Hi think i have a misunderstanding
I bought a temperature sensors with the following specifications
temperature sensors PCA 1.1505 5: Platinum chip temperature sensor, leaded, 500 Ohm at reichelt elektronik
-70 degree Celsius = 20 ohm
550 degree Celsius = 5000 Ohm
so with a litle math the sensor has at
20 degree Celsius = 743 Ohm
that means from my understanding that the value that I read from the pin A0
for -70 degree Celsius = 20 ohm should be 0
for 550 degree Celsius = 5000 ohm should be 1023
and for 20 degree Celsius = 743 ohm should be 132
so far the theory.
Then I put in place this circuit
And this script to read.
void setup(){
Serial.begin(9600);
}
void loop (){
int TemperaturRead = analogRead (A0); // read the value „TemperaturRead“
Serial.print (" ---> TemperaturRead: ");
Serial.println (TemperaturRead);
delay(100);
}
But the value the that I get is 50 and not 132.
Do I have a misunderstanding of how the Arduino interoperates the different Ohm?
from which ohm value the Arduino returns 0 ?
from which ohm value the Arduino returns 1023 ?(ok no connection is = 1023)
I hop it is understandable
What value is that resistor?
-70 degree Celsius = 20 ohm
550 degree Celsius = 5000 Ohm
so with a litle math the sensor has at
20 degree Celsius = 743 Ohm
Can you show your workings?
The analog pin measures a voltage and not a resistance.
I see you have a voltage divider with the sensor as the to ground leg.
Do you go by the rating of the resistor you use or by measured value?
Also when I looked at the datasheet I see versions that cover different temperature ranges at least at the high end.
Perhaps put a meter across the leads and see what it tells?
Hi and thank a lot for your answer.
first Nick
The resistor in the circuit is a 220 ohm resistor.
but does an other resistor at this place make a different according to the output of the analogRead (A0)?
Ok
means a voltage between 0 and 5V divided in 1023 steps --> right?
That makes me think about
So i have to use the ohm law to get the right thing
0 V at 550 degree Celsius
and 5 V at -70 degree Celsius
GoForSmoke:
I see you have a voltage divider with the sensor as the to ground leg.
Do you go by the rating of the resistor you use or by measured value?
I mesur the value with the with analogRead (A0)
GoForSmoke:
Perhaps put a meter across the leads and see what it tells?
If i put the meter across the legs it chages the ohm Value for instance the ohm value at room temperatur is 514 Ohm.
now I will think about ohms law :
To measure a resistance you need a voltage divider:
http://www.raltron.com/cust/tools/voltage_divider.asp
Now with a 220 ohm resistor, and measuring 512 ohms on your sensor you should get:
512 / (220 + 512) * 5 = 3.5 volts on A0
Therefore the analogRead would give you:
3.5 / 5 * 1024 = 716.8
This is not what you observe, so something is wrong.
Cool and 1000 Thanks
you are really good amassing
I will look in detail
and again tanks
So now I have reconstruct this by math and by measuring.
and what you say is 100% correct. 
I mesur on the temperature sensors around 540 Ohm, output voltage 4,89V
resistor is 220 ohm and i mesure input Voltage 3,51V
by calculation
540 Ohm /(540 Ohm +220 Ohm )*4,89 V output =3,47V input
In that case to get at -70 degree Celsius the full output voltage on A0 I don't need a the second resistor
to remember -70 degree Celsius = 20 Ohm
20 Ohm / (20 Ohm + 0) * 4,89 V output = 4,89
But can i do it like this or is it a short circuit?
Yes you need a second resistor. Where does the voltage come from otherwise? You have +5v at the top of the voltage divider and 0v at the bottom of it. Without the second resistor (to give a middle point) there is nowhere to measure from. With just the sensor, either you will have 5v at one end of it or 0v at the other end, and it will also get warm from the current flow, which is hardly what you want.
Yes you are right
The thing is that with this sensor the change of the Ohm value is to soft.
To get a bigger amplitude I think I need a sensor where the Ohm difference between low and high is less.
Thanks a lot for your help
i will now search for a other ntc sensor