If one had a power mosfet with really hunky source and drain connections, would the thermal conductivity from source to drain depend on the gate voltage?
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John Larkin Highland Technology Inc
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OK I'll play... but I know squat about power fets! So there is some channel resistance as a function of gate voltage. You could do some Wiedermann-Franz type law and estimate the thermal conductance of those electrons. I think it's something like a factor of 10^5 at room temperature. So 1 ohm (volt/amp) is ~10^-5 deg-K/watt. (That seems small?? Maybe I've got the numbers wrong.) Now if it's two dimensional rather than 3D then the ratio may be different.
What kind of numbers are you seeing? And what's the channel resistance?
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Most common metals seem to have about 180,000 K/W thermal resistance per ohm. But metallic conduction is clearly a special case: consider an insulator, like diamond maybe.
Haven't tried it yet. Got other things to do right now.
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John Larkin Highland Technology Inc
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Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
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Multichannel arbitrary waveform generators
I believe that thermal conductivity in metal is primarily through the free electrons, while thermal conductivity in insulators is carried by the lattice. Guessing further, I'll bet that the thermal conductivity of metals and similar materials is the thermal conductivity of the crystal lattice in parallel with the thermal conductivity of the free electrons, and that the free electrons dominate.
But I am just guessing here.
This is interesting:
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It mentions semiconductors, but I couldn't make out what it means in your proposed case.
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Oops I got the ratio upside down... What's a factor of 10^10 between friends? :^)
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Yup, electrons or lattice vibrations. Say I wonder if there's a speed of sound vs thermal conductivity ratio for insulators? (Well, for crystal insulators.)
George H.
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Where do you find the speed of sound in diamond, alumina, glass, AlN?
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John Larkin Highland Technology, Inc
jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com
Precision electronic instrumentation
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Mobility decreases with increasing temperature, hence the notion of no thermal runway with FETS. So the carriers don't carry heat, but do slam the lattice and each other due to the increased temperature. With conservation of energy, I suppose this translates to higher conductivity.
I would think it would have to do with the dissipation factor for the phonons (more == less conductivity), and the allowable energy levels for the phonons (more == more conductivity).
But I'm waaaay out on a limb, here.
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My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?
Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
In principle it should work, justified by the Wiedemann-Franz argument others have already stated. My colleaque Jan v d K had suggested the idea to me when we met in the WOLTE workshop a few years back.
I actually made an experiment some time ago, using an FDV301N in our sub-K fridge. My thermometry was lacking, however, so the result was inconclusive, and I haven't had an opportunity to re-try.
A crude thermal conductivity estimate for a 0.2mm x
0.2mm x 0.1mm Si chip is 15 mW/K edge-to-edge, and the electrical resistivity required for equal or better Wiedemann-Franz conductivity would be < 500 u-ohm at room temperature. Not likely for a FET on such a small substrate, but then again that is not so far, and the through-substrate thermal leak is probably much smaller, owing to various thermal on-chip interfaces. It would be simplest just to try.
The situation improves a lot at low temperatures, which I'm interested in. Phonon mediated thermal conductivity goes down ~T^3 but W-F only ~T, provided that the FET on-state resistance does not change (which seems roughly to be the case). Various Kapitza resistances complicate theoretical estimation so much that I decided just to try it in practice. But with an inconclusive result, so far.
Well I was reading in Kittel?s ?Intro to SS physics? last night (3rd ed.) (Is that cheating?) The thermal conductivity will go as,
K = 1/3 (heat capacity) x (velocity) x (mean free path)
If this link works you can find the pages on amazon... the 8th edition.
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6X#reader_047141526X
But things are more complicated than the ?simple? electron case. There?s this interesting question of how local thermo equilibrium is established. Apparently it was Rudolf Peierls who figured it out. The answer is that you have to have phonon scattering that also involves a reciprocal lattice vector. (The dreaded Umklapp process.)*
George H.
*Sorry this is solid state ?speak?. It just means scattering out of the first Brillion zone.
If there were a commercial need for thermal switching you could probably optimize the die for it. Heck, clever laser trimming may even do it, if you get a grant, or have a crying need.
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My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?
Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
I would think the answer is "some, but the S-D current flow would also be a factor. Delve into "Transistors and Active Circuits" by John G Linvill and James F Gibbons (McGraw Hill 1961).
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