Thevenins method and the dependent sources - Page 1
Forum section: electronics / beginners

author: lastwarrior
author_rank: Contributor
timestamp: April 09, 2020, 02:13:32 pm
content: Hello everyone,I have some problem with developing intuition on this topic. So, straight to the subject. Suppose a circuit with only independet sources. If I want to apply thevenins theorem to investigate how changes will affect the power delivered to the load, I have to disconnect the load and start with finding the Rth. I have to disable all sources and calculate the Rth between the terminals where the load previously where connected. Easy.If some dependent source appears in the network, this method of finding Rth cant be used anymore. All references that I have checked just mention it as a law, without any kind of explanation. It is said that you cant, it will generate wrong results. Instead of explanation, the authors jumps to modifications that you need to do to the original thevenins theorem to accomplish correct answer.Disabling the independent source will automaticaly set to off the dependent source. That according to the reference manual can be replaced with short circuit equivalent.So, the Rth between A and B will be only R1. And this is mentioned as the wrong answer.I can understand in superposition that if I suppress one of the sources from controlling-controlled pair, the circuit will have no sens anymore as a model of something. It will be completly changed. But in case of thevenins theorem I cant find any explanation why I cant use resistance reduction with all sources off as I've done with independent sources.

author: lastwarrior
author_rank: Contributor
timestamp: April 17, 2020, 08:40:30 am
content: Seams that it is very noob quation, the problem is that I cant find any explanation of it in the web. One reason could be, because of superposition theorem convention. As I found out the superposition theorem is used to prove the Thevenin's theorem. Also some strange info like, "suppresing a dependent source is like disconnecting some electronic components from the network, the responce will never be the same" but not any kind of explanation to prove or the dependent source rules. Maybe there is some conventions taken abot the dependent sources or some rules about how they are used to model a network or a portion of the network.

author: RoGeorge
author_rank: Super Contributor
timestamp: April 17, 2020, 09:14:34 am
content: I'm not sure I understood the question.Superposition principle can be applied only for linear systems. To be more precise, the validity of the superposition principle and the linearity of the system are equivalent.In case of electrical circuits, a controlled source most probably will make the circuit non linear. Therefore, the usual superposition techniques (power sources passivization followed by the summation of the effects given by each power source, one by one) can not be applied any more.Is this what your question was about?

author: kjpye
author_rank: Contributor
timestamp: April 17, 2020, 11:55:58 am
content: A dependant or controlled source won't make the circuit non-linear.Given the second circuit where the independent current source is 0 (open-circuit). Suppose you tried to measure the resistance by forcing 1mA from A to B. What would the voltage be between A and B. What is the measured resistance?

author: RoGeorge
author_rank: Super Contributor
timestamp: April 17, 2020, 01:07:37 pm
content: Quote from: kjpye on April 17, 2020, 11:55:58 amTry to apply superposition in a circuit that contains a voltage controlled source, e.g. Vout = V2in.Superposition won't work any more.Numerical example:Two series connected voltage sources V1 = 1V and V2 = 2V are feeding the input of a voltage controlled voltage source Vout = Vin * Vin. At the output of the controlled source is a 1 ohm resistor. What is the current through the resistor?If we apply the superposition we getI caused by V1 = 1AI caused by V2 = 4AI final through R = 5A according to superposition, which is a wrong answer, because the superposition can not be applied to a non linear circuit.The correct answer is 9A, not 5A.

author: lastwarrior
author_rank: Contributor
timestamp: April 17, 2020, 04:09:33 pm
content: Thank you all for your opinion guys, realy appreciate this.The thing that I'm trying to clarify is :"Why I cant use general thevenin procedure (where all sources are killed and the thevenins impedance is found by solving the network without sources) to networks where a dependent source is present with its controlling variable in the same network?"I dont get it. The thevenin procedure was modified from the general one (that aslo can be applied to DC networks), only because of dependet sources with its controlling variable in the same portion of the circuit to which Thevenins has to be applied.All references that I have checked mention that thing as mondatory law without any clarification.

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 17, 2020, 05:43:19 pm
content: I suggest you consider a simpler circuit to help you realize why Thevenin reductions will not work.

author: lastwarrior
author_rank: Contributor
timestamp: April 18, 2020, 11:43:40 am
content: In this case I can suppress the dependent source only if Vo is 0. It will be if I short the output terminals. By shorting them I can suppress the dependent source and substitute it by a short. As the result I will get a closed loop with 1 resistor and the resistance Rth will be equal to resistance of the short places between output terminals.Now returning to my original example posted in post#1. I can see why it will generates the wrong result only and only if I suppress the dependent voltage first by shorting out the R1 and then kill the current source.If I kill the current source first -> the current through the network is 0 -> as the result, the voltage drop across the R1 is 0. If Vr1 is 0 -> The dependent source can be substituted by its short-circuit euivalent.