I have following circuit reading an externally connected PT1000 sensor. The Sensor is on a cord, the side we are looking at in the schematic uses a a female connector. So the pins are not easily accessible. The device will however be used in a bio lab setting. Moisture (up to 100% rel.) and temperature (might go up to 50°C) might be an issue. Corrosion will probably not be an issue.

I want to foolproof the device for non-technical personel with an academic background.

What are important scenarios I would need to failsafe the circuit against, and - if a suggestion is easy - how would I do it?

I have already checked following conditions:

  • Short-Circuit

For people wanting to reuse: The circuit specs to 0-50°C with about 0.1°C resolution on a 10bit ADC. Make sure to use the right Opamp, I've been burned by that.

enter image description here

Datasheet for diodes: SMAJ5.0A

  • \$\begingroup\$ Can they break the alumina substrate of the thin-film sensor? \$\endgroup\$ Feb 28 '20 at 19:59
  • \$\begingroup\$ @SpehroPefhany No, I dont think so. The sensor is in a metal housing. (It is an external product and very well manufactured). Only the device side is made by me. The sensor cord might be cut by accident. \$\endgroup\$ Feb 28 '20 at 20:01
  • \$\begingroup\$ Not on topic, but it's almost never worth it to build an instrumentation amp if CMRR is important. Instead, just buy an inamp \$\endgroup\$ Feb 28 '20 at 22:13

You would need a low pass filter on each input of the instrumentation amplifier. Then you could use some diodes with extremely low leakage current to protect the input of the instrumentation amp. inputs.


simulate this circuit – Schematic created using CircuitLab

Not sure if your design is the best suited for RTD conditioning, but this is up to you: I_rtd at 0C = 2.5V/2000ohm= 1.25mA, it's on the limit of usability. Usually 100 to 200uA constant current source is used.

If you want a bullet proof, then you would need a galvanic isolation between MCU and PT1000, by using a complete DAQ front end and then using some SPI, I2C,...galvanic isolators.

  • \$\begingroup\$ Why the filter? \$\endgroup\$ Feb 28 '20 at 22:14
  • \$\begingroup\$ The filter is a must, since the Pt1000 has a very big impedance, it will act as an antenna and it will pick all the mains noise. You'd better go with constant current source and Sallen-Key filter, yet better to use Pt100 instead. See: ww1.microchip.com/downloads/en/appnotes/00687c.pdf \$\endgroup\$ Feb 28 '20 at 23:35
  • \$\begingroup\$ That's not protection per se. Also, with respect to CMRR, better to filter after the in amp. \$\endgroup\$ Feb 29 '20 at 0:29

I think if you give it a unique connector that can't easily be used to apply external voltage to the sensor you'll be okay. Pt sensors are inherently isolated so they're not as fussy as thermocouples.

Your circuit is applying an awful lot of voltage to the sensor, so it is not the best. Usually for a Pt1000 we like to see 100uA, give or take, so about 100mV. Putting 12.5x the voltage across it means 150x the self-heating, among other things.

Using a Pt100 sensor at 0.3mA would be less sensitive to moisture that might have conductive ionic contaminants, but you'd have much less output. It depends what kind of resolution and accuracy you are going for, but the self-heating is going to play havoc with readings if you are trying for high resolution.

  • 1
    \$\begingroup\$ Thats very good advice, I have not thought about this. Will see if I can improve. \$\endgroup\$ Feb 28 '20 at 20:22
  • \$\begingroup\$ Given that the usage of a voltage source is a must, how would I go about reducing voltage to the sensor? I could use higher resistors in series (say 10k), but my calculations show that the sensor's swing is the highest when its resistance is matched. I could use a 1.0V source instead, but that would merely be a reduction by 50%. \$\endgroup\$ Mar 2 '20 at 9:23
  • \$\begingroup\$ One method is to use a constant-current source of, say, 100uA. Or much higher values of resistors. If you use a high value resistor you can give it a taste of positive feedback to linearize the output (exact at 3 points) though with digital linearization that's no longer that exciting. Have a look at suggested circuits- I think LTC has a number of them. P.S. maximizing the output is usually not the main consideration- there is a trade-off between output and self-heating. \$\endgroup\$ Mar 2 '20 at 15:26
  • \$\begingroup\$ Thank you for your considerations. Unfortunately I had to decide which answer to accept, and with regards to the question I felt the other answer as more on topic. Your Input as very much appreciated, however \$\endgroup\$ Mar 2 '20 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.