How do acceptor energy levels form in p-doped semiconductors?

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In summary, with p-type doping in semiconductors, the impurity can be modeled as a negatively charged ion with one less electron than necessary to form covalent bonds. This creates an electron deficiency, or hole, which is weakly bound to the impurity. The impurity creates a repulsive potential that pushes one electron out of the continuum and into the acceptor state, forming a covalent bond. This reorganizes the states in the valence band, resulting in a completely filled band and one electron per impurity atom in the bound state above the band. This process is explained by Jenö Sólyum as the impurity potential reorganizing the states in the valence band. The exact mechanism
  • #1
SchroedingersLion
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How are acceptor levels formed? Doped semiconductors.
Greetings,

I have troubles at getting a good intuition on how acceptor energy levels are formed above the valence band in p-doped semiconductors.

Ashcroft describes the following scenario:
With n-type doping, the added impurity can be modeled as a positive ion with an additional electron. The ion has enough valence electrons to perfectly form the typical covalent bonds with neighboring atoms, as if it was part of the intrinsic semiconductor. Now, due to the attraction of the electron and the ion, the energy state of the electron is not part of the conduction band but slightly below that level: It would take a bit of energy to break the attraction to the ion.
With p-type doping it is the same, only now the impurity can be modeled as a negatively charged ion that forms the covalent bonds with neighboring atoms, coupled to a weakly bound positively charged particle (hole).

Now, I have troubles understanding the p-type model. How can the impurity form covalent bonds if one electron is missing? This should be impossible. The very naive picture is that electrons from other covalent bonds can 'jump in', but this would already correspond to the state where the hole entered the valence band, which is initially not true as it should still be bound to the impurity.

I have taken a look into a more advanced book by Jenö Sólyum who writes:
"On the other hand, the outermost shell of elements of group IIIA is one electron short compared to silicon. With spatially well localized covalent bonds in mind this implies that in samples doped with such impurities the number of electrons per impurity atom is one less than what would be necessary to form valence bonds. The formation of the chemical bond requires an additional electron, creating an electron deficiency, an apparently positively charged hole. Such impurities are called acceptors.
One would expect that the electron deficiencies brought about by the covalent bonds in the presence of acceptors give rise to empty states, holes, in the valence band
[the naive picture I described above and which many undergraduate courses introduce]. This is not the case, as the states in the valence band are reorganized by the impurity potential. When the bonds are formed, the acceptor can be considered as a negatively charged ion, which therefore repels other electrons. The repulsive potential pushes a state outside the continuum, above the band. Since the impurity does not change the total number of possible electron states, there remain one less state in the band as without the impurity. Thus,in spite of the electron deficiency, the valence band is completely filled in the ground state, and there is one electron per impurity atom in the bound state above the band. Since this level could accommodate two electrons, one may say that acceptors add weakly bound holes to the system."

So how exactly can the acceptor impurity form covalent bonds if there is an electron missing? I need four electrons, but I only have three.
Sure, I can act as if I had 4 electrons (=negatively charged ion) and an additional hole and then do the maths, but in reality, there are still only 3 electrons, so it is impossible to form the covalent bonds.
He said "the states in the valence band are reorganized by the impurity potential". What exactly does this mean? Does this mean the typical covalent bond structure does not apply to the impurity and it can 'somehow' form bonds with one electron less? Or that one of the three electrons or neighboring electrons does twice the 'work' and thus the impurity can be considered as negatively charged since the number of electrons that reside in its proximity to form the bonds is greater than the actual number of impurity valence electrons and, to compensate for this, we have to introduce the positively charged hole?

And even if I accept the model of negatively charged impurity + attached hole:
One electron is pushed out of the continuum into the acceptor state. Which electron is it? Is it one of the valence electrons that are part of the bonds of the impurity?

Please help me understand this conceptually.
SL
 
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  • #2
SchroedingersLion said:
Now, I have troubles understanding the p-type model. How can the impurity form covalent bonds if one electron is missing? This should be impossible. The very naive picture is that electrons from other covalent bonds can 'jump in', but this would already correspond to the state where the hole entered the valence band, which is initially not true as it should still be bound to the impurity.
In fact it is the common understanding, that electrons from surrounding covalent bondings 'jump in' to the unbound electron and form a new covanelt bonding leaving the former bounded electron unbounded. This is the situation around room temperature. The vancancy of the 4th covalent bonding (within a Si or Ge host crystal) is regarded as an additional hole. As the energy of this unbounded hole state is some meV higher than the bond state, it layes above the valence band edge.
 
  • #3
BPHH85 said:
In fact it is the common understanding, that electrons from surrounding covalent bondings 'jump in' to the unbound electron and form a new covanelt bonding leaving the former bounded electron unbounded. This is the situation around room temperature. The vancancy of the 4th covalent bonding (within a Si or Ge host crystal) is regarded as an additional hole. As the energy of this unbounded hole state is some meV higher than the bond state, it layes above the valence band edge.
Sorry, but you didn't answer all the questions... and your answer is pretty vague.
The 'unbounded hole state'. What do you mean by that?` You mean the state where the hole is freely moving through the valence band? Than its state is within the valence band, and not above it.
It is the acceptor state that lies above the valence band. At T=0, the hole is bound there and only released to the valence band, if an electron is thermally excited from there.

The question was: How exactly does this acceptor state come to be? How can one understand it intuitively as with donor levels (see first part of my post)?
 

Related to How do acceptor energy levels form in p-doped semiconductors?

1. How do acceptor energy levels form in p-doped semiconductors?

In p-doped semiconductors, the acceptor energy levels form due to the introduction of impurities into the crystal lattice. These impurities, known as acceptor atoms, have one less valence electron than the atoms in the semiconductor crystal. This creates a "hole" in the crystal structure, which acts as a positive charge carrier. The acceptor atoms also have a higher ionization energy, meaning they require less energy to remove an electron from their outer shell. This leads to the formation of acceptor energy levels within the band gap of the semiconductor.

2. What is the role of acceptor energy levels in p-doped semiconductors?

The acceptor energy levels in p-doped semiconductors play a crucial role in the conduction of electricity. As mentioned before, the acceptor atoms create "holes" in the crystal structure, which act as positive charge carriers. When an external electric field is applied, these holes are able to move through the crystal lattice, allowing for the flow of current. The acceptor energy levels also contribute to the formation of the depletion region at the p-n junction in a diode, which is essential for its functioning.

3. How do the acceptor energy levels affect the band structure of p-doped semiconductors?

The acceptor energy levels affect the band structure of p-doped semiconductors by creating a new energy level within the band gap. This energy level is known as the acceptor level and is located closer to the valence band. Since the acceptor atoms have a higher ionization energy, this energy level is typically closer to the valence band edge compared to the donor energy level in n-doped semiconductors. The presence of these energy levels alters the band structure, making it easier for holes to be created and increasing the conductivity of the material.

4. How are the acceptor energy levels determined in p-doped semiconductors?

The acceptor energy levels in p-doped semiconductors are determined by the energy difference between the valence band and the acceptor level. This energy difference can be measured using various techniques such as photoluminescence spectroscopy, capacitance-voltage measurements, and deep-level transient spectroscopy. Additionally, the acceptor energy levels can also be calculated using theoretical models and simulations, taking into account the properties of the acceptor atoms and the crystal lattice of the semiconductor.

5. Can the acceptor energy levels be controlled in p-doped semiconductors?

Yes, the acceptor energy levels in p-doped semiconductors can be controlled by varying the concentration and type of acceptor atoms introduced into the crystal lattice. Different acceptor atoms have different ionization energies, which affect the position of the acceptor energy level within the band gap. Additionally, the temperature and the external electric field can also influence the position and behavior of the acceptor energy levels. By carefully selecting the impurities and controlling the external conditions, the acceptor energy levels can be tailored to meet the specific requirements of a semiconductor device.

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