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First of all, in considering thermoelectric effects, we have to realize that we 
are dealing with a non-equilibrium situation.  |  
 
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A general theory of non-equilibrium is beyond our means, suffice it to say that 
Lars Onsager, with a paper entitled "Reciprocal 
relations in irreversible processes" induced some fundamental 
insights as late as 1930; he received the Nobel price for his contribution to non-equilibrium thermodynamics in 1968 - for chemistry, of all things. |  
 
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However, what we should be aware of, is the essential statement of non-equilibrium 
theory: |  
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As long as there is no equilibrium, we always have currents 
of something trying to establish equilibrium by reducing a gradient in something else that is the actual cause of the non-equilibrium. 
A gradient in the electrical potential, e.g., causes our well-known electrical currents, and a gradient in a concentration 
causes diffusion currents. |  
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But we must abstract even more, and consider things like entropy currents as well as all kinds 
of combinations of gradients and currents. |  
 
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While Onsager discovered some quite general relations between gradients and currents, 
we will not delve into details here, but only look a bit more closely at what causes the Seebeck effect. |  
 
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For that, we still treat the thermoelectric effect with equilibrium thermodynamics, 
simply assuming that locally we are not very far from equilibrium and thus can still 
use band structure models with a Fermi energy (which is only a well defined quantity for equilibrium) and resulting carrier 
distributions. |  
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In the simplest possible case, what we will get for a long bar of metal, hot 
at one end and cold a the other, is something like this: |  
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At the hot end, the Fermi distribution is "soft", and we have a noticeable 
concentration of electrons well above the Fermi energy. At the cold end, the Fermi distribution is sharp, and we have fewer 
electrons above the Fermi energy. |  
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The drawing, of course, grossly exaggerates the real situation. Note also that the total concentration 
of electrons a both ends is the same - even so the drawing does not show this because the holes below the Fermi energy are 
not included. |  
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Note too, that the Fermi energy is constant throughout the material (we neglect any possible 
effects of the temperature on the Fermi energy, as we have it, for example, in doped 
semiconductors). |  
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As always, electrons go to where the energy is lower; the electrons would tend to move from 
the hot end to the cold end, thereby transporting energy and thus equilibrating the temperature eventually. Equilibrium, 
with a constant temperature everywhere will be achieved. |  
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An equally valid alternative interpretation just looks at the concentration gradient of the 
electrons in energy space, which would automatically drive a kind diffusion current until the concentration (and thus the 
temperature) is equalized. |  
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Yet another way of looking at it is to consider that the average momentum of the electrons 
at the hot end is larger than that of the electrons at the cold end. They would therefore "run away" faster (taking 
energy with them) than the electrons from the other end would "run in". |  
 
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However, since we keep the temperature difference 
constant, all this cannot happen. We will have to maintain constant but different temperatures 
and therefore different energy distributions at both ends of the metal bar. |  
 
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If nothing happens, we will loose the electrons with large momentum faster than we gain electrons 
with smaller momentum; and a temperature gradient cannot be maintained. The only way to change that, is to lower the potential 
at the hot end somewhat, i.e. make the ends positively charged, and to raise it at the cold end. |  
 
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The potential difference must built up until it is large enough to exactly counteract the 
net loss of "hot" electrons due to momentum imbalance. |  
 
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This is essentially the reason why we find a thermovoltage. 
 |  
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Note that the junction is not directly essential. However, if you just plug a wire from one 
material into your Voltmeter and heat up the middle part, leaving the two ends cold (and at the same temperature), your 
potential along the wire may change, but at the two ends you have the same potential, and it is the potential difference 
between the two ends you measure |  
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Plucking the hot end into your Voltmeter is a bit unpractical, so you necessarily end up with 
a junction to some other material. The other material now will also have a hot end and a cold end, and thus develop a potential 
difference. |  
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Since the potential at the other end can only have one 
value, you will now get a potential difference between the two cold ends which depends, 
of course, somehow on the choice of materials. |  
 
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Still, there is a potential difference between the hot and cold end of one piece 
of material, and even so it cannot be measured directly, we can measure it indirectly somehow and tabulate the values. |  
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We can do this, with somewhat more involved but similar reasoning not only for metals, but 
also for semiconductors. The table below gives some absolute values and shows that semiconductors are good candidates for 
actual thermocouples, because their Seebeck voltage is fairly large. The values are 
for about room temperature, or about 700 oC for the last three materials |  
 
 |   |   
  | Material |  Al |  
Cu |  Ag |  W |  
(Bi,Sb)2Te3 |  
Bi2(Te,Se)3 |  ZnSb |  
InSb |  Ge |  TiO2 |     
Seebeck voltage [µV/K]  (Vhot - Vcold) |  
-0,20 |  +3,98 |  
+3,68 |  +5,0 |  +195 |  
-210 |  +220 |  -130 |  
-210 |  -200 |    
 
  |  
 
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Surprise! There are positive (as expected) 
and negative values of the voltage. What does it mean? |  
 
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Simply that you are looking at positively charged carriers 
being reponsible for the Seebeck effect - holes, in other words. |  
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Not so surprising for semiconductors, perhaps, but somewhat unexpected for Al. But, 
as we should know, conduction in Al relies heavily on holes, as evidenced, e.g., in its 
positive Hall coefficient while most other metals have a negative one |  
 
 |   |    |  
 
 
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Understanding qualitatively the Seebeck effect does not help much to understand 
the Peltier effect. |  
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Again, lets look at some simple junction, this time an ohmic conduct of a metal 
to a semiconductor. |  
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Shown is an equilibrium situation, where the Fermi energy is constant throughout, and the 
flow of electrons across the junction must be equal in both directions. Note that only the high-energy end electrons oft 
the Fermi distribution in the metal makes it across the junction, whereas all electrons of the semiconductor can flow into 
the metal. |  
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The electrons of the metal thus also transport some thermal energy out of the metal, but in 
equilibrium exactly the same amount is gained by the semiconductor electrons, which are high-energy electrons in the metal. |  
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Now consider some external voltage driving some net current through the junction 
in either direction. |  
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If this current is an electron current flowing from the metal into the semiconductor, it still transports some thermal energy out of the metal, but since it is now much larger than the 
electron current flowing back, we have a net transport of thermal energy out of the 
metal, which therefore must cool down. The semiconductor part heats up, of course. |  
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If the current is reversed, the flow of thermal energy reverses, too, and now the semiconductor 
part cools down. |  
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It is conceivable then (also far from clear) that the total effect in terms of temperature 
differences is proportional to the current I flowing. |  
 
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Note, however, that as a completely independent process, you always have ohmic 
heating (or Joule heating) which is simply given by the total power P dumped into the system via  |  
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with U = voltage applied, R = total series resistance of the system. |  
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Since this general heating of the whole device is proportional to I2, 
it can easily overwhelm any cooling effect that might be there. |  
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If you want to use the Peltier effect as an elegant way of cooling something, you must not 
only choose your materials very carefully, but also optimize your system design and working points. |  
 
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That this is possible is evidenced by the successful marketing of Peltier cooling elements, mostly for scientific applications. Here is a table with technical data from 
a major supplier (EURECA Messtechnik GmbH, Am Feldgarten 3 D-50769 Köln, GERMANY ) |  
 
 |   |    |  
  |   |  
   Micro Peltier Elements 
  | Module  |  
Imax   [A]   |  
Qmax   [W]   |  
Umax  [V]   |  
dTmax  [K]   |  
Dimensions  |  
Unit Price  [Euro]   |     
| A [mm]   |  B [mm]   |  
C [mm]   |  D [mm]   |  
H [mm]   |     
| TECM-4-4-1b/69  |  1,4  |  0,7  |  
0,9  |  69  |  4,3  |  4,3  |  
4,3  |  4,3  |  2,95  |   
 28,75  |     | TECM-4-5-1/67  |  
0,7  |  0,4  |  1,0  |  67  |  
3,4  |  3,4  |  3,4  |  5,0  |  
2,30  |    29,50  |     
| TECM-5-7-1/67  |  0,7  |  0,9  |  
2,2  |  67  |  5,0  |  5,0  |  
5,0  |  6,6  |  2,30  |   
 38,50  |     | TECM-7-8-2/67  |  
0,7  |  1,7  |  3,9  |  67  |  
6,6  |  6,6  |  6,6  |  8,3  |  
2,30  |    52,50  |     
| TECM-9-12-4/67  |  0,7  |  3,5  |  
8,0  |  67  |  9,1  |  9,9  |  
9,1  |  11,5  |  2,30  |   
 66,25  |     | TECM-12-6-4/69  |  
1,7  |  4,4  |  4,3  |  69  |  
6,0  |  12,0  |  6,0  |  12,0  |  
2,75  |    57,50  |    
 
   In this class you will find elements with various geometries and electrical parameters. For this reason, these elements 
are suited for very different and partly exotic requests as you have in the research. Our support will help you with the 
selection and the startup of the elements in consideration of your particular requests.   
  Generally you will receive these elements in small quantities from stock.      |  
 
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© H. Föll (Advanced Materials B, part 1 - script)