pt100 thermistor and analog input module

--- Are you sure it's a thermistor?

"PT100" is a designation usually used when referring to a 100 ohm platinum resistance thermometer.

---

--- I don't think so. The wiring you're describing is consistent with a

3-wire RTD:

---

--- You have been, but first things first.

Is the device a platinum resistance thermometer or a thermistor?

JF

Dear JF, thank you for you reply. it is exactly what you have described in your post. a platinum resistance thermometer.

I think I understand now that this device works with a very low voltage, but i have no idea on exactly what voltage it is needed. What do i have to do in order to read successfully the small voltage variations that are provoked by the temperature variations?

Thank you, Vasilis.

--
The usual approach is to buy an already built meter like:

http://www.omega.com/ppt/pptsc.asp?ref=DP460-T&Nav=temm07

If you want to build your own then it gets complicated.

What do you want to do?

JF

I would like to read with my 0-5V analog input device the values of the thermometer. Applying a very small voltage on the two wires of the thermometer ( 18 mV ) and heating it, the voltage between the two wires of the termometer increases. Problem is that i cannot get a temperature value ( a resistance value ) just from reading the voltage change, or can I?

Thank you, Vasilis.

--- Yes, you can.

Consulting the platinum resistance tables at:

we find that a 100 ohm sensor with an alpha of 0.00385 exhibits a resistance of 100.00 ohms at 0C and a resistance of 138.51 ohms at 100C

That's a difference of 38.51 ohms, and if you want to use an analog input device to indicate zero volts at 0C and 5V at 100C, then we must convert that 38.51 ohm resistance change into a 5V change in voltage.

To start with, we must excite the RTD electrically, but we mustn't force too much current through it or it'll self-heat to the point where its accuracy will be degraded.

Assuming that 1mA is OK and that we have a 10V excitation supply, we can start like this:

10V Vin | [R1] | +--->Vout | [RTD]R2 | 0V

With 1mA through the RTD at 0C, we'll have, as Vout:

Vout = Irtd R2 = 0.001A * 100R = 0.1V

Now, in order to limit the current to 1mA, we must calculate the value of R1:

Vin - Vout 10V - 0.1V R1 = ------------ = ------------ = 9900 ohms Irtd 0.001A

So, we now have:

10V E1 | [9900]R1 | +--->0.1V E2 | [100]R2 | 0V

At 100C we'll have:

10V E1 | [9900]R1 | +---> E2 | [138.51]R2 | 0V

And the output voltage from the divider will be:

E1 * R2 10V * 138.51R E2 = --------- = ------------------ = 0.138V R1 + R2 9900R + 138.51R

The voltage difference, then, for the temperature change from 0C to 100C will be:

dV = V100 - V0 = 0.138V - 0.1V = 0.038V

which means that for your analog device to read 0V at 0C and 5V at 100C, there must be something with a gain of:

Vout 5.0V Av = ------ = -------- = 131.58 Vin 0.038V

interposed between the output of the divider and the analog device.

A difference amplifier would do it, and the circuit would look like this:

+V>---+---------+---------------+ |R1 | | [9900] [R3] +----|--[R6]--+-->Vout | | | | | | +--[R5]----+---|-\\ | | | | >------+ +---------|--[R7]----+---|+/ |R2 | | | [RTD]R2 [R4] [R8] | | | | | 0V----+---------+----------+----+

Working out the values of the resistors and selecting the opamp is left as an exercise for the student. ;)

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