Simplify

We now make a number of simplifications:

• 
  

  The conductors are perfect, so no charge will penetrate into the volume of the conductors, i.e., they are surface charges.

  
  
• 
  

  We divide each conductor surface into much smaller surface segments where the charge is constant: see Figure 6.1.

  
  
• 
  

  There is only one dielectric medium, so no dielectric boundaries need to be considered.

  
  
• 
  

  We will ignore the complication of self-interaction where a given charge impacts the voltage on its own segment (there is a singularity). This is an important effect and not that difficult to deal with, but for illustrative purposes we will not address this problem here.

Figure 6.1 Conductors in three dimensions where the surface is divided up into smaller areas within which the charge is constant.

Solve

Let us now define the problem by looking at the potential from a charge on a surface segment jj, at a point rr. We have from equation (4.37)

Vr=∫Surface_jρjr′r−r’dr’.

We now calculate this voltage at a point in the center, r_iri, of another segment ii and sum over all segments. We find

Viri=∑j∫Surface_jρjr′ri−r’dr’=∑jρj∫Surface_j1ri−r’dr’.

This can be written as a matrix equation:

From the boundary condition we know the value of VV on all surfaces, and we can calculate GG and solve for ρρ. To get to capacitance, let us be a little more careful with notation. Let us assume we have N_segNseg conductor segments, each of which is subdivided into N_subNsub charge segments. We have a total number, N_tot = N_segN_subNtot=NsegNsub of charge segments. Let us denote by ρ_ijρij charge segment jj on conductor segment ii. In order to find the capacitance between conductor segment ii and jj, we need to ground all segments except ii, which has a voltage of V_i = VVi=V. We solve (6.1) and calculate the capacitance from (4.48)

Cij=1V∑k=1Nsubρjk.

With this procedure one needs to go through all conductors and set each voltage to VV one at a time, keeping all others grounded, solve the equation, and calculate the resulting capacitance. It sounds easy enough and it sort of is, but the difficulty lies in the details. The integrand has a divergence when i = ji=j that is fairly strong and the accuracy required in the numerical integration can be excessive. Furthermore, the grid is critical and in order to set it up properly one has to have a deep understanding of how charge will distribute itself on the surfaces beforehand. However, the pseudocode is straightforward.

Pseudocode

Subroutine build_G(Mesh)

                    * Loop over Mesh to build Matrix G.

                    For i = 0,N do

                              For j = 0,N do

                                        Integrate_element(r[i],r[j])

                              End for

                    End for

                    End build_G

                    Subroutine CapacitanceSolver

                    Create_mesh(Mesh)

                    G = Build_G(Mesh)

                    Ginv = Invert_matrix(G)

                    For i = 1,N do

                              V = Define_voltages(Mesh)

                              rho = Multiply(Ginv,V)

                              C[i] = sum(rho) // Assuming the voltage V = 1.

                    End for

                    End CapacitanceSolver

Verify

Simulators are traditionally verified by having them solve known problems and comparing the solutions. We leave it as an exercise to the reader to implement a code and gain experience by learning by actually doing.

Evaluate

We are not going to go to the trouble of actually building codes here, but simply state that although this may sound really simple and in some meaning of the word it is, the problem comes when you want accuracy. The charge tends to want to sit at the edges of the conductors, similar to the current distributions we examined in Chapter 5. This means that around those edges the surface elements must become really small to resolve the charge distribution, adding up to quite a few elements for large geometries. Some serious tricks need to be employed to get to the bottom of this and make the problem manageable. The most successful simulators have in common that they have found clever ways to manage the grid-generating algorithms. This has made some inventors very wealthy indeed. The construction of the grid and the speed with which the matric, GG, is calculated are critical parts of building simulators. If this is done poorly it will take a long time to find a solution for a given accuracy.

Simplify

We assume we are dealing with one long wire of a certain thickness and width that are much smaller than the conductor length. We chop the wire into small conductor segments where we assume several things:

• 
  

  The magnitude of the current in each conductor segment is the same, so there is no capacitive or other loss.

  
  
• 
  

  The current is uniform in a cross-section of the wire, in other words skin depth is not important. Each conductor segment can then cover the full cross-section of the wire.

  
  
• 
  

  These is only one magnetic medium, so no boundaries need to be considered.

  
  
• 
  

  We will ignore the complication of self-interaction where a given current segment impacts its own magnetic field (there is a singularity). This is an important effect and not that difficult to deal with, but for illustrative purposes we do not address this problem here.

Solve

Let us now define the problem by looking at the vector potential from a current on a conductor segment ii, at a point rr. We have from (4.36)

Air=∫Segment_iIir′r−r′dr′

With this expression we can now calculate the partial inductance matrix, L_{i, j}Li,j, between segments i,   ji,j, using (4.46) and

We find

Li,j=1IiIj∫Segment_jIjrj⋅Airjdrj=1IiIj∫Segment_jIjrj⋅∫Segment_iIirirj−ridridrj≈Ij⋅IiIiIj∫Segment_j∫Segment_i1rj−ridridrj=ej⋅ei∫Segment_j∫Segment_i1rj−ridridrj.

We refer to the case i ≠ ji≠j as the mutual inductance and i = ji=j as the self-inductance, similar to what we discussed earlier. Here we see explicitly the convenient fact that when the two currents are perpendicular to each other their mutual inductance is zero. If we decide not to divide the conductors – perhaps they are really thin, or frequency is low – we will be done here. We merely have to sum up all the elements to get the total inductance of the wire.

L=∑i,jLi,j