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Re: So let's double check this.....

""BUT, if the cable is symmetrical, each junction has a partner "reverse junction" at the other end of the cable. This junction exhibits exactly the same Peltier effect, but reversed.
This leads to a Seebeck effect potential that is exactly opposite the first junction.""Peter

And, I agree, the shield should have the exact same things happening..

What I'm interested in is the non perfect conversion from current to heat, and back again...As, that will cause an energy dissipation.

Peltier: Heat = (A-B) * I (A,B) being peltier coeff..

Heat can be + or -

So, one heats, the other cools. Note the amount of heating does not depend on temp.

Seebeck: V = (C-D) * (T2-T1) C,D seebeck coeff...T being temps

Now, since the temp diffs T2, T1 are so very small, the seebeck return is very small..Seebeck depends on temp diff, where peltier does not.

So, not too much is returned by seebeck..

Any that is, has to be the local stuff..Remember, some of the heat and cold (yah, I know) is travelling away from the junction never to return..so seebeck will not have all the energy available to return.

What still gets me is how the peltier effect is related only to current, whereas a resistor is current times voltage..I still think that will remove energy in a non linear way..and in the way JC has measured.

Cheers, John



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