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The accuracy of a measurement system can be significantly reduced by loading effects in the system. A thermocouple is used together with an indicator to measure temperatures. The thermocouple has a sensitivity of 50µV/°C and an output resistance of 500Ω. The indicator gives a measured temperature of \$T_M = 20V_{Load} K\$ and has an input resistance of 10 kΩ.

  1. How do you sketch this system using Thevenin equivalent circuits? (I'm thinking something like this, but I'm not sure.)

schematic

simulate this circuit – Schematic created using CircuitLab

  1. How do you determine the accuracy of this measurement system? I'm thinking I need to calculate \${T_M}/T\$, but I don't know how to determine the voltage across the load (\$V_{Load}\$).
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  • \$\begingroup\$ \$T\$ is the true temperature, and \$T_M = 20*V_{Load}K \$ is the measured temperature by the indicator. This will be significantly reduced by the effects of the load, with resistance \$R_{Load}=10kΩ\$ and where the voltage across the load is given by \$V_{Load}=V_{out}{R_{Load}/({R_{Load}+R_{out})}}\$. \$\endgroup\$ – JompaTormann Oct 22 '18 at 12:09
  • \$\begingroup\$ You are missing cold junction compensation for the thermo couple. Without it this will greatly decrease the accuracy. \$\endgroup\$ – vini_i Oct 22 '18 at 12:49
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    \$\begingroup\$ Welcome to EE.SE! This appears to be a homework question. As such, you need to show us your work so far, and explain which part of the question you're having trouble with. For future reference: Homework questions on EE.SE enjoy/suffer a special treatment. We don't provide complete answers, we only provide hints or Socratic questions, and only when you have demonstrated sufficient effort of your own. Otherwise, we would be doing you a disservice, and getting swamped by homework questions at the same time. See also here. \$\endgroup\$ – Dave Tweed Oct 22 '18 at 13:11
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In this problem you are asked to consider an old-fashioned (D'Arsonval, perhaps) mechanical meter movement attached to a thermocouple. (as an aside, with a bit of bimetal or a manual trim to do the cold-junction twiddle that was once a common type of thermocouple meter- aka "pyrometer").

The meter is fitted with a scale that indicates 20,000 degrees per volt of input. So it would read exactly correctly (assuming the 50uV/°C is correct) if the meter did not load the thermocouple.

In answer to your questions, you have the equivalent circuit correct. The error will be proportional to the temperature reading in this simplified problem, so Tm/T is a reasonable way to describe it.

The voltage across the meter is a simple voltage divider, which you should be able to calculate. By inspection the error is of the order of -5% (but not exactly -5%). I'll leave that to you.


In actual practice, this type of thermocouple meter would be designed for a specific resistance of probe + wiring (often something more like 10 ohms), and that requirement would be marked on the meter face. So a known amount of loading would be compensated for in advance via the factory calibration.

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Two hints:

  • A voltage source and two resistors is called a "voltage divider". Calculation is straightforward.

  • \$K\$ is a "calibration constant". You adjust its value until the reading matches the temperature exactly.

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