Nodal analysis

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In electrical engineering, nodal analysis, node-voltage analysis, or the branch current method is a method of determining the voltage (potential difference) between "nodes" (points where elements or branches connect) in an electrical circuit. It is used to solve for the voltages and currents at any point in a circuit without working through many individual KCL or KVL rules.

[edit] Method

Label all the nodes in the circuit (e.g. 1, 2, 3...), and select one to be the "reference node." It is usually most convenient to select the node with the most connections as the reference node.
Assign a variable to represent the voltage of each labeled node (e.g. V₁, V₂, V₃...). The values of these variables, when calculated, will be relative to the reference node (i.e. the reference node will be 0V).
If there is a voltage source between any node and the reference node, by Kirchhoff's voltage law, the voltage at that node is the same as the voltage source's. For example, if there is a 40 V source between node 1 and the reference node, node 0, V₁ = 40 V.
Note any voltage sources between two nodes. These two nodes form a supernode. By Kirchhoff's voltage law, the voltage difference at these two nodes is the same as the voltage source. For example, if there is a 60 V source between node 1 and node two, V₁ − V₂ = 60 V.
For all remaining nodes, write a Kirchhoff's current law equation for the currents leaving each node, using the terminal equations for circuit elements to relate circuit elements to currents. For example, if there is a 10 Ω resistor between nodes 2 and 3, a 1 A current source between nodes 2 and 4 (leaving node 2), and a 20 Ω resistor between nodes 2 and 5, the KCL equation would be (V₂ − V₃)/10 + 1 + (V₂ − V₅)/20 = 0 A.
For all sets of nodes that form a supernode, write a KCL equation, as in the last step for all currents leaving the supernode, i.e. sum the currents leaving the nodes of the supernode. For example, if there is a 60 V source between nodes 1 and 2, nodes 1 and 2 form a supernode. If there is a 40 Ω resistor between nodes 1 and the reference node, a 2 A current source between nodes 1 and 3 (leaving node 3), and a 30 Ω resistor between nodes 2 and 4, the KCL equation would be (V₁ − 0)/40 + (−2) + (V₂ − V₄)/30 = 0 A.
The KCL and KVL equations form a system of simultaneous equations that can be solved for the voltage at each node.