RTD excitation current source circuit

Thread Starter

mah115

Joined Sep 11, 2009
5

beenthere

Joined Apr 20, 2004
15,808
Did you try reading the article from National Semiconductor?

You might also want to read up on Kelvin four wire sensing.
 

Thread Starter

mah115

Joined Sep 11, 2009
5
Yes, I did. I still don't see what advantage the National Semiconductor circuit (figure 1) has, except that you can ground one terminal of the RTD directly to ground. But why would this matter if we are reading the differential voltage across the RTD anyway, if your differential amplifier has a good CMRR?
 

beenthere

Joined Apr 20, 2004
15,808
Why do you expect to have a differential voltage to amplify? - http://www.omega.com/prodinfo/rtd.html

Put another way, what is the advantage of not having one of the RTD's terminals at ground? It's simply a very well defined resistor. The means to a good measurement is a stable current source and a quiet amp.
 

Thread Starter

mah115

Joined Sep 11, 2009
5
We would want to measure the voltage differentially because the actual RTD may be some distance away from the circuitry, which means you can have many feet of wire between the driver circuit and the RTD. Current flowing through the wire will produce a voltage drop, and so we want to measure the voltage across the resistor with high-impedance inputs.

The question is what benefits such a fancy excitation circuit provides.
 

rjenkins

Joined Nov 6, 2005
1,015
My first thought is that the common mode range of many opamps includes 0V, but rarely includes V+ or even close to it.

Using a current source that works with the load to ground makes reading the true RTD voltage simpler.
 

Thread Starter

mah115

Joined Sep 11, 2009
5
But in terms of accuracy, would not having two op-amps instead of one introduce more sources of error (bias, input offset, etc.)?
 
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