0
\$\begingroup\$

I'm looking for the correct temperature compensation equation to use on our project.

We are measuring the output of a detector who's signal is very sensitive to temperature drift. Any external temperature drift is reflected in the output. A cycle due Day/night variation is typically what we see.

So, we apply a formula to the detector's output to try to compensate for the changes in temperature.

The formula we use is: RITC = ((RT - AT) × TC) + I

Where: RITC = temperature corrected output RT = Reference Temperature, AT = Actual Temperature as Read on Instrument, TC = Temperature Coefficient, I = Input (Actual reading from the detector).

This equation works if the temperature variation is constant, say (±5°C) from day to day. However, if the variation isn't constant and it changes from day to day, the equation doesn't work well. Also if there is sudden changes due to maybe air conditioning or Fans turning On/Off, again the equation does not hold up.

So what is the solution? How can the affects of temperature be removed from the detector signal? What is the correct method to deal with sudden dynamic changes and also to deal with the slower variations? I feel a better equation is needed.. just not sure what it is!

\$\endgroup\$
3
  • \$\begingroup\$ I worked with two companies making temperature sensors -- expensive ones. Thousands of US dollars per measurement point. They were accurate down to 5 mK and had repeatable precision well below that. And yes, without compensation they would definitely be impacted by an air conditioner turning on and they would allow you to see day vs night. And more. Compensation is a process that involves studying your complete sensor system and uncovering, one by important issue after another, the reasons involved and then working through the related sensitivity equations. One resolved, it's on to the next. \$\endgroup\$
    – jonk
    Commented Apr 29, 2021 at 16:48
  • \$\begingroup\$ The process ends when you decide it does. Sometimes, you uncover surprising things. It can be something as unlikely as changes in a thin film filter (if there is an optical system using one.) Each thing you find teaches you something knew about the world you hadn't seen before, or hadn't considered in this case. And they each have their own elements of correction and behavior vs temperature. There's a simple story I like to tell that gets the point across. I don't tell it the same way twice. Depends on the audience. \$\endgroup\$
    – jonk
    Commented Apr 29, 2021 at 17:15
  • \$\begingroup\$ So, as it may help improve your question, please provide specific details about complete system, your applied theory and why it is taken as true and the boundary conditions within which that theory is supposed to apply, and all your data on the variations. Without a lot of detail, it's worse than guesswork trying to avoid an ovenized situation. I don't know how anyone can do much more than that without knowing a lot of detail. And I don't even know what sensor you are using, though I might guess about it!! \$\endgroup\$
    – jonk
    Commented Apr 29, 2021 at 17:25

1 Answer 1

2
\$\begingroup\$

So you have a simple linear compensation equation.

The "variation isn't constant" doesn't really mean much, if it's constant it's not a variation. Maybe you mean that the correction does not work for variations that are not specific amounts.

In any case, it seems that dynamic changes are a problem, which indicates that whatever you are using as a temperature sensor is not really sensing the temperature closely enough to the part of the detector that needs correctly.

There is probably not a simple cheap solution to this. You can get a better sensor, which is probably the best solution.. other options..

Perhaps you ovenize the sensor so that the temperature does not vary significantly.

Or you could calibrate the sensor in an environmental chamber with many points and then create a polynomial fit equation that cancels out the errors-- but that won't help with dynamic effects.

We did have a compensation situation where dynamic effects were partially corrected by assuming a certain response curve, but in some cases it made the errors worse.

\$\endgroup\$
1
  • \$\begingroup\$ If the external temperature did not change from day to day yes we'd have a have constant. As it does, need a method of adapting to these changes and apply appropriate levels of compensation. The equation we've used is too simple, does not adapt. There must be an equation that does?! Placing sensor in an Oven is an alternative option yes. With the correct method of compensation, hoping it won't be necessary. \$\endgroup\$
    – Conor
    Commented Apr 29, 2021 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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