Okay, sounds like a nice approach.
You must have skilled hands ! I never used smd for a home project because at my age i can't even see those tiny parts.
If i take your words exactly as you wrote them, here's a problem.
When you measure in circuit your meter shows resistance between its probes . The other three legs of your bridge will let some current pass which makes the total resistance between the probes smaller.
Is this your circuit, where R1 R2 and R3 are each 100 ohms??
View attachment 114378
How to analyze this :
Assuming there is no source connected to V+ and V- ,
Connect a meter across any leg of the bridge and you'll read:
(resistance of that leg )
in parallel with (sum of resistances of other three legs)
Do you know how to calculate resistances in parallel? It's too easy to believe, just remember this phrase:
"Reciprocal of sum of reciprocals"
Example:
Connect your meter across R
Pt100 leg. If it's a DMM it probably applies one mlliamp of testing current and measures the voltage drop.
Meter will push its small testing current through R
Pt100 . Draw it on the above circuit perhaps in red.
But some of that test current will sneak around R
Pt100 through R3, R2, R1 loop. Draw that too, maybe in orange
So the meter will report the parallel resistance of those two paths.
Resistance of the R
Pt100 path is just R
Pt100 and that has reciprocal 1/R
Pt100
Resistance of the R1 R2 R3 path is (R1 + R2 + R3) which is 300 ohms and that has reciprocal 1/300
Let's just guess that R
Pt100 is 110 ohms at your workbench temperature, which has reciprocal 1/110 .
Now take your pocket calculator and add 1/110 to 1/300. That's the sum of the reciprocals. I get 0.0124242... What did you get ?
Take reciprocal of that, that's what the meter should read. I get 80.49 ohms . What did you get ? Is that about what you measured?
Remember ,
"Reciprocal of sum of reciprocals" .
Just too easy, isn't it? That 1/X key on pocket calculators is handy for resistances in parallel, 1/X then M+ ; M accumulates the sum of the reciprocals for you. MR and 1/X again, presto you're there.
You're not the first guy to get fooled by in-circuit versus out-of-circuit measurements.
Now ---
What do you calculate your ohmmeter should read if you connect it across 1 to 2 in above bridge? Try it, what does it indicate ??
......
Power = V
2 /R . I suspect that one watt would be enough to hurt your Pt100 in a matter of seconds. . That'd be 10 volts.
5 volts is too much. Your 100 ohm smd resistors should be plenty hot, feel them.
We always use something to set current through a Wheatstone bridge . Simplest is just a resistor in series with supply.
With Pt sensors I like to set for one milliamp in each leg at balance.
That's because one milliamp will not heat your sensor above ambient temperature.
The effect is called "Self Heating" and should be mentioned in your sensor datasheet. It makes your sensor report high.
Please calculate how much power one milliamp develops in your 100 ohm sensor.
Multiply that by your sensor's "°C per Watt self heating" coefficient, that's the self heating error.
Repeat for 5 volts across your sensor. Do you see the evils of high current through a platinum temperature sensor?
Now to your bridge:
One milliamp through a 200 ohm bridge leg would be 200 millivolts , 0.2 volts, across your bridge.
That means your series resistor needs to drop the 5 volt supply by 4.8 volts.
4.8 volts / 2 milliamps = 2400 ohms.
Add a 2400 ohm resistor in series with your supply and you'll have a high quality temperature measurement instrument.Use all your senses when troubleshooting. A fingertip will tell you if something is being overheated.
A Pt sensor dying from too much current will crackle and pop as the insulation inside it boils and burns away, and it'll be hot to the touch.. Smoke is always a bad sign.
Did you touch test your 100 ohm smd's ? They should be plenty warm at 1/4 watt.
old jim