In this video, we would discuss pn Junction under forward bias. Now let's recall the pn Junction of at equally region and the band diagram looks like this there is a built-in electric field across the depletion region. And in the energy band diagram that built-in electric field produces potential energy difference, and so it shows up as the energy band bending, producing an energy difference between P-side and in the N-side. And the energy difference of course is related to the built in potential. Now when you apply reversal or forward bias, then it changes the band bending and our approximation is that the entire potential drop. The voltage drop due to your applied bias is entirely dropped across the depletion region and there is no voltage drop in the neutral region, quasi-neutral region outside the depletion region. Now in general, the resistance in the quasi-neutral semiconductor region is not exactly 0. So there will be a finite voltage drop. But usually that voltage drop is negligibly small, especially in the modern devices. And even in the case that you have to include it, you can always include it as a series of resistance to your analysis at the very end. So in this analysis, we will assume that there is no voltage drop in the quasi-neutral region and your entire, applied voltage v sub r, the reverse bias. Or the v sub a you're applied forward bias, is dropped entirely across the depletion region here. So in that assumption, the energy band bending is given by replacing your built in potential with phi sub i plus Vr in the case of reverse bias or phi sub i minus Va in the case of forward bias. Okay, so now at equilibrium, diffusion and the electric field produced by diffusion, produced as a result the of diffusion of carriers. They balance each other out and so there is no net current. And obviously, when you don't apply any voltage, you shouldn't expect any current. And in pn junction that balance is established the tendency of diffusion due to the carrier concentration difference between the P side and the N side. And the built in electric field across the junction due to ionized owners and ionized acceptor. When they cancel each other out you have equilibrium. Under forward bias, you are lowering the energy barrier, or reducing, bending, or decreasing the built-in electric field. Therefore, diffusion takes place, diffusion prevails. And the diffusion moves the majority carriers to the opposite side. So there are a lot of carriers that can move, therefore you can expect a rapid increase in current. That's what happens in the forward bias. On the reverse bias however, you increase the electric field. So the electric field prevails, the drift current will win. However, the direction of the electric field is in such a way that it tries to move minority carriers to the opposite side, but the minority carrier concentration is very small, and therefore you don't really have any substantial current on the reverse bias. So this is the origin of the rectifying behavior, current voltage or behavior of the diode. And to analyze the forward bias current we adopt the assumption of low level injection. What does that mean? That means under forward bias majority carriers move to the opposite side. When they move to the opposite side they become minority carriers, they are minority carriers there. And so the minority carrier density will increase. And that increase the amount of minority carriers due to this forward bias current is still small compared to the equilibrium majority carrier concentration. So this is the assumption of low level injection. And the boundary condition will be that the majority carrier concentration and sub n naught, so n here is electron concentration obviously. Subscript n represents the n type side and o represent equilibrium. So this here is the majority carrier concentration on the n type side at equilibrium. At the edge of the depletion region, which is causing neutral reason, that should be equal to just the doping density. This is a standard result for intrinsic semiconductor; likewise the moderated carrier concentration add equilibrium on the p side should be equal to the doping density there, on the p side. Now let's estimate the carrier concentration, minority carrier concentration using these majority carrier concentrations. Now if you recall, the carrier concentration difference between two location is related to the potential difference between the two locations, okay? So the n sub p naught at negative xp, so what is this? Equilibrium minor retailer concentration on the p side, okay? At the edge of the depletion region edge on the p side, okay? That should be equal to the majority carrier concentration on the n side at equilibrium times the exponential factor containing the potential difference. But at equilibrium, the potential difference between n-type and p-type side is simply the built-in potential. So the potential difference is simply phi sub i here. Now the equilibrium majority concentration on the n side is simply equal to the doping density, and we can derive a similar equation for the minority carrier concentration, whole concentration on the n side in terms of the doping density on the p side and the built in potential. Under bias we can simply replace phi sub i with phi sub i minus Va because entire voltage, applied voltage has dropped across the depletion region and therefore all of your applied voltage is used to reduce the built in potential, okay? So these are the two equations for the minority carrier concentration under bias. Now combine those two equations and you can write out the excess minority carrier concentration, n prime at the edge of the depletion region. So n prime is the excess minority carrier concentration at the edge of the depletion region on the p side. By definition it is the minority carrier concentration minus the equilibrium minority carrier concentration and that is given by this equation here exponential factor minus 1. Similarly you can derive a similar expression for the excess hole concentration on the inside. Which is given by the same exponential factor here. So what does that tell us? As you apply a forward bias voltage Va, you're excess minority carrier concentration increases exponentially. Now, we can calculate the carrier concentration as a function of x, as a function of position in the neutral n type region and causing neutral p type region. How do we do that? By setting up and solving continuity equation. So in the quasi-neutral region there is no electric field. So the drift current components, drift current terms in the quasi-linear equation is zero. So you are left only with the diffusion current term here and the generation and the recombination term. There is no generation so you only have the recombination term. So that's the right-hand side and the left-hand side is the rate of change of your carrier concentration. So this is your continuity equation. And if you consider a steady state, then the left-hand side, the rate of change time derivative is zero. And so, and if you further assume that doping is uniformed then your equilibrium carrier concentration is uniformed. And therefore the x derivative of your carrier concentration is simply the derivative of the excess carrier concentration only. And the recombination term had the excess carrier concentration to begin with. So you have a very simple second order differential equation for excess minority carrier concentration. And you can solve this equation and the solution to this equation is an exponential function. And in general, you get an exponentially decreasing function and an exponentially increasing function.