In this video, we will continue our derivation of current in diode on the forward-biased and derive the ideal diode equation. So we left off with this solution to the continuity equation. So the excess minority carrier concentration on the n-side has this general solution which is basically a linear combination of two exponential functions, exponential with decreasing function and exponential with increasing function. Now we need to determine these two constants A and B. And these are to be determine by the boundary conditions and we already know one boundary condition. That is the excess carrier concentration at the edge of the depletion region, x equals x sub n. We need to know the other boundary condition which will be the other end of the quasi-neutral region, which is the end of the semi-conductor. That's where you typically put an ohmic contact non resistive and non rectifying metal contacts, so you can apply voltage and extract current. So we have to make some assumption on where it is located, and that consideration gives us these two cases here, long base diode and short base diode. Long base diode is where the end of the quasi neutral region, or the width of the quasi neutral region is very, very large. So you can effectively consider that your semiconductor p type side, or n type side, is semi infinite. That is the long base case, and the short base diode is this very close to the junction, close to the depletion region x. Now, how close is close and how far is far? You have to compare that with this L sub p quantity that goes into this exponential factor. And from the definition that we have used from the derivation of the continuity equation. The L sub p is given by square root of D sub p times tau sub p, diffusion coefficient times the carrier lifetime. Now, this quantity has a dimension of length and we call that the diffusion length. It is the average length your carrier travels before they recombine and disappear. So if the width of the quasi-neutral region is much larger compared to your diffusion length, you have a long-base diode. And if it is smaller, much smaller than the diffusion length, you have a short-base diode. Now before we move on, I just want to reiterate that in the quasi-neutral region, we have no electric field. Our assumption was that the resistance in the quasi-neutral region is negligible and, therefore, there is no voltages drop. Entire voltage drop occurs within the depletion region, therefore, in an ideal p-n diode, your current is entirely diffusion current and there is no drift current. In a long-based diode, the end of the depletion region, the quasi-neutral region, is located far, far away. You can consider the semiconductor as a semi-infinite material. And if I may go back to the previous slide here, the second term in your general solution contains an exponentially increasing term. Now, if x is allowed to become very large, this exponential term will blow up causing an infinitely large excess carrier concentration which is clearly non physical. So your constant B here must be 0. Now physically, what happens is that as the x becomes large, as the carriers moved far, far away from the depletion region, so deep into the crossing nature of a region, it experiences more and more re-combination process. So the fewer and fewer excess carriers survive and eventually by the time the carriers reach the end of the quasi neutral region, all of the excess carriers have recombined and your excess carrier concentration become 0. Your total carrier concentration reaches the equally reinvalue. So in either case, in any case, you must set the constant B as 0, and you are left with only one constant A. And that is to be determined by the other boundary condition which is the excess carrier concentration at the edge of the depletion region, which we derived in the previous video. So plug that in. You get the full solution of the excess carrier concentration as a function of x here. And plug that into the definition equation for diffusion current and you derive this expression here. So what does it look like? The diffusion current decreases exponentially as a function of x. So if you look at the graph, if you look at how the excess carrier concentration change as a function of x. So this is the depletion region, this is the edge of the depletion region, x sub n, and this is the boundary condition set by the applied voltage here. And they decrease exponentially as we move away from the depletion region deeper into the quasi-neutral region. And diffusion current for holes also decrease exponentially according to the equation that we derived in the previous slide. Now, you have to wonder, Kirchhoff's law says that unless there are branches, your current should remain the same as a function of position. So how come you have an exponentially decaying hole current here? Well, you have to think about what causes this hole current decay? The hole current decay arises from the exponentially decaying carrier concentration here. And there exponentially decaying carrier concentration arises from the fact that holes continue to recombine with electrons. That's why the hole concentration decrease. What that means, is that there has to be electrons that recombine and disappear. Recombine with holes and disappear as well. And there has to be electron current, electron current must be provided from the other side. From the right that has to flow in in order to replenish the electrons disappearing due to recombination with hole. So this current here represents the electron current. This represents the electron flux from right to the left. And these are the electrons that are replenishing those electrons that disappeared due to recombination with hold. So adding the electron recombination current to the whole diffusion current gives you a total current constant as it should according to the Kirchhoff's Law. Now, you can do the similar analysis for electrons in the p-side, you get a similar equation for exponentially decaying decreasing diffusion current for electrons. And the total current then is simply the summation of the electron current and the hole current and you get a very familiar exponential qVa over (kT- 1) times some saturation current. And the saturation current is given here, full expression is given here, this is called the ideal diode equation. Now, let's summarize. The total current in a forward-biased p-n junction consists of an injected minority carrier current, that's J(1). So on the n-side, you have an injected hole current which decreases exponentially. And on the p-side, you have injected electron current which decreases exponentially as well. And then, on both sides, you have majority carrier recombination current, here in j sub n 2 and j sub p 2 here. And then of course, you have to have a current here, majority carrier current, which provides these excess holes. So where do these holes come from? They come from the p-side. So this is the current that provides excess injected holes on the n-side, and same thing, the electric current here, j sub n, is the one that provides these excess electrons showing up on the p-side. And then, this is the ideal part, 1, 2, and 3 are the ideal part. And in general, there is some non-ideal current, which we will discuss later. Now, we consider short-base diode, and this is the opposite extreme. Your quasi-neutral region is extremely short. Short compared, again, to the diffusion length. So what that means is that the injecting minority carriers don't have enough time or don't have enough distance to travel to get recombined. So you can ignore recombination. So the continuity equation becomes even simpler. So the original recombination part goes away, so you only have the diffusion turn on the right hand side. Left-hand side is the time derivative for steady-state, that's 0. So you simply have a situation equation that says the second derivative of your carrier concentration is 0. What does that mean? That means your carrier concentration varies linearly. So the general solution will be A plus Bx and we have an only contact at the end of the quasi-neutral region. And the only contact is by definiti on a contact that allows free exchange of carriers. So if there are any excess carriers at the only contact that excess carriers are free to exit the semi conductor and goes into metal counter. At the ohmic contact, you always have 0 excess carrier concentration. So excess carrier concentration at x=WB end of the quasi-neutral region, the location of your ohmic contact should be 0. And again, use the boundary condition for x sub n, the edge of the depletion region. You then have the full solution, complete solution for p sub n. And once again, it is a linear function of x, okay? And it has the same familiar exponential term, depending on your applied voltage. So the diffusion current now, if you plug this into your diffusion current equation, the diffusion current now becomes constant. It is independent of your x, independent of the position. And you can do the similar analysis for electrons injected into the p-side and combine that into the total current. You basically get the same equation, and the exponential term here that contains the voltage dependents remain the same. The saturation current term in front of these exponential term also largely remain the same. The only difference is that the diffusion length that was originally located here in the denominator is now replaced by the W prime B and W prime E. This is the length of the quasi-neutral region on the n-side and W prime E is the length of the quasi-neutral region on the p-side. So if you plot the excess carrier concentration, it varies linearly from the edge of the depletion region to the end of the quasi-neutral region. Where the ohmic contact is located, and if you take a derivative of this, you get a constant. So your current is constant, and because there is no variations in current, there is no need for majority carrier recombination current. There is no recombination, therefore, there shouldn't be any recombination current anyway. So your minority carrier injection current here, Jp constant. And on the p-side, there has to be the same amount of current that provides these access carriers on the n-side. Likewise, the electron injection current on the p-side remains constant throughout the neutral region and there has to be the same amount of electron current on the n-side that provides these excess carriers. So this is the composition of current in a short-base diode