1. EMLAB 1 6. Capacitance

2. EMLAB 2 Contents 1.Capacitance 2.Capacitance of a two-wire line 3.Poisson’s and Laplace’s equations 4.Examples of Laplace’s equation 5.Examples of Poisson’s equation : p-n junction

3. EMLAB 3 Capacitance

4. EMLAB 4 Due to Coulomb force, positive charges rush to the capacitor. As the amount of charges increases, the voltage increases. If the voltage difference between the terminals of the capacitor is equal to the supply voltage, net flow of charges becomes zero. Charging capacitor

5. EMLAB 5 Potential distribution near parallel plates

6. EMLAB 6 Capacitance The magnitude of an electric field is proportional to charges, and voltages are proportional to electric field. Hence, charges are proportional to voltages. This proportionality constant is called capacitance.

7. EMLAB 7 Capacitor

8. EMLAB 8 Variable capacitor

9. EMLAB 9

10. 10 Example: Capacitance of a parallel plate capacitor

11. EMLAB 11 Capacitance from electrostatic energy Example : parallel plate capacitor

12. EMLAB 12 Capacitance of 2-wire transmission line PEC In this problem, capacitance between two parallel wires needed to be obtained whose radius is b. The distance between the center of the wires is 2h. - + + + + + + + + +- - - - - - - - With finite radii, the distribution of charges on the wires are non-uniform, which prevents the application of Gauss’ law. - To simplify the problem, the potential field is approximated by that of two wires with infinitesimal radii. - - - - - - - - - - - - - - - -

13. EMLAB 13 Example : transmission line

14. EMLAB 14 Capacitance : 2-wire line (a,0,0) (-a,0,0) P(x,y,0) Equation for equi-potential surfaces If the surfaces of wires are fitted to those equi- potential surfaces, boundary conditions on PEC’s are satisfied. With the surfaces fitted to equi-potential contours, the relation between the voltage and the charge density is

15. EMLAB 15 Capacitance of a wire above a PEC plane (a,0,0) (-a,0,0) If image method is adopted, the PEC plane is replaced with a wire with negative charges. Then the problem geometry becomes that of the previous example. In calculating capacitance, it should be noted that the voltage difference is V as compared with 2V of the previous example. (a,0,0)

16. EMLAB 16 Poisson’s and Laplace’s equations

17. EMLAB 17 Derivation of Poisson’s & Laplace’s equations for homogeneous medium Laplace operator has different forms for different coordinate systems.

18. EMLAB 18 Laplace’s equations The differential equation for source-free region becomes a Laplace equation. (rectangular coordinate) (cylindrical coordinate) (spherical coordinate)

19. EMLAB 19 Uniqueness theorem Solution of a differential equation The solution that satisfies the differential equation and its boundary condition is unique regardless of the solution procedure. To prove the uniqueness theorem, we assume that there exist two distinct solutions that satisfy the same boundary condition. Then, the difference of the those two solutions will have zero value at the boundary and will have non-zero value in the interior region. This situation is contradictory to the original assumption that there exist two distinct solution that satisfy the same boundary condition.

20. EMLAB 20 Example 1 : Laplace eqs. Unlike the procedures in the previous chapters, the potential V is first obtained solving Laplace equation. Then, using the potential, E, D, , Q, C are obtained. If the plates are wide enough to ignore the variation of electric field along x and y directions Using the boundary conditions on the two plates,

21. EMLAB 21 Example 2

22. EMLAB 22 Example 3

23. EMLAB 23 Scanning tunneling microscope probe Bi-conical antenna Examples

24. EMLAB 24 Intrinsic semiconductor

25. EMLAB 25 Semiconductor doping

26. EMLAB 26

27. EMLAB 27 Example : Poisson eq. Diode (simple model) P-typeN-type P-type region N-type region

28. EMLAB 28 Method of separation of variables The original equation on the left is split into three equations containing single variable. The equations are related to one another through variables  and . ,  can be fixed using boundary conditions. If the boundary condition becomes complex, three dimensional Laplace equation is very difficult to solve. This is because the Laplace equation is a partial differential equation that contains three variables x, y and z. If the boundary is parallel to coordinate axes, the solution to Laplace equation can be represented as a multiplication of three functions that contain only one variable. This method is called as a method of separation of variables.

29. EMLAB 29 Example : Separation of variables If the boundary extends to infinity in the z-direction, the derivative with respect to z becomes zero.

30. EMLAB 30 Result-Matlab code % potential_demo.m % 변수 분리 방법에 의해 푼 전압을 그림. clear,clf; hold on; imax = 30; jmax = 30; a=2.5; b = 1; for i=1:imax+1 for j=1:jmax+1 x(i,j) = (i-1)*a/imax; y(i,j) = (j-1)*b/jmax; z(i,j) = Vseries(x(i,j),y(i,j)); end colormap jet; surf(x,y,z); shading interp; %caxis([0,100]); colorbar('vert'); view([0,90]); %Vseries.m % 변수 분리 방법에 의한 전압 계산 function f=Vseries(x,y) V=100; f=0; a=2.5; b=1; for i=1:10 n=2*i-1; f=f+sinh(n*pi*x/b)/sinh(n*pi*a/b)*sin(n*pi*y/b)*(1/n); end f=f*4*V/pi;

31. EMLAB 31 Solving Laplace equation through Finite difference method Electrostatic potentials can be found from the Laplace equation. Numerical Laplace equation

32. EMLAB 32 The unknown voltages on the lattice points are set to V(i,j). Using the numerical Laplace equation, the unknowns are related to one another. If as many equations as the number of unknowns are generated, a set of simultaneous equations is formed that has unique solution. Matrix formulation