Dependent sources - what to do with them? - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: December 31, 2017, 06:19:04 pm
content: Hello,If I use superposition theorem and remove a branch which controlls a dependent source, then the dependent source becomes a short circuit. Right?But If in some circumstances the dependent source stays in the circuit, do I treat him like any other source?Because in superposition theorem I remove only independent source, dependent sources stay in the circuit. Is it this all right till now?Second thing:Why do we need them? for modelling a transistor? A transistor has own symbol and it's nonlinear.

author: MrAl
author_rank: Super Contributor
timestamp: January 01, 2018, 12:39:29 am
content: Quote from: nForce on December 31, 2017, 06:19:04 pmDependent sources are very useful. You can simplify models of op amps and transistors and more.Superposition works with dependent sources, but it's not exactly the same as with non dependent sources. There are techniques you can use. For example, if you model an op amp as a dependent source you can short the output to ground to kill the source.It would be best though if you post the problem circuit as then we can discuss how this works.

author: nForce
author_rank: Frequent Contributor
timestamp: January 01, 2018, 11:48:08 am
content: Are my first three statements correct?Because then I can solve any linear circuit with superposition. I just want to know for general purpose.

author: orolo
author_rank: Frequent Contributor
timestamp: January 01, 2018, 01:27:59 pm
content: On superposition with dependent sources, take a look at thisinteresting article. Lots of examples there.

author: nForce
author_rank: Frequent Contributor
timestamp: January 01, 2018, 02:03:17 pm
content: Quote from: orolo on January 01, 2018, 01:27:59 pmThanks for the article, can you just confirm these three statements? Because I think I am right about it, but just yes/no will be the best answer.Thank you all.

author: MrAl
author_rank: Super Contributor
timestamp: January 01, 2018, 03:55:01 pm
content: Quote from: nForce on January 01, 2018, 02:03:17 pmHi,I dont think your questions should be answered exactly the way they are presented. Here's the first one:"If I use superposition theorem and remove a branch which controlls a dependent source, then the dependent source becomes a short circuit. Right?"What do you mean 'remove a branch'. You dont remove a branch when you use superposition. You "Kill" sources, that's it. You kill all sources then unkill them one at a time and sum the responses. There's nothing about removing a branch in that.If you sill wish to ask this question anyway then show a circuit where it applies. That's the best way so we can understand each other better.If i answered your first question directly it would not make sense. If you simply "remove a branch" from any circuit you potentially change the entire circuit so it will never work right. Superposition may do that indirectly, but that depends on the circuit at hand and i dont think it is something that should be generalized because that question is more of a topological question not a question about superposition.Post a circuit in which you think would be relevant and we can take a look.The link looks interesting but i am not sure if we have to stick to that rule about the controlling signal or not in every case. If one source is killed another may provide the signal for example, but we can look at that more too later.

author: nForce
author_rank: Frequent Contributor
timestamp: January 01, 2018, 04:16:38 pm
content: Ok, I wasn't clear enough:If I have a current source, which in superposition theorem I remove it from the circuit. That means the branch disappears. But what if the current in that particular branch controlls a dependent source elsewhere. So is it true that, then I make short ciruit of that dependent source?I think other statements are clear enouqh.

author: orolo
author_rank: Frequent Contributor
timestamp: January 01, 2018, 04:35:34 pm
content: QuoteRight.If you are using the standard method of superposition, and you have linearly dependent sources, then when you remove a controlling branch then you zero the dependent sources from that branch: they turn open (current source) or short (voltage source).For an example, take a look atFigures 7 and following here.QuoteIf a dependent source remains, you don't remove it, but use nodal analysis to deal with it, and solve for that part of the superposition problem.QuoteYes. You only eliminate independent sources. The dependent ones, if you are lucky, go away when you remove some sources. If you aren't lucky, you can't recursively use superposition and remove a dependent source, that will lead to error. Use node/cycle analysis and the dependent current will just become a constraint.An attractive alternative to this scheme is the article by Leach I linked above, where all sources, dependent or not, are treated equally. It's quite elegant. A further discussion of Leach's method can be foundhere.

author: MrAl
author_rank: Super Contributor
timestamp: January 03, 2018, 11:35:44 am
content: Quote from: nForce on January 01, 2018, 04:16:38 pmHello again,You should be able to follow the method for the technique and get reasonable results.If you have a current source you kill it by open circuiting it, and a voltage source is killed by short circuiting it.The only catch given in that link was that if you kill a dependent source and it also kills the controlling variable then it may not work. So if when you open circuit a dependent current source and it ends up making the controlling variable also go to zero, then it may not work.However, we always have to deal with human errors in circuit analysis anyway. There are all sorts of reasons for errors when we do circuits even with a simulator. What this means is we need to check the result in some way, that uses a secondary method to test the solution. Often all this requires is to use a basic rule like the sum of currents into a node equals zero. Even if we have all dependent sources we need to do this so we should also do it with independent sources to make sure everything worked the way it should have.This gives YOU a way to check your own work. If you open that current source and do the circuit the way you think you should and then check the result and see a problem with a current or voltage, you know something is wrong.Just to recap...1. There's only that one catch with dependent sources.2. You should always check your work using a second method.

author: nForce
author_rank: Frequent Contributor
timestamp: January 08, 2018, 07:18:40 pm
content: I have another question.Why is power of harmonic signals 1/2*U*I*cos(phi)?

author: bson
author_rank: Unknown
timestamp: January 09, 2018, 12:02:51 am
content: Quote from: nForce on January 01, 2018, 04:16:38 pmNo, the controlling current becomes 0 and you calculate dependent source's response to that. If it's say V=Ai it becomes 0V, which is a short. If it's I=Ai it becomes open. If it's something else, like V=A(1-i) then it becomes A.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: January 09, 2018, 02:52:43 am
content: Quote from: nForce on December 31, 2017, 06:19:04 pmJust to address this. There are uses for dependent sources in transistor circuit analysis, especially the VCCS.One method of circuit analysis that involves transistors is to look at the large and small signal operating conditions, where large signal conditions are mainly concerned with transistor biasing and steady-state operation. Once steady state conditions have been determined, a hybrid-pi model can be established (and swapped in place of the transistor) for small signal analysis.This is especially useful if you want to determine the system transfer function is performed in the Laplace domain (s-plane).https://en.wikipedia.org/wiki/Hybrid-pi_modelNote: This is one method of circuit analysis that involve transistors, not the only one. More accurate models of transistors do indicate that they are non-linear and thus it is known that this method effectively "linearises" its behaviour around an operating point.

author: nForce
author_rank: Frequent Contributor
timestamp: January 09, 2018, 12:35:43 pm
content: Ok, thanks for these notes.But I stil don't know where does 1/2 come for power of harmonic signals.

author: bson
author_rank: Unknown
timestamp: January 09, 2018, 07:52:13 pm
content: Because it's averaged over 1/f of the fundamental?