Hi folks!

I'm trying to thoroughly understand P/N junctions from a physical standpoint. However, I'm confused on a couple things that I keep seeing in videos and docs.

First, when _P-type_ silicon is created, it's (usually) doped with Boron, which is missing an electron (has a hole), and therefore should have a positive charge (more protons than electrons), yet we call them negative ions, why is that?

 

In it's neutrally charged state (the same number of electrons as protons), Boron is missing one valance electron (i.e., has an available hole). This does NOT make it positively charged, this is it's normal neutrally charged state.

When the hole is filled (which it wants to be), then the net charge of the ion will be negative. It still has five protons, but now it has two inner shell electrons and all four valance electrons.

 

Hi folks!

I'm trying to thoroughly understand P/N junctions from a physical standpoint. However, I'm confused on a couple things that I keep seeing in videos and docs.

First, when _P-type_ silicon is created, it's (usually) doped with Boron, which is missing an electron (has a hole), and therefore should have a positive charge (more protons than electrons), yet we call them negative ions, why is that?

That is your fundamental error. Boron (element) has a neutral charge, as do all elements. It has 5 protons and 5 electrons. There are potentially three, 2-p orbitals. That means that of the three p-orbitals, one is not completely filled. That vacancy is called a hole. When it accepts an electron in that 2p orbital, it becomes negatively charged. That gives it very interesting chemical properties. It is a Lewis acid, but not everyone considers it a Bronsted acid (see this discussion: https://en.wikipedia.org/wiki/Talk:Brønsted–Lowry_acid–base_theory).

Its action as a Lewis acid is often represented as this:



Thus, unlike something like sulfuric acid (H2So4) that ionizes to HSo4^- + H^+ to "liberate" a proton, boric acid liberates a proton from water by adding the hydroxyl.

@WBahn
Sorry, I had a problem with attachments and took a bit longer to post than I anticipated.

 

That is your fundamental error. Boron (element) has a neutral charge, as do all elements. It has 5 protons and 5 electrons. There are potentially three, 2-p orbitals. That means that of the three p-orbitals, one is not completely filled.

I don't follow this. There are five electrons total, but two are in the 1-s and two are in the 2-s, leaving just 1 for the six possible electrons in the 2-p orbitals.

 

I don't follow this. There are five electrons total, but two are in the 1-s and two are in the 2-s, leaving just 1 for the six possible electrons in the 2-p orbitals.

Sorry for the bad description. For comparison, take carbon with 2, 2s and 2, 2p electrons that can hybridize to give 4 bonding sp^3 orbitals and a neutral charge. Boron has 2, 2s and 1, 2p electron (formally). Which after hybridization allows up to 4, sp3 orbitals (in theory) or 3, sp^2 orbitals (e.g., B(OH)3, which is effectively planar). When that "potential" sp^3 orbital is occupied you get a tetrahedral boron with a negative charge.

As an aside, one of boric acid's useful characteristics is that it readily and reversibly forms esters with hydroxyl groups. At high pH (e.g., pH ≥ 9.2), those ester are negatively charged as the sp^3 hybrid orbital is occupied. That allows for, among other things, use of boron-based acids to aid in the electrophoretic separation of anything with vicinal hydroxyls, e.g ., otherwise neutral polysaccharides. Boronic acids (i.e., boric acid attached to an organic radical instead of a hydroxyl) are also used in electrophoretic separations and can also be used as solid phases for affinity chromatography.

 

Ha! Much more information than I needed, but thanks for the awesome answers!

Seriously, these are fantastic responses, they completely clear up my question and provide substantial, interesting, supporting information.