Scribd
Upload
44 views15 pages

Capacitor Modeling

The document compares the capacitance of parallel plates calculated using the simple parallel plate equation to that calculated using a finite difference model that accounts for fringing fields. The model calculates capacitance by solving Laplace's equation on a grid, determining the electric field and stored energy distribution, and integrating over the grid. It finds the capacitance predicted by the model is higher because it accounts for energy in the fringing fields not included in the simple equation. As plate separation increases or dielectric constant rises, the simple equation becomes more accurate since less energy is in fringing fields.

Uploaded by

Bill White
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
44 views15 pages

Capacitor Modeling

The document compares the capacitance of parallel plates calculated using the simple parallel plate equation to that calculated using a finite difference model that accounts for fringing fields. The model calculates capacitance by solving Laplace's equation on a grid, determining the electric field and stored energy distribution, and integrating over the grid. It finds the capacitance predicted by the model is higher because it accounts for energy in the fringing fields not included in the simple equation. As plate separation increases or dielectric constant rises, the simple equation becomes more accurate since less energy is in fringing fields.

Uploaded by

Bill White
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 15

Ignoring fringing fields, the

capacitance is
Area S
+Q A
C   0 r
 d
d
-Q
For our example, let the plate dimensions
be 1 m x 1m, separated by 1 m of air.
2
C   8.854 1012 m  1.0 
1 m
F
 8.854 pF
Area S 1m
+Q
Is this right? To test, the device was

d analyzed using the finite-difference
-Q
method.
Cnum  18.96 pF

The model predicts a higher capacitance


because there is energy in the fringing
fields that was not accounted for
previously.
The Model (1 of 4)
Step 1 – We put a device on a 2D grid.
The Model (2 of 4)
Step 2 – Compute the scalar potential by solving    r  V    0
The Model (3 of 4)
Step 3 – Calculate the field E  V

We already see
that the field is
not uniform
between the
plates due to
fringing.
The Model (4 of 4)
Step 4 – Calculate the electric flux density D   0 r E

Step 5 – Calculate the total stored energy W  1   D  E  ds


2 S

Step 6 – Calculate capacitance C  2W2


V0

Cnum  18.96 pF
Effect of Separation

w
1
d
53% Error w
2
d
36% Error

w
 10
As w becomes much larger than d, the field within the d
gap is more uniform and less energy resides in the 6.7% Error
fringing fields. Our simple equation is more accurate.
Effect of Dielectric Constant, r

r  1
w
2
d
36% Error r  2
w
2
d
15% Error
r  5
As r becomes larger, a greater fraction of energy resides w
2
between the plates and the zero-fringing fields d
approximation becomes more accurate. 4% Error
How Does the Model Work?
We construct separate grids for the dielectric distribution and the distribution of metals.
How Does the Model Work?
We approximate Laplaces’s equation using finite-differences (or finite elements, etc.)

 2V  0
 2V  2V
 2 0
x 2
y
V  i  1, j   2V  i, j   V  i  1, j  V  i, j  1  2V  i, j   V  i, j  1
 0
 x   y 
2 2

We collect common terms.

 2 2  1 1 1 1
   V  i , j   V  i  1, j   V  i  1, j   V  i , j  1  V  i, j  1  0
     
   
  
  
  

2 2 2 2 2 2
x y x x y y

This equation must be satisfied at each point in our grid.


How Does the Model Work?
We write our finite-difference equation at each point on the grid. This large set of
equations can be written in matrix form as

Lv  0

 V 1,1 
L
 
 V  2,1 
v   V  3,1 
 
 
  x y 
V N , N 

This equation is not yet solvable


because
v  L1 0  0
How Does the Model Work?
We must incorporate a “source” by enforcing the known potentials.

 #   #  #  #   #   #   V1   0 
  V2   0 
 #   #  #  #   #   #    
    
   V metal   
 0 0 1 0 0   m   Vapplied 
    
    
 #   #  #  #  #  #  VN x N y 1   0 
 # 
  #  #  #  #  #   VN x N y   0 
L v b
How Does the Model Work?
Calculate the potential

v  L1b
How Does the Model Work?
Calculate the E Field

E  V
How Does the Model Work?
Calculate the D Field

D   0 r E
Calculate total-energy stored

0
1
 
2
W   D  E ds    r E ds
2 grid 2 grid

Calculate capacitance

1 2W  0 2
W  CV02
2
 C 2  2
V0 V0  
grid
r E ds

You might also like