I'd like to talk about now, what happens when we apply a bias to a Schottky junction. I've left the band bending diagram in equilibrium up here just for reference. So, you know that there is this built-in potential in the semiconductor, and we know what happens here if we apply forward bias in a pn junction, this barrier becomes less. It's the same thing that happens here. So, if I apply forward bias, then essentially now my potential barrier is reduced and this becomes qVb-i minus V. So, essentially we've reduced the barrier, and so, the net current flow is this way. So, essentially what happens is that jn plus increases due to the lower barrier and nothing happens with jn minus. So, jn minus is unchanged, because, remember that when I apply a bias, it does not change the barrier between the metal and the semiconductor. Now, let's consider applying reverse bias. So, if we apply reverse bias we can draw the same diagram. So here, I have my vacuum level, here I have my barrier, this is q phi B, and then here, I essentially have my conduction band for my semiconductor. So, this is qVb-i. If I now apply a reverse bias, we know what happens in the case of the pn junction. What happens is that we increase that potential, and so what we find is we got something that looks like this. So here, this is for the case of reverse bias, and so this becomes qVb-i plus V. What happens now is this actually flips the direction of the current flow. So now, the net current flow is going into the metal. So, jn minus, so this is net current, and so, essentially, we can think of jn minus essentially goes down or jn plus, excuse me, goes down due to the barrier height, and jn minus is unchanged, because remember it's independent of the barriers. So, it's independent of bias, because it's set by the barrier between the metal and the semiconductor that does not change when I apply bias. So, if we look at this, we essentially now have something that asymmetric and we essentially have a rectifying current voltage characteristic. This is why it's referred to as a rectifying junction. So, if you look at the current voltage characteristic, actually it looks very similar to what you got from a pn junction. But, the problem is that the mechanisms that cause this are actually very, very different. So, we can go ahead and write down an expression for the total current from this kind of junction. It's going to be equal to Jn is equal to Jn plus, plus Jn minus, and so, this is nothing more than essentially JST, and then the exponential of q times V over k sub BT minus one. So, this looks almost identical to that that you get from the pn junction. Remember that JST is equal to A star, T squared, and an exponential of, and then here we're going to have minus q phi B over k sub BT. So, this is similar to the pn junction but it's got a couple of key differences. JST is a function of T indicating that the current is due to the thermionic emission of electrons, and JST depends on the barrier height. So, it's really material specific. So, if you want to change JST, you really need to change your metal and your semiconductor. Remember, that the barrier height, phi B is effectively equal to q and then phi m minus chi, and those are fundamental material properties. So, hopefully now you understand how a metal semiconductor Schottky junction works.