Nodal-Voltage Analysis & Supernodes

Nodal analysis solves for node potentials and obtains branch currents from element laws. It naturally handles current sources and produces a sparse conductance matrix; voltage sources require direct constraints or supernodes.

Nodal analysis makes node potential the primary unknown. Every resistor current becomes a voltage difference times conductance, so KCL produces one equation per non-reference node. The method scales well because each equation touches only neighboring nodes.

Conductance stamping #

For a resistor RabR_{ab} between nodes aa and bb with conductance Gab=1/RabG_{ab}=1/R_{ab}, current from aa to bb is

Iab=Gab(Va−Vb).I_{ab}=G_{ab}(V_a-V_b).

KCL at node aa collects all incident branches:

(∑kGak)Va−∑kGakVk=Ia,\left(\sum_k G_{ak}\right)V_a-\sum_k G_{ak}V_k=I_a,

where IaI_a is the net independent current injected into the node. Stamping all nodes gives

GV=I.GV=I.

Diagonal entries are sums of connected conductances; off-diagonal entries are negative mutual conductances.

Voltage sources and supernodes #

A voltage source between a node and reference sets that node voltage directly. A voltage source between two unknown nodes does not give its current, so enclose both nodes and the source in a supernode:

  1. write one KCL equation for the supernode boundary;
  2. add the source constraint Va−Vb=VsV_a-V_b=V_s.

Dependent voltage sources are handled the same way, with their controlling relation added to the system.

Choosing the reference #

Any node may be ground, but a good choice simplifies source constraints and reduces awkward supernodes. The physical branch voltages do not depend on the choice.

Verification #

After solving, compute every branch current from the original element law, not from rounded intermediate values. Check KCL at the omitted reference node and calculate a power balance. These checks catch the two dominant errors: a source injection with the wrong sign and a conductance stamped between the wrong pair of nodes.

Two-node nodal-analysis example A ten-volt known node connects through two kiloohms to unknown node v, which connects to ground through three kiloohms and receives one milliampere. 10 V2 kΩv3 kΩ1 mA(v−10)/2k + v/3k − 1mA = 0 → v = 7.2 V
Each KCL term is a conductance times a node-voltage difference

Sources

  1. docs6.01SC Course Notes: Chapter 6, Circuitspdf

Searches titles, aliases, headings, summaries, and full text.↑↓ move ↵ open esc close