Video Transcript
A doped semiconductor that contains
acceptor ions and is at thermal equilibrium is modeled using three variables. The density of free electrons in
the semiconductor is represented by 𝑛. The density of acceptor ions in the
semiconductor is represented by 𝑁 subscript 𝐴 superscript minus. The density of vacancies in the
semiconductor is represented by 𝑝. Which of the following formulas
correctly represents the relationship between these variables in the
semiconductor? (A) The density of vacancies equals
the density of acceptor ions plus the density of free electrons. (B) The density of vacancies equals
the density of free electrons minus the density of acceptor ions. (C) The density of vacancies equals
the density of acceptor ions minus the density of free electrons. (D) The density of vacancies equals
the density of free electrons divided by the density of acceptor ions. (E) The density of vacancies equals
the density of acceptor ions divided by the density of free electrons.
This question is asking about a
doped semiconductor. But we’re gonna begin by first
thinking about a pure undoped sample. Recall that for an undoped sample,
free electrons and vacancies are created in pairs whenever an electron receives
enough thermal energy to break its atomic bond, become freed, and leave behind a
vacancy or electron hole in its place. Recall that in order to have
thermal energy, the temperature of the semiconductor must be above absolute
zero. If we had a pure sample at absolute
zero, there would be no vacancies and no free electrons. So, throughout the course of this
question, we’re going to assume the temperature and, therefore, the quantities 𝑝
and 𝑛 to be greater than zero.
So, for a pure semiconductor, we
know that for each vacancy in an electron shell, there’s a free electron somewhere
in the sample that used to occupy a space in that shell. Therefore, in a pure semiconductor,
the density of vacancies must be equal to the density of free electrons, or 𝑝
equals 𝑛.
Keeping this in mind, let’s move on
to think about our doped sample, which contains acceptor ions. So we know that the concentration
of acceptor ions is greater than zero. Recall that acceptor ions are first
introduced to the atomic lattice with one fewer outermost electron than the
surrounding atoms that made up the original undoped sample. So there’s room in its outermost
shell for one more electron. And that room for one more possible
electron constitutes one vacancy. Therefore, doping with these
acceptor ions has increased the concentration of vacancies in the semiconductor
without creating any new free electrons. So, once the sample is doped, we
can no longer say that 𝑝 equals 𝑛.
So now we need to figure out which
of these formulas actually does correctly model our doped semiconductor. Let’s start down here, where answer
choice (E) says that the density of vacancies equals the density of acceptor ions
divided by the density of free electrons. And answer choice (D) is pretty
similar, saying that the density of vacancies equals the density of free electrons
divided by the density of acceptor ions. To determine whether either of
these makes sense, let’s consider the process of taking a doped sample and somehow
making it more and more pure, as if we’re removing the acceptor ions along with the
vacancies they came with. If we were to do this, we would be
reducing the density of acceptor ions down closer and closer to zero.
Let’s see how these two formulas
would react to this process. First, looking at (E), if the
density of acceptor ions goes to zero, that means the numerator would go to zero,
causing this entire quantity and, therefore, the density of vacancies to go to zero
as well. Now, looking at (D), if the density
of acceptor ions goes to zero, the denominator would be going to zero, causing this
entire quantity and, therefore, the density of vacancies to increase without limit
to ∞. But because for a pure
semiconductor 𝑝 equals 𝑛, we know that as a doped sample becomes more and more
pure as the density of acceptor ions goes to zero, the density of vacancies should
just become the density of free electrons, which as we established earlier is not
zero and it’s definitely not ∞. So we know (E) and (D) are
incorrect.
Option (C) says that the density of
vacancies equals the density of acceptor ions minus the density of free
electrons. The subtraction of 𝑛 implies that
the concentration of free electrons somehow counts against the concentration of
vacancies. This isn’t true. Each free electron had to have
created a vacancy. And therefore, the density of free
electrons contributes to, not takes away from, the density of vacancies. Answer choice (C) is wrong as
well.
Answer choice (B) says that the
density of vacancies equals the density of free electrons minus the density of
acceptor ions. This implies that adding acceptor
ions would decrease the density of vacancies. But we’ve already established that
adding acceptor ions does the opposite. It increases the density of
vacancies. Therefore, (B) is incorrect as
well.
And we’re left with answer choice
(A), which says that the density of vacancies equals the density of acceptor ions
plus the density of free electrons. In a doped sample, the total amount
of vacancies must equal the amount of vacancies before it was doped plus the amount
of vacancies contributed by the acceptor ions. This agrees with the formula. And therefore, in our doped sample,
the density of vacancies equals the density of acceptor ions plus the density of
free electrons.
