ADN8834 PT100 calculations

Hello

Is it possible to use the chopper circuitry with a PT100 thermistor?

I'm trying to get those resistor values, namely Rx, with the equiation provided by the datasheet.

Problem is that the resistor result is negative. The divisor parcel is negative, so the resistor result is negative.

The document states that with Rlow(resistance at low endpoint), Rhigh(resistance at high endpoint), and Rmid you can calculate the Rx to get a linearization of the impedance/temperature ratio along the desired range.

So, let's say I want a range of 0ºC~20ºC. Rlow is 100 Ohm, Rhigh is 107.79 Ohm. And Rmid, I assume it's the resistance for 10ºC (in datasheet it states that Tmid "is the average") is 103.9 Ohm.

These are standard PT100 values.

So, I use this provided formula in "THERMISTOR SETUP" section, to get Rx, and it is negative. Always. No matter the values I  choose.

Clearly, I'm missing something.

So please, help me.

Thanks in advance.

  • A couple of thoughts:

    The equation for Rx is aimed at linearizing the response of a thermistor with a negative temperature coefficient, but a PT100 is an RTD (as opposed to a thermistor) which has a positive temperature coefficient.  See an RTD resistance example here:  http://www.micropik.com/PDF/pt100.pdf

    The resistance change of a PT100 is VERY flat compared to that of a typical thermistor.  You might want to consider whether or not you need to add a linearization scheme for your application.

    Hopefully this helps somewhat.

  • I'll add to what LMinCv just said.

    The PT100 is sufficiently linear for your application,  Rx is used with NTC thermistors to generate a linearised operating region around approximately half voltage.

    Platinum sensors on the other hand vary resistance linearly with temperature, and your circuit needs to exploit this.

    Substituting a PTC sensor for an NTC will result in the circuit failing to work at all, you need to put a PTC sensor where R is , and a fixed resistor where (Rx + Rth) are, this will result in a circuit that at least operates, even though nonlinear.

    The 100 ohm resistance is quite low,  so difficult to get more than a few mV out of the sensor, before self heating is significant, and power dissipation in the op-amps becomes significant, the 1000ohm variants (PT1000) are more suited to direct inclusion in an op-amp circuit.

    --------------- case 1 --- a general example of using a PT100 in a half bridge -- where Vtempset is always at Vref/2---

    (a) However, referring to the schematic of figure33,  when using just another fixed resistor R  , the voltage at "in1P" will not be linear w.r.t. temperature.

    (b) In a typical application, if  R is a potentiometer, then the controlled temperature will be linear with rotation ,   As R and Rth form one arm of a wheatstone bridge Wheatstone bridge - Wikipedia, the free encyclopedia .  with the Vref/2 divider forming the other arm.

    (c) If for example R (of figure3)  was  a 100.00 ohm resistor in series with a 40ohm potentiometer, and the pot dial was calibrated with 100 marks from 0 to 100 , then the AD8834 would control temperature from 0C to 100C , according to the pot calibration.

    (d) the above pot value is a bit awkward, better to use a PT1000 sensor and a 400 ohm pot (500 ohm is more likely)

    (e) if you change the voltage at Vref/2 to some other reference voltage, this will allow standard values for the potentiometer in (d)

    --------- case2   --- where you want Vout1 to represent the actual temperature.

    You would need to re-configure the amplifier "chopper1"  to operate with a constant current through the PT100 .  There are a couple of ways of doing this

    (a) use a constant current source driven into the grounded PT sensor (a PT1000 will work better than a PT100) , then configure chopper1 as a non-inverting amplifier.

    (b) with the figure33 circuit, use a  PT1000 as Rfb   ,   set R to infinity , set (Rx + Rth) to 1000ohm  ,  then Vout 1 will be 0.5Vref for 0deg C and approx  0.7vref for 100degC.

    Basically you need to google Pt100 application schematics, and just adapt the components around chopper1 to match those schematics.

  • Thanks for your replies.

    I understand than, that I don't need Rx because PT100 is pretty much linear. OK.

    I understand that PT1000 has a bigger variation in impedance. The problem is that PT1000 generally has less power dissipation maximum standard. That would be in the inverse proportion when comparing to the PT100 resistance. So, if you use 10x less current on it, you will get the same voltage variation, and that voltage will represent the temperature.

    About case1:

    - Can I just ignore Rx and use that circuit has temperature sensing?

    - Can I change the chopper 1 arm with VRef/2 at In1P, so I get a different offset voltage? I was thinking about this, because I want to increase the gain, but when I do it, the influence of In1P is too big at the output. If it is possible, what is the Vout formula?

    About case2:

    - I already figured a way of doing a current source circuit

  • Hello again.

    The self heating effect of a PT1000 will be the same as a PT100  when supplied with either 3.15 times  the voltage  or 1/3.15 times the current. It is possible your circuit could tolerate more self heating. 

    You should do an error budget to see the magnitude of self heating error is comparable to other noise sources, like offset  error in the opamp, current source errors and thermal EMF's.

    Re "about case 1, ignoring Rx" ,  while you can ignore Rx, you can't use the same topology, because the PTC has an INCREASE of voltage/resistance with temperature, while the thermistor has a DECREASE , so you need to invert the polarity somewhere.

    Re "about case 1, Vref/2" ,  It's very unlikely you will get 1.2v out of a PTC , so you wiil need to lower the reference voltage, e.g. with a pot.  A better approach would be to use a 100ohm reference resistor, so you are basically making a wheatstone bridge. the reference arm would be current source + 100.000ohm resistor , the sensing arm would be current source + PT100,  so the bridge difference would be zero at 0degC.

  • OK. I will try the second approach.

    About the polarity:

    I think that circuit is really for an INCREASE of voltage/resistance type of temperature sensing.

    Let's place a scenario:

    - Vout2 ranges from  0 to 2.5V, when you use the full range of the output voltages and have no limits(only to keep it simple).

    - According to the formula given at  "MOSFET DRIVER AMPLIFIERS" a VOut2 increase will decrease VLdr.

    - At this time, the voltage at Vout1 is minimum (2V). With a PT100 this means the temperature is really high, because of the inverting characteristics of chopper1.

    - We place at Tempref a voltage of 2.5V, because we want the temperature to decrease.

    - Looking at chopper 2, it will increase it's output voltage (let's assume the gain is exactly 1) until the difference is zero, i.e., until it reaches 2.5V. This means LDR voltage will increase.

    This only works, if increasing the LDR voltage efectivelly DECREASES the sensor temperature, because the chopper2 loop is expecting an voltage INCREASE on it's negative input, which is what happens.

    When you say "invert the polarity somewhere", it also applies to inverting the PT100 position on the system? I mean, I could place PT100 on one of the TEC's sides...