Advantages of a Wheatstone Bridge compared to a simple voltage divider

hi guys!

I want to measure the resistance of a Resistance Temperature Device(RTD) using the analog input of my arduino. Within the temperature range i want to measure the resistance varies from 1700 Ohms to 2300 Ohms. Aditionally i have to restrict the current to 200*10-6 A.

Ideally i would like to measure 0V if the resistance of the RTD is 1700 Ohms and 5V at 2300 Ohms.

Now my question:

What is the advantage of using a wheatstone bridge over a simple voltage divider?

It seems that there is no way of getting around the use of a OA? Is that right?

Greetings

TheGreatAlex: hi guys!

I want to measure the resistance of a Resistance Temperature Device(RTD) using the analog input of my arduino. Within the temperature range i want to measure the resistance varies from 1700 Ohms to 2300 Ohms. Aditionally i have to restrict the current to 200*10-6 A.

Ideally i would like to measure 0V if the resistance of the RTD is 1700 Ohms and 5V at 2300 Ohms.

Now my question:

What is the advantage of using a wheatstone bridge over a simple voltage divider?

Certain error terms cancel - its pretty much a requirement for high accuracy strain guages for instance, and high accuracy measurements in general. You don't necessary need it for this application, but you haven't said what accuracy you want...

It seems that there is no way of getting around the use of a OA? Is that right? Greetings

If you want to stretch the output to a 5V range, then no way round - 200uA will only get a fraction of a volt in the resistors. The current limit is of course to reduce self-heating effects.

I have been pondering about this during the day.

What you could do:

5V -----------
        |
        22 K resistor
        |
        |---- to analog pin
        |
        RTD (between 1.7K and 2.3K)
        |
GND -------

The 22K will limit the current to small amounts. However, a change in RTD values will only result in a difference of about 0.1 Volts, which is about 24 steps on Arduino ADC when Vref = 5 Volts.

Now I'm wondering, if you set Vref to say 0.15 Volts, then you might get all 1024 ADC steps across what you wish to measure.

Perhaps someone can confirm this idea.

Cheers,

Jack

if you set Vref to say 0.15 Volts,

Then you would be working outside the limits of the arduino's A/D which needs a minimum Vref of 1.0V

Telecommando:

CaptainJack: Now I'm wondering, if you set Vref to say 0.15 Volts, then you might get all 1024 ADC steps across what you wish to measure.

Except with a 0.15 V reference, the largest voltage you could measure would be 0.15 Volts. With those resistor values, the input voltage would range from about 0.36 to 0.47 Volts, both above the range of the reference.

A 0.5V reference would work. The analog values should range from about 734 to 969, a span of 235 values.

Or to take the silly math out, one could do this:

5V ---------------------------
        |                    |
        R1                   R2
        |                    |
        |---- to Vref pin    |---- to analog pin
        |                    |
        R3                   RTD (between 1.7K and 2.3K)
        |                    |
GND --------------------------

Where R1 == R2 (sorry, been on the programming board too long, the double equals is habit...) and R3 == Max resistance of the RTD.

Grumpy_Mike:

if you set Vref to say 0.15 Volts,

Then you would be working outside the limits of the arduino's A/D which needs a minimum Vref of 1.0V

Oh, bother. There goes Grumpy_Mike actually reading the spec like we should have. ;-)

I guess the next step would be to throw in an OpAmp to raise the low voltage of the RTD (to maintain the low current). While doing that, may as well set the gain of the OpAmp so the resulting signal is 0-5VDC to keep the ADC of the Arduino happy. So much for trying to keep this simple.