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Apr 26, 2000, 3:00:00 AM4/26/00

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Heterojunctions can convert heat into electric energy

Gerhard Kainz E-Mail: gerhar...@siemens.at

Gerhard Kainz E-Mail: gerhar...@siemens.at

This paper shows that a special semiconductor heterojunction acts like a

homogenous semiconductors which is illuminated: Both produce additional

electron-hole pairs which can be separated and so produce electric energy.

However, while a normal solar cell needs light as energy source, a special

kind of heterojunctions just absorbs heat energy of the lattice in order to

produce electric energy. Most importantly, there is no need to have a

temperature difference. This is shown by only using the common drift-

diffusion equation. Additionally, this effect can be simulated by

"SimWindows", a Semiconductor Device Simulator from the University of

Colorado.

Fig.1 shows a homogenous semiconductor, which is illuminated only on the

left part. Photons generate additional electron-hole pairs. Because of

these additional charges, the charge concentration is higher than the

intrinsic charge concentration. Therefore these charges diffuse in the

right part of the semiconductor, where they recombine after a certain time.

. photons

. |

. ********|*******************************************

. * \|/ *

. * x ----> e -----> e ------> x *

. * ----> h -----> h ------> *

. * *

. ****************************************************

. generation diffusion recombination

.

Fig. 1: A homogenous semiconductor is illuminated on the left part.

Electron-hole pairs are generated and this additional charges diffuse to

the right part, there they recombine.

This kind of diffusion is the base of the photovoltaic effect. However it

is possible to generate such a charge diffusion in an other way. Fig. 2

shows two undoped semiconductor with the same work function and band gap,

but medium A has higher effective masses of electrons and holes than medium

B. Therefore A has also a higher intrinsic carrier concentration.

. semiconductor A semiconductor B

. ************************** ***************************

. * e h e h e * * e h *

. * e h e h h * * h e *

. * h e h e h * * e *

. * h e h e h e * * h *

. * e h e h e h * * h e *

. ************************** ***************************

.

Fig. 2: Both undoped semiconductors have the same work function and band

gap, but A has higher effective masses of electrons and holes and

therefore a higher intrinsic charge concentration compared to B.

What happens if the two semiconductors are brought into contact? Since both

media have the same work function and band gap, the charges can cross the

junction without any restriction or energy emission or absorption. To be

more specific, there will be no depletion zone nor interface spike nor

quasi-electric field due to discontinuities in the band-edge energies.

However the concentration of electrons and holes are in medium A much

higher therefore there is a diffusion of charges from A to B, see Fig. 3.

. A B

. ****************************************************

. * e h e h * e h *

. * e h e h e -----> e e *

. * h e h e h -----> h e *

. * h e h e h * h *

. * e h e h e * h e *

. ****************************************************

. generation diffusion recombination

.

Fig. 3: The semiconductors are in contact. Both media have the same

work function and band gap, but A has a higher charge concentration,

therefore charges diffuse into B. Thus in B there are more

recombinations than generations and in A there are more generations

than recombinations.

This diffusion would stop if there were the same charge concentration in

both media. However in B there is (because of diffusion) a higher charge

concentration than in the intrinsic case, therefore more electron-hole

pairs will recombine and so the diffusion will not stop. So there will be a

special thermodynamic equilibrium: All the time, electrons and holes which

are generated in medium A will diffuse to B and recombine there. Fig. 4

shows this situation in the band diagram.

. e -------------------> e

. /|\ |

. ----------------|----------------------|------------------ Ec

. | |

. ----------------|----------------------|------------------ Ef

. | |

. ----------------|----------------------|------------------ Ev

. | \|/

. h -------------------> h

. generation diffusion recombination

Fig. 4: The band diagram shows the generation of electron-hole pairs

in medium A, the diffusion and recombination in medium B. The Fermi-

level Ef remains flat.

So in medium A there are more electron-hole pair generations than

recombinations because of the diffusion of charges into medium B. Since

every generation absorb heat energy and every recombination emit this

energy, medium A will cool down a little bit. In medium B there are more

recombinations therefore medium B will heat up. So there will be

established a temperature difference between media A and B only because of

the flow of charges and without any external force or energy. However this

temperature difference will be quite small because of heat transfer from B

to A.

Please note the difference to a normal p-n junction. At a p-n junction

electron and holes diffuse in DIFFERENT directions. Moreover the internal

potential prevents a higher rate of diffusion. However in the

heterojunctions described in this paper, most importantly the electrons and

holes diffuse in the SAME direction. And since they always recombine in

semiconductor B, there cannot be an end of the diffusion.

Lets make an example:

Semiconductor A and B have a bandgap of Bg=0.5 eV

Semiconductor A has effective electron and hole masses m*=1

Semiconductor B has effective electron and hole masses m*=0.5

The total carrier concentration of A is therefore 1.6*10^15

The total carrier concentration of B is therefore 5.6*10^14

So in semiconductor A, there 2.9 times more charges than in semiconductor

B.

On the other hand, the kinetic energy E_kin of an electron must be constant

at the same temperature. Since the charges in A have more mass, they are a

little bit slower:

V_th=(3kT / m*)^0.5

At 300 degree: The carriers in A have a velocity v_th=1.17*10^3 m/s

The carriers in B have a velocity v_th=1.65*10^3 m/s

So the carriers in A are only 1.4 times slower than in A.

So there are in A many more charges which are only a bit slower than in B.

The pressure of the electron gas is

p= 1/3 n m v^2

Since m v^2 is the kinetic energy and constant, the relevant part is only

the number of charges n. Therefore the pressure of the electron gas in

material A is 2.9 times higher than in B therefore electrons move into B.

Exactly the same happens with the holes. Therefore there is a diffusion of

electrons and holes in the same direction, from A into B.

Every effect which uses the increased charge concentration because of

photon absorption can be also done with this special heterojunction, like

the photovoltaic or the photomagnetoelectric effect.

A possibility to use this effect is to convert heat into electric energy.

Fig. 1 and 3 shows the similarity between an illuminated semiconductor and

this special heterojunction: Both produce a higher charge concentration

than in the normal case. So it is easy to construct a "heat cell" with such

a heterojunction, see Fig. 5.

. p-doped undoped semiconductor A undoped n-doped

. *****************************************************************

. * * * * * *

. * * * * * *

. * h <------- h <-------- h e -----------> e --------> e *

. * * * * * *

. * * * * * *

. *****************************************************************

. diffusion generation diffusion

.

Fig. 5: This arrangement is similar to a PIN solar cell, however in

this case additional charges are produced from semiconductor A which

have a higher charge concentration. Everything else is like in a PIN

solar cell: These charges are separated which produce an electric

potential.

This arrangement is quite similar to a solar PIN cell: In the middle is an

undoped semiconductor. In case of a solar cell, photons produce in the

middle of the cell additional charges which are separated. However in this

case there is instead a special semiconductor A in the middle. This medium

has a higher charge concentration and therefore charges diffuse in both

directions out of the medium. Because of the internal electric field of the

p and n area, the different charges will diffuse in different directions

and the electric potential will be produced.

The only difference between a PIN-solar cell and this "heat cell" is in the

middle, how additional electron-hole pairs are produced.

However the "heat cell" produces electric energy without illumination. It

just absorbs heat energy because of the higher number of generations than

recombinations of electron-hole pairs. This energy will be used to produce

electric energy.

This effect can be shown with SimWindows, the Semiconductor Device

Simulator from David W. Winston, University of Colorado.

http://www-ocs.colorado.edu/SimWindows/simwin.html

The following device file is used:

grid length=2.1 points=1000

structure material=gaas length=2.1

Band_gap length=2.1 value=0.5

ELECTRON_DOS_MASS length=1.0 Value=0.5

ELECTRON_COND_MASS length=1.0 Value=0.5

HOLE_DOS_MASS length=1.0 Value=0.5

HOLE_COND_MASS length=1.0 Value=0.5

ELECTRON_DOS_MASS length=0.1 Value=1.0

ELECTRON_COND_MASS length=0.1 Value=1.0

HOLE_DOS_MASS length=0.1 Value=1.0

HOLE_COND_MASS length=0.1 Value=1.0

ELECTRON_DOS_MASS length=1.0 Value=0.5

ELECTRON_COND_MASS length=1.0 Value=0.5

HOLE_DOS_MASS length=1.0 Value=0.5

HOLE_COND_MASS length=1.0 Value=0.5

doping length=1.0 Na=5e18

doping length=0.1

doping length=1.0 Nd=5e18

This means that there is a structure of 2.1 micrometer based on GaAs.

However the bandgap is reduced to 0.5 eV. There is in the left and right a

p and n doped material. In the middle there is a 0.1 micrometer undoped

material with the double mass of electrons and holes.

The short circuit current is about 1.8 mA. (In this case, SimWindows

doesn’t solve the drift-diffusion equation for its own since it supposes an

equilibrium case. In this case it helps e.g. to define a VERY small

illumination of e.g. only 10^-100 mW/cm^2. For comparison: For solar cells,

the often used 1.5AM illumination of the sun is 10^2 mW/cm^2.)

The open circuit voltage of this "heat cell” is about 1mV. The maximum

power delivered by this circuit is about 0.0005 mW/cm^2.

This is not very much, however many of this circuits can be put of one

another and put together. Moreover there it is very possible to find

semiconductors with much better values for this effect.

The "heat cell" will cool down a little bit, if it converts heat energy to

electric energy. However from the environment heat will flow to the heat

cell.

Although this is a theoretical case, it shows that this effect exists, but

it is hard to find good semiconductors, and especially all the data needed

for a device simulator.

This "heat cell" would contradict the 2nd law of thermodynamics. However it

does not contradict the 1st law of thermodynamics, since the whole energy

will be conserved, only the distribution will be changed.

Apr 26, 2000, 3:00:00 AM4/26/00

to

Gerhard Kainz wrote:

>

> Heterojunctions can convert heat into electric energy

> Gerhard Kainz E-Mail: gerhar...@siemens.at

>

> This paper shows that a special semiconductor heterojunction acts like a

> homogenous semiconductors which is illuminated: Both produce additional

> electron-hole pairs which can be separated and so produce electric energy.

[snip]

Peletier Effect. You are 100 years late. Gather up ye filled

skutterudites.

--

Uncle Al

http://www.mazepath.com/uncleal/

http://www.ultra.net.au/~wisby/uncleal/

http://www.guyy.demon.co.uk/uncleal/

(Toxic URLs! Unsafe for children and most mammals)

"Quis custodiet ipsos custodes?" The Net!

Apr 26, 2000, 3:00:00 AM4/26/00

to

Uncle Al wrote:

> Gerhard Kainz wrote:

> >

> > Heterojunctions can convert heat into electric energy

> > Gerhard Kainz E-Mail: gerhar...@siemens.at

> >

> > This paper shows that a special semiconductor heterojunction acts like a

> > homogenous semiconductors which is illuminated: Both produce additional

> > electron-hole pairs which can be separated and so produce electric energy.

> [snip]

>

> Peletier Effect. You are 100 years late. Gather up ye filled

> skutterudites.

Nah, just a perpetual motion machine. To be Peltier he would have to be getting

cooling from an imposed potential difference. He thinks he can get the cooling

and produce a current at the same time with no free energy input. Who wants to

break the bad news to OPEC?

Ben Buckner

Apr 27, 2000, 3:00:00 AM4/27/00

to

Uncle Al schrieb:

> Gerhard Kainz wrote:

> >

> > Heterojunctions can convert heat into electric energy

> > Gerhard Kainz E-Mail: gerhar...@siemens.at

> >

> > This paper shows that a special semiconductor heterojunction acts like a

> > homogenous semiconductors which is illuminated: Both produce additional

> > electron-hole pairs which can be separated and so produce electric energy.

> [snip]

>

> Peletier Effect. You are 100 years late. Gather up ye filled

> skutterudites.

>

Sorry, this is not the Peltier effect.

For the Peltier effect, you need an electric current, which produces at a

heterojunction heat or coldness.

The Seebeck-effect produces electric energy, if one junction has another

temperature than the other.

But the effect described in this paper makes another thing:

It can convert heat (without a temperature difference) into electric energy.

This is neigther Peltier nor Seebeck.

Gerhard

Apr 27, 2000, 3:00:00 AM4/27/00

to

Gerhard Kainz wrote:

>

> Heterojunctions can convert heat into electric energy

> Gerhard Kainz E-Mail: gerhar...@siemens.at

>

>

> Heterojunctions can convert heat into electric energy

> Gerhard Kainz E-Mail: gerhar...@siemens.at

>

As with all violations of the 2nd Law, this is based on an erroneous

analysis. (Note to Uncle Al: he is claiming isothermal energy conversion,

not the inverse Peltier effect.) The error is a very subtle point in

semiconductor heterostructure physics.

...

> This kind of diffusion is the base of the photovoltaic effect. However it

> is possible to generate such a charge diffusion in an other way. Fig. 2

> shows two undoped semiconductor with the same work function and band gap,

> but medium A has higher effective masses of electrons and holes than medium

> B. Therefore A has also a higher intrinsic carrier concentration.

>

...

> What happens if the two semiconductors are brought into contact? Since both

> media have the same work function and band gap, the charges can cross the

> junction without any restriction or energy emission or absorption. To be

> more specific, there will be no depletion zone nor interface spike nor

> quasi-electric field due to discontinuities in the band-edge energies.

> However the concentration of electrons and holes are in medium A much

> higher therefore there is a diffusion of charges from A to B, see Fig. 3.

>

This is the error. There are four terms in the drift-diffusion equation

for a heterostructure: (see http://www.utdallas.edu/dept/ee/frensley/technical/hetphys/node14.html)

drift due to the classical electric field, drift due to the quasi-electric

field due to a band-edge gradient, classical diffusion, and a term

of the form (elementary charge) (Diffusivity) (carrier density) 3/2

grad (effective mass) / (effective mass). This term (called the

"entropy force" by some authors) provides the "force" which prevents

the claimed diffusion from happening.

The heterostructure DD equation is usually derived by imposing the

equilibrium condition that current density = 0 in equilibrium. So,

if one is inclined to believe in 2nd Law violations, one may not accept

the validity of the fourth term. We can get the same result from a

microscopic analysis. This involves, first, using the right effective

Hamiltonian. The simplest one will have akinetic energy term of the form

-grad (1/m*) grad psi, where m* is the effective mass that is varying with

position. This is the simplest manifestly Hermitian form for the effective

Hamiltonian. (One has to be very careful to make sure that the effective Hamiltonian is Hermitian.

Otherwise, continuity will be violated and the

model will predict spontaneous generation of electrons. You can certainly

get a 2nd Law violation out of those models!) You then use the WKB

approximation to find the semi-classical dynamics, and insert the result

into a simple derivation of the drift-diffusion equation (Boltzmann equation

with displaced Maxwellian). (I worked all this out in about 1984, but the

details are in a notebook retained by my former employer.)

>

> This effect can be shown with SimWindows, the Semiconductor Device

> Simulator from David W. Winston, University of Colorado.

> http://www-ocs.colorado.edu/SimWindows/simwin.html

>

Well, that just means that Dave left out the entropy force term in his

implemenation. Since he has gone on to other things and Russ Hayes has

retired from CU, I don't think it is very likely that this will be fixed.

- Bill Frensley

Apr 27, 2000, 3:00:00 AM4/27/00

to

Thanks for your very interesting response!

"William R. Frensley" schrieb:

> This is the error. There are four terms in the drift-diffusion equation

> for a heterostructure: (see http://www.utdallas.edu/dept/ee/frensley/technical/hetphys/node14.html)

> drift due to the classical electric field, drift due to the quasi-electric

> field due to a band-edge gradient, classical diffusion, and a term

> of the form (elementary charge) (Diffusivity) (carrier density) 3/2

> grad (effective mass) / (effective mass). This term (called the

> "entropy force" by some authors) provides the "force" which prevents

> the claimed diffusion from happening.

Well, I know your page and these 4 terms. However I thought that the 4. term

has a force contrary to the diffusion of charges, but does not cancel

it out completely.

Please look at this example:

Lets make an example:

Semiconductor A and B have a bandgap of Bg=0.5 eV

Semiconductor A has effective electron and hole masses m*=1

Semiconductor B has effective electron and hole masses m*=0.5

*) It can be caluclated, that in material A, there are 2.9 times more charges

than in B.

*) The termal velocity of the charges in A are only 1.4 times slower than in B.

*) The kinetic energy of the charges must be equal in both semiconductors

(all at equal temperatures).

So in A there are 2.9 more charges, which are only a little bit slower than

in B and importantly they have all the same kinetic energy. Therefore the

pressure of the electron gas is in material A 2.9 times higher than in B.

So do you understand, why the charges in A should not diffuse into B?

> > This effect can be shown with SimWindows, the Semiconductor Device

> > Simulator from David W. Winston, University of Colorado.

> > http://www-ocs.colorado.edu/SimWindows/simwin.html

> >

> Well, that just means that Dave left out the entropy force term in his

> implemenation. Since he has gone on to other things and Russ Hayes has

> retired from CU, I don't think it is very likely that this will be fixed.

Well, in Winston's thesis, he wrote that he uses a DD equation INCLUDING

the forth term, the entropy force (Page 21, Eq. 21). And in spite of this,

I get these results!

Gerhard Kainz

Apr 28, 2000, 3:00:00 AM4/28/00

to Gerhard Kainz

Gerhard Kainz wrote:

>

> Thanks for your very interesting response!

>

>

>

> Thanks for your very interesting response!

>

>

> Well, I know your page and these 4 terms. However I thought that the 4. term

> has a force contrary to the diffusion of charges, but does not cancel

> it out completely.

>

> Please look at this example:

>

> has a force contrary to the diffusion of charges, but does not cancel

> it out completely.

>

> Please look at this example:

>

> Lets make an example:

> Semiconductor A and B have a bandgap of Bg=0.5 eV

> Semiconductor A has effective electron and hole masses m*=1

> Semiconductor B has effective electron and hole masses m*=0.5

>

> Semiconductor A and B have a bandgap of Bg=0.5 eV

> Semiconductor A has effective electron and hole masses m*=1

> Semiconductor B has effective electron and hole masses m*=0.5

>

> *) It can be caluclated, that in material A, there are 2.9 times more charges

> than in B.

> *) The termal velocity of the charges in A are only 1.4 times slower than in B.

> *) The kinetic energy of the charges must be equal in both semiconductors

> (all at equal temperatures).

>

> So in A there are 2.9 more charges, which are only a little bit slower than

> in B and importantly they have all the same kinetic energy. Therefore the

> pressure of the electron gas is in material A 2.9 times higher than in B.

>

> So do you understand, why the charges in A should not diffuse into B?

>

The simplest way to approach this is to combine the diffusion and entropy> than in B.

> *) The termal velocity of the charges in A are only 1.4 times slower than in B.

> *) The kinetic energy of the charges must be equal in both semiconductors

> (all at equal temperatures).

>

> So in A there are 2.9 more charges, which are only a little bit slower than

> in B and importantly they have all the same kinetic energy. Therefore the

> pressure of the electron gas is in material A 2.9 times higher than in B.

>

> So do you understand, why the charges in A should not diffuse into B?

>

force terms into

q Dn Nc grad (n / Nc), where Nc is the thermal effective density of states

in the conduction band (see my equation 28). Now, in your example of an

intrinsic semiconductor with equal masses in the conduction and valence

bands, the intrinsic Fermi level is precisely at the middle of the gap,

and thus the electron density equals Nc exp (-Eg / 2kT). This

is true in each material, and the difference in Nc's leads to your factor

of 2.9 ratio of electron densities (actually 2^{3/2} is 2.83).

Inserting this intrinsic density into the above expression gives

grad [ exp (-Eg / 2kT) ], which is equal to zero, as the bandgaps are assumed

to be equal. Thus there is no current flow.

We can look at the condensed form of the diffusion current in terms of the

quasi-Fermi level. The fundamental form of the transport equation is

J = - mu n grad Ef (mu is the mobility). In the nondegenerate approximation,

Ef = Ec + kT ln (n / Nc).

Therefore,

J = - mu n grad Ec - kT mu grad ln (n / Nc)

= - mu n grad (Ec_unscreened - q V) - q D Nc grad (n / Nc),

where V is the electrostatic potential, and we use the Einstein relation

qD = kT mu. The first term above is the quasi-electric field, the second

term is the ordinary electric field, and the third term is the combined

diffusion and entropy terms used in the previous paragraph. The conclusion

is simply that in equilibrium the Fermi level is everywhere constant and

thus the current density is zero.

- Bill Frensley

Apr 30, 2000, 3:00:00 AM4/30/00

to

"William R. Frensley" wrote:

> The simplest way to approach this is to combine the diffusion and entropy

> force terms into

> q Dn Nc grad (n / Nc), where Nc is the thermal effective density of states

> in the conduction band (see my equation 28).

Yes, I think we are now at the main point of this matter:

Does the the diffusion depends on grad (n / Nc)

or on grad (n)

If it is grad (n / Nc) then there is of course no diffusion in my circuit

(equal masses and equal bands) since the term (n / Nc) is constant over the

hole circuit.

However the density of states Nc just comes because of the entropy term of

the DD-equation in order to prevent diffusion in such a case.

I'm not sure, if this correct. In material A there are appr. 3 times more

charges which can all cross the junction to material B.

So if you just observe the junction, then from material A there are comming 3

times more charges, which cross the junction in order to go into B

than in the reverse direction.

So in this case, Nc has no importance.

Gerhard Kainz

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