# Why 0V when thermocouples are placed in an ice bath

Discussion in 'Electronic Design' started by cheese9988, Mar 8, 2006.

1. ### cheese9988Guest

Why is it that when you place both junctions of a thermocouple in an
ice bath, the output is zero volts. I know it has something to do with
it canceling out but why? Wont the dissimilar metals still produce
different voltages even though they are at 0 deg c?

2. ### Guest

The voltage is produced at the junction between the dissimilar metals,
not by their individual bulk. So if your setup is

- A - B - A -

And both the AB and BA junctions are at the same temperature they
should produce the same magnitude of voltage. Since they are wired
opposite each other, the voltages in your circuit will be opposite and
cancel.

Ice bath isn't magic - for this purpose anything that insures
consistent temperature might work. But ice bath is a good reference to
leave one junction in and compare the other against, because as long as
the bath contains both ice and water it's temperature is going to
depend on the purity of the water and not much else (ambient temp,
pressure, etc)

There's also a nifty circuit out there that behaves like a reference
junction in a virtual ice bath. I forget how it works, but omega's
temperature measurement handbook might tell you.

3. ### cheese9988Guest

I tell you exactly what I was doing. I have a type s thermocouple I was
doing a quick reference check on. I hooked copper wires from a hp 3458,
which has copper inputs, I hooked the copper wires to the thermocouple
and placed the junction in the ice bath. I placed the thermocouple in
the ice bath. It read 6uV so I know its good. I'm just trying to
understand why. The thermocouple should produce its own emf of x,
having platinum and rodium. But wouldn't the platinum to copper and the
rodium to copper produce a different emf on each junction?

4. ### Spehro PefhanyGuest

The EMFs cancel out since the temperatures at all three junctions are
the same (well, your 6uV indicates an error of perhaps 1 degree C, a
mismatch with extension leadwire alloy, or some combination of the
two).

Best regards,
Spehro Pefhany

5. ### Glen WalpertGuest

Aack! The old thermocouple junction myth rears it's ugly head again!
You have this exactly backwards - thermocouple voltage is produced by
the temperature gradients in the bulk of the wires, not by the
junction. The junction merely provides an electrical connection which
allows the bulk gradients in between the measurement junction and
reference junction to be measured. In the above example with the
reference junction (the connection of the thermocouple wires to the
copper meter leads) at the same temperature as the measurement
junction (both in the ice bath), the net temperature gradient between
the junctions is zero degrees, regardless of any different
temperatures in the middle of any of the wires.

There is no known possible physical mechanism which can produce
thermocouple voltages at the junction. Voltage gradients and
temperature gradients are associated because the high thermal
conductivity of good electrical conductors is due to transport of high
energy electrons from the hotter side to the colder side, balanced by
lower energy electrons returning to hot side. Same current both
directions, but the hotter electrons "see" more resistance, hence
higher voltage drop in the hot to cold direction, amount differing

http://www.electronics-cooling.com/Resources/EC_Articles/JAN97/jan97_01.htm

http://www.chem.cornell.edu/fjd3/thermo/intro.html

http://uhavax.hartford.edu/~biomed/gateway/Thermocouple.html

6. ### Guest

Oops, you're right, and I actually had heard that before, but managed
to forget. It's crossing the thermal gradient with different metals
that gives rise to the difference...

In terms of what the correct theory clarifies that the flawed one
misses, the most obvious practical implication would seem to be that
you can use a mix of metals in your voltmeter as long as the meter
comes close enough to being at a uniform (if arbitrary) temperature
throughout - or do I have that wrong as well?

7. ### John PopelishGuest

The junction of dissimilar metals produces no voltage, regardless of
the temperature of the junction. It is temperature change along the
length of a conductor that produces voltage, and each alloy has a
different amount per degree, and a different curve of voltage per
degree versus temperature. So any number of junctions of any
combination of alloys produces no voltage, as long as the whole thing
is at the same temperature.

8. ### Robert BaerGuest

Not quite correct; the platinum to copper connection was "at" room
temperature, and the rodium to copper connection was "at" room
temperature - the temperatures could be reasonably different, but fairly
"close" to each other.
Only the platinum to rodium connection was in the ice bath (*read*
what he said).

9. ### Spehro PefhanyGuest

Your interpretation is incorrect, or the measured voltage would have
been about 115uV, not 6uV. He clarified in his second post that all
three junctions were in the ice bath when the voltage measurement was

"I hooked the copper wires to the thermocouple and placed the junction
(sic) in the ice bath. I placed the thermocouple in the ice bath."

Best regards,
Spehro Pefhany

10. ### Glen WalpertGuest

You have that correct; no temp gradient, no voltage. Of course it is
difficult to find/build a decent meter with no power dissipation,
which would be required for no temperature gradients, so in practice
you try to make sure that temperature gradients are as small as
practical and occur in the same materials so they cancel.

11. ### cheese9988Guest

But why is there a voltage difference then when you heat the ends. Is
it because a temperature gradient in the middle would null itself out?

12. ### John PopelishGuest

Because the two different metals passing from the junction temperature
to some other common temperature (where the temperature is measured)
generate different potentials when passing between those two
temperatures.

To have a calibrated temperature measurement that ignores the
temperature at the meter, you subtract the difference produced when a
similar couple passes from the reference ice temperature to the meter
temperature.