Question

Multi-Layer Capacitor Use MATLAB to plot the capacitance of a parallel-plate capacitor versus the separation distance d e [0.01, 1]mm between its two plates (take a suitable discretization step value). The capacitor has an area S = 1 cm-. The dielectric material between the two plates has relative permiativity Er that gradually increases from the positive plate along the distance d, until reaching the negative plate. The equation for Er is given by Er = 1+x where x is the distance of the layer from the positive plate (x = 0 refers to thelayer touching the positive plate, while x = d refers to that touching the negative plate). Provide a proof for the capacitance equation you are plotting. Hints and notes: • It might be helpful to discretize the dielectric material into N layers, each with relative permiativitiy Er, then take the limit on N. • Alternatively, you can use the integration approach. • The total grade is 20 points. You will be graded based on the following: o 7 points: Derivation of the capacitance equation and any needed schematic diagram(s) to support your way of proof. o 7 points: MATLAB code o 6 points MATLAB plot (axes have to be labeled with units)

Answer 1

x=d area = A dx x+dx x=0

If dx is taken as the infinitesimal width of each layer, the capacitor can be imagined as d/dx parallel plate capacitors in series. The total capacitance can be obtained using the formula for equivalent capacitance of n capacitors in series.

$\small \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ...+\frac{1}{C_n} = \sum_{i=1}^n\frac{1}{C_i}$

Capacitance of ith layer with width dx, cross sectional area A and relative permittivity [for ith strip, x =distance from positive plate= idx] is given by:

Eri = 1+ ide

where i = 1, 2, ... n=d/dx [dx is chosen such that n=d/dx is an integer]

C;= _EQEriA dr

Equivalent resistance :

1- ΣΑ dr Ce

MATLAB code:

e0 = 8.854*10^(-12); % permittivity of vacuum
dx = 1e-7; % distance in meter between two consecutive parallel areas
x = 0;
a = 0.0001; % area of parallel plate in m^2
Cinv=0;
while x <= 1e-5
Cinv = Cinv+dx/((1+ x)*e0*a); % equivalent capacitance calculation for
% capacitors in series
x = x +dx;
end
C(1)=1/Cinv;
d(1) = 1e-5; % lower limit of d in meter
i=2;
while x < 0.001 % upper limit of d in meter
d(i) = x+dx;
x = x+dx;
Cinv = 1/(C(i-1))+dx/((1+ x)*e0*a);
C(i) = 1/Cinv;
i = i+1;
end
figure;plot(d*1000,C/1e-12,'r','linewidth', 2) % Show d in mm and capacitance in pF
xlabel('d (mm) --->')
ylabel('C (pF) --->')
xlim([0.01 1])

C (PF) ---> 0.2 0.4 0.6 d (mm) ---> 0.8