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Theorem

***TODO Add Figure***

Consider a network as shown in Figure 1.

Millman's theorem states that the voltage at the common junction 0', is given by:-

$V_{00'}={\frac {\displaystyle \sum _{i=1}^{n}{V_{i}Y_{i}}}{\sum {Y}}}$

Where V_i is the value of each voltage source, Y_i is the branch admittance (given by Y_i=1/Z_i) and ΣY is the sum of all the admittances.

This can also be arranged in terms of branch impedance:-

$V_{00'}={\frac {\displaystyle \sum _{i=1}^{n}{\frac {V_{i}}{Z_{i}}}}{\displaystyle \sum _{i=1}^{n}{\frac {1}{Z_{i}}}}}$

The theorem can be remembered as the voltage generated by the sum of the short circuit currents of all branches, flowing through an impedance equal to all the branch impedances in parallel.

Derivation by superposition

A standard method of solving such problems is to set all sources to zero, then enable them one at a time and sum the contribution.  Voltage sources are replaced with a short circuit when set to zero.  Therefore the superposition is given by the sum of the voltage sources, each scaled by the appropriate potential divider ratio:-

$V_{00'}=V_{1}{\frac {Z_{k1}}{Z_{k1}+Z_{1}}}+V_{2}{\frac {Z_{k2}}{Z_{k2}+Z_{2}}}+V_{3}{\frac {Z_{k3}}{Z_{k3}+Z_{3}}}+{...}=\displaystyle \sum _{i=1}^{n}{V_{i}{\frac {Z_{ki}}{Z_{ki}+Z_{i}}}}$ (1)

where Z_ki is the parallel combination of all resistances apart from Z_i.  Let the admittance Y_i=1/Z_i.  As admittances in parallel add, it follows that

$Z_{ki}={\frac {1}{(\sum {Y})-Y_{i}}}$ (2)

where ΣY is the sum of all the admittances (a constant).  Substituting (2) into (1) gives:-

$V_{00'}=\displaystyle \sum _{i=1}^{n}{(V_{i}{\frac {\frac {1}{(\sum {Y})-Y_{i}}}{{\frac {1}{(\sum {Y})-Y_{i}}}+Z_{i}}})}=\displaystyle \sum _{i=1}^{n}{(V_{i}{\frac {1}{1+Z_{i}((\sum {Y})-Y_{i})}}})=\displaystyle \sum _{i=1}^{n}({\frac {V_{i}}{Z_{i}\sum {Y}}})={\frac {\displaystyle \sum _{i=1}^{n}{V_{i}Y_{i}}}{\sum {Y}}}$ (3)

Derivation by conversion of each branch to a Norton equivalent circuit

The V_iY_i term in equation (3) represents the short circuit current of the Thévenin generator consisting of V_i and Z_i.  This suggests a more intuitive derivation by conversion of each of these Thévenin generators to their Norton equivalents.

The Norton equivalent circuit of each generator is a current source of value V_iY_i in parallel with an admittance, Y_i.  The circuit can therefore be redrawn as shown below:-

Current sources in parallel simply add.  Admittances in parallel simply add.  The voltage is given by the total current multiplied by the total impedance. This argument yields the same expression as equation (3).

Generalisation to include impedances and current sources

Figure 1

***TODO: This needs much work to align properly with the above. The designator subscripts are confusing as they can be mixed up with the index numbers. The figure also needs updating. It's simplest to rewrite and redraw from scratch.***

The theorem can be generalised to include impedance and current source branches.

Let V_k be the voltage generators and I_m the current generators.

Let Z_i be the impedances on the branches with no generator.

Let Z_k be the impedances on the branches with voltage generators.

Let Z_m be the impedances on the branches with current generators.

The voltage at the ends of the circuit is given by:

$V_{00'}={\frac {\sum _{}{\frac {V_{k}}{Z_{k}}}+\sum _{}I_{m}}{\sum _{}{\frac {1}{Z_{k}}}+\sum _{}{\frac {1}{Z_{i}}}}}$

Proof is by extension of the derivation by superposition given above. R_m does not feature in the equation because current sources become open circuit when set to zero.

Alternatively the proof by conversion to Norton circuits can be extended. The Norton equivalent circuit of the branches with current sources are simply the current source; The resistor does not feature.

Dual Theorem

The network above contains parallel branches, each with series impedance and voltage source.

The dual network of this is the series cascade of