The structure we’ve just shown you – the basic pn junction – is the very fundamental structure of modern semiconductors. Without it, none of what we have today would be possible. Now that we know that this is the critical structure, the next step is to demonstrate how exactly it is useful electrically. Before we can do that, though, we need to learn just a bit more about the complex interactions that take place inside a pn junction.
Let’s look again at the electric fields that are created when we combine p-type and n-type regions. Because opposite charges are left on either side of the depletion region (positive left on the n side, and negative on the p side), the electric field lines all point in the same direction! In other words, the field lines point from the positive, to the negative – the electric fields from each side have combined to form a very strong, concentrated electric field within the depletion region.
http://media.hardwareanalysis.com/articles/small/10757.gif" alt="Semiconductor Physics">Fig. 8 - An electric field is established inside the depletion region. Both the donor and acceptor ions left behind contribute to the strength of the field.
Something very interesting takes place here. Remember what we have when there’s an electric field over a region in space? We have a voltage! In fact, if we were to stick a meter on either end of our sample above, we would see that the n-type side is at a higher voltage than the p-type side! The diffusion of electrons and holes has resulted in a strong electric field, which means there’s a voltage across our sample. This is known as the built-in voltage
, and is typically on the order of 0.6 to 0.8V for most junctions. It is important to note, however, that the voltage differential occurs only
in the depletion region. The rest of the n- and p- type material has no voltage. If we were to plot the voltage of our sample, it would look something like this.
http://media.hardwareanalysis.com/articles/small/10758.gif" alt="Semiconductor Physics">Fig. 9 - Where there's an electric field, there's a voltage difference, and here we see that the internal electric field causes a voltage differential across the depletion region.
The n-type region is at a certain voltage, and there’s a constant increase in voltage across the depletion region due to the internal electric field. As a result, the n-type side is at a noticeably higher voltage than the p-type side. (Remember that Voltage is a potential, and is relative, so the absolute values of the voltages are not of consequence, just the difference between them. In other words, we could say the p-type side is at 10.0V, while the n-type side is at 10.7V. It would mean the exact same thing.)
Now, our pn junction, as it sits there, has a very interesting property: it does not conduct! The p-type sample, on its own, conducted, as did the n-type sample, but when we combined them and allowed this internal diffusion/drift equilibrium to form, we would discover that the sample no longer allows current to pass! Think back to our earlier discussion of bandgaps. Recall that when we drew bandgaps, the unit on the y-axis of our plots was electric potential! Therefore, if we have some potential difference across the depletion region, then the bandgaps must be at a different level on either side. This is precisely the case – the built-in voltage of the junction essentially drives the bandgaps apart, so that they look something like this.
http://media.hardwareanalysis.com/articles/small/10759.gif" alt="Semiconductor Physics">Fig. 10 - The voltage differential means that the bandgaps have now been pushed apart, due to the internal electric field.
Now think of a charge carrier – an electron, say – in the n-type region. Current is the movement of charge carriers, so for the sample to conduct, that electron would need to be able to flow across the depletion region, and into the p-type region. But the depletion region contains a very strong electric field which forces the bandgaps apart. The electrons are stuck in a lower energy level, and lack sufficient energy to break through the depletion region. Most students find it easiest to visualize this as a hill: the established built-in voltage causes bandgap separation that acts like a hill which, in order to cross, the electron would have to climb. Most of the electrons simply do not have the energy to climb the hill, therefore no charge carriers can cross the depletion region, so the sample cannot conduct.
http://media.hardwareanalysis.com/articles/small/10760.gif" alt="Semiconductor Physics">Fig. 11 - The difference in the bandgaps acts like a 'hill' that electrons (or holes) must climb if they wish to cross the depletion region to aid in conduction.
The same principle applies to holes attempting to move from the p- to n-type region; they’re held back by the energy difference caused by the electric field. Electrons will not move in the other direction (from the p- to n-region) simply because their concentration is so low in the p-type area, and so high in the n-type area. It would be like the cream in your coffee mystically confining itself to one area. Again, the same holds true for holes in the opposite direction.
So, all of a sudden, we now have a useless, non-conductive block of wood again, right? Not quite. Suppose we wanted to make this thing conduct. We might try to give the electrons more energy, so that they can make it up the hill; a logical, brute force solution. Or we can think outside the box – remove the hill!