In this video we will conclude our discussion of pn Junctions Under Reverse Bias and we will concentrate on the junction capacitance. The junction capacitance appears even as part of the MOS transistor because it does have pn junctions, and it limits the speed of MOS transistors, so it is important to understand the basics about it. So, we'll be talking about the small-signal capacitance of a pn junction under reverse bias. Here we have the junction as we've seen it before. It is a step junction with uniform doping on the n side and uniform doping on the p side. And we have applied a reverse bias, V sub R. And certain charges have uncovered themselves on both sides and formed a depletion region as we have seen. Now, we're going to augment the reverse bias by a small amount, delta V sub R, like this. And the question is, what will happen to the charges? Of course, the total reverse bias is larger, so the charges on each side of the depletion region will increase. So on the n side, additional donor atoms will be uncovered and contribute to an extra positive charge. And on the p side additional acceptor atoms will be uncovered and contribute to the total negative charge. Now, where do these charges come from? Clearly, because these are immobile charges, for example, the donor atoms shown with the plus sign inside the circle are immobile atoms. So it could not be that somehow those charges move to that position. They were already there, but they were covered by electrons. So what happened is, the electrons now moved away from that part of the depletion region and left the donor atoms uncovered, and therefore with a net positive charge. So delta Q1 is an immobile charge on the n side and delta Q2 is an immobile charge on the p side. But it was not those charges that moved externally and moved to those positions. The electrons that moved out of where donor atoms are now, they moved in this direction and because they are negative charges, negative charge going up is equivalent to positive charge going down as I show. So we can think of a positive charge delta Q as corresponding to the negative charge going out of the depletion region. Similarly, on the other side, we had holes that were covering the acceptor charges. Those have moved out, but holes moving down being of positive charge is equivalent to a negative charge minus delta Q going in. The total structure is still neutral, so therefore the magnitude of these two charges, + delta Q and- delta Q, is the same. But they have the opposite sign. Now this is the total picture, talking about total charges and total voltages. If we instead concentrate on the effect of the change in voltage, delta V R to the change in charges, delta Q and- delta Q, then we can construct an equivalent circuit. An equivalent capacitance, C sub j, driven by some voltage. The voltage would be only the change of the total reverse bias. And therefore on the plates of the capacitor, we will have + delta Q, and- delta Q. So I emphasize that this small signal equivalent circuit only relates changes of charges, to changes in the voltage. It does not talk about the total charges and the total voltages. Now Cj is assumed to be a linear capacitor. So it can be defined as the charge on one of its plates divided by the corresponding voltage. So it's delta Q over delta VR. And of course, more properly we have to allow delta VR and delta Q to approach 0. So eventually Cj will be defined by a derivative. Now, what is the relation between delta Q2 here and- delta Q? They are equal, right? Because the amount of charge that was uncovered, delta Q2corresponds to the amount of holes that moved out or equivalently the amount of negative charges going in. So therefore I can write delta Q =- delta Q2. And I can replace this in here. So now we have the junction capacitance, in fact I will divide by the cross sectional area as I see it from above, from the top of the structure. And I will have the junction capacitance per unit area Cj prime I will call it, so it will be delta Q prime over delta VR, where Q prime is the charge per unit area. That, as we have already seen in this here, is equal to- delta Q2 so I replace delta Q prime by- delta Q2 prime, and I let the deltas go to 0, so we have a derivative which defines Cj prime. Now depending on the details of the junction, you have to use different expressions for Q2 prime. But we will concentrate on the step junction, which has uniformly doped p and n regions, and in fact the one sided step junction which we have called n + p. And that is the case where you have much higher doping on the n side than you have on the p side. For that situation, we have already derived In the previous video an expression for Q2 prime as a function of the reverse bias, so if you plug that expression in and you take the derivative with respect to the reverse bias you end up with this expression. So this expression relates the junction capacitance per unit area to the doping of the p side. The doping on the n side doesn't matter, as long as it is much higher than the doping on the p side. And it relates also this capacitance to the square root of the reverse bias. So the larger the reverse bias, the smaller the junction. One way to kind of see this also intuitively is that initially you had this distance, and now you have this distance. So it's as if you have made charges appear, which are farther and farther away from each other as you increase the reverse bias. Now if you look at this expression here, you can write it in this form, if you like. So, it's very easy to do an algebraic manipulation and end up with this form where instead of square root you have a power of sub j, which is equal to one half. And from the form of this equation you can see that when V sub R is equal to 0, Cj prime = Cj0 prime. In other words, Cj0 prime is nothing but the zero-bias value of the junction capacitance. This formula applies for zero bias and for reverse bias. It does not apply for forward bias. Now if you have more general doping profiles, it turns out that you can use the same form of the equation, only the values for phi bi and alpha j will be different. In particular, if you have a linearly graded profile, as we call it, so the doping varies gradually from the p side towards the n side, then alpha j is equal to one-third. For general doping profiles, typically alpha j is between one half and one third. The quantities alpha j, and phi bi, we will assume are given to us for a given fabrication process. At this point, we have finished our review of basic facts about semiconductors and pn junctions. As I promised, this was a rather fast review. I was not able to prove everything I gave you. But I consider this as a basic background for the main part of the course which will begin next week. See you then.