I'd like to discuss the current response of an APD. The current response really depends on the number of secondary electron and hole pairs that are generated by the original photo generated carriers. In a typical APD, we eliminate from one side usually the p side, and then the photo generated holes are immediately swept away by the built-in field while the electrons are accelerated through the depletion region. If you have sufficiently strong bias, the E field depletion region is large enough for the electrons to attain the threshold energy and go through the impact ionization process. The original electrons donated by these solid circles are called parent electrons are initiating electrons and then the carriers generated by impact ionization are daughter carriers or secondary carriers. The secondary carriers can also go through impact Ionization and move across the depletion region towards the collecting contact. So, really what happens here is you get a cascade of secondary carriers from each single photo generated carrier. The electron current is opposite to the flow of the electrons because current is a flow of positive charge and P the hole current flows in the minus x direction and also undergoes impact ionization. So, we can actually write down some expressions for the current in these devices. Let's define a couple of key variables. So alpha, this is the electron ionization rate. We can define a similar quantity for holes. So for holes it's beta, and then G is essentially the photo generation rate. So, the current will be made up of two components, one from the holes and one from the electrons. So, we can go ahead and write down the electron current. So, we're going to write down the derivative of the electron current as a function of position and this is equal to alpha of x, Jn of x plus beta of x Jp of x plus qG of x. We can write down a similar expression for the holes. So, this will be dJp dx equals alpha of x Jn of x plus beta of x Jp of x plus qG of x. In steady state we know that the current J should be constant and this should be equal to the sum of Jn plus Jp. So, that essentially allows us to rewrite our expressions for Jn and Jp and we could write the expression for the electron current just in terms of the total current and the electron current and similarly for the holes. So, let's go ahead and do that. What we get is we get that dJn dx equals alpha of x minus beta of x Jn of x plus beta of x J plus qG of x. Then here we can write down an expression for the hole current. This is dJp dx and this is going to be equal to alpha of x minus beta of x and here this will be Jp of x minus alpha of x times J plus or minus here this should be a minus sign qG of x. So, that's essentially what we have for these two current component. But what we really are after here is not the derivatives. We're really after the total current. So, in order to get that we really need an integral. So, the total current J is going to be equal to the integral from x equals zero to x equals w of the equations here which are number one and two. So, that's what needs to happen. So, in order to do this we can define things to make our lives easier. So, we will define that the exponential here of minus integral of alpha minus beta dx this integral we're going to define as exponential of minus phi of x. So, let's go ahead and write down our expression for J. J is equal to Jp of w and then here exponential of minus phi of w plus Jn of zero plus q, and then the integral from zero to w of g of x e to the minus phi of x dx. We only have half the expression written here because we also have to write the denominator. So, the denominator here is just going to be one minus and then here zero to w alpha of x exponential here of the integral from x to w of alpha x prime minus beta x prime and then this dx prime here and then this whole thing multiplied by dx. So, this is a pretty big mouthful and it's also not as analytic expression as we'd like. So, we can look at some special cases. So, let's look at a special case where essentially we have the G the photo generated current equal to zero and we only have one carrier. Either electrons or holes. So, in this case let's assume that Jn of zero equals zero for a hole injection and then essentially here that Jp of w equals zero for only electron and injection. So now, we can calculate what we really want to know. Is we really want to know the electron hole multiplication factors. So, if I look at m I can write down essentially my total current compared with the hole current at essentially the beginning W. So, in this case we're going to end up with the one over and then here we'll have one minus the integral from zero to w of beta of x exponential and then here we'll have again the integral from x to w of alpha x prime minus beta x prime, this whole thing multiplied by dx prime and then the whole thing times dx because it's part of a larger integral. We can write down a similar expression here for the electrons, it's very similar. So, let me go ahead and do that. So, m here is going to equal to J over J and 0. So, these are the two cases of either hole or electron injection and here I'm going to end up with one over and then essentially all have a very similar expression this is one minus the integral from zero to w of beta of x. Excuse me, this should be alpha of x. So, alpha of x and then we're going to have exponential, and then here this will be the integral from x to w of alpha x prime minus beta x prime dx prime times dx. Those are really the two expressions that we got in these limiting cases. So, we can obtain alternate expressions for the electron multiplication factor Mn and Mp using other equations for Jn and Jp. It's also important to note that Mn and Mp have singularities that can cause them to go to infinity. If that happened, then essentially you get Avalanche breakdown and the device is unstable. So, really we want to restrict the APD operation to regions below Avalanche breakdown.
