This slide shows the current flow in non-uniform doped semiconductor under equilibrium, and this slide is very important and very difficult. The reason that we are learning this is to understand the basic concept of the band diagram and the other thing is we want to learn this slide to derive Einstein relationship. First thing to understand is the semiconductor band diagram of the right side image. This is very important. The first thing we need to know here is the Fermi energy is the constant. In chapter 3.5, says that, "If Fermi energy is the constant, then this is the equilibrium." They solve this with the equation but I will not go over this. All you need to do is that you memorize this. If Fermi energy is the constant, this means this is under equilibrium. Equilibrium means that there is no voltage apply, there is no optical excitations occur, totally under equilibrium. This semiconductor is equilibrium because Fermi energy constant. Then what band diagram looks like this? This band diagram means then left side of semiconductor Fermi energy is close to the E_c, means very highly doped reason. Right side image is the Fermi energy is close to the E_i, very lightly doped or n_i. Then what's going to happen? This is the highly doped region, this is the less doped region. Carriers will move from the left to the right, from the high concentration to the less concentrated area by diffusion. Then what's going to happen? Electron move from the left to the right. There is positive As' fixed charge in left side. Then this is the positive charge and electron will be accumulated here, then there is the plus n, minus n. Electric field generate from the left to the right. Then electron then move from the left to the right by the diffusion will go back to the original position by this electric field drift current. How much drift current will flowing when diffusion current and drift current is equal? Think about that, if you don't applying voltage then there shouldn't be any current flowing. If you can design the device that you don't apply voltage and huge current flowing and then device can be operating like LED, there'll be exciting thing, but it's not going to happen. Under equilibrium, current is zero. How becomes zero in this case? Highly concentrated electron will be moved to the less concentrated area by diffusion then there will be fixed charges produced and positive charge left side and electron is moving back to original position by the drift. Towards drift current equal to the diffusion current, then current will be zero. This is the same thing for the hole, they're applying in independently. Electron case drift electron current plus the diffusion, electron current is zero by the diffusion and drift. Same thing for the hole for drift current plus the diffusion current should be zero. Where is the hole here? Minority hole is located in left side, more higher hole located in right side. Then higher concentration of the hole move to this side by the diffusion. An electric field is formed then same thing, holes are going back, how much? Same equal number of the electron current density occurs and total current should be zero. In this [inaudible] , another thing you need to know, how much electric field is generated. Electric field is voltage change per distance. So voltage is energy per queue. Therefore, electric field is proportional to the E_i, the changing by dx. So slope of the E_i is electric field. So one more time. Highly-concentrated electron carrier will be moved to the less concentrated area by diffusion. The fixed charge of the positive charges produced here and electric field is generated. How much d E_i per dx? Then because of this electric field, the electron will be moved back. With the equal number of the diffusion current and diffusion current cancel. Therefore, total current is zero for electron and hole. Electron drift current plus the electron diffusion current is equal and drift current is queue mobility, electron concentration, electric field plus qDn concentration gradient, which is that the d_n per d_x. Electron carrier concentration is n_i exponential, E_f minus E_i per KT. So dn per dx is the derivative of this one over kT and i exponential E_f minus E_i kT times dEi per dx, because E_i is function of the x. N equal this. So there's the same thing here. d_n per dx is Q. Q per kT and electric field, because the electric field is one over Q dEi dx. So if we put d_n per d_x, this equation two here, final equation becomes like this mobility diffusion factor. Then conclusion is the d_n per mobility equal kT per Q. This is the Einstein relationship for electron. Same thing for the hole, d_p per mobility hole equal to kT per queue. This is the Einstein relationship for hole. So let's use this equation and doing interesting thing to find out the diffusion constants. So in silicon, and you dope donor with a ten to the 14. From the graph of the carrier concentration versus doping concentration, you can find out that if you're doping ten to the 14, then carrier mobility of electron is 1400. Then from the Einstein relationship, at room temperature, kT equals 0.026, diffusion constant becomes 35.2 centimeters cubed for a sec. So we can find out the diffusion constant from the Einstein relationship when you know carrier mobility. Another concept we need to know to solve steady-state condition, excess carrier condition is the continuity equation. Continuity equation by concept from the equation is looks like this. It seems very difficult to concept. However, if you understanding the original concept, then it's much easier. Let's assume that there is some box, and the hole accumulation inside this box which is the d_p per d_t, as time goes by, is proportional to the hole comes in minus holes comes out plus hole generation minus hole recombination. So comes in is generation positive, recombination minus, exit is to minus. To express this by equation is the d_p per d_t hole buildup inside this box can be expressed by the holes comes in minus holes comes out in within a box, Delta x, and minus recombination. You might think that there is generation. However, in steady-state injection or steady-state excess carrier condition, we ignore the generation. So assumption of the steady-state injection is the constant minority excess carrier supply. Then the excess minority carrier is much higher than minority thermal generation. Let's say the excess minority carrier supply is 10_14, and then thermal generation, let's say the 10_3. So these will be generation is ignore, so we do not consider the generation. Another steady-state injection assumption is there is no build-up inside the box. As I said in previously, steady-state is the meaning that condition is maintained as a steady-state. There shouldn't be any build-up. Build up doesn't make sense because in steady-state, if you placed in the applying voltage in LED, LED light comes a steady-state not increasing the power in same voltage. So no builds up of the carriers, no builds up means this builds up come d_p d_t equaled 0. That's the two steady condition assumption. Therefore, d_p d_t is the flux in minus flux out minus recombination term. So this case also go to the conclusion first. This is the x-axis and this is the minority carrier hole. If you're constantly supplying the minority excess carrier, then there will be huge minority carrier concentration at x equaled 0. They will go to the inside of a semiconductor because there is concentration gradient. However, this minority carrier go inside the semiconductor, will then recombine with a majority carrier electron. Therefore, carrier concentration is decreasing until the carriers go to the original equilibrium conditions. In this condition, d_p per d_t equal to the flux of the minority carrier in and out minus the hole recombination. The only current generating in steady-state condition is diffusion current because the minority carrier is dominant in diffusion current. The hole cases, this is the minority carrier, let's say the 10_3. Then drift current is ignore because the drift current is dominant by the carrier concentration, then dominant by the majority carrier, not minority carrier. So majority carrier, in this case, is the electron and drift current. This is the minority carrier and dominant by the diffusion current, and drift current is ignore. Therefore, the minority carrier is diffusion current. So this is the p, and this is the n, same thing. But the major contribution of the current of minority excess carrier is the diffusion current, so you put diffusion current equation qD_n d Delta n per dx. So final conclusion is Delta n per dt equal D_n, second derivative of excess carrier per dx square minus recombination. Here says that current carried by diffusion of the minority carrier, so negligible of the drift current because the small portion of the minority carrier concentration in here. We're almost there. Let's start with the steady-state carrier injection. As I said, steady-state carrier injection, there is no buildup. Buildup means doesn't make sense if you put the LED voltage. LED will comes out constantly. No building up or more power of LED light, it does not comes out. Therefore, buildup will be zero, this equation becomes zero. Then, second derivative of excess minority carrier equal to the dp per diffusion constant and the recombination time. Here, diffusion constant time recombination time can be defined with the minority carrier diffusion constant L_p. Therefore, L_p equal root to D_p per recombination time. Let's start with this derivative equation. Possible solution for the minority excess carrier is Delta px equal C_1 exponential x per Lp plus C_2 exponential minus x per Lp. Because this is the second derivative, other than this equation, cannot satisfy this equations. We have two boundary condition where the x equal infinite, then excess carrier will be 0. Another boundary condition is x equaled 0, then excess minority carrier will be Delta p. So first boundary condition x equal infinite, then dp equaled 0, then C_1 should be 0. Second boundary condition, x equaled 0, excess carrier will be Delta p, then C_2 should be Delta p. So excess carrier concentration dpx equal to the Delta p exponential minus x per Lp means that excess carrier can be decreasing in exponentially as distance go deep inside of the semiconductor. So minority carrier expression by equation is px equal original p_0 plus minority excess carrier, which is the Delta p times exponential minus x per Lp. Now, you know everything. This is the steady-state carrier injection, where you supplying the minority excess carrier constant consistently from the left side. Then, choose minority excess carrier is supplied in the x equaled 0. Then there will be concentration gradient occurs. These minority excess carrier will go to inside of semiconductor by diffusion. Those excess carrier will be recombined with the majority carrier of electron and reducing the carrier concentration. Mixed up minority hole distribution versus x equal p_0 plus Delta p exponential minus x per L_p. Now, if you know the concentration equation versus distance, then you can calculate diffusion current, because the diffusion current is minus q diffusion constant dp per dx. Then you do derivative over minority carrier hole, which becomes like this. One more time. Excess carrier injection supplying left side, and then these are then recombined with the inside semiconductor, and there is carrier concentration gradient occur. Carrier concentration gradient also makes the minority excess carrier diffuse to inside of semiconductor. In conclusion of the lecture 4, there's a two condition of the one time exposure and steady-state illumination. In one time exposure, at t equals 0, excess carrier is generated and then after positive time, there's no optical excess carrier is generating, therefore, excess carrier of minority carrier can be decreased exponentially by recombination. This is a nanosec region, they go back to the original carrier concentration, which in this case is 10 to the five. The equation derivation is starting, dn per dt is thermal generation by recombination, and final conclusion of the minority excess carrier in one time exposure is Delta n exponential minus x per recombination time. So the x-axis is the time. For steady state illumination, at t equals 0 and positive time, there's always consistent optical excitation or excess carrier supply is existing from the left side. There is the consistent excess carrier is supply, and they will recombine with the majority carrier of electron and carrier distribution becomes decreasing. Because of this concentration gradient, there is a diffusion current is formed. Minority carrier hole in steady state illumination can be expressed by the p_0 plus Delta p, of excess carrier, exponential minus x per L_p. Here, x is not the time but the distance. This is the excess carrier concentration equation. If you know the carrier concentration equation in steady state condition, then you can calculate the minority carrier diffusion current by the concentration gradient equation. Comes like this. In lecture 5, we studying the p-n junction carrier concentration, starting from the steady state diffusion current without this equation. The derivation of this equation is dp per dt. The carrier comes in, carrier goes out, minus recombination. This dJ_p per dx, J_p is the diffusion current, so comes like second derivative term, and steady state condition low buildup of the carrier, so this becomes zero, and then equation there satisfying the second derivative is this format. Then excess carrier is Delta p. You can put this one to here. That's it for the excess carrier. In next lecture, we are going to learn the real p-n junction. Thank you.
