Wheatstone bridge solve unknown RTD?

Sdyke4898

Is it possible to solve for resistance of RTD coupled with a wheatstone bridge when 3 resistors are fixed?
I know the resistor values and voltage in and out.
I do not know current flow on RTD arm nor do I know the temperature.
I want to determine the temperature on a portion of an air compressor at a distance of approximately 75 yards.

evanh

It's not normal to use a bridge construction when temperature measuring. Usually, for long run transducer wiring, a four wire connection is used that just involves two wires for current supply and the other two wires for sense of the voltage across the PT100 element. Three wire also works for long runs. Two wire is normal for short distance. The PT100 element is a single platinum wire on a ceramic base. Fragile construction.

It's more common to use a themocouple or thermistor for such an application. Accuracy is not a priority.

Sdyke4898

I have a three wire RTD that I plan on using.
I would like to use a current source, however I am having a difficult time finding a circuit that can be built with through hole components. I don't have the knowledge so I have to copy from others hard work.
You are correct in that accuracy is no priority. Although it was my understanding that thermocouple usage was not recommended due to low voltage and lead wires acting as an antenna for noise.
Thank you for your time.

kwinn

Sdyke4898 wrote: »

I have a three wire RTD that I plan on using.
I would like to use a current source, however I am having a difficult time finding a circuit that can be built with through hole components. I don't have the knowledge so I have to copy from others hard work.
You are correct in that accuracy is no priority. Although it was my understanding that thermocouple usage was not recommended due to low voltage and lead wires acting as an antenna for noise.
Thank you for your time.

A constant current source can be built with a small signal transistor, a resistor, and a couple of signal diodes, all of which are available as through hole components. What current level do you want?

Alexander (Sandy) Hapgood

You could place a known resistance in series with the sensor and measure the voltage drop across that to determine the current flowing in the circuit. Measure the voltage drop across the sensor and using the current value you got previously you can determine the resistance of the sensor.

Sdyke4898

Thank you Kwinn and Alexander. I have seen simple circuits containing the components you mentioned. I just didn't know enough to be able to count on them; I figured I needed some expensive ic to do the job.
I need to keep current low such as 100 micro-amps so as not to cause significant heating of RTD.

DigitalBob

Use an LTC1043 look at the data sheet where it shows an RTD signal conditioner. I've used this one before and it works pretty good. Range is 0-400 C (0-4V out) but you can convert it in your code. The schematic is missing a connection. The 887R, pin 14, 5 and the RTD are tied together.

kwinn

Sdyke4898 wrote: »

Thank you Kwinn and Alexander. I have seen simple circuits containing the components you mentioned. I just didn't know enough to be able to count on them; I figured I needed some expensive ic to do the job.
I need to keep current low such as 100 micro-amps so as not to cause significant heating of RTD.

I'm afraid I may not have been of as much help as I had hoped to be. Never had to build any circuits using RTD's, so I thought it was a typo and that you meant TDR. It may be that if the change in resistance is so small that the circuits I suggested will not do the job. OTOH building a constant current source to provide 100uA is simple and inexpensive so it may be worth trying it out.

Tracy Allen

Is it a 100 ohm RTD? That is the most typical value, 100Ω at 0°C. The temperature coefficient is typically 0.385Ω per °C. If you pass 100µA through it, that gives a total voltage of about 10mV, and a change of 38.5 microvolts per °C. So, you see, the measurement circuit has to be pretty sensitive. What ADC do you have that can measure at that level of precision? I've used the ADS1115, a 16 bit converter that can be set up to do a ratiometric conversion in relation to a precision fixed 100Ω resistor. I think you can pass 1mA without worry of self-heating. Even with 1mA, the self-heating power is only 0.1 mW. Then you get ten times as much signal, 385µV per °C.

That said, you might be better off with an an AD592, or an LM334, or an LM34, all of which can be set up to transmit a current over a pair of wires with much less trouble.

Sdyke4898

Thank you so much Tracy Allen. If 1mA will not cause much self heating then I will take that path, the math will be much simpler.
I was unsure due to the fact I sent 15mA through the little guy and picked the probe up and it literally burnt my fingers.
Again thank you all for the direction.