# Creating a thermomether with a pt1000 (Help with circuit)

1. Oct 17, 2016

### TimeLordo123

I have been tasked with creating a thermomether using a RTD, more specifically a pt1000, also using a wheatstone bridge. Conceptually I think I have figured how to do it. I build the wheatstone bridge, that works as a voltage divider, with the pt1000 integrated in it. Then we use 2 amp ops, one first to amplify the signal, and one after to substract the offset voltage of the pt1000 (i think), or as my professor calls it, to create a differential step. The problem is that I lack actual eletronic knowledge, so creating a schematic circuit is proving to be a problem. I have come across some in the internet, like this: https://i.gyazo.com/573ecbc4d7d5ec284f3a7b72b2bad398.png (though this one uses a pt100), but I think that the amp ops are in a different order, and I cannot even spot the wheatstone bridge. Could someone give me some insight?

2. Oct 17, 2016

### Nidum

Last edited by a moderator: May 8, 2017
3. Oct 17, 2016

### jim hardy

4. Oct 18, 2016

### dlgoff

5. Oct 18, 2016

### jim hardy

A Wheatstone bridge is at its simplest just two voltage dividers with a common supply.

I'll explain a little more further down.
First though, about that circuit you found:
i think it has these shortcomings

1. It's not really a Wheatstone bridge because one leg, R5-R6 is from +5V supply while other leg uses an active current source to drive almost constant current through R1 the platinum sensor. So supply voltage variance affects the two legs unequally.
2. It doesn't account for resistance of the wires going to the sensor . How long are they ?
3. Differencing amp tracking is dependent on matching of R3-R4 and R7-R8,
3a. and i'm not convinced it's drawn correctly. Seems to me R4 should be tied to pin 4 not 6, but i may have missed some exotic detail .

Here's a wheatstone bridge at its simplest

Vleft = Vexc X R2/(R1 + R2)
Vright = Vexc X RX/(R3 + RX)
Vr (output) = Vlefft - Vright

Often we make R1=R3 so that at balance both legs have equal current

for yours maybe choose 4 kohm and 5 volt excitation? That'd give about a milliamp when Rx is 1000 ohms, provided you set R2 to balance the bridge there.
I suggest you make a spreadsheet and study Vr versus Rx over expected range of sensor temperature , for several different Vexc's maybe 5 volts plus and minus ten percent. You'll see the differencing effect of the bridge reduces sensitivity to imperfect power supply regulation.

That's one advantage of a bridge.

Electronic Thermometers are often located some distance from the point where it's desired to measure temperature. The resistance of the interconnecting wires becomes significant and causes error in reported temperature.
So we use what's called "Three Wire" connection to the bridge, shown here:

Since all 3 leads are same length their resistances should be very nearly equal.

That's how we measured temperatures around the reactor where i worked. Bridges were in the control room, platinum sensors were in pipes around reactor connected by a couple hundred feet of copper wire in between.

Okay , so now to the differencing amplifier
in my day they were discrete component chopper stabilized monstrosities, a testament to tenacity of circuit designers in the late 1950's.
Nowadays you should use a specialty interface from somebody like Analog Devices
or build one. TI makes INA106 that'd be a good candidate for your learning exercise.

I myself would balance the bridge for the temperature where you want most precision in your measurement, select R2 = Rsensor at that temperature.
Or if it's a general purpose thermometer you're after, balance it at bottom of range you want to measure. That simplifies calibration.

Now to the temperature versus ohms of your sensor.
The platinum sensor is fairly linear, but only fairly . If you want digital precision you'll have to linearize output by either a microcomputer or analog .
I prefer analog but that's just me.

Platinum sensors are used to define the "International Practical Temperature Scale"
which at its simplest is a quadratic curve fit of ohms versus temperature.
Search IPTS68 and IPTS90
https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19870019452.pdf

I prefer the Callendar-Van-Dusen calculation of '68 but everybody has gone to simpler quadratic . You can prove them equivalent with twenty minutes of algebra.

Anyhow you can use your spreadsheet to calculate resistance at any temperature , then Vlaft and Vright and output. You'll get a fairly linear output which will lend itself well to a least squares fit. A quadratic can fit to very nearly 0.1% over a 700 degree range.

If you add a tiny bit of positive feedback you will add a square term that is an analog quadratic adjustment. I once built an analog one that hit within 1 degree over range 0 to 700 degrees F

Here's how we did it

That alfa X Vr block was just an operational amplifier i think LM324. The bridge only needed a couple milliamps which is well within capability of a garden variety opamp,
we let it drive the bridge just summed Vexc+(Alfa XVr )

We included a post-amplifier to give output of 1 millivolt per degree . It was great fun to dial in on our test equipment resistance equivalent to some temperature and watch the digital voltmeter indicate within 1/10 degree over our range of interest, 500 to 600 deg F, and within 1 degree on down to 0F.

So,
there's some ideas for a "better" home-made RTD thermometer .

You'll want a "decade resistance box" to simulate the RTD for calibrating your thermometer

and temperature stable precision(0.1%) resistors for your bridge. Buy a few extras and hand pick them for a good match.

Have fun, guys. You should learn a lot from this project.

old jim

Last edited: Oct 19, 2016
6. Oct 20, 2016

### jim hardy

i guess we lost him.