Modified nodal analysis

All you want to know about Modified nodal analysis

In Electrical Engineering Modified Nodal Analysis or MNA is an extension of the nodal analysis method that not only determines the circuit's node voltages (as in the classical nodal analysis), but also some branch currents.

Method

The MNA uses the element's Branch Constitutive Equations'or BCE Lecture ,i.e., their voltage - current characteristic and the Kirchhoff's circuit laws. According to ^[1] the method is done in four steps, but it can be reduced to three.

Step 1

Write the KCL of the circuit. At each node of an electric circuit one writes the currents coming in and out of the node. Take care however in the MNA the current of the independent voltage sources is taken from the "plus" to the "minus". See Figure 1. Also note that the right hand side of each equation is always equal to zero. So that the branch currents that come inside the node are given a negative sign, whereas the branch currents coming out are given a positive sign.

Step 2

Use the BCE in terms of the node voltages of the circuit to eliminate as many branch currents as possible. Writing the BCE's in terms of the node voltages saves one step. If the BCE's were written in terms of the branch voltages, one more step, i.e., replacing the branches voltages for the node ones, would be necessary. In this article the letter "e" is use to name the node voltages, while the letter "v" is used to name the branch voltages.

Step 3

Finally, write down the unused equations.

Example

The figure shows a RC series circuit and the table shows the BCE of a linear resistor and a linear Capacitor. Note that in the case of the resistor the admittance $G$ i, $G = 1 / R$ , is used instead of $R$ . We now proceed as explained above.

RC Circuit

Element	Branch equation
Resistor	$I R = G V R$
Capacitor	$I_C = C\frac{dV_C}{dt}$

Step 1

In this case there are two nodes, $e 1$ and $e 2$ . Also there are three currents: $i_{V_s}$ , $i R$ and $i C$ .

At node e1 the KCL yields:

$i_{V_s} + i_R = 0$

and at node e2:

$- i R + i C = 0$

Step 2

With the provided BCEs in the table and observing that:

$V s = e 1$

$V R = e 1 - e 2$

$V C = e 2,$

the following equations are result: $G(e_1 - e_2) + i_{V_S} = 0$

$C\frac{de_2}{dt} + G(e_2 - e_1) = 0$

Step 3

Note that at this point there are two equations but three unknowns. The missing equation comes from the fact that $e 1 = V s$ .