stable trimmed resistor

You can set a single-turn or multiturn trimpot - they're about the same, really - to about 0.1% before it gets really annoying. They are stable to a bit better if not shocked. 0.01% might be pushing your luck.

It's better to extend the range with fixed resistors...

--------+---------+ | | R1 | | | | P ------+--R3---->O | T | | R2 | | |

--------+---------+

Note that here the pot is still acting like a pot, not a 2-terminal rheostat. They have better TCs that way.

Big old true 10-turn (spiral wirewound) pots with shaft locks are very stable. There are wirewound-plastic hybrids that are stable and have essentially infinite resolution. But I'd still use the extender circuit above, with some good low-TC fixed resistors. 0.01% is just

John

--R4--------+---------+ | | R1 | | | | P ------+--R3---->O | T | | R2 | | | -R5--------+---------+

I think John has covered it all... If you are tuning only over a small range you can add two 'good' resistors in the arms (as shown) So the pot only has to work over the limited range you desire. (By good I mean good temperature stability.) You want the same type of good resistor in the arm that mirrors the thermistor.

George H.

Actually 0.01% is a little better than necessary, but it should not be worse than 0.05%. Well, actually I mean 0.01C and 0.05C, so maybe the resistors should really be stable to 0.01% so there is room for other errors.

I don't think thermal deviations are a problem as long as they are precise (repeatable). I would keep the whole bridge with the sensor so it will be subjected to the temperature to be measured.

What I'm worried about is that the slider of the pot might change position irreversibly due to thermomechanical motion. Also, the thermometer (including the pot in a casing) will likely be put on a hard surface thus experiencing shock or it will possibly be dropped a few centimeters. And really it would be good if it survived somewhat higher drops without decalibrating.

Hmm, I don't understand your circuit. I thought of something like this:

Bernhard

It is getting clearer ... :)

Bernhard

If you turn the pot to one end, it shorts R3. So you can only really use a fraction of the pot's rotation. Tweak is very nonlinear on rotation, and only goes in one direction. And the pot TC, 100 PPM/K maybe, changes its effective value.

My circuit allows the entire sweep of the pot to almost-linearly tweak the main divider ratio by a controlled amount, above and below the native R2:R3 ratio. That tweak amount can be changed by changing R3, rather than having to buy a new pot that only comes in stock values. R3 could be 10M, but it's hard to buy a 10M pot. This circuit is just has less constraints and is more orthogonal.

John

Actually, this isn't bad in your situation

except that it's a little bit harder to scale the pot influence.

John

Opps, I made a mistake on my previous post. (Just ignore it.)

I use a circuit like this for temperature control of laser diodes. I've put the thermistor in the lower part of the arm. (I'm not sure this makes much difference but I feel more comfortalble when the NTC is touching ground.) I replaced your ADC with an instrument amplifier.

Then Johns original suggestion looked like this, (I think)

Which looks like a loaded pot, except I'm not sure what R4 is doing. (calls for a bit of analysis.)

George Herold

R4 reduces the effect of the pot to a small trim of the (stable) voltage set by precision resistors R2 and R3.

I put it on top because that gives me a non inverted voltage response to temperature changes. Not actually important, though, since the voltage is going to a uC. Hmm, but if there are leak currents between case and NTC it might be better on ground/case potential.

I'll be using an in-amp, too. It has to arrive, yet. Currently I only have normal op-amps.

Bernhard

Yeah, I was thinking of this last night, at least one reason I like to have part of the temperature sensor grounded is that the sensor is (ussually) stuck into a grounded piece of metal. And sometimes I like to attach one leg of the sensor to what it's measuring for heat flow reasons. There's no better thermal coupling route than through the leads.

George H.

If you weld a thermocouple to the metal you're measuring, *both* leads have ideal thermal coupling.

John

rd

Thanks, I had to sit down and write an approximate equation. I assumed R4 >> (X10) any other resistance. And then I(R4) is some small perterbing current.

George H.