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Are you sure you can use this Seebeck/Peltier cause and effect model?

we're jumping posts again....

you said:

Now, drop that current...(like an audio signal does twice frequency rate).. The already established heat gradient will now produce electricity, (well, at least according to Seebeck...but what does he know??))

Are you sure that's correct? I suspect the two effects exist in parallel at the same instant, rather than in a circular fashion.

When you "drop the current" and reverse it, you immediately begin to reverse the heat flow that happened previously, with a net of zero.

Also, the heat gradient has to result in a temperature gradient for Seebeck to apply. Given the high conductivity of these metals, the temperature gradient must be very small, since dQ/dt = -kdT/dx, right?

So, at 30mV signals into a 600ohm load, we have a current of 50uA. Signal freq = 5kHz, so one half cycle = 100uS

Total heating on one half cycle = integral of (diff in Peltier coeffs) x 50uA peak current over 100uS interval.

Temperature gradient dT/dx can now be determined from the heat flow (dQ/dt) and the materials' thermal conductivity.

Given that the thermopower coefficients of metals are down around 10-5 to 10-6 V/K, and we're depending on the heating caused by less than 10-9 AmpSeconds of signal....

Don't have enough data in front of me to complete the math, but these are looking like very, very small numbers......

Peter



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