Pt100 with wheatstone bridge

1. Mar 10, 2017

asherraph

currently i working on RTD sensor using wheatstone bridge...but after i complete the circuit where Ra=Rb=Rc=100ohm wiith RTD pt100...i measure the resistance through multimeter...the value resistance is drop to 78 - 84 ohm..its normal?...or anyone has explanation..

i am also has a problem with RTD..where it cannot reach high resistance...i already test with water at 80*C(130.9ohm on datasheet)..but practically its not..it only can reach 120ohm (50*C)..

2. Mar 10, 2017

jim hardy

We have no idea what you measured.

To what did you connect your multimeter probes?

What kind of meter is it ? Does it read zero ohms when you connect its probes together ?

First reading of your question suggests either your meter is inaccurate or the RTD is no longer 100 ohms at ice point.
How much voltage have you placed across the RTD in prior testing? More than just a very few volts will overheat its internal wire and ruin it by shorting adjacent turns, reducing its resistance.
What voltage did you use for supply to your bridge? How much resistance in series with it ?

3. Mar 11, 2017

asherraph

i dont think our meter is inaccurate..before i solder the smd resistance..i measure it..and its show 100ohm...after i connect into wheatstone bridge...i try to measure using multimeter on one of the resistance...its show 80ohm something...Voltage input is 5V..supply with voltage precision..should i lower the voltage supply again??

i m create bridge using pcb board with smd resistance..

i am using normal multimeter...yes it show 0ohm when the probe is connect together..

Last edited: Mar 11, 2017
4. Mar 11, 2017

tech99

I believe from a quick search that the max current for the Pt100 is 1mA. In such case the power supply to your Wheatstone Bridge would need to be less than 200mV. 5 volts is too high and will cause damage.

5. Mar 11, 2017

asherraph

yes it true...but there is no specific voltage recommended...if that so.. i will give a shot with lower voltage..

6. Mar 11, 2017

SlowThinker

You didn't just "measure" the resistance, did you? You need to use a formula to get the resistance from the voltage across the bridge. Might be something like
$$R=100\frac{1-2V_G/V_s}{1+2V_G/V_s}$$
where $V_G$ is the voltage across the bridge and $V_s$ is the supply voltage.

7. Mar 11, 2017

asherraph

no...i measure the known resistance..but for RTD i m derive formula by myself from Vg = [ (r1/ (r1+r2)) - (rtd/ (rtd+r3)) ] Vs.. its correct this formula?? i give a try from your formula...

8. Mar 11, 2017

asherraph

actually.. i can measure the RTD..but the RTD itself can not read high temperature as it should be....there is no problem when i test it with cold water and it can read accurately except for high temperature...is there a problem with circuit or RTD itself??

9. Mar 11, 2017

asherraph

i try with 3.3V...still same... nothing change...it cannot read high temperature...i test it wtih 60*C of water...but the RTD show resistivity at 40*C...and test with cold water (15*C)..it read accurately around 14.5*C..

10. Mar 11, 2017

jim hardy

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??

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 RPt100 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 RPt100 . Draw it on the above circuit perhaps in red.
But some of that test current will sneak around RPt100 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 RPt100 path is just RPt100 and that has reciprocal 1/RPt100
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 RPt100 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 = V2 /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?

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

Last edited: Mar 11, 2017
11. Mar 11, 2017

asherraph

okay..its make sense now...yeah its true..
..............................................................

i am using arduino 3.3V as power supply...i measure the output current is 0.2mA, while if i am using 5V, the current output is 0.4mA..when i connect directly to the bridge...the pcb become hot...in other way.i m using voltage precision...

NOTE:i measure current output by using ammeter...positive probe connect to output power..while negative probe to ground..there is no circuit...technically in series

12. Mar 11, 2017

jim hardy

Well, i'm sorry but those words do not paint in my mind a picture of :
what is a arduino 3.3v ?
from where and to where in your circuit does 0.2 ma flow ?
where does your 5V come from and where is it connected ?
from where to where does 0.4 ma flow while you are using 5V ?

When you connect what directly to the bridge ?
What on the pcb gets hot ?

I don't mean to be harsh, and i know English is probably not your first language .
That you are able to communicate at all in a second language places you ahead of me.
But a question well stated is half answered.
You have to paint an image in my brain(if i have one) . A word picture is okay , but an actual image usually replaces a thousand words.

Try right clicking that image in post #10, copy, then open Paint and paste it in. Draw in where you connect your meter and your power. Than save it someplace on your computer(i use a folder named PF) and put it up here by clicking UPLOAD button. It'll ask you where is that picture you just saved, just tell it.

............

Ohms law predicts 3.3 volts / 0.2 ma = 16,500 ohms. So i'm totally confused by what you have said about current. There's no resistors that high in your circuit unless you've burnt them open by applying overvoltage.

And your meter indicated 0.2 ma? Was it set to DC milliamps , or AC ? What range?
Does it have a separate test prod jack for measuring current ? Check its internal fuse by selecting ohms and touch red probe to the metal inside that current jack. You should read an ohm or less.

.......................................................................

I'm happy to see you are experimenting with real parts instead of computer simulation. Look at the practical hands-on skills you learn this way .!

old jim

13. Mar 11, 2017

asherraph

i m sorry for my poor explanation. I m still new to this forum interface.. i m not sure whether i upload it or not.
...........................

Arduino UNO is one of the most easiest learning microcontroller and its open-source code( meaning your code can be sharing). There are 3.3V and 5V pin from Arduino UNO board. I m try to find out what is the output current from Arduino UNO board through internet, but i could not verify it. So that why, i am using multimeter to identify the output current.

There are two ways i measure current, first series with bridge (Untitled.png) and the other way, directly from 3.3V or 5V to GND pin of microcontroller(Current_reading.)..for 3.3V the current output is 0.2mA while 5V the current output is 0.4mA.

is it correct the way i am taking the current measurement??

Attached Files:

File size:
7.6 KB
Views:
42
• Arduino UNO.jpg
File size:
10.6 KB
Views:
37
14. Mar 12, 2017

SlowThinker

The first measurement should show around 33 mA, unless there's a resistor you didn't show.
The second measurement will, with great probability, destroy the Arduino. You can measure voltage this way but never current.
If you measure the voltage, does it still show 3.3V ?

15. Mar 12, 2017

jim hardy

Looks great, CONGRATULATIONS ! You deserve a wings medallion like they give to new pilots.

Okay. Slow Thinker said it, something in your observed currents defies laws of electricity.

What i think really happened is first time you connected 5V to common through your ammeter , the fuse inside the meter opened and that's what saved the Arduino.

A working ammeter will have a very small voltage across it. It's almost a short circuit.
If you connect an ammeter across a voltage source a lot of current will flow through it.
That's why they are fused, so there'll not be an electrical explosion in your hand and close to your eyes.

something is limiting current to less than a milliamp.
It might be the burnt out fuse in an ammeter .
It might be the DMM probe is not plugged into the current measuring hole in the meter.
It might be the PCB is mis-wired so your circuit is not what the schematic shows.
It might be the resistors on your PCB are failed to high resistance state by too much voltage.

5V/0.4 ma yields 12,500 ohms. My guess is that's the burnt out fuse in your multimeter.
Do you have another meter? Use it , set for ohms, to measure the resistance of yours when set for current.
Do it on all your current ranges.
More than just a few ohms is too much. Maybe 100 ohms on lowest range like a milliamp.

That'd be my next step.

Then disconnect the PCB and measure across its V+ and V- terminals .

16. Mar 12, 2017

asherraph

nope..i connect directly from 3.3V to GND its show 0.2mA...while i connected to bridge also same...i not sure whether the way i connect ammeter might destroy the Arduino or not.

17. Mar 12, 2017

asherraph

if the fuse was broken...it should be no reading in ammeter right?. the fuse for my multimeter is MAX 400mA...

if the reading of ammeter show the correct value..why my pt100 cannot read high temperature and become hot?
if the voltage is high.. i already try voltage divider to reduce the voltage to 1.7V but the current is to low.
do you have any suggestion for this problem??

i have only one DMM and i m sure fuse is not broken yet.

i m using this kind of DMM. the proble is attached to the DMM

18. Mar 13, 2017

jim hardy

Well , I'm not sure.

Since you said you've violated this warning in your meter's instruction manual, specifically number 2,

http://www.sanwa-meter.co.jp/prg_data/goods/img/PH41381900473.pdf (Red added by me - jh)

Don't you think it would be wise to take a look at the fuse ?

Do you think that black vaporized metal deposited inside the glass tube might conduct just a trickle of current between the ends?
Are you are seeing just a trickle of current ?

Pull your fuse and measure it with the ohms scale. I bet you see about ten thousand ohms.

Whatever you see it'll resolve any question as to condition of your fuse.
That's how you troubleshoot, eliminate possibilities one at a time.
There's a big difference between "I'm sure" and "I verified by test" .

And you're far from alone in blowing multimeter fuses. I keep a pack of ten on hand.
That's the reason better multimeters make you move a wire before you can measure current. It's too easy to roll that knob around into milliamps position.

That's why it's the very last one, to avoid getting there by accident.

old jim

Last edited: Mar 13, 2017
19. Mar 13, 2017

asherraph

since you said that and show the warning in the manual...i already checked and verified it.

the fuse already blow but the DMM still work i guess except for the current measurement. i check the continuity of fuse since the fuse body not in transparent glass using same DMM.

20. Mar 13, 2017

jim hardy

My Beckman came with a spare inside .
Look around in yours?