Node voltage analysis

Node voltage analysis is the holy grail for circuit analysis. It’s not easy, but the method is always the same. This analysis method allows us to solve and analyse complex circuits.

So how does it work? It all starts with KCL. Lets take the last circuit from the Ohm’s law lab as a starting point.

Figure 1. The circuit from the last Ohm’s law quiz question

Reference node

We are first going to define some nodes (points in the circuit where various components are connected). For starters we have to define a reference node. This node will be defined as zero volt. It can be any place in a circuit without affecting the analysis. However, in practice we usually use the negative terminal of the power supply as reference and call this ground. Why do we choose that as a reference? Because in many circuits that is the node that has the most connections. Having that node be ‘zero’ volts makes calculations a lot easier.

Figure 2. Circuit with reference node ground

The reference node has a symbol which we call ground.
The ground symbol only indicates this is our zero volt reference. It is not some magical drain and does not have any special properties. It does mean that all nodes connected to the same reference are connected together. So the above circuit is the same as this one:

Figure 3. Same circuit, but drawn with multiple reference (ground) symbols

Labeling unknown nodes

Now that we defined the reference node we will define all unknown nodes. In this circuit there are 3 nodes:

• The node connecting \(V_1\), \(R_1\) and \(R_2\)
• The node connecting \(R_1\), \(R_3\) and \(R_4\)
• The node connecting \(R_2\), \(R_3\) and \(R_5\)

For the first node the voltage is known. Since we know the negative terminal of the voltage source is at zero volt, the node on the top on the voltage source must be zero plus the voltage of the source, or 9V. This leaves us with two yet undefined nodes. We are going to call those \(V_A\) and \(V_B\).

Determine the KCL equations

The next step is to use a combination of KCL and Ohm’s law to define the voltages. Let’s start at node \(V_A\). Using KCL we know that the three currents leaving the node must add up to 0. So we can state:
$$I_{R_1} + I_{R_3} + I_{R_4} = 0$$
With Ohm’s law we know that the current through a resistor is the voltage across the resistor divided by the resistance. So for \(I_{R1}\) the current is the voltage across it \(V_A – 9\) divided by the resistance value \(R_1\), or $$\dfrac{V_A – 9}{R_1}$$

Note that we are assuming the currents leave each node. This assumption might be wrong, but the solution will tell us if we are correct or not.
Because we assume the current is leaving the node we always calculate with the node voltage (in this case \(V_A\)) minus the voltage of the other node.

To complete our equations for the other currents we get:

$$I_{R_3} = \dfrac{V_A – V_B}{R_3}$$
$$I_{R_4} = \dfrac{V_A – 0}{R_4}$$
We can now substitute those currents in our main KCL equation
$$\dfrac{V_A – 9}{R_1} + \dfrac{V_A – V_B}{R_3} + \dfrac{V_A – 0}{R_4} = 0$$

After completing the equation for node \(V_A\), we can do the same for node \(V_B\). We first start again with our general KCL equation:
$$I_{R_2} + I_{R_3}+I_{R_5} = 0$$
The next step is to calculate the equations for the currents.
Remember, we chose all current flowing out of the nodes.
So \(I_{R_2} = \dfrac{V_B – 9}{R_2}\), since we know the top node is 9V.
For the other currents we can state: \(I_{R_3} = \dfrac{V_B-V_A}{R_3}\) and \(I_{R_5} = \dfrac{V_B – 0}{R_5}\).
Combined this gives us:
$$ \dfrac{V_B – 9}{R_2} + \dfrac{V_B-V_A}{R_3} + \dfrac{V_B – 0}{R_5} = 0$$

Separate the variables

At this moment we have written down our equations for the nodes. We can now use linear algebra to solve for the two unknown values \(V_A\) and \(V_B\). I’ll show you how that works.

First we rearrange the equations:

For the first node we get:
$$V_A: \dfrac{V_A – 9}{R_1} + \dfrac{V_A – V_B}{R_3} + \dfrac{V_A – 0}{R_4} = 0$$

Separating the variables:

$$\dfrac{V_A}{R_1} + \dfrac{V_A}{R_3} + \dfrac{V_A}{R_4} + \dfrac{-V_B}{R_3} = \dfrac{9}{R_1}$$

Combining factors:
$$V_A \left( \dfrac{1}{R_1} + \dfrac{1}{R_3} + \dfrac{1}{R_4} \right) +V_B \left( \dfrac{-1}{R_3} \right) = \dfrac{9}{R_1} $$

For the second node we get:

$$V_B : \dfrac{V_B – 9}{R_2} + \dfrac{V_B-V_A}{R_3} + \dfrac{V_B – 0}{R_5} = 0$$

Again separating the variables:

$$\dfrac{V_B}{R_2} + \dfrac{V_B}{R_3} + \dfrac{V_B}{R_5} + \dfrac{-V_A}{R_3} = \dfrac{9}{R_2}$$

And combining the factors:
$$V_B \left( \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_5} \right) +V_A \left( \dfrac{-1}{R_3} \right) = \dfrac{9}{R_2} $$

Now we can put both equations easily in a matrix. If you are unfamiliar with matrices a short recap is available here.
Once our matrix is filled, the circuit can be solved.

\begin{gather} \begin{bmatrix} \dfrac{1}{R_1} + \dfrac{1}{R_3} + \dfrac{1}{R_4} & \dfrac{-1}{R_3} \\ \dfrac{-1}{R_3} & \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_5} \\ \end{bmatrix} \begin{bmatrix} V_A \\ V_B \end{bmatrix} = \begin{bmatrix} \dfrac{9}{R_1} \\ \dfrac{9}{R_2} \\ \end{bmatrix} \end{gather}

Solving the system

To solve the system we input the values for the resistors from the exercise in the previous lab into our matrix:
$$R_1 = 10k\Omega$$
$$R_2 = 20k\Omega$$
$$R_3 = 1k\Omega$$
$$R_4 = 30k\Omega$$
$$R_5 = 5.6k\Omega$$

\begin{gather} \begin{bmatrix} \dfrac{1}{10k} + \dfrac{1}{1k} + \dfrac{1}{20k} & \dfrac{-1}{1k} \\ \dfrac{-1}{1k} & \dfrac{1}{20k} + \dfrac{1}{1k} + \dfrac{1}{5.6k} \\ \end{bmatrix} \begin{bmatrix} V_A \\ V_B \end{bmatrix} = \begin{bmatrix} \dfrac{9}{10k} \\ \dfrac{9}{20k} \\ \end{bmatrix} \end{gather}

It is good practise to enter the values of components at the very last step. This will help you with circuit understanding later on. 
The next step is to simplify (multiply everything by 1k, and find common denominator) the matrix:

\begin{gather} \begin{bmatrix} \dfrac{17}{15} & -1 \\ -1 & \dfrac{43}{35} \\ \end{bmatrix} \begin{bmatrix} V_A \\ V_B \end{bmatrix} = \begin{bmatrix} \dfrac{9}{10} \\ \dfrac{9}{20} \\ \end{bmatrix} \end{gather}

Followed by multiplying row 1 with \(\dfrac{15}{17}\) and adding it to the second row we get:

\begin{gather} \begin{bmatrix} \dfrac{17}{15} & -1 \\ 0 & \dfrac{206}{595} \\ \end{bmatrix} \begin{bmatrix} V_A \\ V_B \end{bmatrix} = \begin{bmatrix} \dfrac{9}{10} \\ \dfrac{423}{340} \\ \end{bmatrix} \end{gather}

Now we can solve for \(V_B\) which is
\(\dfrac{206}{595}V_B = \dfrac{423}{340}\)
giving us:
\(V_B = 2961/824 = 3.59V\)

Now that we know $$V_B$$ we can input it in one of our previous equations: $$dfrac{17}{15} V_A – V_B = \dfrac{9}{10}$$
Solving this gives us \(V_A = 3.96V\).

Lastly if we want to solve the current \(I_1\) all we have to do is to use Ohm’s law to calculate it.
\(I_1 = \dfrac{V_A-V_B}{R_3} = \dfrac{3.96 – 3.59}{1k} = 370µA\)

Recap

Ok phew…. that was a lot of work wasn’t it? So to recap, what have we done?

1. First we selected a reference node (commonly known as ground)
2. Next we labeled all the nodes of which the voltage is unknown
3. Then we wrote down all the KCL equations of each node, with the currents flowing out of the node
4. Now we separate all the variables (nodes)
5. Solve the network

Node voltage equations

We won’t solve all circuits we encounter completely. Mainly because this is a lot of work to do by hand. In my opinion it is much more valuable that you can correctly setup the equations for a network, which is what we are going to do in this next example.

Take the circuit above, where step 1 and 2 have already been done. Step three is to write done the KCL expressions for every node and rearrange them. For node \(V_A\) this will be:

$$\dfrac{V_A – 5}{R_1} + \dfrac{V_A – V_B}{R_2} + \dfrac{V_A – V_C}{R_3} = 0$$

$$V_A \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \right) + V_B \left( \dfrac{-1}{R_2} \right) + V_C \left( \dfrac{-1}{R_3} \right) = \dfrac{5}{R_1} $$

Next up is node \(V_B\):

$$\dfrac{V_B-V_A}{R_2} + \dfrac{V_B – V_C}{R_4} + \dfrac{V_B – 0}{R_5} = 0$$

$$V_A \left( \dfrac{-1}{R_2}\right) + V_B \left( \dfrac{1}{R_2} + \dfrac{1}{R_4} + \dfrac{1}{R_5}\right) + V_C \left( \dfrac{-1}{R_4} \right) = 0 $$

And the final node \(V_C\):

$$\dfrac{V_C – V_A}{R_3} + \dfrac{V_C – V_B}{R_4} + \dfrac{V_C – 0}{R_L} = 0$$

$$V_A \left( \dfrac{-1}{R_3}\right) + V_B \left( \dfrac{-1}{R_4} \right) + V_C \left( \dfrac{1}{R_3} + \dfrac{1}{R_4} + \dfrac{1}{R_L}\right) = 0 $$

Plugging this all into a matrix gives us:

\begin{gather} \begin{bmatrix} \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} & \dfrac{-1}{R_2} & \dfrac{-1}{R_3} \\ \dfrac{-1}{R_2} & \dfrac{1}{R_2} + \dfrac{1}{R_4} + \dfrac{1}{R_5} & \dfrac{-1}{R_4} \\ \dfrac{-1}{R_3} & \dfrac{-1}{R_4} & \dfrac{1}{R_3} + \dfrac{1}{R_4} + \dfrac{1}{R_L} \\ \end{bmatrix} \begin{bmatrix} V_A \\ V_B \\V_C \end{bmatrix} = \begin{bmatrix} \dfrac{5}{R_1} \\ 0 \\ 0 \\ \end{bmatrix} \end{gather}

If we wanted to solve this we can always plug all the number into a computer and let the computer do the math or visit Matrix solver where you can enter the values and have the matrix solved.

Now you need to exercise with some circuits and see if you can solve them. In the next lesson we will see what happens when we add multiple voltage and current sources to a circuit.