Newsgroup: it.scienza.fisica
Titolo: Vorrei opinioni su msg da sci.engr.semiconductors [inglese]


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 sci.physics:96911 sci.engr.semiconductors:3418
     Poole-Frenkel ionization can defeat the 2nd law of thermodynamics
             Gerhard Kainz E-Mail: 
gerhar...@siemens.at
Abstract - Poole-Frenkel ionization of charges is due to electric-field-
enhanced thermal excitation of trapped electrons into the conduction band
and holes into the valence band. This means that there are _more_ mobile
charges in the conduction and valence band than the fermi-dirac would
determine. If this happens inside a p-n junction (e.g. through a special
doping of deep impurities) these additional charges will be separated
through the internal electric field and produce an electric potential, like
a solar cell. However in this case, more donors and acceptors are ionized
than the fermi-dirac distribution would determine. So the donors and
acceptors try to recombine with a charge, which means that there are more
generations than recombinations in this region. So heat energy (without a
temperature difference) will be converted in electric energy.
The fermi-dirac equation specifies the distribution of electrons in every
energy level at a certain temperature. According to this distribution the
donors and acceptors in a semiconductor are only partly ionized - the
higher the temperature the more are ionized. If this fraction is disturbed
in whatever direction, the charges will try to establish the normal
distribution.
However this distribution can be changed by a static electric field - the
Poole-Frenkel effect. J. I. Pankove writes in "Optical Processes in
Semiconductors":
"An electric field can exert a stronger force on the electron than the
local forces binding the electron to an impurity or to an excitonic state.
In this case, the center is ionized and the carrier is free to move in the
appropriate band. If donors are ionized, free electrons appear in the
conduction band. If acceptors and ionized by the field, free holes appear
in the valence band. The ionization of excitons produce both free holes and
free electrons. The free carriers are subsequently accelerated by the
applied field. The resulting high kinetic energies allow the "hot" carriers
to interact with the lattice ..."
So there are two important facts for the Poole-Frenkel effect:
1) There are more donors and acceptors ionized than it would be
determinated by the Fermi-Dirac distribution. So this part behave like it
would have a higher temperature, since there are more free charges than the
intrinsic number. In other words, the charges have a "higher temperature".
2) The additional charges will be separated by the electric field.
How can this be used to convert heat into electric energy?
There is a e.g. a p-i-n device, where the small i-region is doped
symmetrically with both deep level donors and acceptors.
..      p-region                     i-region             n-region
..                            with deep impurities
..   +-----------------*********************************-----------------+
..   |     -   -   -   *                               *   +   +   +     |
..   |       -   -   - *         <--- + - --->         * +   +   +       |
..   |     -   -   -   *   Poole-Frenkel effect:       *   +   +   +     |
..   |       -   -   - *   additional donors and       * +   +   +       |
..   |     -   -   -   *   acceptors are ionized, so   *   +   +   +     |
..   |       -   -   - *   more charges are generated  * +   +   +       |
..   |     -   -   -   *                               *   +   +   +     |
..   +-----------------*********************************-----------------+
So in the p and n region almost donors and acceptors are ionized, however
in the i region there are only a few according to the fermi-dirac
distribution. However, because of the electric field, additional donors and
acceptors ionized, these charges will be separated and flow in opposite
directions in the p- and n-region and produce an electric potential like in
a solar cell.
In this situation, there happens an interesting thing in the i-region. The
donors try to establish the normal fraction of ionized impurities e.g.
through absorbing an electron of the conduction band or through electrons
which move "up" from the valence band through thermal motion. This can only
be done, if there is one generation of an electron-hole pair more than a
recombination. The same happens with the acceptors.
And this try to restore the fermi-dirac distribution enables again the
electric field to ionize the donors and acceptors.
So every time the higher amount of generations of electron-hole pairs
absorb kinetic energy and therefore heat of the lattice - thus the junction
will cool down a little bit. This energy will be used to create additional
charges which will be separated in order to generate an electric potential
and electric energy. So the electric field is only the initiator, but does
not give any energy to the charges, it is therefore just a "catalyst".
It is logical that the higher the electric field, the more additional
impurities can be ionized. This can be done, if the i-region is very small.
However any electric field, also a small one, ionizes impurities and so
should have an effect. This effect can be enhanced, if more of this
circuits are put together.
This situation using the Poole-Frenkel effect will contradict the 2nd law of
thermodynamics since heat without a temperature difference will be
converted in electric energy. However this does not contradict the 1st law
of thermodynamics, since the whole energy will be conserved, only the
distribution will be changed.
--- Gerhard Kainz

 wrote :
| This situation using the Poole-Frenkel effect will contradict the 2nd law
of
| thermodynamics since heat without a temperature difference will be
| converted in electric energy. However this does not contradict the 1st
law
| of thermodynamics, since the whole energy will be conserved, only the
| distribution will be changed.
Ad occhio, prima che l'effetto vada a regime, il silicio fonde, o almeno si
altera la struttura cristallina che da' la semiconduttivita', aumentandone
resistenza e di riflesso la temperatura, impedendo per di piu' il passagio
di cariche. Di conseguenza serve un campo + forte, e vai in feedback.
Ma non ho le equazioni di F-P sotto mano, ne' so come dovrebbe essere
drogato il silicio secondo l'articolo