pn junction in equilibrium

Imagine taking two separate bars of silicon, one doped to be n-type (N_D) and the other doped to be p-type (N_A). Push these bars together so that there is a single bar where the middle abruptly changes doping types and levels.

At the moment of contact, there is a huge concentration gradient for both holes and electrons. The p-type region with a majority of holes is right next to the n-type region with very few holes and vise versa.

This large gradient causes the holes at the edge of the p region to flow via diffusion to the low concentration side and electrons at the edge of the n region flow via diffusion the opposite direction. Because of the opposite signs for charge polarities and movement direction, the net diffusion current is in the same direction, which is to the left in Figure 3.

After a long time, this diffusion will eventually stop because the electron and hole densities have become uniform:

Think about the Group IV dopant atom (say phosphorus) on the n side of the junction. That extra electron (and extra proton!) is in a region of many free electrons and next to a region with few. So it will tend to move (diffuse) to the right towards the p region.

This particular free electron came from the 5^th outer valence electron of the phosphorus dopant atom. Therefore, when it diffuses away, the 15^th proton in the nucleus is left without a charge-balancing mate. We now have a +1 positively-charged phosphorus ion.

Similarly, when the hole contributed by the Group III dopant (say boron) on the p side diffuses to the left, we remember that the hole was filled with an electron. This electron fills the 4^th outer valence “slot” and completes the covalent bonding structure in that area. Don’t forget that boron is short a proton that matched the now-filled hole. Therefore the region has a net −1 negatively-charged boron ion.

Both of these types of ions cannot move because the charges are either a) part of the nucleus, or b) a bound electron. This means that we have:

or called the depletion region because the volume is eventually depleted of free charge carriers.

Now that there are separated charges, Gauss’s law, \(\nabla \!\bullet\! \left(\epsilon \vec{E}\right) = Q\), implies that there is a non-zero E-field inside the depletion region. This field is maximum at the boundary between the n and p type doping and, for this example, points to the right.

Recall that the diffusion currents are to the left and the drift current from the E-field is opposite, to the right. Because this semiconductor bar is not connected in a circuit, KCL forces the current at every x location to be zero. The junction is in equilibrium and the drift and diffusion currents must be equal and opposite.

Take the free electrons equation and substitute their definitions:

Integrate and divide both sides by n

The left side is the potential difference developed across the depletion region and is given the symbol \(V_0\). Also, recall Einstein’s relation, \(\frac{D}{\mu} = \frac{k_B T}{q}\), and substitute

We usually are only concerned with the magnitude of V₀; notice that swapping the fraction so the majority carriers term (n_n) is in the numerator cancels the leading negative sign. Recognize, for the next section, that the higher potential is on the n side, at location b, given the E-field direction. A similar process can be done to find V₀ from the hole currents.

One final set of substitutions using our (good) approximations of majority and minority carrier concentrations

where \(V_T\) is called the thermal voltage.

The book derives the built-in potential \(V_0\) from book equation (2.62).

Look back at book Figure 2.18. There are two regions of the semiconductor bar that suddenly change their doping types and levels at the middle dotted line.

Focus on the curve for one type of charge carriers, e.g. holes. The concentration starts low on the left since this is the (majority-)n-type region, then then increases until it is the majority type of carrier in the right side.

Now, go back to the calculus used to derive the built-in potential. This needs the E-field (or voltage) versus position (x, here), and needs charge carrier concentration.