# Thermocouple junctions

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I am learning about how thermocouples work. I am trying to understand why it is necessary to have a junction for you to accuratly measure a thermocouple. I understand from another topic that with any wire, when heated produces am emf due to energy being given to that particular section on the wire. Now, do you need the disimilar metal with a separate cold junction to produce what would be two separate emfs, and through some math be able to determine the temperature from that? What is it about the dissimilar metals with the extra cold juntion that makes it easier to measure?

• posted

Circuits involve a loop. How do you propose to connect your proposed homogenous wire with a temperature gradient to a voltmeter without effecting another temperature gradient which precisely cancels out the voltage resulting from the first one?

Best regards, Spehro Pefhany

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• posted

Any piece of conductor creates voltage when it passes through a temperature gradient. Unfortunately, you can't complete a circuit to measure that voltage, except by having another conductor also pass through that same gradient. If you complete the circuit with the same metal, you get two equal and opposite voltages that cancel, and you get nothing. But if you use two different metals to pass in and back out through the gradient, their different voltages produce some net total.

In order to have that voltage represent the total temperature change from a known temperature requires that that couple be in series with another to a known temperature (usually at ice temperature) or in series with a temperature controlled active voltage source that simulates this second couple.

• posted

Another way to look at it is that you need an asymmetry in the circuit. If you used a single metal type to make a wire loop, intended to measure temperature, what would be the polarity of the result? Why would one output be positive or negative relative to the other? OK, it wouldn't.

The thermocouple uses different metals so that when you create the voltage gradients as a result of temperature gradients, the two legs are different so you get something with a predictable voltage (and polarity!) that you can measure.

John

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So if I understand this right, the other conductor creates a different emf, and becuase of the difference a small current is allowed to flow?

• posted

So if I understand this right, the other conductor creates a different emf, and becuase of the difference a small current is allowed to flow?

• posted

Here's a type K thermocouple:

chromel ________________________ / B / T1 A * T2 \\ \\________________________ C

alumel

The wires are made of chromel and alumel, and are welded together at A. The junction is at temp T1 and the other ends are at T2. We call T2 the "reference junction temperature". Call the absolute circuit voltages A, B, and C.

Each wire generates a voltage that depends on the temperatures at T1 and at T2. These voltages are A:B and A:C. Note that these voltages depend on both T1 and T2, *not* (accurately) on the difference T1-T2.

For a given T1 and T2, the chromel generates some voltage A:B, and the alumel makes A:C, and we get to measure the difference B:C.

Now add a voltmeter, with copper-wire probes, to measure B:C...

chromel B ________________________ copper / *---------------+ / | T1 A * T2 meter \\ | \\________________________ | *---------------+ alumel C copper

Heat up junction A, and the voltmeter indicates positive voltage, around 40 microvolts per degree C.

We've created two new thermocouples * and *, but they cancel out, so we still measure voltage B-C.

We don't want current, usually, we just want to measure the voltage with the meter. If we know temperature T2 and the metered voltage, we can calculate T1. Thermocouple wire has high resistance relative to copper, so we usually want a very high impedance voltmeter to avoid loading the loop and making errors.

Sometimes you do want current, like for holding in the gas solenoid on a water heater, where the pilot light heats the thermocouple. But in that case you're not trying to measure temperature.

I'm writing this same stuff today, in a manual for a thermocouple simulator. Good practice.

John

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So the difference between ab & ac is what your looking for. The two produce two different voltages. The copper wire really has nothing to do with the circuit? So long as it is either kept in an ice bath as a reference or temperature compensated, because if you know what the voltage produced at the reference would be, then you can figure out the voltage per degree of the thermocouple would be from that reference?

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No. Ideally, the current approaches zero. A metal passing through a thermal gradient is a thermal battery (different metals are batteries that produce different millivolts per degree), but you get a most accurate measure of the difference between the two battery voltages at zero current (so wire resistance doesn't get involved).

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It completes the circuit to a voltage measurement. It will have no effect on the voltage, as long as the temperature is the same at both points where the copper connects to the thermocouple leads and also the same temperature (though possibly, a different temperature) at the far ends where the voltage measurement is made. That is, any voltage generated by one copper wire will be balanced by an equal and opposite voltage from the other copper wire, as long as the pairs of ends share temperatures.

There may be a second thermocouple in series with the measurement one that is kept at ice temperature, so you know what temperature to start with when you calculate the degrees from the volts. But many systems just measure the temperature of the terminals where the thermocouple wires connect to copper, and either generate a voltage based on that temperature (to simulate the voltage generated between an ice bath and the terminal temperature) or communicate this temperature to the measurement process so it can calculate the couple temperature starting at that temperature.

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