Lecture 7: Electron Emission

Transcription

1 Lecture 7: Electron Emission Solid state physics of metals E_f > E_c --> Many conduction carriers E_f - Fermi level E_c - minimum conduction band energy A realistic potential well in a metal crystal An approximate potential well, where phi_w is the work function. Here the conduction band bottom looks approximately like a square well. n - number of conduction band carriers g_c(e) - density of states function (a function of energy) f(e) - Fermi function

2 So, for a metal at room temperature 1) Fermi level is in the conduction band 2) Tail of g_c(e)*f(e) extends to high kinetic energy 3) if kt raised enough, the tail of g_c(e)*f(e) > phi_w, and electrons will escape the metal Thermionic emission Examine particle momentum momentum of particles just escaping over the work function number of escaping particles escaping current Note that <... > means average over all carriers The density of states function, as a function of momentum is where h is Planck's constant

3 And the Fermi function as a function of momentum is So we can find the current density of carriers with enough energy, and momentum in the correct direction to overcome the work function, and exit the metal we can integrate using the identity which can also be integrated, and we can substitute back in for the just escaping momentum Richardson-Dushman Equation

4 In this expression: J_th -- thermionic current density [A/m^2] e -- electron charge [C] m -- electron mass [kg] h -- Planck's constant [J*s] k -- Boltzmann's constant [J/kg] T -- metal temperature [K] phi_w -- metal work function [ev] Note: work function and Fermi energy were expressed in Joules here. They are more commonly tabulated in electron-volts. to convert, multiply quantity in ev by the electron charge. 1 ev = E-19 J We typically determine the constants in the Richardson-Dushman equation empirically, since it is hard to know exact work functions, other losses, etc. A [A/(cm^2 K^2)] phi_w [ev] Cu W Ta W(Th) BaO SrO We can measure these paramenters for a given thermionic emitter by construction a Richardson-Dushman plot. We measure current in [A] as a function of temperature in [K]. If we know the area of the thermionic emitter we can calculate J=I/area. With current density versus temperature, we can make the following plot.

5 We can make a couple of corrections to improve our accuracy of this solution Quantum mechanical transmission over a barrier -- we should include a QM calculation of the probability of transmission over the barrier. Since we usually use the form, and empirically determine the constants, this is not so important. Schottky Effect -- in the presence of and electric field, the height of the barrier (the effective work function) is lowered. This we do need to account for in the R-D equation in these circumstances. There is an image charge within the metal, of opposite sign and equal distance, which produces and attractive force The work function barrier height (expressed in Volts) is lowered by the image charge effect. If an electric field is also applied, the total barrier height in Volts is given by the following. image charge electric field original barrier

6 to find V_max, take dv/dx = 0 approximate thermofield emission solution Field Emission + by the Schottky effect, in the presence of an electric field, the barrier has a finite width + if the field is high enough, particles can tunnel through the barrier + this is called field emission Fowler-Nordheim Equation

7 Assumes + T = 0 K, at high T, there is a significant thermofield effect + 1-D smooth planar barrier + image force and applied field as in the Schottky effect As with R-D, we can write out a more functional, empirical form of the Fowler-Nordheim equation phi_w -- work function [ev] beta -- field enhancement factor [1/cm] V -- voltage applied [V] note, the electric field at the metal surface is E=beta*V Fowler-Nordheim plot

8 Field emitter arrays - microfabricated cathodes metal insulator - emitter tip radius ~ nm, so high field enhancement - tip J > 1E8 A/cm^2 - cathode an array of tips Triple point insulator Photoemission vacuum metal - high field enhancement at the insulator-vacuum-metal junction triple - can cause field emission from the point cathode - leads to secondary emission avalanche on insulator, potentially + if photon energy, h*nu, is larger than the work function, and electron is emitted + J is a function of the intensity and temperature + not very efficient cathodes but they have a low thermal spread Secondary Emission secondary emission yield curve + and electron or ion collides with a surface, releases a secondary electron + important in electron multipliers, magnetrons, and RF or pulsed flashover and breakdown. + the SE yield curve, for many materials has a peak > 1. Hence 1 electron or ion in yields > 1 electron out. + for energies EC1 < E < EC2 the surface charges positively + otherwise charges negatively

9 Secondary electron emission avalanche metal insulator + an electron is emitted from the cathode, K due to field emission + gains enough energy from the applied field to strike insulator with EC1<E<EC2 + insulator charges positively, pulling the released secondary electron toward insulator + if enough positive charge, secondar electrons have EC1 < E < EC2, pushing the positively charged region toward anode + discharge walks down insulator + generally supposed that adsorbed gases play a role at late times, forming plasma bridge Theory W_o = electron emission energy from K v_o = electron emission velocity say v_o along the x-direction there is some E_x electric field from charging, and some E_y from the applied electric field electron travels vertically a maximum distance from the insulator returns to insulator within

10 energy upon return if a cosine distribution of initial angles if the return energy is greater than EC1, get more secondaries than primaries and get positive charging on the insulator, amplifying E_x Divide the electric field strength at the insulator surface into one with charge and one without. The field E_o = V/d (the applied field), near the surface of the insulator is bent from theta by the presence of the insulator. So with no charge on the surface, we get the following with surface charge on the insulator

11 define, using the returning energy above To determine the charging necessary to result in an avalanche, take the return energy to equal EC1 for the material, W_o as the average secondary energy, and E_o as the applied field.

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