# Precision voltage reference divider vs. constant current source to drive an RTD sensor

I need to measure temperature with an RTD between 0-300 C. I believe I have two options to make this work:

1. A constant current source feeding the RTD
2. A precision voltage reference and a resistor forming a voltage divider with the RTD.

It seems like I could use a precision voltage with an op-amp constant current source circuit to complete option 1. I could also use something like an LT3092.

This approach seems more complicated than option 2, where I could connect an ADR02 5V reference in line with a 5k 0.01% resistor to form a divider with the RTD.

I plan on using a 1k platinum 2 terminal RTD. Which is the better option, and why?

I've come up with this circuit after reading through the various suggestions.

Does it look like this will work? Should this reduce noise reasonably well?

The 10V clamp is for protection of the ADC. I think I also need an inline resistor on the output before the clamp, but not sure how to size it.

• option 3 is the best, use a non-precision source together with a Wheatstone bridge to compare the RTD with a precision low drift 1k resistor Jun 15 '16 at 17:15

A current source ideally has infinite resistance. A resistor has positive resistance. There's a third choice- use current source with a large but negative output resistance which will linearize the RTD so that it has an 'S'-shaped residual error curve. Probably not worth it these days if you are going into the digital domain anyway.

Since an platinum RTD will only change from 1K to about 2.14K for 0°C~300°C you will lose about half your resolution if you use a simple resistor. A bridge can offset the voltage at 0°C and give you approximately full resolution. In any case your resistor should be connected to the same reference as your ADC so that reference voltage error cancels out and only the resistor values matter.

Something like this (you'd want to add other parts in a real implementation typically for EMI and so on).

simulate this circuit – Schematic created using CircuitLab

The output voltage is:

Vo/5V = R4/(R4+R1) - R3/(R2+R3) = R4/(R4+20K) - 1/21

So the output voltage would be 0 to 245.05mV for 0 to 300C.

• If I want to increase the magnitude of the output voltage, what is the best approach? Lower the values of R1 and R2 to let's say 1-5k, or use the gain of a difference amplifier? I'm specifically worried about noise here. My ADC is an external NI 9205 unit. Jun 16 '16 at 15:44
• Lowering R1/R2 will increase self-heating. Looks like the NI 9205 has a 200mV range, so if you increase the resistors you can use that. For example, 24.90K each. Or use an amplifier. Jun 16 '16 at 20:01

RTDs are non-linear so let's just say that using a resistor to feed it adds a bit more non-linearity you have to cope with. But if you are feeding the RTD signal to an ADC then it makes sense to tie the top of the resistor to a reference voltage that is also used by the ADC - this is called a ratiometric measurement and, to a significant extent that voltage reference can drift this way or that without numerically affecting the ADC conversion number.

Using a current source may make things a bit more linear but you lose the benefit of a ratiometric measurement.

Using three more resistors to make a bridge isn't worth the hassle in my opinion - that's three resistors that have to be precision instead of 1 and any quarter bridge circuit will be non-linear anyway.