Troubleshooting RTD Sensor Resistance Drop with Wheatstone Bridge

In summary, you are measuring the resistance of an RTD in circuit, but the RTD itself is not reading as high as it should. You think there might be a problem with the circuit or the RTD.
  • #1
asherraph
15
0
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)..
 
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  • #2
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 ?
 
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  • #3
i don't 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..
 
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  • #4
asherraph said:
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)..
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
tech99 said:
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.

yes it true...but there is no specific voltage recommended...if that so.. i will give a shot with lower voltage..
 
  • #6
asherraph said:
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..
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
SlowThinker said:
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.
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
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
asherraph said:
yes it true...but there is no specific voltage recommended...if that so.. i will give a shot with lower voltage..
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
asherraph said:
i m create bridge using pcb board with smd resistance..
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.

asherraph said:
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...

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??

Bridge.jpg

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?

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
 
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  • #11
jim hardy said:
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 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?

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
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
asherraph said:
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...

Well, I'm sorry but those words do not paint in my mind a picture of :
what is your circuit ?
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 ?

asherraph said:
when i connect directly to the bridge...the pcb become hot...
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.

asherraph said:
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
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
 
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  • #13
Untitled.png
Untitled.png
jim hardy said:
Well, I'm sorry but those words do not paint in my mind a picture of :
what is your circuit ?
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

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??
 

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  • #14
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
asherraph said:
i m not sure whether i upload it or not.
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.

asherraph said:
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.
If that reading is correct,
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
SlowThinker said:
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 ?[/QUO
SlowThinker said:
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 ?

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
jim hardy said:
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.

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

jim hardy said:
If that reading is correct,
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.

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??
jim hardy said:
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 .

i have only one DMM and i m sure fuse is not broken yet.
mKM3W1BmnOFN3o_VHJWVGcQ.jpg
i m using this kind of DMM. the proble is attached to the DMM
 
  • #18
asherraph said:
i have only one DMM and i m sure fuse is not broken yet.
Well , I'm not sure.

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

DMMsanwa2.jpg

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 ?
the-blown-glass-fuse.jpg


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.
DMM_Sanya.jpg

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

old jim
 
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  • #19
jim hardy said:
Well , I'm not sure.

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

View attachment 114481
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 ?
the-blown-glass-fuse.jpg


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.
View attachment 114482
That's why it's the very last one, to avoid getting there by accident.

old jim
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
asherraph said:
the fuse already blow but the DMM still work i guess except for the current measurement.
My Beckman came with a spare inside .
Look around in yours?
 
  • #21
jim hardy said:
My Beckman came with a spare inside .
Look around in yours?

do you mean, you got another fuse when you buy the DMM??

unluckily ,they did not give any spare for me,. just user manual with multi language.

i will buy new fuse tomorrow..
 
  • #22
asherraph said:
do you mean, you got another fuse when you buy the DMM??
Yes. Inside my meter there is a round indent in the plastic case, deep enough to hold a replacement fuse. The factory put a spare fuse in there for just this situation.

I keep a pack of them in my electronics toolbox. Use probably one or two a year when doing a lot. Mistakes do happen.
 
  • #23
Hi everyone. I'm setting this project so i need some help. I read all comments and I have a question. If i choose R3 and R1 to 5000 ohms and R2 is 100 ohm. Thus 5V / (5000+100) = 0.98 mA so Arduino will fine and Multimeter don't blow up i think.
 
  • #24
That would be safe.

However: You WILL have to calibrate the whole circuit. Please consider the tolerance of those 5000 Ohm resistors and what values may have and still be within tolerance. Even with 1% resistors one could be 5050Ω and the other 4950Ω. That could put the bridge out of balance by the total resistance of the PTD!

For the majority of uses/designs, the "best" design/operation of a bridge circuit is when all the resistors are the same value. Then in your case, you would put another resistor in series with the supply voltage to keep the current down in the bridge.

Here are some excerpts from Post #10 above:
jim hardy said:
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.
...
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.

Cheers,
Tom
 
  • #25
Tom.G said:
Here are some excerpts from Post #10 above:
Old Jim :oldcry:
 
  • #26
Tom.G said:
That would be safe.

However: You WILL have to calibrate the whole circuit. Please consider the tolerance of those 5000 Ohm resistors and what values may have and still be within tolerance. Even with 1% resistors one could be 5050Ω and the other 4950Ω. That could put the bridge out of balance by the total resistance of the PTD!

For the majority of uses/designs, the "best" design/operation of a bridge circuit is when all the resistors are the same value. Then in your case, you would put another resistor in series with the supply voltage to keep the current down in the bridge.

Here are some excerpts from Post #10 above:Cheers,
Tom
You mean like this?
9A3E8837-6732-4538-82F3-E3EAF5E55B12.jpeg
 
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  • #27
This thread got me thinking back when I was working for a pharmaceutical company doing calibrations. IIRC it uses a platinum thermocouple. They're kind of expensive but this is what I used. Check out it's users manual I've attached. Very nice specs.
 

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  • #28
Thanks but i decided to use MAX31865 with PT100 on Arduino. It gives better results.
 

FAQ: Troubleshooting RTD Sensor Resistance Drop with Wheatstone Bridge

1. What is a Pt100 with wheatstone bridge?

A Pt100 with wheatstone bridge is a type of temperature sensor that uses a platinum resistance thermometer (Pt100) in a wheatstone bridge circuit to measure temperature. The resistance of the Pt100 changes with temperature, allowing for accurate temperature measurement.

2. How does a Pt100 with wheatstone bridge work?

In a wheatstone bridge circuit, the Pt100 is one of the four resistors. As the temperature changes, the resistance of the Pt100 changes, causing an imbalance in the bridge. This imbalance is measured and converted into a temperature reading.

3. What are the advantages of using a Pt100 with wheatstone bridge?

A Pt100 with wheatstone bridge offers high accuracy and stability, as well as a wide operating temperature range. It also has a linear response to temperature changes, making it easier to calibrate and use in various applications.

4. What are the applications of a Pt100 with wheatstone bridge?

Pt100 with wheatstone bridge sensors are commonly used in industries such as aerospace, automotive, and manufacturing for temperature measurement and control. They are also used in scientific research, medical equipment, and environmental monitoring.

5. How do I choose the right Pt100 with wheatstone bridge for my application?

When selecting a Pt100 with wheatstone bridge, consider the temperature range and accuracy required for your specific application. You should also consider the materials and construction of the sensor to ensure it is suitable for your environment. Consulting with a technical expert can also help in selecting the right sensor for your needs.

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