In this video, we will discuss pn junction under Reverse Bias. Now, let's first consider pn junction at equilibrium. At equilibrium, pn junction passes no current and this is a result of a balance between the diffusion of carriers, which is caused by the carrier concentration difference between the n-type side and the p-type side, and the built-in electric field which is a result of carrier diffusion which leaves ionized donors and acceptors behind. So, at equilibrium, the electric field builds up as shown here across the junction within the depletion region, and the direction of the electric field is in such a way that it opposes diffusion. So, if this electric field completely balances out the diffusion, then you reach equilibrium, there is no net current, the built-in electric field produces potential energy difference. So, in terms of the energy band diagram, it produces energy band bending as shown here, and that characterizes the energy difference between the n-side and the p-side, the amount of energy band bending is characterized by the built-in potential. Now, when you apply voltage, then you're tipping the balance, you're increasing the electric field or decreasing the electric field depending on the polarity of your bias. The convention is that the reverse bias voltage here, shown in the middle figure here, is in the direction that it increases the potential of the n side and it lowers the potential of the p side. So, that produces a greater electric field at the junction, it adds to the built-in electric field, and that means that the energy band bending across the junction becomes even larger. So, in this case, the carrier motion due to electric field prevails over diffusion. But if you look at the direction of this electric field, the electric field tries to drive holes from n-type side to the p-type side, but there are no holes or there are very few holes on the n-side. Likewise, the electric field direction tries to drive electrons from p-side to n-side but there are very few electrons in the p-side, they are minority carriers. So, even as you increase your electric field, there are no carriers that these electric field can drive to produce the essential currents. So, under reverse bias, you really don't have much current. On the other hand, under forward bias, you reduce the electric field, reduce the built-in electric field. In terms of energy, you are lowering the energy barrier, and now diffusion prevails. Now, the diffusion current drives majority carriers towards the other side so electrons from n to p, and holes from p to n, and there are plenty of these carriers, there are majority carriers, so the current is very large. This is really the origin of these asymmetric current behavior or what we call rectification behavior of the diode which makes this device very useful. To quantitatively describe the situation, the major change of the energy band structure under reverse bias or forward bias is the change in the band bending. So, at equilibrium, your band bending is characterized by the built-in potential V sub i, on the reverse or forward Bias, your band bending is reduced or increased by the amount of your applied bias. So, you simply replace V sub i with V sub i minus V_a, then use the same equation that we derived for the equilibrium situation. So, for example, if you want to calculate the depletion region width, all you have to do is change V sub i with V sub i minus V_a. Forward bias means V sub a is positive, reverse bias means V sub a is negative. Likewise, inlet maximum electric field which is the electric field at the junction is given in terms of the depletion region width, and so you just again substitute V sub i minus V_a for V sub i, you get this equation, and this is the maximum electric field on the bias. Now, under reverse bias as I mentioned a few minutes ago, you don't expect much current, you don't get a lot of current. So, the response of the diode pn junction is primarily capacitive. It increases and decreases the stored charge across the junction, this is the main response of the diode under reverse bias. By definition, your capacitance is the change in charge, dQ, in response to the change in applied voltage. Now, the charge stored at the junction is the space charge within the depletion region, the ionized donor and ionized acceptors within the depletion region. So, the change in storage charge is due to the increase and decrease of the depletion region in response to your bias voltage change. So, say that this was the depletion region width here, X sub n at certain voltage V sub R, reverse bias voltage, if you increase it by dV_R, then the depletion region width on the n side will increase by this much, dx sub n. Likewise, the depletion region width on the p side will slightly increase as well and we call that dx_p. So, if you know how much change in depletion region width on either side in response to the change in the reverse bias voltage, then you can calculate your capacitance. So, we already have an expression for the depletion region width x sub n and x sub p. And once again, you substitute V sub i minus V_a for V sub i in the equilibrium equation, then you get the equations for x sub n and x sub on the bias and you take the derivative with respect to the applied voltage. Then, from the definition of your C, capacitance in the previous slide, you derive this equation here and it simplifies into this very simple expression of epsilon s , the permitivity of the semiconductor divided by x sub d, the total depletion region width. So, this capacitance looks very similar to the parallel plate capacitor separated by distance x sub d and filled with a dielectric material with a dielectric constant epsilon sub s. Now, the capacitance, C, is proportional to the square root of one over the applied voltage. So, if you plot the applied voltage with respect to one over C squared, then you get a straight line curve as shown here in the red curve, and the slope of this curve is related to the doping density, and the x intercept is the built-in potential. So, this is a very standard technique that are used the capacitance voltage characteristics, is a standard technique used to characterize the semiconductor material in diode, and is commonly used to determine the built-in voltage and the doping density. Now, in this case, the doping density is kind of convoluted, you get one over N sub A plus one over N sub D. So, you don't get the individual doping densities on the n-side and p-side, but you get something that contains both. If you make a one sided junction, that is, if the doping density on one side is much greater than the other side, for example, if your acceptor density is much greater than the donor density, then the depletion region width on the p-side, the heavily doped side is negligible, the entire depletion builds on the lightly doped side, n-side, and your capacitance depends solely on the doping density of the lightly doped side. So, in this case, the slope, if you plot one over C squared versus the applied voltage, then the slope is inversely proportional to the doping density of the lightly doped side, and once again, the x-intercept is the built-in potential.