 =9.19	  T =298.47	  Temperature: 25.32 C
  Vo =2074	 R2 =9744.46	  logR2 =9.18	  T =298.50	  Temperature: 25.35 C
  Vo =2079	 R2 =9696.97	  logR2 =9.18	  T =298.62	  Temperature: 25.47 C
  Vo =2095	 R2 =9546.54	  logR2 =9.16	  T =299.03	  Temperature: 25.88 C
  Vo =2092	 R2 =9574.57	  logR2 =9.17	  T =298.95	  Temperature: 25.80 C
  Vo =2095	 R2 =9546.54	  logR2 =9.16	  T =299.03	  Temperature: 25.88 C
  Vo =2096	 R2 =9537.21	  logR2 =9.16	  T =299.05	  Temperature: 25.90 C
  Vo =2101	 R2 =9490.72	  logR2 =9.16	  T =299.18	  Temperature: 26.03 C
  Vo =2102	 R2 =9481.45	  logR2 =9.16	  T =299.20	  Temperature: 26.05 C
  Vo =2103	 R2 =9472.18	  logR2 =9.16	  T =299.23	  Temperature: 26.08 C
  Vo =2106	 R2 =9444.44	  logR2 =9.15	  T =299.30	  Temperature: 26.15 C
  Vo =2102	 R2 =9481.45	  logR2 =9.16	  T =299.20	  Temperature: 26.05 C
  Vo =2106	 R2 =9444.44	  logR2 =9.15	  T =299.30	  Temperature: 26.15 C
  Vo =2119	 R2 =9325.15	  logR2 =9.14	  T =299.63	  Temperature: 26.48 C
  Vo =2099	 R2 =9509.29	  logR2 =9.16	  T =299.13	  Temperature: 25.98 C
  Vo =2112	 R2 =9389.20	  logR2 =9.15	  T =299.46	  Temperature: 26.31 C
  Vo =2108	 R2 =9426.00	  logR2 =9.15	  T =299.35	  Temperature: 26.20 C
  Vo =2113	 R2 =9380.03	  logR2 =9.15	  T =299.48	  Temperature: 26.33 C
  Vo =2119	 R2 =9325.15	  logR2 =9.14	  T =299.63	  Temperature: 26.48 C
  Vo =2115	 R2 =9361.70	  logR2 =9.14	  T =299.53	  Temperature: 26.38 C
  Vo =2116	 R2 =9352.55	  logR2 =9.14	  T =299.56	  Temperature: 26.41 C
  Vo =2115	 R2 =9361.70	  logR2 =9.14	  T =299.53	  Temperature: 26.38 C
  Vo =2113	 R2 =9380.03	  logR2 =9.15	  T =299.48	  Temperature: 26.33 C
  Vo =2114	 R2 =9370.86	  logR2 =9.15	  T =299.51	  Temperature: 26.36 C
  Vo =2105	 R2 =9453.68	  logR2 =9.15	  T =299.28	  Temperature: 26.13 C
  Vo =2097	 R2 =9527.90	  logR2 =9.16	  T =299.08	  Temperature: 25.93 C
  Vo =2114	 R2 =9370.86	  logR2 =9.15	  T =299.51	  Temperature: 26.36 C
  Vo =2113	 R2 =9380.03	  logR2 =9.15	  T =299.48	  Temperature: 26.33 C
  Vo =2115	 R2 =9361.70	  logR2 =9.14	  T =299.53	  Temperature: 26.38 C
  Vo =2123	 R2 =9288.74	  logR2 =9.14	  T =299.73	  Temperature: 26.58 C
  Vo =2117	 R2 =9343.41	  logR2 =9.14	  T =299.58	  Temperature: 26.43 C
  Vo =2090	 R2 =9593.30	  logR2 =9.17	  T =298.90	  Temperature: 25.75 C
  Vo =2058	 R2 =9897.96	  logR2 =9.20	  T =298.09	  Temperature: 24.94 C
  Vo =2021	 R2 =10262.25	  logR2 =9.24	  T =297.17	  Temperature: 24.02 C
  Vo =1991	 R2 =10567.55	  logR2 =9.27	  T =296.42	  Temperature: 23.27 C
  Vo =1970	 R2 =10786.80	  logR2 =9.29	  T =295.90	  Temperature: 22.75 C

I found this code also but haven't been able to make it work either:

byte NTCPin = A0;
#define SERIESRESISTOR 10000
#define NOMINAL_RESISTANCE 10000
#define NOMINAL_TEMPERATURE 25
#define BCOEFFICIENT 3950

void setup()
{
Serial.begin(9600);
}
void loop()
{
float ADCvalue;
float Resistance;
ADCvalue = analogRead(NTCPin);
Serial.print("Analog value ");
Serial.print(ADCvalue);
Serial.print(" = ");
//convert value to resistance
Resistance = (1023 / ADCvalue) - 1;
Resistance = SERIESRESISTOR / Resistance;
Serial.print(Resistance);
Serial.println(" Ohm");

float steinhart;
steinhart = Resistance / NOMINAL_RESISTANCE; // (R/Ro)
steinhart = log(steinhart); // ln(R/Ro)
steinhart /= BCOEFFICIENT; // 1/B * ln(R/Ro)
steinhart += 1.0 / (NOMINAL_TEMPERATURE + 273.15); // + (1/To)
steinhart = 1.0 / steinhart; // Invert
steinhart -= 273.15; // convert to C

Serial.print("Temperature ");
Serial.print(steinhart);
Serial.println(" oC");
delay(1000);
}

Even after changing the value to 4095 .

Is this the way your thermistor is connected?

VCC for an ESP32, with it's 3.3volt logic, is of course the 3.3volt pin.

Resistance = (1023 / ADCvalue) - 1; // that sketch also needs it's 10-bit value changed to a 12-bit value

Thermistor code contains two parts.

  1. calculating the resistance of the thermistor.
  2. converting that resistance to temperature, with a Steinhart-Hart equation.

To test the first part, you should see about 10000 ohms printed if the thermistor is 25C.
For other temps, you could look for a "10k thermistor resistance table" online.
Leo..

Hi,

Resistance = (1023 / ADCvalue) - 1;

Should be;

Resistance = (4095 / ADCvalue) - 1;

Also use your DMM and see what the thermistor resisitance measures at your 22C, then compare it with the controller monitor display.

Tom... :slight_smile:

@ JCA34F, I have it connected as per the attached diagram

What is interesting (I think) is that with this code:

//---------------
byte NTCPin = A0;
#define SERIESRESISTOR 10000
#define NOMINAL_RESISTANCE 10000
#define NOMINAL_TEMPERATURE 25
#define BCOEFFICIENT 3950

void setup()
{
Serial.begin(9600);
}
void loop()
{
float ADCvalue;
float Resistance;
ADCvalue = analogRead(NTCPin);
Serial.print("Analog value ");
Serial.print(ADCvalue);
Serial.print(" = ");
//convert value to resistance
Resistance = (4095 / ADCvalue) - 1;
Resistance = SERIESRESISTOR / Resistance;
Serial.print(Resistance);
Serial.println(" Ohm");

float steinhart;
steinhart = Resistance / NOMINAL_RESISTANCE; // (R/Ro)
steinhart = log(steinhart); // ln(R/Ro)
steinhart /= BCOEFFICIENT; // 1/B * ln(R/Ro)
steinhart += 1.0 / (NOMINAL_TEMPERATURE + 273.15); // + (1/To)
steinhart = 1.0 / steinhart; // Invert
steinhart -= 273.15; // convert to C

Serial.print("Temperature ");
Serial.print(steinhart);
Serial.println(" oC");
delay(1000);
}
//-------------

I get these results :

Temperature 32.32 oC
Analog value 1730.00 = 7315.01 Ohm
Temperature 32.21 oC
Analog value 1735.00 = 7351.70 Ohm
Temperature 32.09 oC
Analog value 1728.00 = 7300.38 Ohm
Temperature 32.25 oC
Analog value 1728.00 = 7300.38 Ohm
Temperature 32.25 oC
Analog value 1730.00 = 7315.01 Ohm
Temperature 32.21 oC
Analog value 1728.00 = 7300.38 Ohm
Temperature 32.25 oC
Analog value 1741.00 = 7395.92 Ohm
Temperature 31.95 oC
Analog value 1726.00 = 7285.77 Ohm
Temperature 32.30 oC
Analog value 1732.00 = 7329.67 Ohm
Temperature 32.16 oC
Analog value 1743.00 = 7410.71 Ohm
Temperature 31.90 oC
Analog value 1726.00 = 7285.77 Ohm
Temperature 32.30 oC
Analog value 1729.00 = 7307.69 Ohm
Temperature 32.23 oC
Analog value 1739.00 = 7381.15 Ohm

Which are almost double the ones I get with:

int ThermistorPin = A0;
int Vo;
float R1 = 10000;
float logR2, R2, T, Tc, Tf;
float c1 = 1.009249522e-03, c2 = 2.378405444e-04, c3 = 2.019202697e-07;


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


void loop() {


  Vo = analogRead(ThermistorPin);
  Serial.print ("  Vo =");
  Serial.print(Vo);
  R2 = R1 * (4095.0 / (float)Vo - 1.0);
  Serial.print("\t R2 =");
  Serial.print(R2);
  logR2 = log(R2);
  Serial.print("\t  logR2 =");
  Serial.print(logR2);
  T = (1.0 / (c1 + c2 * logR2 + c3 * logR2 * logR2 * logR2));
  Serial.print("\t  T =");
  Serial.print(T);
  Tc = T - 273.15;
  // Tf = (Tc * 9.0)/ 5.0 + 32.0;


  Serial.print("\t  Temperature: ");
  Serial.print(Tc);
  Serial.println(" C");


  delay(1000);
}
  Vo =1728	 R2 =13697.92	  logR2 =9.52	  T =289.92	  Temperature: 16.77 C
  Vo =1734	 R2 =13615.92	  logR2 =9.52	  T =290.07	  Temperature: 16.92 C
  Vo =1731	 R2 =13656.85	  logR2 =9.52	  T =290.00	  Temperature: 16.85 C
  Vo =1728	 R2 =13697.92	  logR2 =9.52	  T =289.92	  Temperature: 16.77 C
  Vo =1724	 R2 =13752.90	  logR2 =9.53	  T =289.83	  Temperature: 16.68 C
  Vo =1728	 R2 =13697.92	  logR2 =9.52	  T =289.92	  Temperature: 16.77 C
  Vo =1724	 R2 =13752.90	  logR2 =9.53	  T =289.83	  Temperature: 16.68 C
  Vo =1727	 R2 =13711.64	  logR2 =9.53	  T =289.90	  Temperature: 16.75 C
  Vo =1731	 R2 =13656.85	  logR2 =9.52	  T =290.00	  Temperature: 16.85 C
  Vo =1742	 R2 =13507.46	  logR2 =9.51	  T =290.27	  Temperature: 17.12 C
  Vo =1730	 R2 =13670.52	  logR2 =9.52	  T =289.97	  Temperature: 16.82 C
  Vo =1723	 R2 =13766.69	  logR2 =9.53	  T =289.80	  Temperature: 16.65 C
  Vo =1745	 R2 =13467.05	  logR2 =9.51	  T =290.34	  Temperature: 17.19 C
  Vo =1737	 R2 =13575.13	  logR2 =9.52	  T =290.15	  Temperature: 17.00 C
  Vo =1731	 R2 =13656.85	  logR2 =9.52	  T =290.00	  Temperature: 16.85 C
  Vo =1735	 R2 =13602.31	  logR2 =9.52	  T =290.10	  Temperature: 16.95 C
  Vo =1734	 R2 =13615.92	  logR2 =9.52	  T =290.07	  Temperature: 16.92 C
  Vo =1733	 R2 =13629.54	  logR2 =9.52	  T =290.05	  Temperature: 16.90 C
  Vo =1733	 R2 =13629.54	  logR2 =9.52	  T =290.05	  Temperature: 16.90 C
  Vo =1732	 R2 =13643.19	  logR2 =9.52	  T =290.02	  Temperature: 16.87 C
  Vo =1731	 R2 =13656.85	  logR2 =9.52	  T =290.00	  Temperature: 16.85 C
  Vo =1729	 R2 =13684.21	  logR2 =9.52	  T =289.95	  Temperature: 16.80 C

Wawa:
VCC for an ESP32, with it's 3.3volt logic, is of course the 3.3volt pin.

Resistance = (1023 / ADCvalue) - 1; // that sketch also needs it's 10-bit value changed to a 12-bit value

Thermistor code contains two parts.

  1. calculating the resistance of the thermistor.
  2. converting that resistance to temperature, with a Steinhart-Hart equation.

To test the first part, you should see about 10000 ohms printed if the thermistor is 25C.
For other temps, you could look for a "10k thermistor resistance table" online.
Leo..

you mean to use a medium of a known 25C and then measure the resistance of the thermistor?
The manufacturer has provided me with this (attached) table , that states @25C Rt / R25 Nominal : 1.00000 and
Temp. Coef (% / ºC) 4.32. Can this be implemented in the sketch?

Best regards,
John

Thermistor connection diagram.jpg

Thermistor connection diagram.jpg

Hi,
Have you compared the measured resistance with a DMM and the calculated resistance with the code?

Thanks.. Tom... :slight_smile:

If your thermistor was 32 degrees C, then the first code was right (~7300 ohm).

If your thermistor was 18 degrees C, then the second code was right (~13700 ohm).

See this table.
Leo..

float c1 = 1.009249522e-03, c2 = 2.378405444e-04, c3 = 2.019202697e-07;

I don't know where you got these parameters from, but c3 is pretty far away from other values I have seen.

For example

Did you use a calculator like this to get the parameters?
https://www.thinksrs.com/downloads/programs/Therm%20Calc/NTCCalibrator/NTCcalculator.htm

TomGeorge:
Hi,
Have you compared the measured resistance with a DMM and the calculated resistance with the code?

Thanks.. Tom... :slight_smile:

Hmmm, you know ...no I should have, and now that I did I read on DMM 11.40K while serial monitor shows : Vo =1710 R2 =13947.37 logR2 =9.54 T =289.48 Temperature: 16.33 C

The above is with the values I had copied from the link I posted in the beginning of this thread.

cattledog:

float c1 = 1.009249522e-03, c2 = 2.378405444e-04, c3 = 2.019202697e-07;

I don't know where you got these parameters from, but c3 is pretty far away from other values I have seen.

For example
https://www.skyeinstruments.com/wp-content/uploads/Steinhart-Hart-Eqn-for-10k-Thermistors.pdf

Did you use a calculator like this to get the parameters?
SRS Thermistor Calculator

After changing the values per first link, the returned result (compared to the the values I was using) is:
Vo =1710 R2 =13947.37 logR2 =9.54 T =290.73 Temperature: 17.58 C
After using the calculator per second link (attached screenshot) I entered the values given by the manufacturer (attached) and adjusted the code:

int ThermistorPin = A0;
int Vo;
float R1 = 10000;
float logR2, R2, T, Tc, Tf;
float c1 = 1.145226154e-3, c2 = 2.296821079e-4, c3 = 1.194648779e-7;


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


void loop() {


  Vo = analogRead(ThermistorPin);
  Serial.print ("  Vo =");
  Serial.print(Vo);
  R2 = R1 * (4095.0 / (float)Vo - 1.0);
  Serial.print("\t R2 =");
  Serial.print(R2);
  logR2 = log(R2);
  Serial.print("\t  logR2 =");
  Serial.print(logR2);
  T = (1.0 / (c1 + c2 * logR2 + c3 * logR2 * logR2 * logR2));
  Serial.print("\t  T =");
  Serial.print(T);
  Tc = T - 273.15;
  // Tf = (Tc * 9.0)/ 5.0 + 32.0;


  Serial.print("\t  Temperature: ");
  Serial.print(Tc);
  Serial.println(" C");


  delay(1000);
}

Returned result is : Vo =1710 R2 =13947.37 logR2 =9.54 T =290.62 Temperature: 17.47 C


UPDATE: I measured with a DMM between A0 and board's GND and got 1.52v and I see in the latest result above a 1.71v, is this normal? Also please correct me if I am wrong but based on the sketch's coding shouldn't R2 be (3.27v / 1.71v) - 1 * 10000 = 9122 (Instead of 13947.37). Again though, 9122 based on the manufacturer's specs for the thermistor 9122 is about 27C (between 25C and 30C) and I have a 22C. OMG...

3.27v is what I measure with my DMM between the board's 3.3v (where the thermistor is connected) and pin GND.


UPDATE2: OK, maybe this thing is solved but I would like your thoughts: after doing more reading and through trial and error I have calculated Steinhart-Hart model coefficients as: c1 = 1.174407533e-3, c2 = 2.248274514e-4, c3 = 1.324430127e-7; based on the manufacturer's "extreme" readings and middle one at 105C. Thermistor's specs declares at -40C : 31.44855 , at 105C : 0.05931 and at 250C : 0.00263 . I entered the values on this calculator (provided by cattledog) : [SRS Thermistor Calculator](https://www.thinksrs.com/downloads/programs/Therm Calc/NTCCalibrator/NTCcalculator.htm) as -40C : 314485, at 105C : 593 and at 250C : 26 . In the PDF (file name: DG103395) that I have attached above it shows -40C : 31.44855 , 105C : 0.05931 and 250C: 0.00263.

Am I correct to assume 31.44855 is 314485 Ohms , 0.05931 is 593 Ohms, and 0.00263 is 26 Ohms at 250C?

DG 103395.pdf (227 KB)