Nodal Analysis: Correct/Standard Method?

Question

Nodal analysis is mostly quite straightforward and most people that explain/teach it do a lot of things similarly, except for this one thing that seems pretty important: assuming the direction of current over every non-source element or not assuming the direction and just assuming that all currents will flow outwards from any given node.

For example; for the same circuit, Educator #1 will pick a voltage reference node (GND), pick the nodes the voltages of which are unknown, then start to assign polarities to all non-source elements, thus assuming the direction of current before getting to the maths part.

Educator #2 will pick a reference node, write down the nodes the voltages of which must be found then go straight to the maths part, assuming that current will only flow out from any given node (unless if specified otherwise).

I tend to mix the sources that I study from, and I can not tell if any method is superior to the other/'healthier' to use in the long term, and I couldn't settle on any one yet. Does anyone have any experience with this?

Answer 1

I tend to mix the sources that I study from, and I can not tell if any method is superior to the other/'healthier' to use in the long term, and I couldn't settle on any one yet. Does anyone have any experience with this?

Nodal analysis is just minimization and it relates to relaxation and probability, where similar principles result in similar solutions. In this case, nodal minimizes \$E\left(\vec{x}\right)=\vec{x}^T\left[A^T C A\right]\vec{x} \$, where the bracketed section is the weighted Laplacian describing the circuit and \$\vec{x}\$ is the node-potentials vector. (Tellegen's theorem says that in single sink-source circuit, this minimum will take on the value of the effective conductance of that circuit. See also dirichlet's principle, which is a dual view to Thomson's principle. Weierstrass found an error in Dirichlet's proof. A correct proof came later from Kneser.)

Each nodal (KCL) equation is a weighted average of the adjacent surrounding nodes based on the connected conductances.

To illustrate, consider the following simple schematic. It has 3 boundary potentials (knowns) and 1 harmonic potential (unknowns). So only one KCL equation is needed:

^{simulate this circuit – Schematic created using CircuitLab}

Let's start with the usual pedagogy for the above circuit:

$$\frac{V-V_1}{R_1}+\frac{V-V_2}{R_2}+\frac{V-V_3}{R_3}=0\:\text{V}$$

While technically correct, it buries the lead (which is \$V\$.) And one has to keep track of signs. In more complex circuits, the chances of making mistakes increase (even when one is meticulous.)

I prefer to start this way:

$$\begin{align*} \frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}&=\frac{V_1}{R_1}+\frac{V_2}{R_2}+\frac{V_3}{R_3}\tag{1}\label{eq1} \end{align*}$$

The lead, \$V\$, now clearly stands out and is entirely the focus on the left, which is where one begins when writing. The right side almost looks like an average. Still better is the fact that signs are kept simple.

A little more massaging exposes the fact that this actually is just a weighted average of the surrounding nodes:

$$\begin{align*} V\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)&=\frac{V_1}{R_1}+\frac{V_2}{R_2}+\frac{V_3}{R_3}\tag{2}\label{eq2} \\\\ V\left(G_1+G_2+G_3\right)&=V_1\: G_1 + V_2\: G_2 + V_3\: G_3 \tag{3}\label{eq3} \\\\ V&=V_1\: \frac{G_1}{G_T} + V_2\: \frac{G_2}{G_T} + V_3\: \frac{G_3}{G_T}, \tag{4}\label{eq4}\\\\&\text{ where }G_T=G_1+G_2+G_3 \end{align*}$$

In eq. \$\ref{eq4}\$, each adjacent node contributes to the resulting node voltage by the ratio of its path-conductance as compared against the sum of all path-conductances to the node. That just makes sense.

There's something else that may be intuitively obvious -- \$V\$ cannot be larger than the largest of the surrounding nodes and cannot be smaller than the smallest of the surrounding nodes. And abstracting outward from eq. \$\ref{eq4}\$, this leads to a maximum principle and a minimum principle -- the largest node value and the smallest node value are always found in the boundary potentials.

Eq. \$\ref{eq1}\$ keeps the focus on the node-at-hand when developing the equation, avoids minus signs, and ultimately promotes a larger and better viewpoint. It also doesn't come into conflict with more advanced ideas that may come later.

Answer 2

They are not assuming a direction of current flow. They are placing a reference arrow, so that when the maths finally gives an answer, they will know what the direction means. If the current turns out to be positive, it's in the direction of the arrow. If it's negative, it's opposite to it.

It doesn't matter which way the reference arrows point, as long as they keep pointing the same way as you set up and solve the equations, and decide what the results mean. Be consistent.

Some authors try to set the reference directions so that the equations have the fewest minus signs (I hate minus signs, always lose one of them!). Others emphasise a nice geometric regularity to make it easier to check the equations against the schematic. There is no wrong way, as long as you are consistent.

Answer 3

Consistency makes it easier but it is not necessary to be consistent in your choice of current reference directions. All that is necessary is to write KCL correctly AND to correctly express the currents in terms of the node voltages.

(You can easily verify that if you change the reference direction of the current that there will be a sign change when expressing the currents in terms of node voltages).

Having said that- a consistent choice, and doing everything the same way every time, is least likely to lead to errors. Writing the equations is a mechanical process- otherwise, how could a circuit analysis program obtain a system of equations from a netlist?

Answer 4

I'm not sure whether this will help, but I am educator #3, a guy who advocates for not using reference nodes at all, be it ground or some other known potential. KVL doesn't care about absolute potentials, only potential differences. This isn't necessarily a better way, it's just a different, and potentially less ambiguous way.

In defence of #1 and #2, there are circumstances where it's advantageous to write a KVL equation in terms of some absolute potential, and with practice it becomes clear why one might do so:

^{simulate this circuit – Schematic created using CircuitLab}

On the left, there are a few things you can state immediately, without resorting to strict and complete application of KVL. For instance, since we know that emitter potential is \$V_E=0V\$, and we know from experience that in this configuration current \$I_2\$ will flow downwards through R2, making base potential \$V_B\$ negative with respect to \$V_S\$, but positive with respect to \$V_E\$. We know also that \$V_{BE}=0.7V\$, and therefore, without having to employ KVL or KCL in a strict manner, we can instantly state some potentials and currents:

$$ \begin{aligned} V_B &= V_E + V_{BE} \\ \\ &= 0\rm{V} + 0.7\rm{V} \\ \\ &= +0.7\rm{V} \end{aligned} $$

$$ \begin{aligned} I_2 &= \frac{V_S-V_B}{R_2} \\ \\ &= \frac{12\rm{V} - 0.7\rm{V}}{10\rm{k\Omega}} \\ \\ &= 1.1\rm{\ mA} \end{aligned} $$

This simplicity is only possible because we are familiar with the arrangement. Another engineer reading these conclusions will agree because he too is familiar with this setup. However, this is lazy (justifiably so in this case), and even though we are still applying KVL it's not really the full KVL as described in the textbooks. To do this the textbook way, you'd have to go around the entire loop SBES, as shown above-right. Starting at S, and going clockwise:

$$ \begin{aligned} -V_2 - V_{BE} + V_1 = 0 \end{aligned} $$

The educator-#3 point I want to make is that nowhere in that equation is there any reference to an absolute potential, ground or otherwise. If you wished, you could convert the equation to be in terms of absolute potentials \$V_S\$, \$V_B\$ and \$V_E\$, but this is still only potential differences, all parenthesised here:

$$ \begin{aligned} -(V_S-V_B) - (V_B - V_E) + (V_S - V_E) = 0 \end{aligned} $$

That's still correct, but it has lost its KVL "roots", and requires great care to simplify. Sure, you can plug in known values such as \$V_S = +12\rm{V}\$ and \$V_E=0\rm{V}\$, and find \$V_2\$ eventually, but why go that route when \$-V_2 - V_{BE} + V_1 = 0\$ is more direct? You can always find absolute potentials later, but I am advocating for sticking with potential differences, at least until you've really grokked KVL. KVL doesn't deal with absolute potentials, it relates changes of potential as one visits nodes around a loop.

As further motivation for avoiding absolute potentials until the very last stage, consider a circuit that is unfamiliar, about which we can't make instant and short-hand assumptions about potentials anywhere:

^{simulate this circuit}

In the true spirit of KVL, the equations I write will all be in terms of potential differences across the various elements, \$V_1\$, \$V_2\$ etc., with not an absolute potential in sight. I ignore \$V_G=0\$ altogether, and at the very last stage I'll derive absolute potentials.

KVL for loop ABGDA:

$$ -V_3 - V_5 - V_6 + V_1 = 0 $$

KVL for loop BCDGB:

$$ +V_4 -V_2 +V_6 + V_5 = 0 $$

Usually you'd write those equations directly in terms of resistances and currents (using Ohm's law) to save a step, but I'm trying to stick to the fundamentals, so I'll write the Ohm's law equations separately:

$$ \begin{aligned} V_3 &= I_1 R_3 \\ \\ V_4 &= I_2 R_4 \\ \\ V_5 &= I_3 R_5 \\ \\ V_6 &= I_3 R_6 \\ \\ \end{aligned} $$

KCL at B:

$$ I_1 + I_2 - I_3 = 0 $$

To keep things simple, I've obeyed passive sign convention for every resistor, with current labelled to enter a resistor terminal marked as being more positive. If you don't do this, then you will find that current polarity will not match voltage polarity, and while the values may be correct, it becomes ambiguous (and easy to forget to correct later) which one has the correct sign, and which must be negated. Stick with passive sign convention.

In the solution to all those simultaneous equations, we get:

$$ \begin{aligned} V_3 &= -3.6\rm{\ V} \\ \\ I_1 &= -1.2\rm{\ A} \\ \\ \end{aligned} $$

Clearly I incorrectly guessed current direction through V1 and R3, and voltage polarity across R3, but that changes nothing. I forgive my mistake because it isn't at all obvious. The solution is still correct, though, I just need to remember that in reality, conventional current is flowing left through R3, and down through V1, and that the right end of R3 actually has the higher potential.

Two more results from solving those equations:

$$ \begin{aligned} V_5 &= +1.6V \\ \\ V_6 &= +8.0V \\ \\ \end{aligned} $$

This is where we can finally find some absolute potentials, the last stage. We can do some "lazy" KVL now, and say things like "potential \$V_B\$ is higher than \$V_G\$, by amount \$V_5\$:

$$ \begin{aligned} V_B &= V_G + V_5 \\ \\ &= 0\rm{V} + (+1.6\rm{V}) \\ \\ &= +1.6\rm{V} \end{aligned} $$

Potential \$V_D\$ is lower (look at the labelled polarity of \$V_6\$) than \$V_G\$:

$$ \begin{aligned} V_D &= V_G - V_6 \\ \\ &= 0\rm{V} - (+8.0\rm{V}) \\ \\ &= -8.0\rm{V} \end{aligned} $$

Knowing \$V_B\$ and \$V_3\$, it's easy to find \$V_A\$ in the same lazy way, but I will also stress the importance of sticking to your labelled polarities. On my schematic I assumed that current flowed to the right, and that \$V_A > V_B\$, which we now know to be false. However, I stick to my schematic's labelling, and trust the algebra to work. My schematic says "Node A is higher in potential than node B (see labelled \$V_3\$ polarity), by amount \$V_3\$", so that's what I write algebraically:

$$ \begin{aligned} V_A &= V_B + V_3 \\ \\ &= +1.6\rm{V} + (-3.6\rm{V}) \\ \\ &= -2.0\rm{V} \end{aligned} $$

The sign of \$V_3\$, being negative, took care of the "correction" for me.

Answer 5

Example of nodal analysis of a resistive network, based on the analysis of the network topology. (the elements of the individual resulting current and voltage vectors are positive if they agree with the directions of the corresponding branches, otherwise they are negative) (the drawback of the TABLEAU method is that it is limited to small-sized networks, otherwise the TABLEAU exceeds all limits and the calculation requires very long times,in such case you should use the modified node analysis or MNA):

Topology rules: