Need help understanding Thévenin’s Theorem - Page 1
Forum section: electronics / beginners

author: Spork Schivago
author_rank: Frequent Contributor
timestamp: February 21, 2018, 05:57:44 pm
content: I've started reading The Art of Electronics, Third Edition, and I'm a bit confused with what they say about Thévenin's Equivalent Circuit. So I head to Google, and find this site:https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/thevenins-theorem/I was hoping someone could answer a question for me. If you scroll down to the fourth picture, you'll see a circuit with two resistors (R1 = 4 ohm, R3 = 1 ohm), and two power supplies, (B1 = 28V, B2 = 7V).Underneath the circuit there's a chart where they show various values that they calculated using Ohm's Law. So, I want to calculate the voltage drop of R1. First, I calculate the current of the entire circuit, using I = V / R.What I don't understand is their current of 4.2 amp. The only way I can get 4.2 amp is to subtract the 7V from the 28V, to get 21V. Then, 21 / (R1 + R3 = 5) = 4.2 amp.Why do we subtract the 7V from the 28V? I would think we would add the voltages together, not subtract them. I'm a bit confused here and was hoping someone could clarify what's going on in that image so I have a better understanding.Thank you.

author: lordvader88
author_rank: Frequent Contributor
timestamp: February 21, 2018, 06:17:19 pm
content: Well using loop current on the 2 small loops, making 2 KVL loops going clockwise in both loops, means 1 loop current goes down Rload and the other goes up through it. Thats the superpostion of the 2 loops. The total is their sum, but watch the signs as usual.

author: rstofer
author_rank: Super Contributor
timestamp: February 21, 2018, 06:23:59 pm
content: You subtract because the batteries are opposing each other. Write a KVL clockwise around the loop starting in the lower left corner without the load:+28 - (4*I) - (1*I ) - 7 = 05*I = 21VI = 4.2AThen follow along where the author plugs the current back in and calculates the voltage drops across the resistors to arrive at 11.2V across the load resistor. That is the Thevenin voltage.To get the Thevenin internal resistance, simply evaluate the two resistors in parallel: (4 * 1) / (4 + 1) = 0.8 Ohms.That is the Thevenin resistance.

author: lordvader88
author_rank: Frequent Contributor
timestamp: February 21, 2018, 06:49:38 pm
content: Oh right, the 1st thing of Thevenin's rule is find loop resistance without the load, and sources shorted. I forgot the basics by not doing things in order.No wonder I drew a blank on actual Th. theorem.

author: Spork Schivago
author_rank: Frequent Contributor
timestamp: February 21, 2018, 06:59:35 pm
content: Quote from: rstofer on February 21, 2018, 06:23:59 pmI'm sorry for such a dumb question, but because I'm just starting out, with KVL, you mean Kirchhoff's Voltage Law, right? With the math, I think I understand.Code:+28 - (4*I) - (1*I ) - 7 = 0+28 - 5I - 7 = 0+21 - 5I = 0+5I +5I+21 = 5IBut how do we know the batteries are opposing each other? Oh! Because of their schematic symbol. I didn't pay attention there, I see now. Thank you!!!!!Does this mean the voltage going over R3 is supplied solely by B2?

author: Spork Schivago
author_rank: Frequent Contributor
timestamp: February 21, 2018, 07:02:52 pm
content: Quote from: lordvader88 on February 21, 2018, 06:49:38 pmI read your first post but to me, it didn't really explain much, but I think that's because you're much further ahead than I am. I'm just starting out, and I'm trying hard to understand how everything works here. With sources, you mean the supply, right? Ie, the batteries or voltage sources?I need to print some circuits and then work on calculating the voltage drops for the individual components, the total circuit, the current , etc. Kinda like the table they have, but having it blank and me figuring out the answers.I'm hoping I can post them here to have them double checked, just to make sure I'm doing it right.Thank you guys!!!!!!

author: daqq
author_rank: Super Contributor
timestamp: February 21, 2018, 07:20:23 pm
content: QuoteGenerally, yes. It should be noted though that you can have (in theoretical ideal terms):Voltage sources - maintains a constant voltage across its terminals, regardless of current being drawn (used in Thevenin)Current source - maintains a constant current through itself, regardless or the voltage it needs to supply to the outside circuit. (Used in Nortonhttps://en.wikipedia.org/wiki/Norton%27s_theorem)QuoteYup, practice, practice. If you are confused, you might want to look up some youtube lecture on the topic, some are pretty good.

author: rstofer
author_rank: Super Contributor
timestamp: February 21, 2018, 08:01:51 pm
content: Quote from: Spork Schivago on February 21, 2018, 06:59:35 pmWe know the Thevenin voltage is 11.2V and we know that the internal resistance is 0.8 Ohms. So, we add in the 2 Ohm load resistor and we get the current through the center resistor11.2V / (2.8Ohms) = 4 Amps. So, 4 amps flows down the center resistor, out of the node. Working back up through the 2 Ohm load resistor, we find that the center point is 8V (4A * 2 Ohms).If the center point is 8V and B2 is 7V then there is 1V dropped across R3 (we have to get from 8 down to 7) so the current flow is INTO the battery - it is getting charged! The current flow is out of the node.Well, we know that there is 1A flowing to the right out of the center node toward the battery and we know there is 4A flowing out of the center node down through R2 so there must be 5A coming into the center node through R1. That means the drop across R1 is 20V (5A * 4Ohms) which means the center point should be 28V - 20V (=8V) and that's good because we already established that point.The circuit balances!The current through R1 is 5A going into the center node.The current through R3 is -1A going into the center node (which means it is actually flowing out of the node)The current through R2 is -4A going into the center node (again, the current is flowing out)The sum of the currents going into and leaving the center node is zero - good thing!This nodal analysis is the easy way to figure out how the circuit works without resorting to Thevenin.The current going into the node from the left is (28-V1)/4The current going into the node from the bottom is V1 / 2The current going into the node from the right is (7-V1)/1Realizing that the sum of the currents going into the node must equal zero, we just add them up. What we wind up with is the node voltage and from there it is easy sailing. Note that the current from the right will be negative as will the current coming up from the bottom. This is a sign convention issue. We assumed the current was flowing into the node, the minus sign just says we were wrong and the current is actually flowing out of the node.The equation for the nodal analysis is(28-V1)/4 + (7-V1)/1 + (0-V1)/2=0 Note how the 3rd term is convoluted to force current INTO the node.28-V1+28-4*V1-2*V1=0 -- multiply through by 456-7*V1=0V1=8VIt seems silly to assume the load current flows into the node and you will get the same answer if you assume it flows out because that current will be subtracting (V1/2) which is just what happened above!Just be consistent in how you draw the arrows and write the matching equations. Everything will work out!

author: rstofer
author_rank: Super Contributor
timestamp: February 21, 2018, 08:09:44 pm
content: Here is a good lecture series along with problems and worksheets:https://learn.digilentinc.com/classroom/realanalog/

author: rstofer
author_rank: Super Contributor
timestamp: February 21, 2018, 08:42:00 pm
content: Regarding sign convention

author: rstofer
author_rank: Super Contributor
timestamp: February 21, 2018, 09:10:28 pm
content: Quote from: Spork Schivago on February 21, 2018, 06:59:35 pmWhen I write a loop equation, I draw the loop starting at the reference and run it up clockwise through B1. Since I enter at the (-) end of the battery, I gain voltage. I drop voltages through the resistors and then I hit B2. Note that I enter that source at the (+) end so I have to subtract.I could start in the lower right corner and run up through B2 which gains 7V then drop voltages through the resistors and enter B1 on the (+) terminal where I would subtract that value.7 - I*R3 - I*R1 - 28 = 0-I(R1+R3) = 21VI = -21 / (R1+R3) = -21 / 5 = -4.2A OOPS! The assumed counterclockwise current flow is wrong. The actual current flow is clockwise - as in my first solution.Nevertheless, it doesn't matter where I start or which direction I travel, it will all work out. As the circuits become more complex, we will often find that our assumed directions were wrong. It's unimportant as long as we are consistent with our direction arrows and the equations.You may have noticed that I posted a nodal analysis when I solved this problem a couple of replies back. I have not posted a mesh analysis! Why?There is only one node - the common connection point of all the resistors and I assigned that a voltage V1. There is only one node, therefore only one equation and only one unknown (V1).Were I to post a mesh solution, it takes two meshes to cover the circuit. One around the outside and one around the B1-R1-R2 loop. I need two meshes to cover all the branches. As a result, I need two equations in two unknowns (the loop currents) and that requires a little more arithmetic.When circuits get more complex, nodal analysis always takes less equations than mesh analysis. When you have to solve them by hand, the difference between a 2x2 and 3x3 matrix solution is substantial. Get out to 4x4 or even 6x6 like we were talking about yesterday and a computer is about the only way to do the job.Watch the Modified Nodal Analysis video in Reply 39 and my computer generated solutions for the 6 equations in replies 46 and 47.https://www.eevblog.com/forum/beginners/using-laplace-to-analyse-circuits/This will show you a bit about where you are headed. The MNA approach came out after I graduated and I had never heard of it. It is a very powerful method that combines KCL and KVL to analyze circuits.

author: Spork Schivago
author_rank: Frequent Contributor
timestamp: February 21, 2018, 10:48:16 pm
content: Quote from: daqq on February 21, 2018, 07:20:23 pmThis is something the book I'm reading talked about, but I don't understand it. I thought Ohm's Law shows us the current flowing in a circuit is proportional to the voltage. So, with voltage sources, for instance, if the current being drawn goes up, wouldn't the voltage that the load use drop?I have been using the book and the internet, which includes Youtube videos, among other sites (eevblog, spark-fun, etc) if the book doesn't explain it in a way that makes sense or if I need more information. Even with the internet at my disposal, I felt I needed to ask here for help with understanding Thévenin's Theorem.Thanks.

author: Spork Schivago
author_rank: Frequent Contributor
timestamp: February 21, 2018, 10:59:52 pm
content: Quote from: rstofer on February 21, 2018, 09:10:28 pmI haven't learned about mesh analysis yet, but thanks for the video! I will watch it. It's slow learning, because I have a daughter and I can only work on this stuff when she's sleeping or when my wife is awake and watching her (which isn't often right now). But things are starting to click. Like someone said, practice, practice, practice. I think that's the key, just to make sure it's set in stone, and then moving towards more complex stuff.

author: lordvader88
author_rank: Frequent Contributor
timestamp: February 21, 2018, 11:01:42 pm
content: Quote from: Spork Schivago on February 21, 2018, 07:02:52 pmYeah its all in that web site tho, Mesh analysis/loop current analysis is what I used.Yeah the 1)voltage source(s) get replaced with a perfect wire, 2)the load is removed /left open 3)capacitors are ignored/ removed ,if u added them into the circuit.

author: rstofer
author_rank: Super Contributor
timestamp: February 21, 2018, 11:03:06 pm
content: I decided to work out the mesh analysis. I wrote two loops: the first is around the lower left circuit starting from the lower left corner and traversing B1, R1 and R2, clockwise. The second loop starts in the lower right corner and traverses B2, R3 and R2, counter-clockwise. When writing the voltage for R2, it is important to remember that both currents are flowing through it. In this case, they add because both currents are flowing top to bottom.The first loop looks like: 28 - 4*I1 - 2*I1 - 2*I2 = 0 -- see how I2 got involved at R2?This cleans up as 28 - 6*I1 - 2*I2 = 0The second loop looks like: 7 - 1*I2 - 2*I2 - 2*I1 = 0 -- see how I1 got involved at R2?This cleans up as 7 - 2*I1 - 3*I2 = 0Solve the first equation for I2 in terms of a voltage and I1: I2 = (28 - 6*I1)/2Plug that into the second equation to get 2*I1 + (84 - 18*I1)/2 = 7Solve for I1 and get 5A - that is the current leaving battery B1Plug the value into the second equation to get I2 = -1A -- we assumed the wrong direction, the current is going into the battery.Now, since we have 5A going into the node and 1A leaving to the right, the other 4 amps are flowing down through R2.Mesh analysis is a little more difficult. I make it easier by jamming the equations into MATLAB and watching the screen. See attached

author: Spork Schivago
author_rank: Frequent Contributor
timestamp: February 22, 2018, 02:52:27 am
content: So I'm looking at the website,https://learn.digilentinc.com/Classroom/RealAnalog/circ1/Text/RealAnalog-Circuits1-Chapter1.pdfAnd I'm at the exercises. Is this the right place? Am I just supposed to make up resistor values for Exercise 1? Tomorrow, I'll read the first chapter and maybe it'll make more sense.I'd like to concentrate on the Art of Electronics, 3rd Edition, but I might be able to slip the Real Analog site into my schedule as well and work through both.Thanks.

author: rstofer
author_rank: Super Contributor
timestamp: February 22, 2018, 04:22:32 am
content: They're not asking you to solve anything just to add current arrows and polarities showing your assumed current flow direction.The first problem has one source and from the reference node (the bottom node) you flow up through the source from - to + Therefore, the current arrow adjacent to the battery is pointing up. Once you get to the top, you will see that all the other arrows point in a direction of getting back to the reference node. To the right and down...The other problems get tricky because the sources are pointing in directions contrary to the flow arrows and, as a result, you need to get the polarity correct.But that's just what I was saying when I talked earlier about the nodal analysis. I made the assumption that ALL currents were entering the node. That is as valid a guess as any. As it turns out, the current if flowing OUT of the node for the center branch and the right branch. That's apparent from the (-) signs in the current calculation.The Passive Sign Convention starts on page 5 and this is the heart of the matter. You can make any assumption you want, the arithmetic will take care of the matter. It's helpful to make an intelligent guess and it could be argued that I knew the center branch current was leaving the node. No problem, assume two currents going into the node from the battery branches and one current leaving through the 2 ohm resistor. That is also a valid assumption (and better than the other).Some people advocate assuming all currents are entering (or leaving) a node. I am wishy-washy... I prefer to guess the proper current flow but I also know that assuming all current enter the node or all currents exit the node will also work. Assuming the center leg current enters the node makes the term a little goofy in that the current is (0V - V1) / R2. That looks like I am assuming that ground is more positive than the node and to get current to flow up hill that is exactly what would have to happen. But look again! If I had assumed the current was leaving the node, the node equation would have had a -V1/R2 term and that's exactly what I used (once you junk the 0V part).Take your time, draw circuits, draw current loops (meshes) and, by all means, work through nodal analysis. It's a lot easier than mesh analysis. IMO...It's all pretty easy but it is vitally important. Dave's Op Amp video uses Kirchhoff's Current Law to describe what happens at the inverting input. You can't do op amps without understanding what happens at that input.

author: Nerull
author_rank: Frequent Contributor
timestamp: February 22, 2018, 01:48:59 pm
content: Quote from: Spork Schivago on February 21, 2018, 10:48:16 pm"Proportional" means that both increase by the same factor. If you have a circuit with fixed resistance the only way to increase current is to increase voltage. Doubling voltage will double current.1V/1Ohm = 1A. 2V/1Ohm = 2A. 200V/1Ohm=200A

author: daqq
author_rank: Super Contributor
timestamp: February 22, 2018, 02:38:52 pm
content: QuoteIt should be noted that we are talking about ideal voltage sources - a mathematical concept that has the same voltage even if you draw a million amps from it. You are quite correct, a real voltage source has a drop when you load it. When dealing with simplified models you can ignore this kind of physics.

author: rstofer
author_rank: Super Contributor
timestamp: February 22, 2018, 02:55:30 pm
content: Quote from: daqq on February 22, 2018, 02:38:52 pmAnd this is the entire point of calculating the Thevenin equivalent voltage/resistance. You are finding out how much internal resistance the source actually has.If you were to build up a source that looked like the circuit above, using 2 batteries and 2 resistors and put it in a black box labelled "Battery" with only 2 terminals exposed, you would find that it had an output voltage of 11.2V and an internal resistance of 0.8 Ohms. As a result, if you were to draw 1A from the battery, the voltage would drop by 0.8V and you would only have 10.4V at the terminals.This should also point out just how low the internal resistance of a real battery must be. That's why, for a first approximation, we consider the internal resistance of sources to be 0 Ohms.A little later on, you will find that maximum power transfer occurs when the external resistance is equal to the internal resistance. But that's a discussion that usually comes a little later.

author: rstofer
author_rank: Super Contributor
timestamp: February 22, 2018, 05:05:20 pm
content: I went Googling for 'why is thevenin theorem important' and I got a lot of pretty poor answers. It was almost like a circular definition: "It's important because you can analyze complex circuits.". So?Where I find it to be of value is in calculating the input and output impedance of amplifiers. Although they don't mention Thevenin directly, the authors keep dancing around the procedure:https://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.htmlYou want to look into the input port and find an impedance and then you want to turn around and look back at the output port and calculate an output impedance. This is Thevenin at work!Next up: Norton Equivalent Circuits.

author: daqq
author_rank: Super Contributor
timestamp: February 22, 2018, 05:52:12 pm
content: QuoteI actually used them in my own method:http://www.daqq.eu/?p=1333to make a few simplifications and test the math out.

author: Spork Schivago
author_rank: Frequent Contributor
timestamp: February 22, 2018, 08:17:57 pm
content: Quote from: rstofer on February 22, 2018, 02:55:30 pm

author: rstofer
author_rank: Super Contributor
timestamp: February 23, 2018, 12:46:31 am
content: The problem is that the original model was deficient. It didn't include all of the resistances in the circuit. As a result, the Thevenin resistance is not quite right. Neither is the open circuit voltage. Poor models guarantee poor performance!In class, all things are "ideal". Wires have 0 Ohms resistance, op amps don't have offset voltages or bias currents. I always thought that things that were supposed to be constant tended toward being variable. Linear variables tend to be exponential, etc.You'll get into distributed parameters a bit later on.

author: Spork Schivago
author_rank: Frequent Contributor
timestamp: February 23, 2018, 02:21:19 am
content: So I'm back to the book right now, and I see they are using the voltage divider and says it must have a Thévenin equivalent.Then they show:Code:1. The open-circuit voltage isV = Vin (R2/(R1+R2))2. The short-circuit current isVin/R1So the Thévenin equivalent circuit is a voltage source,Vth = Vin(R2/(R1 + R2))in series with a resistor,Rth = (R1R2)/(R1 + R2)I don't follow this at all. Why is the open-circuit voltage V = Vin (R2/(R1 + R2))? Is it because V, in this example, is equivalent to Vout in the voltage divider circuit? The short-circuit current I don't follow at all. From the website I was looking at (and the ones you guys linked me to), I was understanding the Thévenin circuit, but this book has me confused now.They show a voltage divider circuit (page 9, Figure 1.12). And then the Thévenin equivalent of that voltage divider. I don't understand the second circuit. I understand the voltage divider and how it works. But with the second circuit, it looks like Vin now becomes VTh, R1 becomes RTh, and R2 becomes Rload.Should I put the book on hold for a bit and read the website to learn about the Thévenin Theorem? I keep reading a bit, but when I reach the next exercise in the book that they expect me to be able to solve, I'm lost. So I think I need more info (maybe from that website).**EDIT: I went back to the website and started reading. I'm on page 6 of chapter one and now I'm even more confused. It says:Code:The assumed direction of the current, i, passing through the element is shown by the arrow on Figure 1.1. InFigure 1.1, i is assumed to be positive if it is going into node a.Early, on page 4, it shows current is equal to dq / dt, where q is the charge in coulombs, t is the time in seconds. Page three says, "The charge of one electron corresponds to -1.6022×10-19 C".So, as we know, the electron is negatively charged. Wouldn't the assumed direction for the current be the opposite of what they show on page 6? If a is positive, b is negative, I'd assume those negatively charged electrons where moving towards the positive side, not from the positive side to the negative side.

author: rstofer
author_rank: Super Contributor
timestamp: February 23, 2018, 02:54:25 am
content: Let's look at the voltage first - this implies there is no load because we want the open circuit voltageThe current through the voltage divider is I = Vin / (R1 + R2) and the output voltage is I * R2 which results in (Vin * R2) / (R1 + R2)There are other ways to solve the voltage divider, I decided to get pedantic about the loop current.Looking back into the source with Vin shorted, all we have is R1 & R2 in parallel so the Thevenin Equivalent Resistance is (R1*R2)/(R1+R2)So, our Thevenin Equivalent Circuit is: Vth = (Vin * R2) / (R1 + R2) in series with a resistor Rth = (R1*R2)/(R1+R2)The open circuit voltage is just what you see when you measure across R2 with no load connected. The Thevenin resistance is, in this case, just the parallel combination of R1 and R2.ADD:Pretend Vin is 10V and R1 = R2 = 1 Ohm. It's fairly obvious that the Thevenin voltage is 5VThen calculate the Thevenin resistance as 0.5 Ohm.Your equivalent circuit is a 5V source with a 0.5 Ohm series resistor. Your maximum current is 10 Amps (5V / 0.5 Ohm), your open circuit voltage is 5V. If you go back to the original circuit and short out R2, the maximum current is 10V / 1 Ohm = 10 Amps. Same maximum current and same open circuit output voltage - 5V. Circuits are equivalent.