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Ok so I understand the PN junction, and how when 2 Semiconductor materials are placed together the Electrons will jump into the Holes near the junction creating a Negatively Ionized Atoms on the P-Side (Near the Junction) and Positively Charged Atoms near the Junction on the N-Side.

Makes sense.

HOWEVER, the Donor Atoms Give out Electrons at room temperature and the Acceptor Atoms move around holes at room temperature.

How come these Ionized Atoms aren't Displacing Electrons to create more holes (In the Ionized P-Side) or Accepting more electrons (In the Positively Ionized Region on the N-Side)?

I understand That the Electric Field is causing some resistance....but Regardless at room temperature why aren't the Electrons getting excited out of the Negatively Ionized Atoms on the P-Side near the Junction....Why are they now "fixed" to the lattice? Like why aren't these Atoms near the Junction allowing Electrons to Get Excited and move at room temperature (Like the Electrons that have filled the holes on the P-Side)?

Every Book/Semiconductor Physics book i've read talks about how the Ionized Atoms are fixed to the lattice (Makes sense)....but it doesn't say why THEY themselves aren't changing due to thermal excitations at room temperature.

edit: to rephrase:Why arent the acceptor atoms releasing the electrons they obtained from the N-Biased junction at room temp? Because I know That the electric field at the depletion region is caused by electrons filling holes in the P-region and Lack of Electrons in the N-Region.....im asking whats stopping those acceptor atoms near the junction in the P-region from "releasing electrons (that they just gained)" or whats stopping the donor atoms from being filled? (the atoms that are producing an electric field). I understand the electric field opposes more electrons from filling the n-side, but whats stopping the p-side negative ions from being excited again?

I guess im confused how the electric fields are stable...when room temperature is what excited the electrons in the first place.

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What do you mean by "...the Acceptor Atoms move around holes at room temperature."? All atoms in the system are fixed; that's why it's a lattice. Only electrons can move around. Did you mean to describe the motion of an electron "jumping" from one hole site to the other? Moreover I don't understand what you mean by "How come these Ionized Atoms aren't Displacing Electrons to create more holes..."? Can you please elaborate more on that? I understand the paragraphs after that. But I would like some more info before answering. –  NanoPhys Feb 23 '13 at 21:29
    
@NanoPhys Made a quick edit above^^ –  Mercfh Feb 25 '13 at 6:25
    
Is this a fair rewording of your questions: "In the depletion region dopants must be ionised. That means that donors have lost an electron and acceptors have trapped a electron. But why does the trapped electron stay with the acceptor atom?" –  boyfarrell Feb 26 '13 at 13:57
    
@boyfarrell thats a fair rewording. Because im mainly concerned with why at room temp electrons are being ionized....but still being trapped with the acceptor atoms. But I also suppose I wonder why electrons aren't filling the donor atoms either (but Im assuming thats because of electric field....which makes sense) –  Mercfh Feb 26 '13 at 20:06
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OK. I'm working on an answer. –  boyfarrell Feb 28 '13 at 11:49

2 Answers 2

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How come these Ionized Atoms aren't Displacing Electrons to create more holes (In the Ionized P-Side) or Accepting more electrons (In the Positively Ionized Region on the N-Side)?

It is only possible for the donor and acceptor atoms to de-ionise in the depletion region if they capture a free carrier (electron and hole, respectively). But there are no free carriers in the depletion region because they have all be swept out by the strong electric field (something like 30$-$40kV/cm$^{\textrm{-1}}$!).

So why then do the electron from the n-side stay with the acceptor atoms on the p-side once the junction has formed?

The short answer is because the carrier trapped by the dopant atoms would have to gain almost a bandgaps worth of energy to de-ionize.

De-ionization of dopants required a large amount of energy.

The longer answer. Let's assume an acceptor atom in the p-side depletion region de-ionises by giving up it's captured electron. What happens? The electron is pushed back to the n-side by the field. However, the system is now no longer in equilibrium because the p-side is charged to +1 and the n-side is charged to -1. This is not stable! You can see that if you run this forward in time, eventually an electron from the n-side will have to neutralise the acceptor, bringing the material back to charge neutrality.

When you solve the Poisson equation for the pn-junction this is what you are solving for: the equilibrium distribution for charge neutrality. There are probably carrier dynamics like de-ionisation happening but they only serve to push the system out of equilibrium temporarily, eventually equilibrium will always be restored.

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I don't think it matters what the rate is. No matter what the perturbing rates are in a system a stable equilibrium state will eventually be reached. I think it's best to view this problem from a statistical perspective rather than from considering the microscopic processes. The depletion region is stable once formed. –  boyfarrell Mar 2 '13 at 15:17
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I think the rate will be quite low because the electron (from the p-side acceptor) would have to acquire enough thermal energy to reach the conduction band of the material (about a ~1eV jump). The thermal energy at room temperature is much lower ~25meV, so I imagine that once acceptor captures the free electron (from the n-side) it will be quite stable. –  boyfarrell Mar 4 '13 at 6:48
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Yes, but the opposite is happening on the n-side. The n-side is ionised and wants to capture an electron from the valence band, therefore the energy gap is the same ~1eV. –  boyfarrell Mar 4 '13 at 12:56
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Right! You can think of that way. On the n-side to ionize a donor the electron only needs several ~kT, however on the de-ionise an acceptor (release an electron to the conduction band) on the p-side the electron needs about ~40kT. The same thing happens when you consider it from the "hole perspective". –  boyfarrell Mar 11 '13 at 1:03
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Yay! You got it! My pleasure, every time I am forced to think about the pn-junction I learn something new, so thank you for your question. The pn-junction is actually surprising complex piece of physics (in concept and computationally). Keep on asking question if you get stuck. –  boyfarrell Mar 12 '13 at 2:36

The Boltzmann distribution is a basic distribution as a function of temperature. It states that the probability of finding particles at a given energy decreases as the energy increases, but that at higher temperatures, higher energies are more likely to be found. For a single particle species, the Boltzmann distribution looks like \begin{equation} P(E)\propto e^{\frac{-E}{k_B T}}, \end{equation} where $k_B$ is the Boltzmann constant and $T$ is the system temperature. Generally, this means that the higher the temperature, the more likely you are to find fast moving particles.

If a charged particle is in an electric field, then that field affects makes a contribution to the energy, in addition to the purely kinetic energy it has ($m v^2/2$). Written this way (and ignoring factors of 1/2, etc...),

\begin{equation} P(E)\propto e^{\frac{-(m v^2 - q \phi)}{k_B T}}, \end{equation}

This says that, at higher temperatures, charges are likely to be moving faster, or be in higher potential regions, or a little of both. Charges cross the P-N junction through simple diffusion, which is related to their velocity $v$. The more charges that move across this junction, the higher $\phi$, the electric potential (a measure of the electric field) becomes. This means that as more charges migrate and establish a large potential, the less velocity particles in those regions tend to have, for a fixed temperature $T$. So while there is still significant temperature in the region, it is no longer manifest purely as kinetic energy. With diminished kinetic energy, diffusion takes place more slowly, and so eventually, charges no longer cross the P-N junction (this is when the junction is in equilibrium).

Alternatively, you can look at it like this: charges cross the P-N junction, creating an electric field across the junction. However, the establishment of this field impedes the crossing of more charges. At equilibrium, the drift velocity of charges due to the electric field exactly counteracts the diffusion drift velocity. So, your hot electrons are bouncing out and trying to cross over, but the accumulated charge opposes the crossing, and your electron settles back where it came from.

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