RTD help?

RTD's need to be in a state of very low power dissipation. The lower the better.. Why would you want heat to offset the ambient readings?

Since most are PT100 (100 ohms) at 0C (32F), I guess that would mean a low voltage in the bridge to prevent internal heating.

Depends on the required accuracy and other factors, but modern instruments usually use 1mA or less. It's a trade-off, obviously.

Your allowable self-heating and the RTD construction and what it's stuck into (usually) determine the maximum dissipation. A tiny sensor in still air will be a lot more sensitive to power dissipation than a big honking sensor potted into a tube and stuck into a block of aluminum or running water.

Best regards, Spehro Pefhany

afair 1mA for pt1000, a few 100uA for pt1000

something like this? (no garantees I just hacked it together on the fly)

Version 4 SHEET 1 1684 680 WIRE 256 112 112 112 WIRE 112 144 112 112 WIRE 256 144 256 112 FLAG 256 224 0 FLAG 112 224 0 SYMBOL res 240 240 M180 WINDOW 0 36 76 Left 0 WINDOW 3 36 40 Left 0 SYMATTR InstName R1 SYMATTR Value {THERM} SYMBOL current 112 144 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName I1 SYMATTR Value -1e-3 TEXT 424 72 Left 0 !.STEP param temp -50 100 10 TEXT 422 108 Left 0 !.tran 1 TEXT 424 208 Left 0 !.param A= '3.9083e-3' TEXT 424 232 Left 0 !.param B='-5.775e-7' TEXT 424 264 Left 0 !.param C = IF(temp+0.5,0,-4.183e-12) TEXT 424 160 Left 0 !.param THERM ='R0* (1 + A*temp

-Lasse

Keep the power dissipation down, well below a milliwatt. It helps to measure them at a low duty cycle, like maybe excite+measure for 10 milliseconds, once a second or some such. That's a 100:1 reduction of any self-heating.

Depends on the size, the packaging, the environment. And how much self-heating error you can tolerate.

Here's some data on a 1206 size, 100 ohm Pt RTD:

ftp://jjlarkin.lmi.net/RTD_in_air.JPG

ftp://jjlarkin.lmi.net/RTD_lotsa_copper.JPG

ftp://jjlarkin.lmi.net/RTD_on_board.JPG

As you can see, the PCB pads/traces have a huge effect on self-heating.

John

Because of the the slowness of the thermal response, it's frequently possible to have the best of both worlds. You can turn the RTD excitation off and on periodically (maybe only once, at power-on), and digitize fast enough to watch its thermal transient. Then use that as an estimate of the self-heating error.

It's unusual to need, say, 0.1 K absolute accuracy, but fairly common to need millikelvin stability. Using 1k RTDs with 1-10 V of excitation often makes that much easier, because it relaxes the drift requirements on the rest of the circuit.

Cheers

Phil Hobbs

I do not want heat; nobody gives a nominal drive, much less any drive. No spec, no hints. Yes, want low drive - maybe in the region of 1mW?

• My first off-the-cuff wild guess was 1mA - and then i thought "that is a huge 100mW of dissipation = = way too much".

• That is roughly what the sellers say or indicate. But still NO CLUE as to "preferred" max drive.

• That seems large; 100mW is a lot of heat. I am going to see if i can get away with 1mW.

• Will look at it; thanks.

• Sounds more sensible; 1mW drive and "pulse" (or sample) the beastie, which is better yet.

Excellent documentation..but what does the 40mA/800 ohms stand for?

• Excellent idea!

I would like to get 1 degree absolute accuracy on the basis of the theoretical info (the Minco tables) and the same repeatability. Anything better than that is gravy. See my most recent posting on ABSE; it seems that either SPICE is not up to snuff or that bridges just do not balance the way one would think (based on pure theory).

Maybe i should not use a bridge, and just "measure" the value..

It *is* way too much, because it's wrong! :) I²R, remember?

So that's 100µW and 1mW for 100 and 1k RTDs respectively. (Lets see if the non-ascii symbols there will come through).

They don't know your application. If you can't be bothered to work it out, just use 1mA. Otherwise you need to model it or make some measurements.

Yeah, 100mW _would_ be too much, but I^2*R= 100uW You'd need 32mA to get 100mW from a Pt100 at room temperature. You use less current for a Pt1000. At, say, 200uA you'll see but 40uW, and you'll be getting double the signal (mV/degree) and less sensitivity to lead wire resistance (but more sensitivity to shunt resistance between the wires and input Z).

It's not limited by the sensor element but by what you do with the sensor element, what your requirements are, how good your circuitry is, and, as others have pointed out, whether your excitation is continuous or not. Oh, and the maximum temperature of the element. There's also some (unspecified) pulsed power level, above which you'll blow the Pt element and passivation right off the substrate.

Another ultimate limitation is the Johnson-Nyquist noise in the element at the operating temperature vs. your required bandwidth. At, say, 300°C, a Pt1000 will have ~17nV/sqrtHz so if you want to resolve

People used to use even more than 1mA on a Pt100 way back when inexpensive electronics wasn't as stable as today, but they got away with it in some applications partly because the most common sensors were actually big fat coils of platinum wire on ceramic formers rather than thin films deposited on alumina. It's still a way of making a really stable low-hysteresis but thermally slow sensor. The thin film ones are cheap and good enough for most applications so they dominate.

Maximum power? I've driven more than a watt through a sensor, because that was what worked in my application. Don't try this at home, kids.

Best regards, Spehro Pefhany

RTDs aren't quite linear--their R vs T curves are concave downwards, unlike thermistors, so they can't be linearized by resistors in series-parallel. You can make them very linear by applying a negative resistance of about -25 times the RTD's resistance--maybe 24 or 26 times, depending on the temperature range you want to linearize over. OTOH since you're digitizing it anyway, you might as well use a voltage divider from your reference voltage, do the algebra, and compute the temperature by table lookup and linear interpolation.

I'm a fan of the OPA378 low noise chop amp for this sort of thing--preferably running inside the temperature-controlled zone. Its output isn't that stiff, so if you're driving a capacitive ADC, you might need something like an AD8605 in the loop, or a big cap on the output.

Cheers

Phil Hobbs

The RTD was connected to a power supply through an 800 ohm resistor, and the current was 40 mA. That, and the numbers on the scope photos, is enough info to calculate the thermal dissipation constants (in K/watt) and the thermal time constants. I did that a couple of times and keep losing the numbers.

John

On a sunny day (Mon, 13 Dec 2010 07:58:24 -0800) it happened John Larkin wrote in :

I was thinking 'what a cheap shitty displays those instruments have, a bit better and you could add the explaining text on screen via the keyboard or PC link. My 35 Euro photo frame has a better display than that.'

What is that, TDS2000? I think they are QVGA CSTN displays. The older TDS3054 were much nicer than this, at least VGA I think. I don't know why they felt the need to save the extra $20.

The post-its work fine.

Pictures of your dog? I have useful data about RTDs, and you don't.

John

On a sunny day (Mon, 13 Dec 2010 09:08:02 -0800) it happened John Larkin wrote in :

I do not have a dog, I'm not Joerg :-)

I use thermocouples, and LM135. And Si diodes, quite linear, actually all the way down to 77K.

I was looking at one of the schematics of a Lab instrument once with a RTD input. The bridge source was coming from a basic Si diode clamp in the forward direction to yield ~ 0.6v. The bridge uses 100R so the path for the RTD is 200R at 0C degrees. I guess this puts it in the 3 ma area yielding ~1.8mw

The bridge drives a chopper op-amp

I don't think you need to stick to strict guide lines as long as you keep the dissipation down in the single digit mw range or less.