The diagram shows a lattice of silicon atoms in a semiconductor. The left side of the lattice has been doped with donor ions. This is called the n-side. The right side of the lattice has been doped with acceptor ions. This side of the lattice is called the p-side. The regions on either side of the dividing line are of equal size and the ion concentration is the same on both sides. The semiconductor is at thermal equilibrium. What is the ratio of free electrons on the n-side to vacancies on the p-side?
In this example, we’re told that the left side of our silicon lattice is doped with donor ions. All the atoms in the lattice on this side are silicon atoms except for these two labeled P. P is the atomic symbol for phosphorus. And every neutral phosphorus atom has five electrons in its valence shell. This means that when a phosphorus atom is inserted into the lattice of silicon atoms, the valence shell of each phosphorus atom becomes filled to overflowing, we could say, with electrons, which leads to one electron per phosphorus atom in the lattice being released to move about freely.
These electrons, once they are liberated, are called free electrons. And we see one free electron here and another one right here. As we expected, there’s one free electron for each atom of phosphorus in the lattice. Phosphorus is called the donor, by the way, because it donates these free electrons.
On the other side, instead of the lattice of silicon being doped with phosphorus, it’s doped with this element referred to as B. B stands for boron. Every boron atom has three electrons in its valence shell. That means that when a boron atom is inserted into the silicon lattice and effectively shares four electrons to give it seven in its valence shell, there remains a hole or a vacancy where that eighth electron might go to fill the valence shell. Since vacancies can accept electrons, these boron atoms are called acceptors. Each atom of boron contributes one hole or vacancy to the lattice.
So we see then that on the n-type side of our lattice, on the left side there are a total of two free electrons, one for each atom of phosphorus. Then on the right-hand side, the p-side of our lattice, there are also two vacancies or two absences of an electron. Therefore, the ratio of free electrons on the n-side to vacancies on the p-side is one.
Knowing this, let’s move on to the next part of our question.
What is the difference between the net relative electronic charge in the two regions?
Let’s first consider the net electronic charge on the left, or n-type, side of our lattice. We can start by noting that each of the silicon atoms in our lattice is neutral; they each have zero net electric charge. In addition, each of the phosphorus atoms we’ve used to dope this lattice is also individually electrically neutral. That may seem strange because we have seen that these phosphorus atoms are called donor ions. We should note though that these atoms only become ionized when an electron is stripped away from them as they’re added to the lattice. That is, once one of the electrons in the phosphorus atom becomes a free electron, the atom that is left behind has an overall positive charge.
However, if we consider the entire n-type side of our lattice as a whole, every positively charged ion has a corresponding negatively charged free electron roaming about. Therefore, the net electric charge of the n-type side of our lattice is zero; all the positive and negative charges balance one another out.
The same thing actually happens on the p-type side of the lattice. Once again, all the silicon atoms are electrically neutral by themselves. And the boron atoms we use to dope this lattice are also neutral. It’s only when the boron atoms are inserted into the lattice and develop vacancies that are free to move about like free electrons that we might come to think of the boron atoms as acceptor ions. But just like on the n-type side, where for every free electron there was a positively charged phosphorus atom, on the p-type side for every positively charged hole, there’s a negatively charged boron atom. The net electric charge of the p-type side of our semiconductor is zero.
In this part of our question, we want to know the difference between these net relative charges. Zero minus zero is zero. So there’s no difference between the net relative electronic charge in the two regions.