How the positively charged holes are created inside a p-type semiconductor?

To be more specific, we know that the holes are intentionally created via doping with some trivalent impurities (i.e Boron, indium, aluminum etc.) and We know that the holes are positively charged. Now, in the process of doping we substitute a neutral atom of an intrinsic semiconductor (i.e silicon and germanium) with a 'NEUTRAL' atom of a trivalent impurity (examples as discussed above). Now if these atoms are neutral then where do these positively charged holes come from? They are just the neutral atoms. How can this be possible for a neutral trivalent atom to introduce a positively charged hole? Even though there are none of the extra positive charges to create a net charge on the crystal, the atom is still neutral but the holes are there. How? Found some articles on the internet but still none of them has discussed about this thing.

For every created hole, you also add an additional electron that is fixed to the doping atom.

The crystal donates the electron it in order to have 4 full covalent bonds between the doping atom and the neighboring atoms.

The fixed electron does not participate in the whole semiconductor party so it is not commented much.

But it exists - and it is this fixed electron that makes the whole crystal electrically neutral.

And in advance to the next question - this is also how N-type semiconductors are made. An atom from V group is added (Nitrogen, Phosphorus). It has 5 electrons available for covalent bonding. 4 are used, the 5th is free to wander. When it does, the doping atom is left with a fixed positive charge.

• True, but as you've stated in your first paragraph that "we are adding an additional $e-$ which is fixed to the doping atom". Would you like to explain me this properly? How are we adding an extra $e-$? Boron has 5 $e-$ and the corresponding positively charged protons to make this atom neutral. Where as let's take an example of an intrinsic semiconductor i.e silicon this has 14 $e-$ and 14 protons, if we only consider the outermost shell then si has 4 $e-$ whereas Boron has 3, so what is extra here? In the case of n-type we can easily say that the dopant has an extra $e-$. Support your answer. Commented May 29, 2022 at 20:44
• Only the outermost electron shell is interesting. Boron has 3 e there, Silicon has 4 e. For a good covalent bond one needs two electrons and they are usually, but not necessarily, donated by both partners of the bond. The Boron atom has 4 neighbors and only 3 electrons. This is a "hole" and it can travel - by borrowing an electron from a nearby atom. Commented May 29, 2022 at 20:53
• wait. The electrons are not donated in the formation of covalent bond, they are shared. And sharing does not mean that the atom will possess a positive charge with the effect of the formation of a covalent bond. What is your opinion on this? Commented May 29, 2022 at 21:05
• Electrons are "donated" to the bond. Or are "shared" by forming a bond. Either way, you can form a normal bond with 2 electrons or "half" bond by using one in an orbital where one can fit two. Commented May 29, 2022 at 21:30
• Okay, now this helps a lot what were you actually trying to convey. Now I'm on a conclusion that the silicon which is tetravalent has an e- which is not bonded with anyone but if we introduce a doping atom then the those three valance e- of the doping atom will form the bond but also the one e- which is not even in the outermost shell will form the bond with that neighboring silicon atom which has one non bonded e, and that's how the Boron will share it's 4e with the neighbors which will leave a positive charge because now the 4e of Boron atoms will be busy is making the bonds instead of 3 Commented May 29, 2022 at 21:48

I'm struggling with the exact question lately and my understanding is this: OVERALL the piece of p-type semi conductor is electrically neutral, but LOCALLY the boron atoms "snatch" 1e from a nearby silicon atom to form a stable 8e valence shell (with another electron still belonging to that silicon atom), i.e. that electron now "belongs" to the boron atom and not the silicon atom anymore. Now you have a positive silicon ion and a negative boron ion. I'm only a lowly electrical engineering student so please correct me if I'm wrong!