Calculating the resistance in a compensating circuit for a thermocouple

  • #1
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Homework Statement:: A K-type thermocouple is connected as shown to sense temperature. The temperature zone is at 25°C. The RTD has a resistance of 100 at 0°C and a TCR coefficient of 0.00385. A 5V power supply drives the circuit. Calculate R2 for proper compensation. What is the emf produced by the compensating circuit? The Seebeck coefficient for the K-type thermocouple is 39.4 mV/°C at 0°C:
Relevant Equations:: None given.

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Hi! I'm trying to solve this problem but I'm very confused. So what I know is that the RTD is a thermistor in this case, and according to my teacher ##R_1## must be the resistance of the thermistor at ##0°C##, so it would be ##R_1=100 \Omega##.

Now, I'm not entirely sure of what I'm supposed to do to compensate the circuit. Since it reminds me of a Wheastone bridge, I think that I might need to make ##V_B=V_A##, is this correct?

Also, I don't understand how the thermocouple ##V_A## voltage affects the compensation circuit, would it act as another source?
Hope you can help me understand this with some hints.
 

Answers and Replies

  • #2
Averagesupernova
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A couple of things. You do understand the purpose of this or not? A thermocouple develops a voltage across the open end the the wires based on the temp difference between the ends. Search this forum for some very good threads on thermocouples. Fascinating stuff.
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So knowing the above, we know a thermocouple can't tell you absolute temperature. The thermistor gives you a reference at the open end of the thermocouple. Think of it as you correctly assumed as a wheatstone bridge with a completely separate voltage source (the thermocouple) attached as shown.
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Seeing how this is for school work, the rules say you have to put in effort, so with that in mind I'll throw the ball back into your court. Also, I'll notify this get moved to the homework section.
 
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  • #3
rude man
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Too bad no one picked up on this after February.
I caculated that it is not possible to adequately compensate the thermocouple:
Assume 1 deg C departure from the nominal +25C.
Then the thermocouple voltage drops by 39.4mV.
This drop must be made up for by the differential voltage between the RTD voltage and the R1-R2 divider voltage.
But the change in the RTD resistance is only ## .00385 \Omega/C \times 100 \Omega = 0.385 \Omega/C. ##
To generate the needed 39.4 mV we would need a current ## i ## thru the RTD such that ## 0.385 i = 0.039 \rightarrow i = 101mA. ##
But that is impossible with a 5V reference voltage into ## 104 \Omega. ##


Hope someone else will contribute.
 
  • #4
sophiecentaur
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Now, I'm not entirely sure of what I'm supposed to do to compensate the circuit.
I don't use thermocouples but I find the term "compensate" a bit confusing. {Edit - or perhaps not} The (isothermal cold junction block) block with the thermocouple attached seems to be a reference junction with its temperature measured with the thermocouple. The resistance of the RTD should be 100R at 25C and vary in resistance so that the voltage across AB is only the thermocouple voltage. So what @rude man calculated should be right BUT. . . . .
But that is impossible with a 5V reference voltage into

The data sheets for K type thermocouples seem to have coefficients in Micro, not Milli volts per degree. An easy typo but very relevant if you want the actual answer. That could be the reason for your "impossible" result.
 
  • #5
Lord Jestocost
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The cold end of a thermocouple at a voltage meter should be at ##T=0°C## (the reference point) to use the established thermoelectric voltage tables or polynomials to determine the temperature ##T_2## at the hot end (see circuit in post#1). As this is generally not the case, the circuit in post#1 serves as a “cold junction compensation”; the “missing” thermoelectric voltage is compensated in case ##T\neq 0°C##.
 
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  • #6
Averagesupernova
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The data sheets for K type thermocouples seem to have coefficients in Micro, not Milli volts per degree.
Hmm. I missed that too. I've worked with K type thermocouples enough to feel quite silly about this. A hundred mA through the RTD just didn't sound right. This is why.
 
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  • #7
rude man
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The data sheets for K type thermocouples seem to have coefficients in Micro, not Milli volts per degree. An easy typo but very relevant if you want the actual answer. That could be the reason for your "impossible" result.
I suspected as much but didn't bother to check. Thanks for so doing.
(Even then this circuit compensates to 1st order only but if T is held reasonably stably then it's a good way to go.)
 
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rude man
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The (isothermal cold junction block) block with the thermocouple attached seems to be a reference junction with its temperature measured with the thermocouple.
Spg. of typos ... :smile:
 
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  • #9
sophiecentaur
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Hmm. I missed that too. I've worked with K type thermocouples enough to feel quite silly about this. A hundred mA through the RTD just didn't sound right. This is why.
Micro schmicro - who cares. :wink:
It's usually me who is the sloppy one in these matters.
 

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