In this video, we will discuss current in Schottky contact. Now, the electrons in metal and semiconductor both are subject to thermal motion, and that's what really derive the electron migration from semiconductor to metal and metal to semiconductor. So, at equilibrium, the electrons in the semiconductor side want to move to metal, because metal band structure offers a lower energy. Now, as they migrate over, they leave the ionized donors behind which produce electric field, which then stops the migration of electron by having this built-in electric field. In the energy band diagram, that built-in electric field is represented by this band bending that produces this energy barrier. So, this energy barrier q times V_bi here is basically the V_bi is another notation for V sub i, the built-in potential, and this built-in potential represents the amount of band bending or the magnitude of electric field that perfectly balances the electron migration from semiconductor to metal driven by this energy difference. Now, if you apply a forward bias voltage, you're lowering the barrier as shown here, and what that means is you're reducing the electric field. Therefore, the electron migration will continue, will prevail. So, you have a large forward current flowing across this device. Now, what drives, what allows electrons to overcome this barrier, because even in the forward bias case there is a finite energy barrier, an electron need to overcome this energy barrier, so what allows the electrons to overcome this barrier, that's their thermal energy. So, the electrons in the semiconductor side and metal side have this energy distribution due to their thermal energy according to the Fermi-Dirac distribution. So, at a given temperature, there is a certain fraction of electrons that have high enough energy to overcome this barrier. As the barrier is lowered, there are exponentially larger number of electrons that have high enough energy to overcome the barrier, and that's why your current increases exponentially as you apply forward bias. So, this current driven by the thermal energy of electron is called a thermionic emission current. So, in Schottky contact, the forward current that we're seeing is thermionic emission current. Also, the Schottky contact is a unipolar device, because you now have a metal and a semiconductor which could be either n-type or p-type. Now, in this example, we're discussing n-type semiconductor. So the majority carriers of the semiconductor is the ones that is solely responsible for the current, there are no minority carrier currents or minority carrier currents are negligible. So, only one carrier species contribute to the current in this type of devices called a unipolar device. So, metal n-type semiconductor Schottky contact, only electrons contribute to current, metal p-type semiconductor Schottky contact, only holes contribute to current, and they both are unipolar device. So, now let's calculate the current. So, let's define the electrons flowing from semiconductor to metal, the current density due to those is J plus. So, this is the current flowing from left to right and electrons flowing from right to left. The electron moving from metal to semiconductor represents the opposite directional carrier density, J minus. So, to calculate J plus, we basically go back to the definition of your current density. So, your current density is basically the charge times the velocity times the concentration of electrons, and that's the density. But the concentration of electrons that actually contribute to J plus are the ones that have enough energy to overcome the barrier. So, the n bottom to n infinity is basically represents the electron with energy this much, electron that has a minimum energy equaling this q times built-in potential minus the applied voltage. That and above, anything above will be able to go over. So, that's the integral limit. To calculate dn, the fractional electron density, we go back to the general equation that the electron density and semiconductor is given by the density of state times the probability function, probability of finding an electron in energy E. So, we plug that in, then we turn this integral over n into an integral over E, and this integral runs from the bottom of the conduction band to infinity. The velocity here is only the X directional velocity matters. Y and Z directional velocities don't contribute to the X directional current. So, we consider a non-degenerate semiconductor, and in a non-degenerate semiconductor, the Fermi-Dirac probability function can be approximated with Maxwell Boltzmann distribution function, probability function, which is just a simple exponential function as shown here. Now, in the numerator, we insert plus and minus E sub C, and separate this single exponential term into two exponential term here. When you do that, then there is one term that is independent of energy, so this term can be taken outside the integral, and this term here that does contain energy, we notice that E minus E_c is the kinetic energy of your electron. So, the kinetic energy of the electron can be written by the standard classical kinetic energy equation, one half mv squared, and if you take the derivative of this, then you get an equation for the differential energy related to differential velocity. So, using this, we are going to turn, you're going to change the energy integral into the velocity integral. Also, the other term in the integral which is the density of state, we go back to the expression for the 3-D semiconductor density of state here, which has the square root of the energy, square root of E term. E minus E_c here is once again the kinetic energy. So, substitute one half mv squared into this, write down the density of state in terms of v as well. Now, we turn the dn into an integral of dv. Now, here, we have equation in terms of the velocity magnitude or the amplitude of the velocity, but we need to separate out the X directional velocity from the Y and Z directional velocity because only the X directional velocity is the one that contributes to the X directional current once again. So, to do that, we use the v squared here, is simply the sum of vx square plus vy square, vz square, okay. So, use this to replace the v square here and v square here. To replace dv, we use the fact that 4 pi v squared dv is basically a volume element in a spherical polar coordinate and that should be equal to the volume element in the Cartesian coordinate dvx times dvy times dvz. So, using this, you turn this integral into three separate integral of over vx, over vx, vy and vz. These vy and vz integral or something that you can just integrate out, you can just look up the integral table and this gives you a number. The integral over vx is a little different because you have a vx here in front of the exponential term, but you can also integrate this, it's something that you can integrate. Once again you can find this integral in your integral table, and you do the integration, you get this equation here. You now have an expression that has only one unknown quantity, and that is this v_x0, and v_x0 is the minimum velocity corresponding to the minimum kinetic energy required to overcome the barrier. What is the minimum energy required to overcome the barrier? And that is the barrier height. Which is the q times the built-in potential minus the applied voltage. So now, let one half m v_x0 square equate that equal to this, then you arrive at the final expression for J plus, the current density in the positive x direction and that is this. This entire thing here, the entire quantity here is lumped into this A star, except for this T square dependence, and this A star is given here, and this is called the Richardson constant, it's a material specific constant because it has this effective mass. So, you will find, in the reference table that lists Richardson constant for various semiconductor materials, so you can use those, we also use the barrier height, is simply the E_c minus E_f plus the built-in potential. So, this here is the final expression for your positive directional current density, and now we only need to find the opposite, the current density to get the total current density. So J minus, but J minus is simply equal to J plus at equilibrium. So, J plus at v equals 0, applied voltage equals 0, that should be equal to J minus. So we just take the equation in the previous slide, plug in V_a equals 0 and let that equal to J minus. So now, you add those two, J plus and J minus, and you arrive at the final equation for the current in the Schottky contact, and this familiar exponential qV_a over kBT minus one has the same dependence, same identical dependence as the p-n junction diode. The only difference here is in this pre-factor here, saturation current density, and the saturation current density in the p-n junction case contained the terms like diffusion coefficients and the intrinsic carrier concentration, because the current flowing in the p-n junction is diffusion current. In the case of Schottky contact however, the J saturation J_st is given by this, Richardson constant T-square and the exponential term containing the barrier height. This is due to the fact that the mechanism of current in Schottky contact is not diffusion current but thermionic emission current.