Measure Temperature: Alternate to myDAQ

[Neuropotential]

Hello Experts, 

I am trying to measure temperature from simple thermistor as shown in the link below: 

http://zone.ni.com/devzone/cda/epd/p/id/6408

I do not have a myDAQ. Hence, I am looking for an alternative for the same. I am using USB-6008/9. The myDAQ in this program performs two functions: 

1) Provide excitation current to the thermistor

2) Record resistance directly instead of voltage for temperature calculations. 

For 1), I have a constant voltage regulator. But how much current should I apply across the thermistor? Does any one have any idea about a range? I do not want to damage the thermistor by passing more current

For 2), In the equation relating Temperature with Resistance, what is R? Is R the absolute temperature at any instance in time? If so, R can be easily calculated with voltage recorded using USB-6008/9 and constant excitation current. 

Alternately, R could be change in resistance from Rref (which is 10k, in this case) Does any one have an idea which one is true? 

Thank you in advance, 

Sanket

[Alex.T]

Hi Sanket,

In the article that you have attached, the equation uses the parameter R to represent the current resistance of the Thermistor; this deviation from the Reference Temperature allows for the current temperature to be calculated. This is how we can measure the equivalent temperature by knowing the Resistance of the thermistor. Using the DMM inputs, a resistance measurement can be made by the myDAQ.

The USB-6008/6009 modules do not have DMM input so we can't configure them for a direct resistance measurement; instead what we can do is measure the voltage across the thermistor. Instead, we can apply a voltage across the thermistor whilst it is in series with a known resistance, such as 10K. If the rail voltage used is +5V and we measure 2.5V across the thermistor, that means that there would be 2.5V across the series resistor; this means that the resistance of the thermistor would be 10K too; then we can use the same calculations provided by the example. Therefore we need to make use of the Potential Divider equation. 

You may want to look into other modules if you wish to directly measure Current. 

Merry Christmas!

[Neuropotential]

Hello Alex, 

Thank you for your reply. Based on your reply, I will take parameter R as the absolute resistance of the thermistor. 

For the other question, I am not looking to measure current. Instead, I wanted to use constant current as the excitation source. With a known constant current, the voltage across the thermistor will change based on the change in resistance. Now, since voltage across thermistor can be recorded by usb-6008/6009, I can calculate instantaneous resistance using this voltage and constant current value. This eliminates the use of another resistor as you suggested. Do you think this approach makes sense? If yes, then what value of current should I use to excite the thermistor? 

Merry Christmas 🙂

Sanket

[Alex.T]

Hi again Sanket,

Use of a constant current source is also a good idea, however on the USB 6008/6009 devices you can only source via a voltage output. In this case, you could make use of a Thevenin Equivalent Circuit (We'd need to grab another resistor!). The problem with this implementation is that the change in load would also produce a change in the current which is sank, so we'd need to find some way to prevent coupling between the source and the load.

In order to properly source current independantly of the applied load, you could make use of a Howland Circuit.

For ease of implementation, I'd recommend using the Potential Divider based method.

[Neuropotential]

Ohh. I understand you wanted to source the thermistor using USB 6008 itself. 

I had an external constant current source and was hoping to use that as current source. USB 6008 was used only for sensing voltage. But your idea is great. Just with an additional resistor, I can get rid of an external source. I guess I will use output channels on usb 6008 as constant voltage +5V and then use the potential divider idea you suggested in the first post

[johnsold]

Sanket,

You came to the correct conclusion about the first question in your original post, but never got verification of that.  In the temperature-resistance equation in the DevZone article you linked T is the absolute temperature in kelvins. R is the resistance of the thermistor at temperature T. Rref is the thermistor resistance at the reference temperature, as specified by the thermistor manufacturer, usually 25 degrees C.  A, B, C, D are coefficients for the equation relating resistance and temperature.  They are dependent on the material and construction of the thermistor, although standardized values are often used.  Note also that the temperature formula implemented in the formula node is not the logarithmic function described in the equation listed on the block diagram.  The VI uses a polynomial approximation.  Inside the formula node R is the natural logarithm of R/Rref.  The use of "R" for two quite different things can be misleading.  If you want to use the logarithmic equation, implementing it with LabVIEW Mahtematics palette functions is straightforward.

The amount of current to be applied to the thermistor depends on several things.  The maximum power dissipation rating of the thermistor is one, although other things always produce a lower limit.  Self heating is a major source of error in some applications.  The manufacturer of the thermistor will provide data indicating how much the device heats up due to the power dissipated in its resistance.  This should be kept below the allowable error in your temperature measurement.  A conflicting requirement is the magnitude of the voltage change with temperature. The larger the current, the larger the voltage change. 

Selecting the appropriate current involves making tradeoffs between the self-heating error and the voltage or temperature measurement resolution you want.

An external constant current source is an excellent way to operate a thermistor because the voltage across the device then is directly proportional to the resistance.  A series circuit consisting of a fixed resistance and the thermistor in series excited by a constant voltage is one of the simplest circuits for using a thermistor.  Your software needs to include the calcuation of the resistance from the output/input voltage ratio of the divider, but that is a simple algebraic calculation.

As long as the total current in the voltage divider does not exceed the current sourcing capability of the AO line, using that as an excitation source is fine.  Typical currents for 10000 ohm thermistors are less than 1 mA, so the AO drive will not be a limitation.

Lynn