Rectification at a semiconductor junction
We would like to show next how it is that a $p$-$n$ junction can act like a rectifier. If we put a voltage across the junction, a large current will flow if the polarity is in one direction, but a very small current will flow if the same voltage is applied in the opposite direction. If an alternating voltage is applied across the junction, a net current will flow in one direction—the current is “rectified.” Let’s look again at what is going on in the equilibrium condition described by the graphs of Fig. 14–9. In the $p$-type material there is a large concentration $N_p$ of positive carriers. These carriers are diffusing around and a certain number of them each second approach the junction. This current of positive carriers which approaches the junction is proportional to $N_p$. Most of them, however, are turned back by the high potential hill at the junction and only the fraction $e^{-qV/\kappa T}$ gets through. There is also a current of positive carriers approaching the junction from the other side. This current is also proportional to the density of positive carriers in the $n$-type region, but the carrier density here is much smaller than the density on the $p$-type side. When the positive carriers approach the junction from the $n$-type side, they find a hill with a negative slope and immediately slide downhill to the p-type side of the junction. Let’s call this current $I_0$. Under equilibrium the currents from the two directions are equal. We expect then the following relation: \begin{equation} \label{Eq:III:14:12} I_0\propto N_p(\text{$n$-side})= N_p(\text{$p$-side})e^{-qV/\kappa T}. \end{equation}  You will notice that this equation is really just the same as Eq. (14.10). We have just derived it in a different way. Suppose, however, that we lower the voltage on the $n$-side of the junction by an amount $\Delta V$—which we can do by applying an external potential difference to the junction. Now the difference in potential across the potential hill is no longer $V$ but $V-\Delta V$. The current of positive carriers from the $p$-side to the $n$-side will now have this potential difference in its exponential factor. Calling this current $I_1$, we have \begin{equation*} I_1\propto N_p(\text{$p$-side})e^{-q(V-\Delta V)/\kappa T}. \end{equation*}  This current is larger than $I_0$ by just the factor $e^{q\Delta V/\kappa T}$. So we have the following relation between $I_1$ and $I_0$: \begin{equation} \label{Eq:III:14:13} I_1=I_0e^{+q\Delta V/\kappa T}. \end{equation}  The current from the $p$-side increases exponentially with the externally applied voltage $\Delta V$. The current of positive carriers from the $n$-side, however, remains constant so long as $\Delta V$ is not too large. When they approach the barrier, these carriers will still find a downhill potential and will all fall down to the $p$-side. (If $\Delta V$ is larger than the natural potential difference $V$, the situation would change, but we will not consider what happens at such high voltages.) The net current $I$ of positive carriers which flows across the junction is then the difference between the currents from the two sides: \begin{equation} \label{Eq:III:14:14} I=I_0(e^{+q\Delta V/\kappa T}-1). \end{equation}  The net current $I$ of holes flows into the $n$-type region. There the holes diffuse into the body of the $n$-region, where they are eventually annihilated by the majority $n$-type carriers—the electrons. The electrons which are lost in this annihilation will be made up by a current of electrons from the external terminal of the $n$-type material. When $\Delta V$ is zero, the net current in Eq. (14.14) is zero. For positive $\Delta V$ the current increases rapidly with the applied voltage. For negative $\Delta V$ the current reverses in sign, but the exponential term soon becomes negligible and the negative current never exceeds $I_0$—which under our assumptions is rather small. This back current $I_0$ is limited by the small density of the minority $p$-type carriers on the $n$-side of the junction. If you go through exactly the same analysis for the current of negative carriers which flows across the junction, first with no potential difference and then with a small externally applied potential difference $\Delta V$, you get again an equation just like (14.14) for the net electron current. Since the total current is the sum of the currents contributed by the two carriers, Eq. (14.14) still applies for the total current provided we identify $I_0$ as the maximum current which can flow for a reversed voltage. The voltage-current characteristic of Eq. (14.14) is shown in Fig. 14–10. It shows the typical behavior of solid state diodes—such as those used in modern computers. We should remark that Eq. (14.14) is true only for small voltages. For voltages comparable to or larger than the natural internal voltage difference $V$, other effects come into play and the current no longer obeys the simple equation. You may remember, incidentally, that we got exactly the same equation we have found here in Eq. (14.14) when we discussed the “mechanical rectifier”—the ratchet and pawl—in Chapter 46 of Volume I. We get the same equations in the two situations because the basic physical processes are quite similar.