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In Reply to: Who's arguing? I'm being bashed for trying to explain. posted by jneutron on August 25, 2003 at 11:21:13:
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
Follow Ups:
During the rise of the signal, the conversion is Peltier effect, setting up the thermal gradient..in a non linear current related fashion..After the peak, when the signal is dropping, energy will return as a result of the Seebeck equation. But, not at 100%, due to all the equation crap..(got tired of the verbage).
Then, the other half cycle does the same, albeit the heat flow and voltage swap polarity.
Then toss in the other confounding influence garbage...heat capacity, conductivity, plating integrity....major headache..
Hell, it's easier soldering a thousand joints than it would be figuring it out..
You said difference in peltier coefficients...The peltier efficiency will be a function of slew rate..I've no idea how to generalize or integrate the time dependent efficiency to develop a simple coefficient.
""Don't have enough data in front of me to complete the math, but these are looking like very, very small numbers......""peterIndeed, they are. But so is 120 db below 30mV.
""I suspect the two effects exist in parallel at the same instant,""peter
Yes, they do..but for peltier, the highest efficiency occurs when the sides are the same temp, or the reverse temp, while the seebeck efficiency is the absolute worst when there is almost no gradient. So they don't exactly balance out.
What strikes me the most is the current dependence for heatflow, and not current times voltage of joule heating..
Now, to really open your eyes, Peter...
All along, I asked for technical discussion about the POSSIBILITY OF THIS EFFECT HAVING SIGNIFICANCE for the measured results of JC's..
I did not state that it is what is causing it...I asked if it was possible...to that I got "bull shit".
You are engaging in the technical aspects, which is a pleasant change from what JC did. Thank you ..you are at least willing to entertain those aspects without glibly dissing me..
Cheers, John
Comments:I'm not tracking with your assertion on the other thread that the effect goes away as the amplitude goes up. Shouldn't this result in a greater assymmetrical heating effect during each half cycle, and therefore increased non-linearity?
OK, back to the calcs (check these and tell me if I screwed up):
120dB below 30mV is 3x10 -8 V, so we're looking for a number in that ballpark.
In JC's test, the heat current peaks at (diff in Peltier coeffs) x 50uA, right? I can't find the Peltier coeffs, but they are related to the Seebeck coef by Seebeck = Peltier/T, so at 300K the Peltier coeffs should range from 3x10 -3 to 3 x 10 -4 . If you assume worst case diff is 3x10 -3 , then the heat current is
1.5x10 -7 Watts. Probably much lower....If the junction is 1mm square, the area is 10 -6 m 2 .
So, we have:
1.5x10 -7 Watts = 400W/mK x 10 -6 m 2 x (temp gradient)
(using 400 W/mK for the approximate thermal conducivity of CU and Ag at 300K)So, temp gradient is 1.5/4 x 10 -3 = 3.75 x 10 -4 K/m.
I'm stopping here for now...
If you estimated the thicknees of the boundary over which this gradient occurs, that would give you a temp diff, which you can use to calc the Seebeck effect.
I still think we will come in way below JC's numbers, but I'm not ruling it out just yet...
Peter
I'm at home, no data here..You said Cu at 400w/mk. there's gotta be a length in there somewhere..Cu is 10.2 watts/inch-degree C. (sorry about the units, they say memory is the second thing to go)
Assuming the effect is real...It gets kinda tuff calculating it, because the diffusivity of the materials comes into play..as the gradient formed at audio frequencies is heavily dependent on the speed at which generated heat flows, both away from and towards the junction..so I can't readily calculate the zone or boundaries over which the heat can spread..
I'm not sure I can calculate with any accuracy the total effect, if I include Seebeck, Peltier, Carnot, and Joule..
""I'm not tracking with your assertion on the other thread that the effect goes away as the amplitude goes up. Shouldn't this result in a greater assymmetrical heating effect during each half cycle, and therefore increased non-linearity?""peter..
I'm not sure..that's why I asked my initial question..
But, consider the effect conversion efficiency..When the hot to cold difference is greater, the peltier efficiency goes down..So, for higher amplitudes, less conversion occurs. Yes, for Seebeck, it goes up with difference, but for the lengths of material on either side of the junction being discussed, thermal conductivity plays the great dissipator role.
That modulation of Peltier efficiency will happen at audio rates very near the junction (remember I Wag'd about 10 to 100 microinches??).
Those distances are verifiable for silicon, just look at the transient thetaJc stuff in the IR hexfet books...
Hmmmm..consider this...Assume a junction...Then, instantly apply lots of current....
Immediately, the peltier efficiency is some value (nano time frame).
As heat is absorbed(gradient forming), the peltier efficiency starts to go down.
As the structure continues to settle in, the peltier eff. still goes down..
Eventually, the eff. will stabilize at some structurally related number..But, before that happens, it is dominated by diffusivity of the materials. Before it stabilizes, it is horribly non linear..
But the efficiency is actually tracking in some fashion the slew rate of the current...which is the derivative of the voltage.
So, the dissipative loss is a function of the current slew rate, and that is quite non linear..
If we can really calculate the possible peltier/seebeck losses, we have to compare them to the joule losses of the IC..and those are definitely low..30mV/10Kohm/milliohms....
I wonder if the thermoelectric conversion losses are really that low compared to the resistive losses of an IC..in series with the load resistance of 10K..
You said Cu at 400w/mk. there's gotta be a length in there somewhere..Cu is 10.2 watts/inch-degree C. (sorry about the units, they say memory is the second thing to go)I found a couple of references to this number:
4 Watts per cm per degree C, which agrees with your 10.2 watts per inch per degree. Converting to MKS should give 400W per meter per Kelvin, unless I got the conversion the wrong way round and it's 0.04 W per meter per K....
Let me think this through....
The equation is actually something like:
dQ/dt (watts) = -k dT/dx (degrees/meter)
(the whole assumes unit area, BTW...)since dT/dx is degrees/distance, I think k has to be watts per degree per distance to be dimensionally correct....
Example: a 1 degree per centimeter gradient = a 100 degree per meter gradient, so....
I GOT IT THE WRONG WAY ROUND.... the thermal conductivity if Cu is 0.04W/m/K. Damn! So what effect does that have.....
The right calc is:
1.5x10 -7 Watts = 0.04W/mK x 10 -6 m2 x (temp gradient)
(using 0.04W/mK for the approximate thermal conducivity of CU and Ag at 300K)So, temp gradient is 1.5/4 x 10 = 3.75K/m .
Ok, now I need to go think about your Seebeck effect theory here...
Maybe tomorrow!
Peter
Again, havin fun with the subject line....When you said 400 w/mK, I thought you meant millikelvin...sorry..the guys here use millikelvin numbers in the dewars..ya had me major confused..that and a martini..
Ya gotta admit...it is interesting, isn't it??
Tomorrow, I gotta figure out how an optical encoder at 216,000 counts per revolution is losing 7.5 counts per turn..the magnetic analysis guy is roasting me cause of some harmonics on a serpentine quad I made using a computer thingy.....
I'm running out of things to test to find the drift error..pissin me off..big time..It must be the mechanical engineer's fault..it's always their fault...:-)
Peter, it's been a pleasure..later..
Cheers, John
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