Talk:Millman's theorem

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THis is the more common form of statement:

If any number of admittances Y1, Y2, Y3, ... meet at a common point P, and the voltages from another point N to the free ends of these admittances are E1, E2, E3, ... then the voltage between points P and N is: VPN = (E1Y1 + E2Y2 + E3Y3 + ...) / (Y1 + Y2 + Y3 + ...) VPN = SEY / SY

The short-circuit currents available between points P and N due to each of the voltages E1, E2, E3, ... acting through the respective admitances Y1, Y2, Y3, ... are E1Y1, E2Y2, E3Y3, ... so the voltage between points P and N may be expressed as: VPN = SIsc / SY

--Light current 03:42, 9 April 2006 (UTC)[reply]

What I usually see in text books is even different, but it resembles more the above than the wikipedia entry itself. If someone has a good reference, please update the article. Some examples to illustrate the theorem would be useful too. Aphexer (talk) 19:51, 25 December 2008 (UTC)[reply]

Superposition

why is this better than superposition? CorvetteZ51 13:26, 20 June 2007 (UTC)[reply]

Or perhaps Thevenin's theorem combined with the Superposition theorem. Isn't a constant current source an uncommon occurance? unsigned comment

The article does not say that this is a better method than superposition or that it is not equivalent to other methods. And no, constant current sources are not uncommon. They are used all the time in electronics, eg current mirrors, long tailed pair etc. Bipolar junction transistors are better modelled using constant current than constant voltage sources. Constant current power systems are also possible, but now everything is the dual of normal - eg, you need to short circuit the load to switch it off rather than open circuit it. Finally, if they are that uncommon, why are there so many people selling them? try this google search string [1]. SpinningSpark 21:19, 5 April 2008 (UTC)[reply]

I think it might be better to actually go through the derivation for this. One possible path is indeed the combination Thevenin's theorem and the principle of superposition. Also, the phrase "by considering the circuit as a single supernode" seems misleading; it is actually saying to treat the nodes at either ends as supernodes. Maybe include a figure with the nodes labeled (e.g., node a for the node at the top, and node b for the node at the bottom).Iccruz (talk) 17:35, 11 July 2008 (UTC)[reply]

Try this: http://www.opamp-electronics.com/tutorials/millmans_theorem_revisited_1_10_10.htm seems to be a nice explanation of Millman, based on Norton-Thevenin. Hoemaco (talk) 12:46, 3 December 2009 (UTC)[reply]

Tank's method

Does someone have a reference for this. I can find no mention of it anywhere. SpinningSpark 21:21, 5 April 2008 (UTC)[reply]

I'm about to remove it. Seems like it exists nowhere. Alexius08 (talk) 11:13, 10 March 2009 (UTC)[reply]

Useful Application?

Can someone cite a useful application of this theorem? --96.244.247.130 (talk) 00:51, 15 July 2011 (UTC)[reply]

Node analysis, one Node at a time

I concur that the general statement of Millmans is:

Vp = sigma(En.Yn)/ sigma(Yn) were Vp is the voltage at the node, En are the emf's and Yn admittances, connected to the node

Another way of thinking about it is, its node analysis, one node at a time instead of using a many nodes at a time matrix solution.

Millman's is easily proved from kirchoffs current law, ie sigma(In) = 0 and In = (Vp - En)Yn (ohms law)

hence (Vp - E1)Y1 + (Vp - E2)Y2 + ...... (Vp - En)Yn = 0

therefore multiplying out

VpY1 + VpY2 + .... VpYn - E1Y1 - E2Y2 - .... EnYn = 0

therefore Vp sigma(Yn) = sigma (EnYn)

therefore Vp = sigma (EnYn)/sigma (Yn) QED

It was first published by Millman in the 1950's - he was professor of EE at Columbia University in the USA

It is very useful for circuit analysis, for example a two pole Salen and Key active filter is easily analysed using Millmans.

Regards, Phil Robinson (UK 2014 www.pcrobinson.co.uk)

References

I have added the original publication of Millman which can be found via the IEEE Xplore Digital Library. Thus, the "This article does not cite any references or sources." should be removed130.237.53.229 (talk) 09:21, 26 February 2015 (UTC)[reply]

Yobot has now changed it to "refimprove" which is about right. Millman's paper is the primary source, there are still no secondary sources. SpinningSpark 18:27, 3 March 2015 (UTC)[reply]

Proposed Major Update / Cleanup

Info from A. Jones, 21 Nov 2015: I am planning to contribute an update to this article with the following changes:-

A rewrite of the first section, sticking to Millman's original equation and nomenclature, removing the generalisation to current sources etc, which masks the simple concept.
A couple of sections on ways of deriving the basic theorem
A section on generalising the theorem, presenting pretty much what is on the page today, plus clarifying the terminology. Currently, V_k etc is used, which gets confused with the index numbers.
A section on the dual of the theorem for current sources in series with parallel admittances across each current source.
Some examples of usage
Better references to key text books

Note that I have the original paper, and believe we should stick as close as possible to this. — Preceding unsigned comment added by Amj4321 (talk • contribs) 09:46, 21 November 2015 (UTC)[reply]

Draft text so far is below.Amj4321 (talk) 09:56, 21 November 2015 (UTC)[reply]

Theorem

***TODO Add Figure - Ideally we should try to get the original figure from the paper. If this isn't possible due to copyright, we'll have to redraw as similar as possible.***

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

***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 current sources, each with a parallel admittance.

TODO*

I have taken the liberty of boxing your draft text to keep it distinct from the conversation. I'm glad to see someone is taking this article by the scruff of the neck, it's needed it for a long time. I don't think that you should get too hung up on Millman's original paper. Yes, it should be covered, both for accuracy and historical interest, but the focus of the article should be on the present day usage and conception of the theorem, rather than sticking rigidly to what Millman originally wrote. SpinningSpark 12:10, 21 November 2015 (UTC)[reply]

Ok no problem. My main task now is figure drawing. Do you have a standard tool for circuit drawings on Wikipedia? I didn't think much of the file suggested on the project page. Amj4321 (talk) 22:21, 22 November 2015 (UTC)[reply]

Overcomplicated

I have removed some of the complications in the expression given in this article. I am not disputing that the expression is correct, but that's not how reliable sources handle it. I've looked in a dozen text books and they all treat only the simple case of multiple voltage sources in parallel. The treatment of current sources can perhaps be added as an additional section. SpinningSpark 17:01, 15 December 2018 (UTC)[reply]