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AHA..gotcha

Not really, just wanted to lighten it up a bit, that last post to you gave me a headache..

Read my post to you above, then this one.

""which suggests that the net effect should be zero for AC current. The junction is heated on one half-cycle, and cooled on the other.""peter.

Because of the delta T, the carnot efficiency will be low..and it's Seebeck effect for the opposite way..

I think the most important thing is that because the carnot efficiency is so low, there will be energy lost to the conductor, which will not return..

And because the energy conversion goes as current, the transfer occurs in a non linear fashion..It's that non linear part that got me thinking about the applicability to JC's results.


""However, you're saying that the transfer function is changed during each half cycle, which is a cause of non-linear behavior.""peter

Yes, I'm saying that it is a combination of Seebeck, peltier, and thermal conductivity coupled with finite heatflow speed away from the junction that gives the non linear behaviour.

""In that case, shouldn't the non-linear component increase with increasing signal amplitude? As the current increases, the total heat increases before reversing (the difference in the Peltier coefficients times the current integrated over the half cycle)""peter

Peltier and seebeck coefficients with carnot efficiency..

And don't forget, the levels of energy loss we are talking about are actually small, as is the joule heating loss..But, as the levels increase, the joule loss increases faster than the peltier/seebeck effect, so the effect gets smaller as the drive goes up..

""According to John, the effects he's seeing disappear at higher amplitudes, more akin to a crossover distortion effect.""peter

I know..that's why I've given so much thought to thermoelectric type effects and how they would have to fit into what is observed. It would be silly for me to try to fit it otherwise..

I visualize the problem as a pair of antiparallel diodes in series with a 1 Kohm resistor..at milliamp levels, the diode dominates (say, half a volt plus 1 millivolt of resistor drop)..but at one ampere, the diodes disappear (actually, about one volt out of one thousand)

It's good to actually discuss the technicals for a change. Thanks

Cheers, John




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  • AHA..gotcha - jneutron 11:58:44 08/25/03 (0)


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