Today, I'd like to focus on high-performance integers. High-performance integers use charged coupled devices with metal insulator semiconductor or MIS detector. To understand these, let's start with a simpler case. So, let's start with the case of the metal semiconductor structure. So, here, when you bring a metal and a semiconductor together, there are two types of junctions. One is referred to as a rectifying junction. It's a Schottky diode, but similar to a p-n junction but very different in the way it operates. The second kind of contact that can be produced is an ohmic contact. It's a non rectifying contact, and it produces a very small electrical resistance that's independent of bias. What you're looking at in this case with the Schottky diode, is the band diagram before the metal and the semiconductor make contact. So, in the metal, we have Phi m which represents the work function of the metal. It's the energy, the minimum energy it takes to ionize an electron and extract it from the surface of the solid to the vacuum level. On the semiconductor side, the electron affinity is what it takes to ionize the conduction band electrons. And Phi sub s is the semiconductor work function. It's the the energy required to extract an electron at the Fermi level to the vacuum level. In order to have a Schottky diode, we need to have the work function of the metal, to be greater than that of the semiconductor. If we don't have that, we have a different case. We have the case of an ohmic contact. So, here, Phi m is less than Phi s. And I've drawn for you, the band diagram, before we make contact in an ohmic contact, you can see the metal and it's work function, the semiconductor and it's work function. So, now, let's put the metal and the semiconductor together and draw the band diagram after they make contact. I'm first going to do this for the case of the Schottky diode. So here, the dotted line that I'm drawing for you is actually, the vacuum level. Here, my dashed line here, represents the Fermi level. We are at equilibrium. So, essentially that means the Fermi level should be flat. Now, between the metal and the semiconductor, we actually find we get a potential barrier. This is referred to as Q Phi B. Here's the work function of the metal. So, this is q Phi m. Then here, essentially, we can draw our valence band, and it looks very similar to that of the conduction band. So, let's label a few key quantities. Here, we have Q times the electron affinity Chi. So, that means that my Q Phi barrier is also equal to Q, and then here, we have Phi m minus Chi. We can also define the work function for the semiconductor. It's q Phi sub s. Finally, we can think about defining a built-in voltage for the semiconductor. This is nothing more than essentially here, Q VBI, or Q, and then here this is Phi m minus Phi s. So, the band bending that you see in the Schottky diode is essentially, due to the fact that electrons are flowing from the semiconductor to the metal. So, this essentially leaves a positive region behind, a positive space charge region. You can see that you got this potential barrier. From the metal side, it has a barrier with height of Phi B. It doesn't change as a function of bias, and then, on the semiconductor side, we have a built-in potential VBI, which is similar to that of the p-n junction. Now, let's look at the case for the ohmic contacts. So here, when I bring things together, I'm going to get something like this. This is my vacuum level, and then here is my Fermi level with the dashed line. So, this is EF, and then here, I'm going to end up with, here is EC, and then also, we're going to have EV. So, I can define again, here is Q Phi m and in here, you have Q Phi s. So, you can certainly subtract these two to figure out what's going on between your conduction band and fermi level. So, this is just Q Phi s minus Phi m. The key thing to note here is that we don't, actually have this barrier between the metal and the semiconductor that we do in the case of the Schottky diode. So, when contact is made, now, electrons flow into the semiconductor in order to achieve thermal equilibrium. So, essentially, electrons from the metal can move into the empty states in the semiconductor, and holes from the semiconductor are flowing into the metal. Let's now go back and focus on the Schottky diode. So, as we said, there are these two barriers, Phi B, this is on the metal side, and then here, we have the barrier on the semiconductor side, due to VBI. So, although, it's similar to a p-n junction, there's really some key differences. The current in a p-n junction, is really due to minority carrier diffusion. While the current in a Schottky diode, is due to thermionic emission of electrons that flow over a barrier. The Schottky diode is also a unipolar device in which electron motion dictates the current. There are two current components in these devices. Jn plus, and so this is where electrons, which I'll denote as this, are flowing from the semiconductor to the metal. Then we also have Jn minus, which is basically the opposite current, where electrons are flowing from the metal to the semiconductor. So, the key things to note is that the electron flux that generates Jn plus must overcome the potential barrier in the semiconductor. This can be changed significantly by simply applying a bias. In the case of Jn minus, the electron flux that generates Jn minus, must overcome the potential barrier on the metal side which is unchanged irrespective of external bias.