I am about to design an amplifier to J | K Thermocouples, but I have been unable to find a trustable equivalent circuit|model|specs for them.

I know a J Type Thermocouple has some general specs, such as, Iron-Constantan Junction|Wires, -210 to 1200 C Temperature Range, 50-58 uV/C Sensitivity, -11 to 60 mV Voltage Range.

Thermoelectric Effect expressions for Seebeck-Peltier-Thompson's roughly don't (or do?) give us clues about how to made an equivalent model far from its simplest equivalent circuit as a voltage source with a resistor.

But I would been unable to find other specs. In particular, I am looking for the following figures, or at least some reference|paper about them:

  • The maximum current one could draw from it (like an equivalent internal resistance?),
  • How it discharges with a resistive load (like a capacitor in parallel with a voltage source?)
  • The operating (recommended?) current under which it could operate without deviating (too much?) their "properties" (repeatability? sensitivity?),
  • The maximum current it could sustain without losing these "properties" permanently (solely through Joule Effect?).

2 Answers 2


You can consider a thermocouple in its usual measurement role as a voltage source with a series resistance. The open-circuit voltage is a nonlinear function of two temperatures, the so-called cold junction(s) and the measuring junction. It can be expressed in terms of a nonlinear function with one parameter.

The series resistance is a function of the temperature along the wires but usually low enough it can be ignored. You can calculate the loop resistance by the usual formula R= \$\rho L \over A\$ for the two metals, or simply look it up in a table of thermocouple wire or extension leadwire from the material type, gauge and length. Extension leadwire may be made of a similar alloy to the thermocouple itself in the case of base-metal thermocouples such as J and K (Iron-Constantan and Chromel-Alumel) E and N, obviously that is a much less attractive proposition in the case of types R, S and B, and for different reasons, W, so the thermoelectric characteristics are approximated with different alloys (that are unsuitable for the measuring junction) for the extension leadwire.

The engineer designing the installation may choose a larger gauge of extension leadwire if the runs are long, in order to keep the resistance relatively low. If everything is within a few meters it usually doesn't matter.

The resistance is usually < 100 ohms, often closer to 10 ohms. Usually measuring and controlling instruments introduce a small current into the thermocouple to detect a broken sensor, maybe a few hundred nA. Say 200nA and a 20 ohm resistance will cause the sensor to read about 0.1°C high for a type K thermocouple. That offset will increase somewhat with temperature as the resistivity of the metals increases.

If the thermocouple junction is grounded (which is preferable in many situations) your measuring circuit should accommodate that using a galvanically isolated or differential front end.

In the old days, thermocouples were made with a defined resistance and fed a meter movement in the pyrometer (with some additional complexity we don't need to go into here), while that may still be true in a few cases, it's rare on industrial instruments. You should just assume the sensor has series resistance 0 < Rtc < 100\$\Omega\$ (in most cases).

To answer your questions:

  1. Assume it has Rtc < 100 ohms or whatever makes sense if your situation is not typical. The more current you draw (or bias it with) the more the offset or error. How much error is too much? Up to you.
  2. It behaves like a low-value resistor for practical purposes here.
  3. See 1. the more current you draw or bias, the more offset (and potentially error). Since the resistance varies with the temperature along the leadwire (not just at the measuring junction) any offset will vary with things that are usually beyond your control.
  4. You can short the thermocouple with no damage. You can even force enough current through it that it glows. As long as the alloys don't change there is little effect on the accuracy.
  • \$\begingroup\$ Thanks for your answer and advices 1) Should we assume that, even given a constant temperature gradient (unlimited heat supply), the TC conductors are not able to convert that heat into current at a fast rate, and that imposes a current limit, or that conversion rate is simply too small that in practice this limit goes into the pA range (or lower?). 2) Ok 3) Ok 4) Ok. As long as the alloys and adherence of the TC is not changed, there should be little effect, and hence this value should attend to Joule effect. \$\endgroup\$
    – Brethlosze
    Commented Apr 7, 2020 at 13:57
  • 1
    \$\begingroup\$ The current limit is from the resistance. If you had massive thick conductors you could have arbitrarily large current, but the heat loss would increase. Simple gas safety systems have a thermocouple or thermopile that directly actuates a magnetic gas valve from the heat of the flame. Pilot goes out, gas shuts off and no unwanted earth-shattering explosion. You can operate a fan directly from a thermopile (there are commercial stove-top devices). And there are nuclear powered electricity sources that use a thermopile for space and the remote arctic (RTGs). \$\endgroup\$ Commented Apr 7, 2020 at 14:11
  • 1
    \$\begingroup\$ But to put some numbers to this, a type K thermocouple with reference junctions at 20° and hot junction at 220°C and resistance of 10 \$\Omega\$ will have an open-circuit voltage of 10.87mV and a short-circuit current of 1.087mA. \$\endgroup\$ Commented Apr 7, 2020 at 14:16
  • 1
    \$\begingroup\$ There is a simplifying assumption there that the temperatures are held constant regardless of heat loss. That ignores the Peltier effect as well as the loss down the wires. \$\endgroup\$ Commented Apr 7, 2020 at 14:21

1) The maximum current one could draw from it (like an equivalent internal resistance?)

A thermocouple, at least a measurement thermocouple, is designed to be used voltage output, that is, unloaded, with a zero output current. The themocouple wires have fairly high resistance (at least compared to copper), and it is expected that for remote measurements, you'll add extra length to the wires with extender leads made of the same thermocouple materials. The total resistance of the thermocpouple is therefore not only undefined, it's defined to be undefined. Any current drawn from it will result in an unknown scale error. The more current you draw, the larger the error.

If you have one specific thermocouple, and you're happy to calibrate the scale error, then draw a current. But if you replace it, you'll have to recalibrate, if you extend the wires, you'll have to recalibrate. A significant 'property' of a voltage-sensed thermocouple is that it can be replaced and extended with no calibration required.

2) How it discharges with a resistive load (like a capacitor in parallel with a voltage source?)

It is a resistor, with negligible stray capacitance to ground.

3) The operating (recommended?) current under which it could operate without deviating (too much?) their "properties" (repeatability? sensitivity?),


4) The maximum current it could sustain without losing these "properties" permanently (solely through Joule Effect?)


  • \$\begingroup\$ Thanks for your answer 1) ok. 2) A resistor plus a voltage source, such as in the included reference? 3,4) As per Seebeck, there should be a non zero, probably in the pA, current range, for which, given a constant source of temperature gradient, that current can be delivered. Perhaps depending on the size of the junction and the cable section. 4) As per Peltier, inversely as 3. I would also expect a non zero current as maximum. Also as a load through Joule i would expect the TC not to burn under some nonzero level of power dissipation. \$\endgroup\$
    – Brethlosze
    Commented Apr 7, 2020 at 13:48

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