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- Thread starter planety_vuki
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The reason the effective mass is negative in a p-type system is because of the formation of energy bands in a solid. The relationship between energy and momentum is called the dispersion relationship. For an n-type system, the dispersion is positive, so that as momentum increases, energy increases. But for a p-type system, the charge carriers are near the top of the band, and the bands curve downward, so that as momentum increases, energy decreases.

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I have read that in wikipedia, but how can effective mass be negative ? I am having hard time imagining what exactly happens down there. Is there any real world analogy where effective mass is negative ? something Newtonian ?

The reason the effective mass is negative in a p-type system is because of the formation of energy bands in a solid. The relationship between energy and momentum is called the dispersion relationship. For an n-type system, the dispersion is positive, so that as momentum increases, energy increases. But for a p-type system, the charge carriers are near the top of the band, and the bands curve downward, so that as momentum increases, energy decreases.

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The effective mass is defined as [tex]\frac{1}{m^*} = \frac{\partial^2 E}{\partial k^2}[/tex] where k is the (pseudo-)momentum. For a classical free particle, [itex]E = k^2/2m[/itex] and the effective mass is equal to the real mass.

In a particular very simple toy model, you have an energy band [itex]E = -t cos( a k)[/itex] (a is the lattice constant and t is a parameter). For an n-type system, you have this band as being unfilled, so that your charge carriers are at the bottom of the band near k = 0 and the band curves up, so you have positive mass. For a p-type system the band is nearly filled, so your free carriers are near the top of the band around [itex]k = \pi/a[/itex] and the band curves down, giving negative effective mass.

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Does this mean that in p-type materials the electrons move the same direction as conventional current ?

The effective mass is defined as [tex]\frac{1}{m^*} = \frac{\partial^2 E}{\partial k^2}[/tex] where k is the (pseudo-)momentum. For a classical free particle, [itex]E = k^2/2m[/itex] and the effective mass is equal to the real mass.

In a particular very simple toy model, you have an energy band [itex]E = -t cos( a k)[/itex] (a is the lattice constant and t is a parameter). For an n-type system, you have this band as being unfilled, so that your charge carriers are at the bottom of the band near k = 0 and the band curves up, so you have positive mass. For a p-type system the band is nearly filled, so your free carriers are near the top of the band around [itex]k = \pi/a[/itex] and the band curves down, giving negative effective mass.

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Sorry me but, if in p-type materials electrons' effective mass is negative then why it's negative only to magnetic field but not to electric field ?? Am I missing something. In fact if mass is negative for noth, E field and B field, then hall effect should be the same as for n-types.

Daveyrocket is it true that hall effect reveals opposite results for n-type and p-type? Istarted my question assuming so but I don't know for sure. Have you done such test before? I think I should make a hall experiment for n-type and p-type materials to see it for myself.

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One other thing I forgot to mention, the velocity of particles is [itex]v = \frac{\partial E }{ \partial k}[/itex]. For n-type systems this behaves normally, as k gets larger, v gets larger and has the same sign. But for p-type systems, (using the dispersion I mentioned above [itex]E = -t\cos (ka)[/itex])... At k = pi/a the velocity is zero and effective mass is negative. As k moves away from pi/a, the energy goes down and the velocity picks up the opposite sign you would expect.. You can approximate the cos function for a p-type system near the band edge as [itex]E(k = \tfrac{\pi}{a} + \delta k) = t ( 1 - \delta k^2/2)[/itex]. Velocity will be negative if [itex]\delta k[/itex] is positive. This gives an additional sign flip for the magnetic force, since it comes in as v cross B.

This is definitely an experiment worth doing just to see it for yourself, even if you thought my explanation was so awesome that you are totally convinced.

There's some more detailed explanation here: http://www.fys.ku.dk/~jjensen/SolidState/Week5.pdf

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Well actually I don't understand anything you explained because I don't anything about the formulas and equations you arote. I just want to understand things qualitatively. I really did not understand the reason why electron mass negative for magnetic but positive for electric field. Whatever electrons effective mass is shouldn't it be the same for all kind of fields ?

One other thing I forgot to mention, the velocity of particles is [itex]v = \frac{\partial E }{ \partial k}[/itex]. For n-type systems this behaves normally, as k gets larger, v gets larger and has the same sign. But for p-type systems, (using the dispersion I mentioned above [itex]E = -t\cos (ka)[/itex])... At k = pi/a the velocity is zero and effective mass is negative. As k moves away from pi/a, the energy goes down and the velocity picks up the opposite sign you would expect.. You can approximate the cos function for a p-type system near the band edge as [itex]E(k = \tfrac{\pi}{a} + \delta k) = t ( 1 - \delta k^2/2)[/itex]. Velocity will be negative if [itex]\delta k[/itex] is positive. This gives an additional sign flip for the magnetic force, since it comes in as v cross B.

This is definitely an experiment worth doing just to see it for yourself,even if you thought my explanation was so awesomethat you are totally convinced.

There's some more detailed explanation here: http://www.fys.ku.dk/~jjensen/SolidState/Week5.pdf

Don't get me wrong I am thankful to you for trying to explain but I don't understand.

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If velocity is opposite then the magnetic force is also opposite, then since F=ma, acceleration is proper, then the induced voltage is proper, as for n-type semiconductors.

note: by 'proper' I mean 'non-opposite'.

Also if velocity is opposite doesn't that mean electrons travel opposite way?so electrons travel the way electric field line points.

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No, I mean the velocity is opposite of the momentum. F = ma is not the correct equation for this situation, it is instead that F = dp/dt, the force is the time derivative of the momentum.

Velocity of the electrons is opposite the conventional current as usual. But their momentum is in the direction of the conventional current. To be very loose with terminology, since m is negative, p = mv will go in the direction of the conventional current.

If electrons went in the direction of the conventional current you'd see some very serious violations of the conservation of energy at a classical level. That can't happen.

Velocity of the electrons is opposite the conventional current as usual. But their momentum is in the direction of the conventional current. To be very loose with terminology, since m is negative, p = mv will go in the direction of the conventional current.

If electrons went in the direction of the conventional current you'd see some very serious violations of the conservation of energy at a classical level. That can't happen.

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