Dependent voltage source circuit analysis - Page 1
Forum section: electronics / beginners

author: MikeK
author_rank: Super Contributor
timestamp: March 24, 2012, 11:50:41 pm
content: I've run into a problem in the MIT 6.002x course. We have an exercise that was given two different answers and we're not sure which is correct. And we haven't heard from the staff yet. So I'd like to ask the real electrical engineers a question.I should point out that I'm not looking for the solution to the exercise (I already have it), but clarity about method. And, in fact, please don't provide an overall solution, because there are still students who have not gotten to this exercise yet.The question I have is...Should alpha be included in the calculation of the Thevenin resistance? See the attached circuit diagram. I say the resistance is R1||R2, while others say it is (R1+alpha)||R2.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 12:56:37 am
content: I'm not doing the course, and I'm not an EE, but I believe the essence of Thevenin is that a combination of linear elements must also behave linearly. Linear means "straight line", and you only need two points to define a straight line. So you find two conditions of the circuit that are easy to solve (often by forcing currents or voltages to zero), and then connect these two points. Any simple linear circuit that is equivalent for the two known conditions must by definition be equivalent for every condition.In this case my intuition says that since alpha is a resistance that appears in at least one path between terminals, it must appear in the solution. The only way alpha could not appear in the solution is if it linearly cancels out for all possible circuit conditions. I have not tried to solve the problem, but in general for alpha to cancel out would require all the circuit parameters to be chosen as a particular special case. In the general case cancellation isn't going to happen.

author: Rufus
author_rank: Super Contributor
timestamp: March 25, 2012, 01:39:27 am
content: Quote from: MikeK on March 24, 2012, 11:50:41 pmThe Thevenin equivalent resistance determines the short circuit current. The only solution for short circuit current requires i to be zero which makes alpha irrelevant.

author: amspire
author_rank: Super Contributor
timestamp: March 25, 2012, 03:05:54 am
content: Quote from: Rufus on March 25, 2012, 01:39:27 amThat is true for the short circuit, but to calculate the Thevenin resistance, you have to determine the open source voltage.I don't know if the course is pushing you towards a particular type of solution, but for one approach, define a current iR2.You know VOUT= R2 * iR2and the input current = i1+ iR2. Just come up with another equation that gives iR2in terms of i1. Use this to eliminate i1from the second equation and then you can work out iR2as a function of the current source, and hence VOUTas a function of the current source.Divide VOUTby the short circuit current in terms of the current source and you have the result. Rufus has given a good clue about how to work out the short circuit current, but since you have worked out these beautiful equations, make R20 ohms to find out what happens when the output is shorted.Sorry if that is not great to read, but I was trying to give a method that doesn't actually solve the problem.Richard.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 03:21:48 am
content: It's over thirty years since I was introduced to this kind of subject matter and I haven't really touched it since (electronics is not my field). So I've been racking my brains to see if I can reach back over the intervening years and figure it out from scratch.Clearly you can solve the problem by algebra and nodal analysis, but there should be an "easier" way, a short cut to the solution.The hint says to add another current source at the port and use superposition. Presumably by choosing an appropriate current source you can make something cancel out and leave a simpler problem? But I can't quite see what that magic current value would be?

author: MikeK
author_rank: Super Contributor
timestamp: March 25, 2012, 03:43:36 am
content: Nodal analysis does solve the circuit and give the Thevenin voltage, but it says nothing about the Thevinin resistance. I hope Dave chimes in on this one.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 03:59:37 am
content: You can work out the short circuit current, and then I believe the Thevenin resistance is the Thevenin voltage divided by the short circuit current?But I want to think you are not intended to do a laborious nodal analysis on this circuit. There is meant to be a quick and easy way to arrive at the answer.

author: MikeK
author_rank: Super Contributor
timestamp: March 25, 2012, 04:40:06 am
content: Actually, I think you can't. Because shorting out the terminals causes an invalid circuit (two different sources in reverse parallel). Already tried it and got weird results. Hopefully some EE's will comment.

author: Rufus
author_rank: Super Contributor
timestamp: March 25, 2012, 04:48:11 am
content: Quote from: amspire on March 25, 2012, 03:05:54 amYou don't have to calculate it to know that the calculation won't involve alpha which is enough to answer the OP's question.

author: vk6zgo
author_rank: Super Contributor
timestamp: March 25, 2012, 04:55:45 am
content: How about a "Tech's" comment?Way too long since I did anything approaching this stuff,but,as Thevenin is normally a voltage based form of analysis,I think you need to convert the current sources to voltage sources to apply it.Do they show you how to do that?If you can analyse the circuit that way,you can maybe work your way back to what they actually want you to do.One suggestion is to look at other books,not just the one they specify,as sometimes the explanation of one point may be a bit weak.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 05:00:47 am
content: Quote from: MikeK on March 25, 2012, 04:40:06 amIf you short the output port it is like setting R2 to zero. So now the (alpha)i voltage source is in parallel with R1. Equating voltages between the two nodes across R1 (or summing voltages around the loop) gives i x R1 + (alpha) x i = 0 => (R1 + alpha) x i = 0 => i = 0. If i = 0 then the all the current I0 must flow through the output, so the short circuit current is I0.

author: MikeK
author_rank: Super Contributor
timestamp: March 25, 2012, 05:30:24 am
content: Yep, you beat me to it. I went about it a different way. I added another resistor to the output and solved and then compared the behavior to the Thevenin circuit I had with the new resistor. The behavior was only the same when I used alpha in the Thevenin resistance.So now my questions is...Why is alpha used in the Thevenin resistance? I thought it was just a constant for determining the dependent source voltage. It doesn't actually have a resistance of 8 ohms....I calculated it to be 8.9 ohms (by using it's voltage with it's current). And that 8.9 isn't a rounding error.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 05:39:06 am
content: I don't know yet. I'm still fumbling my way here. I have not actually solved that example problem in a way I can really understand, such that it becomes transparent to me. Doing nodal analysis and algebra just doesn't feel right. I'm in the process of watching Agarwal's lecture on superposition and Thevenin in the hope of gaining further insight

author: MikeK
author_rank: Super Contributor
timestamp: March 25, 2012, 06:04:07 am
content: Oh, I could do algebra all day long. Love it. But your explanation about the short circuit current helped. I was getting confused, not realizing what i=0 really meant. Thanks.I hope someone can help me understand why alpha is used in Rth, so I can be confident in the process.

author: Rufus
author_rank: Super Contributor
timestamp: March 25, 2012, 06:09:39 am
content: Quote from: Rufus on March 25, 2012, 04:48:11 amOops I was wrong. Right about i=0 and alpha being irrelevant for a short circuit, wrong to infer that made it irrelevant to the Thevenin equivalent resistance. We know the short circuit current and can calculate the open circuit voltage so we know the Thevenin resistance. I don't know how to express it algebraically.On current controlled voltages sources - they have a conversion constant in volts per amp which is the same unit as resistance. They don't have resistance.

author: amspire
author_rank: Super Contributor
timestamp: March 25, 2012, 06:26:15 am
content: Ah! I didn't actually read the hint.Looking at it now. it is really easy.First you can take off R2. Get the Thevenin circuit of what is left. Open circuit volts is just Io x R1 + Io * alpha. For resistance, take Io away and add a current source on the port. Lets call it I2. Vout = I2 * R1 + alpha * I2, Divide by I2 both sides and you have the Thevinin impedance (Since I0 had infinite impedance). You can leave I0 in if you like, but you will just find the effects of the two sources just add together which I guess is what they meant by "superposition of sources".Now we have the Thevenin voltage and resistance without R2. Adding R2 back is a simple solving of a voltage divider and parallel resistors.Richard.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 06:27:12 am
content: OK, I've "got" it now. I stopped trying to work it out in my head and resorted to pencil and paper.The Thevenin voltage can be found by nodal analysis and algebra. I think there is an easier way to find it, but that hasn't been covered in the course yet.The algebra is a bit ugly, but the result comes out as:V(Th) = (I0)(R2)[1 - R2/(R1+R2+alpha)] = 2.13 VSuperposition comes in when considering the Thevenin equivalent resistance. We are looking for the resistance "seen" by an external current source. So we can suppress the internal current source for this calculation as it becomes irrelevant. As soon as we suppress I0, we can rearrange the circuit and it immediately becomes obvious by inspection that:R(Th) = (R1 + alpha) || R2This happens because when the current source is suppressed it becomes an open circuit and vanishes from consideration.[Edit: cross-posted with Richard. Got to cross check with that and see if I missed a simplification.]

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 06:36:11 am
content: Quote from: amspire on March 25, 2012, 06:26:15 amI'm lost at "taking R2 away" and then "adding it back in again". Are you saying to work out the Thevenin equivalent of the simpler circuit with R2 absent, and then combine that simpler Thevenin circuit with R2 to get the new Thevenin equivalent with R2 back in? (Goes away to try working that out.)

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 06:47:46 am
content: Quote from: IanB on March 25, 2012, 06:36:11 amYes! Same answerNote to self, bleeding obvious department: if any whole circuit has a Thevenin equivalent, than any sub-circuit will have a Thevenin equivalent too. You don't have to do it all in one step, you can do it in smaller steps and you must by definition get the same result (otherwise the Thevenin equivalent circuit would not be as equivalent as it is supposed to be...)

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 07:36:39 am
content: Quote from: IanB on March 25, 2012, 05:39:06 amI finished watching the lecture, and he does explain most of the points above. It's fast paced, but he covers the idea of suppressing any internal (non-dependent) voltage or current sources and looking at the resistance of the circuit as seen by a current source from outside, and also the idea of breaking a circuit into parts and working out the Thevenin equivalent of just a part of the whole before recombining.

author: MikeK
author_rank: Super Contributor
timestamp: March 25, 2012, 02:51:14 pm
content: What I don't understand is WHY is alpha used in Rth? 8 ohms is not the resistance of the dependent source. If I calculate the voltage across and current through the dependent source, and divide, I get something other than 8 ohms for the resistance.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 03:35:14 pm
content: I think you have something wrong. 8 ohmsisthe resistance of the dependent source (when the current through it isi-- if the current through the dependent source is notithen its effective resistance is something else of course, although resistance is not really the right term to use in such case).

author: MikeK
author_rank: Super Contributor
timestamp: March 25, 2012, 03:50:21 pm
content: "i" is the current through R1, not the dependent source. The dependent source is linear (and defined as such), so it's resistance must be constant. If you calculate it's voltage drop and current you never get 8 ohms, for ANY setting of Io.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 04:15:46 pm
content: But the current "i" depends on the conditions in the rest of the circuit, so the dependent source cannot be considered to have a resistance in the conventional sense. To have a resistance it would need to be both linear and independent. Since it only satisfies one of these conditions it cannot be looked at in isolation as an independent circuit element. (This circuit is arranged to be vaguely like a transistor, where "i" might be the base current. If you considered the voltage and current relationship between the collector and emitter of a transistor you would also find it to have a non-constant resistance, since it depends on the current at the base.)Another way of looking at this is to ask, "What is the Thevenin equivalent circuit of the dependent source alone?" As such it should have a fixed voltage in series with a fixed resistance. But since the voltage depends oniwhich is an external parameter, it is not possible to represent the dependent source as such. There is no Thevenin equivalent circuit for the dependent source alone.

author: MikeK
author_rank: Super Contributor
timestamp: March 25, 2012, 04:28:32 pm
content: Right, I understand all that. But I'm having trouble understanding why the 8 ohms is used to calculate Rth, since 8ohms is not the resistance of the dependent source...and in my mind it's just a conversion factor.I have no problem now getting the short circuit current and using that to get Rth, but it's maddening to see that Rth can be calculated by using alpha and not understand why.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 04:39:48 pm
content: It's because of the principle of superposition. If you suppress the current source I0 and "look back" into the circuit with an external current source, then the current through the dependent source isi, and thus it behaves like a resistor with value alpha.However, if you suppress any external current sources and consider only I0, then the current through R1 is not the same as the current through the dependent source. In this case the dependent source does not behave like an 8 ohm resistor any more.By superposition you can add the results of the two separate analyses together. So the dependent source still behaves linearly, but with a resistive and non-resistive component combined.It might be interesting to solve the circuit with varying values of external load resistor and plot V & I on a graph for the dependent source. You should get a straight line (because it's a linear circuit), but the straight line won't go through the origin (because the source is not a true resistor). I have not done this, but if you do it I would be interested to know what you find?

author: MikeK
author_rank: Super Contributor
timestamp: March 25, 2012, 05:23:39 pm
content: Quote from: IanB on March 25, 2012, 04:39:48 pmAwesome. THIS I understand. So the Thevenin resistance (with a dependent source) isn't just a blind resistance measurement, but the resistance when some external current flows.Quote from: IanB on March 25, 2012, 04:39:48 pmI already did this, but without the graph. As the external resistor goes up in value, the impedence of the dependent source (V/I) also goes up:external resistorvoltagecurrentimpedence18000.0170454550.0028693185.94059405920000.0173778720.0028277666.1454421330000.0184577520.0026927816.85453160740000.0190496350.0026187967.27419680750000.0194233430.0025720827.551602618100000.0202165420.0024729328.17512944150000.0204955370.0024380588.406501027200000.0206379420.0024202578.527168506250000.0207243390.0024094588.601246225300000.020782340.0024022088.651350572350000.0208239680.0023970048.687498276400000.0208552990.0023930888.714807911450000.0208797330.0023900338.736167734500000.0208993210.0023875858.753331129550000.0209153750.0023855788.767424148600000.0209287720.0023839038.779203043650000.0209401220.0023824858.789194549700000.020949860.0023812678.797776816750000.0209583070.0023802128.805228349800000.0209657040.0023792878.811758803850000.0209722340.0023784718.817529017900000.0209780430.0023777458.822664445950000.0209832430.0023770958.8272643741000000.0209879250.0023765098.831408413Note: voltage, current, and impedence are for the dependent source.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 05:45:09 pm
content: The graph is interesting, though.Here's a graph I plotted of V vs I at the dependent source with an external load resistor varying from 100 to 1100 ohms:The formula for the dependent source works out to be:I = 0.005 - V/8so the "resistance" of the dependent source is -8 ohms.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 05:51:40 pm
content: Quote from: MikeK on March 25, 2012, 05:23:39 pmNot only in this particular case, but in all cases.To find the Thevenin equivalent resistance of a circuit you first suppress all internal current sources and voltage sources (set voltage sources to a short circuit and set current sources to an open circuit). Having done that, you now "look back" into the circuit and calculate what resistance it has using the normal rules of parallel and series resistances. Whatever result you get is the equivalent resistance.

author: MikeK
author_rank: Super Contributor
timestamp: March 25, 2012, 06:00:38 pm
content: Right, but this is what I had a problem with. When you remove the independent source and "look back", the dependent source resistance is confusing, because no current flows.If I have 10 ohms in series with 20 ohms, I can say that they equate to 30 ohms even though no current flows. But the above problem doesn't allow the same analysis.By the way, your graph explains why the voltage across the dependent resistor divided by the current through it is not 8 ohms. Because it doesn't pass through (0,0). It has linear behavior, but is not exactly a resistor, because of negative slope and not passing through the origin.

author: IanB
author_rank: Super Contributor
timestamp: March 25, 2012, 06:17:07 pm
content: Quote from: MikeK on March 25, 2012, 06:00:38 pmThat's why the hint in the problem statement suggested to add an independent current source at the output terminals. If you imagine some currentIbeing imposed on the circuit from outside, a voltageVwill be developed across the terminals. If you work out the value ofVin terms ofIand the circuit elements, you will get the equivalent resistance asV/Iin the normal way.(Remember that this imposed current approach is exactly how you derive the formulas for resistances in series and parallel. If you simply memorize those formulas by rote you may forget where they came from.)