Voltage Nodal Analysis
Introduction: This article focuses on understanding how to analyze voltages at nodes within a given circuit.
The Essentials
An important analysis technique is Node-Voltage Analysis. To be clear in the meaning of this, node voltage is the voltage rise from a reference to a non-reference node (this can also mean we compare a voltage node to ground, which would be zero).
The way nodes are identified is a point in the circuit where at least two elements are connected. Each node will have a unique voltage value.
The hard part of this method is to generate the correct equations necessary. We need to analyze the circuit node by node according to the currents entering each node.
We do this by using Ohm's Law which states
where
I = the current in amps
V = the voltage in volts across the element
R = the resistance in ohms of the element
We will be using this equation a lot in our nodal analysis example.
Example
(Note: For each equation, I will include factored-out versions of the node equations for the purpose of solving by matrices. Using matrices is not necessarily required for solving the node voltages, but this method is an efficient way to solve for all variables. You can simply solve for one variable in one equation and plug into the other to solve for each variable.)
Figure 1. contains an image of an example circuit
Figure 1: Example Problem
Here in the equation there are three key voltage nodes: the input voltage (given as 5 V), \( V_{a} \) and \( V_{b} \). We are interested in knowing the values of \( V_{a} \) and \( V_{b} \).
Let's start by writing an equation that represents the voltage at \( V{a} \). We need to represent the currents entering the node \( V{a} \), so we will use equation 1 above.
For the portion to the left of the node, the current coming from there can be represented as
\[\frac{V_{a} - 5}{1000}\]
We can conclude this since the voltage across the element can be found as the voltage node value minus the other node of interest, in this case the 5 V voltage source. The 1000 in the denominator is the value of the resistor between the 5 V source and \( V_{a} \) node. This represents the current through that branch.
Next, the right hand side
\[\frac{V_{a} - V_{b}}{2500}\]
Finally, the current from the bottom
\[\frac{V_{a} - 0}{1500}\]
Compiling this together, we get this equation
\[\frac{V_{a} - 5}{1000} + \frac{V_{a} - V_{b}}{2500} + \frac{V_{a} - 0}{1500} = 0\]
According to the law of conservation of energy, the current entering a node is the same as the current leaving the node, so the sum of all current entering a node equals 0. This equation is has two unknowns, both \( V_{a} \) and \( V_{b} \). We can tidy up this equation so it can look like this
\[V_{a}(\frac{1}{1000} + \frac{1}{2500} + \frac{1}{1500}) + V_{b}(-\frac{1}{2500}) = \frac{1}{200}\]
This format of the equation may be helpful for those who wish to use matrices to solve for the unknown values given that you know how to do those.
Since we only have one equation with two unknowns, we now need another similar equation but in reference to the voltage node \( V_{b} \).
Let's apply the same process as earlier and we'll get this equation
\[\frac{V_{b} - V_{a}}{2500} + \frac{V_{b}-0}{1500} = 0\]
Rewriting this equation again
\[V_{a}(-\frac{1}{2500}) + V_{b}(\frac{1}{2500} + \frac{1}{1500}) = 0\]
Now since we have two equations with two unknowns, solving for \( V_{a} \) and \( V_{b} \) should be simple. After doing algebra, you should get the following:
\[V_{a} = 2.6086 \, \text{V} \, \ \ \ \ \ \ \ \ \ V_{b} = 0.978 \, \text{V}\]
Practice
Problem 1.)
Using Figure 2., analyze the circuit and find values of \( V_{a} \) and \( V_{b} \).
Figure 2: Problem 1
Problem 2.)
Using Figure 3., there is a switch that is involved. Solve for all labeled node voltages (\( V_a, V_b, V_c \)) and calculate the power dissipated through resistor \( R_L \) when the switch is closed.
Figure 3: Problem 2
Problem 3.)
Using Figure 4., solve for \( V_o \). (Part of this problem involves identifying nodes that you need to set up the equations. Choose these nodes carefully, \( V_o \) is already given)
Figure 4: Problem 3
Solutions:
Problem 1 Solution.) \( V_a = 0 \, V \) and \( V_b = 2.74 \, kV \). We follow the same process as earlier in solving these voltages. What makes this easy is that \( V_a \) is tied to ground, so we automatically know that \( V_a = 0 \, \). Then solving for \( V_b \) can be seen as the following
\[ \frac{V_b - 10}{1500} + \frac{V_b - V_c }{10000} + \frac{V_b}{15000} = 0\]
Looking at the current source there may be intimidating, but we can solve this effectively. \( V_c \) represents the voltage directly above the current source, While we do know the current through that branch, we are interested in the voltage at that node. Knowing the physics of charge flow, the node directly above the current source MUST have a higher voltage than \( V_b \). We can represent the relationship between \( V_c \) and \( V_b \) very simply
\[ V_c = 2*10000 + V_b\]
Replacing that into our previous equation and working some algebra
\[ V_b (\frac{1}{1500} + \frac{1}{15000}) = 2 + \frac{1}{150}\]
Solving for \( V_b \) and we get the value of 2.74 kV. This makes sense since the current source in series with the 10k \( \Omega \) will generate LOTS of voltage and will make the 9 V source essentially negligible
Problem 2 Solution.) \( V_a = 24.79 \, V \), \( V_b = 24.79 \, V \), \( V_c = 2.6 \, V \), \( P_{RL} = 32.8 \, mW \). A good first step is simplifying the circuit so it looks a little more manageable. With the switch closed, the charge tends to go the path of least resistance, therefore the charge will only go through the switch and none through the 5 k\( \Omega \) resistor. I rearranged the circuit to look like Figure 5., but all the values are unaffected as if it were the original circuit
Figure 5: Problem 2 Rewritten
Immediately we notice that \( V_a = V_b \) since they share the same node. From this we can make our nodal equations. Firstly for \( V_a = V_b \)
\[ \frac{V_a - 30}{1000} + \frac{V_a - V_c}{20000} + \frac{V_a - V_c}{15000} + \frac{V_a - V_c}{8500} = 0\]
Rewriting this equation
\[ V_a(\frac{1}{1000} + \frac{1}{20000} + \frac{1}{15000} + \frac{1}{8500}) + V_c(-\frac{1}{20000} - \frac{1}{15000} - \frac{1}{8500}) = \frac{3}{100}\]
Next the equation for \( V_c \)
\[\frac{V_c - V_a}{20000} + \frac{V_c}{500} + \frac{V_c - V_a}{15000} + \frac{V_a - V_c}{8500} = 0\]
Now rewriting this one
\[ V_a(-\frac{1}{20000} - \frac{1}{15000} - \frac{1}{8500}) + V_c(\frac{1}{20000} + \frac{1}{500} + \frac{1}{15000} \frac{1}{8500}) = 0\]
Now with these two equations we can solve for the voltage nodes. You should get \( V_a = V_b = 24.79 \, V \), \( V_c = 2.6 \, V \).
Now we would like to solve for the power through the resistor RL. Since we know the voltage on both ends of the resistor, we can calculate the voltage across the resistor and the current through it using Ohm's Law
\[ I_{RL} = \frac{V_a - V_c}{15000} = 1.479 \, mA \]
Since \( P = (V_a - V_c) \cdot I_{RL} \), we should get a power value of 32.8 mW.
Problem 3 Solution.) \( V_o = 2.787 \, V \). Let's set up equations for each node in the circuit. The nodes needed to evaluate this circuit is the node directly after the 8 mA source (\( V_a \)), after the horizontal 6 k\( \Omega \) resistor (\( V_b \)), and where \( V_o \) is identified.
For \( V_a \),
\[ (-8\cdot10^{-3}) + (2 \cdot 10^{-3}) + \frac{V_a}{3000} + \frac{V_a-V_b}{6000} = 0 \]
Rewriting that equation
\[ V_a(\frac{1}{3000} + \frac{1}{6000}) + V_b(-\frac{1}{6000}) = 6 \cdot 10^{-3} \]
Now for the second equation
\[ \frac{V_b - V_a}{6000} + \frac{V_b}{6000}+\frac{V_b - V_o}{2000} = 0 \]
Rewriting
\[ V_a(-\frac{1}{6000}) + V_b(\frac{1}{6000} + \frac{1}{6000} + \frac{1}{2000}) + V_o(-\frac{1}{2000}) = 0\]
Now for the final equation
\[ \frac{V_o - V_b}{2000} - (2 \cdot 10^{-3}) + \frac{V_c}{1000} = 0 \]
Finally rewriting this one
\[ V_b(-\frac{1}{2000} ) + V_o(\frac{1}{2000} + \frac{1}{1000}) = 2 \cdot 10^{-3}\]
Now we have three equations for three unknown variables, so now we can solve for each. \( V_a = 13.5 \, V \), \( V_b = 4.63 \, V \), and our final answer, \( V_o = 2.79 \, V \).