Basic Kirchhoff
- src/model/kirchhoff.h
This is the most basic algorithm. It is not very efficient, and should only be
used for debugging. Suppose we have a circuit with \(b\) branches and
\(n\) nodes. We take one node (the last one, by convention) as reference.
The corresponding system of equations should be written as:
\[\begin{split}\left[
\begin{array}{c}
&\\ % empty row
\text{Constitutive equations} \\
b \times (b+n-1) \\
&\\ % empty row
\begin{array}{c|c}
\hline \\
\hspace{0.5cm} \mathbf{0}_{(n-1) \times (n-1)} \hspace{0.5cm} &
\hspace{0.5cm} \text{KCL}_{(n-1) \times b} \hspace{0.5cm} \\
\\ % empty row
\end{array}
\end{array}
\right]
%############################################################ \cdot
%############################################################
\left[
\begin{array}{c}
V_1 \\ \vdots \\ V_{n-1} \\
I_1 \\ \vdots \\ I_b
\end{array}
\right]
%############################################################
=
%############################################################
\left[
\begin{array}{c}
B_1 \\ \vdots \\ B_b \\
0 \\ \vdots \\ 0
\end{array}
\right]\end{split}\]
KCL = Kirchhoff’s Current Law
Then we just apply one of Eigen’s solvers to do the dirty work for us.
Example
Each row of the upper part of the system matrix corresponds to a branch and its
constitutive equation (see Modelling a branch).
Suppose that the first row of the system matrix is
\[[ \begin{array}{ccc} a_1 & a_2 & b_1 & b_2 & b_3 & b_4 & b_5 \end{array} ]\]
Since the first branch is only connected to nodes 1 and 3, and \(I_1\) is
the current that flows through it, \(a_1=1,\ b_1=-R_1\), and all
other elements are zero. Do the same for all other branches. The lower part of
the matrix is easy to populate based on KCL.
Finally, the system of equations for this circuit is:
\[\begin{split}\left[
\begin{array}{ccccccc}
1 & 0 & -R_1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & -R_2 & 0 & 0 & 0 \\
1 & -1 & 0 & 0 & -R_3 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 \\ \hline
0 & 0 & 1 & 1 & 1 & -1 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 1 \\
\end{array}
\right]
\cdot
\left[
\begin{array}{c}
V_1 \\ V_2 \\
I_1 \\ I_2 \\ I_3 \\ I_4 \\ I_5
\end{array}
\right]
=
\left[
\begin{array}{c}
E_1 \\ 0 \\ 0 \\ I_4 \\ E_5 \\
0 \\ 0
\end{array}
\right]\end{split}\]