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Why 0V when thermocouples are placed in an ice bath

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

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  1. cheese9988

    cheese9988 Guest

    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. cheese9988

    cheese9988 Guest

    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. 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 Walpert

    Glen Walpert Guest

    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
    between metals but not direction. See also Wiedeman-Franz law.

    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. 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 Baer

    Robert Baer Guest

    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. 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
    made:

    "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 Walpert

    Glen Walpert Guest

    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. cheese9988

    cheese9988 Guest

    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. 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.
     
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