# Using Thevenin with Op amp.

Discussion in 'Homework Help' started by sn00py23, Mar 3, 2010.

1. ### sn00py23 Thread Starter Member

Mar 3, 2010
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Am having problems understanding how to find equivalents for this circuit with the ideal op amp associated with it. Without an output to reference, how do I determine what to do with the op amp in this circuit. I can use Thevenin on simple DC circuits, but this is my first with an op amp.

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Remove the 4Ω resistor; then the node where it was connected (the plus input of the opamp) is your output. Calculate the voltage appearing there and that is your Vth.

3. ### sn00py23 Thread Starter Member

Mar 3, 2010
31
0
it's the part to get there.

I was taught to find Rth you have to replace the voltage source with a short and then solve with no sources, but since I never did one with an op amp and am thinking of the DC voltage to bias it (though it wasn't showing) then I don't know what to do with the op amp to find Rth.

Next, finding Vth. I was taught to find that you needed to solve the circuit as it sits without the load and then do a KVL from A to B which is where the load was, and wallah done. I just don't know what to do with the op amp to distribute Vs.

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Let's see if I can't get you started. We'll use KCL.

Number the nodes like this: the junction of the 8, 16 and 32 ohm resistors will be labeled V1; the + input of the opamp will be V2; the - input of the opamp will be V3 and the output of the opamp will be Vo.

Sum the currents at each node; currents away from the node are taken as positive.

The currents leaving node V1 are calculated as follows:

The current in the 16 ohm resistor is (V1-6)/16 amps.

The current in the 8 ohm resistor is (V1-V2)/8 amps.

The current in the 32 ohm resistor is (V1-Vo)/32 amps.

Summing all the currents to zero we have:

(V1-6)/16 + (V1-V2)/8 + (V1-Vo)/32 = 0

In a similar manner write equations for the sum of the currents at node V2 and V3 (remember to delete the 4 ohm resistor). Finally to get a fourth equation, set V2=V3 (because the opamp is taken to be ideal, with infinite gain).

You should then be able to solve the 4 equations for the 4 unknowns.

The voltage at nodes V1 and V2 should end up being the same; this will be your Vth.

To find Rth, replace the 6 volt source with a short (and remember the 4 ohm resistor is still deleted), apply a 1 amp current source at node V2 and solve the network again. You should be able to use the same equations, but the equation for node V2 will sum to 1 instead of 0. The voltage thus determined at node V2 will be numerically equal to the impedance there; that will be Rth.

5. ### sn00py23 Thread Starter Member

Mar 3, 2010
31
0
I understand how to do Nodal and I was right with you until you got to the Op Amp. I'm not sure I understand the concept of Ideal as it pertains to the circuit and how I should look at it. I know what to do with current sources and voltage sources but now you're adding one with 1A and I don't know where you got that from.

Also, there are five unknowns and you say there are four. Are you looking at the two 10 Ohm resistors as being in parallel?

I'm just confused having the op amp in the circuit and it's affect. I can do everything else except know what is going on with the op amp.

I appreciate all your work and I'm sure this is a simple concept I'm missing, but I was only taught basics about an op amp and how it amplified in a circuit and not to know it's affect on anything like this to where it's situated in a circuit with a DC source vice any focus on the bias for it and it's relational output to input.

What we learned was only that you had an input and where the signal was coming in made it invert or not. Then you could sum what was going on and to focus on feedback. So, forgive me, but I'm kinda lost here and I'm thinking I'm missing something.

Is the op amp being treated like a resistor from V2 to Vo? Is is acting as an open? Can I justify flow from the 32 Ohm resistor's path to the two 10 Ohm resistors? It's crazy, I know, and I'm feeling mighty stupid but this is the first time I've seen an op amp in a circuit like this.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I will be assuming that you made the notation on your schematic of the 4 nodes (V1, V2, V3 and Vo) as I described them.

An ideal opamp is considered to have infinite voltage gain, and this leads to the well used notion that the voltage between the + and - inputs is zero. That's why one of your equations is V2=V3.

The opamp is a voltage controlled voltage source; it's a voltage amplifier. The infinite voltage gain of the (ideal) opamp causes the output, Vo, to be whatever it has to be to cause the + and - inputs of the opamp to be the same, when you have negative feedback. That's the important concept that allows you to easily solve these kind of opamp problems.

Imagine that you have some electrical network with some arbitrary number of nodes, one of which is the reference node (ground, usually). Suppose you want to know what impedance (resistance in your case) will be measured by an ohmmeter connected between some node, Vx, and ground. A real ohmmeter does it by applying a current to that node and measuring the resultant voltage; from that, the impedance can be determined.

In circuit analysis, you can achieve the same result by applying a unit current to the node and calculating the voltage that will appear at the node as a result of injecting the current (unit current; 1 amp). When you pass a current of 1 amp through a resistance, the voltage across the resistance is numerically equal to the resistance--simple Ohm's law.

You need to know Rth at node V2; you can find it by applying a 1 amp current to the node (with the 6 volt battery replaced with a short) and calculating the voltage that results at that node.

But, before we try to calculate Rth (for which you short the 6 volt battery and apply the 1 amp source), let's determine the node voltages (V2 will be Vth), for which we leave the 6 volt battery in place and don't apply any extra current sources.

No. The way I have described to solve the network, there are 4 unknowns--the 4 node voltages, V1, V2, V3 and Vo. What do you think the fifth unknown is?

All you need to know is that the voltage at the + input is the same as the voltage at the - input. This is a result of the infinite gain of the ideal opamp and of the fact that there is negative feedback via the two 10Ω resistors.

You assume the voltage at the output of the opamp is Vo and that V2=V3 and it will all work out.

I'll repeat what I said above: the opamp is a voltage controlled voltage source; it's a voltage amplifier. The infinite voltage gain of the (ideal) opamp causes the output, Vo, to be whatever it has to be to cause the + and - inputs of the opamp to be the same, when you have negative feedback. That's the important concept that allows you to easily solve these kind of opamp problems.

The current in the two 10Ω resistors doesn't come from the 6 volt battery; it comes from the opamp output. The opamp has its own power supply (not shown in the schematic, but it's there--otherwise the opamp won't work), and the current in the 10Ω resistors comes from that power supply.

The current in the other resistors can also be partially supplied by the opamp, and partially by the 6 volt battery.

Try writing the 3 equations for nodes V1, V2 and V3, summing the currents at those nodes to zero as I described, and as I gave the equation for node V1.

Remember that your fourth equation will be V2=V3. When you get the four equations, solve them as a system of simultaneous equations. Show this work, which will allow you to determine the voltages at the nodes with the 6 volt battery in place (V2 will be Vth) and we'll go from there to calculate Rth.

7. ### sn00py23 Thread Starter Member

Mar 3, 2010
31
0
Okay, I'm really getting kind of confused.

I was thinking of Rth and finding solutions for the loads, all of them and that's how I was thinking of using the two 10 Ohm resistors.

Anyway, I need to use Thevenin on this circuit and you are using Nodal. I see how and I used the Node voltages but I'm not getting how to solve for V2 and V3, which are equal when I can't figure out how to find V3.

If there is infinite resistance in the Op amp from + to - then I would see that as an open whereby the circuit would have a flaw.

Just to recap. I remove the 4 Ohm resistor which is what I'm solving for, let's skip that I need to find current for the moment and I'm putting together an equivalent Thevenin circuit.

I need to find Rth. I short the voltage source and remove the 4Ohm resistor and place it in the equivalent circuit. Here's where I'm lost. I don't know how to look at the op amp. How can I say that V2 is = to V3 when I can't solve for either if I don't know how it appears in the circuit.

For instance, I replaced the voltage source with a short, do I replace the op amp as a resistor so I can determine Rth? I just can't see the configuration I need to find Rth.

Then, to find Vth, I have to solve the circuit as if there was no 4Ohm resistor and again I'm stumped by the op amp.

I do get the bias voltage and that it is irrelevant but if I were reducing the circuit just how does one reduce the circuit to find the total resistance when I can't figure out how the op amp is situated for flow in the circuit.

I am So grateful to you for what you have helped me through but I'm still not getting the op amp. If you could clarify it's configuration and flow in the circuit then all else should fall into place for me.

Thank you.

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Loads will be connected to the V2 node, not the V3 node. The two 10Ω resistors will not be loads. The will be absorbed into the Thevenin equivalent.

The problem statement first says "Replace everything except the 4Ω resistor by its Thevenin equivalent...". Then it says "...and use the result to find i." The first thing you have to do is find the Thevenin equivalent, then you can use that Thevenin equivalent to find i. We could use the nodal method to find I directly, but that's not what I'm helping you use the nodal method for. I'm showing you how to use the nodal method to find the Thevenin equivalent.

If you set up the four simultaneous KCL equations as I've suggested, solving them will give you all four node voltages, V3 included.

Even if the opamp itself has infinite resistance from + input to - input, there are other circuit elements (resistors) which provide a less than infinite resistance to those two nodes, so that there is no flaw in the circuit. In fact, the other assumption about an ideal opamp that we are using is that the two inputs take no current because of that "infinite" resistance.

In the equations we are setting up it is assumed that no current flows into the opamp inputs.

Go back to post #6 and re-read the part where I talked about an ohmmeter.

To find Rth, you replace the battery with a short (the 4Ω resistor is also removed at this point) and find the resistance that an ohmmeter would measure if you connected it from the V2 node to ground.

When the 6 volt battery is replaced with a short the ohmmeter would be the only source of voltage applied to the network. The ohmmeter would inject a current into the V2 node and measure the voltage that was thereby created at V2. The ohmmeter would know how much current it injected and after measuring the voltage it would use Ohm's law to calculate the equivalent resistance at that node. That resistance is Rth.

When solving a circuit with nodal analysis, you can do what the ohmmeter would do by attaching a 1 amp current source to node V2 (with the 6 volt battery replaced by a short) and calculating the voltage induced by the current. Ohm's law tells you that the voltage is numerically equal to the equivalent resistance at that node.

Go back and read post #4 where I showed how to set up the first equation by calculating the current in each resistor connected to node V1, and then summing those currents to zero.

The opamp is dealt with like the other circuit elements by assuming that no current flows into the inputs, and that the voltage at the two inputs is the same. Then you just sum the currents at nodes V2 and V3 to get equations #2 and #3. Equation #4 is just V2 - V3 = 0.

You have to have 4 equations to find the 4 node voltages V1, V2, V3 and Vo when the 6 volt battery is in place. After you find the node voltages, you just use the fact that the V2 node is the output of the Thevenin equivalent to say "Vth is equal to V2".

The second equation may be a little confusing because with the 4Ω resistor gone, and with the assumption that the opamp + input takes no current, there will be no current in the 8Ω resistor. So, I'll give you the second equation; it's:

(V2 - V1)/8 = 0

It's this equation that results in V2 being the same as V1.

I've told you that the fourth equation is V2 = V3.

Give a try to deriving the third equation which is found by setting the sum of the currents in the two 10Ω resistors to zero. The currents in these two resistors can be found by using the voltages at nodes V3 and Vo.

Don't let the opamp stump you. You give the output voltage a name, Vo, and just use it as an unknown in setting up the KCL equations. Doing that plus using the ideal opamp assumptions that V2=V3 and that no current flows into the opamp inputs, will take care of the opamp.

After you've found Vth, then you can use the same 4 equations with a couple of slight modifications (replace the battery with a short and add a 1 amp current source into node V2) to find Rth.

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Imagine that you removed the opamp, leaving the Vo node that was the output node of the opamp.

Now you would have a network (with the 4Ω resistor also removed) without any opamps, and you should have no trouble solving the network for the node voltages.

In fact, the solution would be:

V1 = 4
V2 = 4
V3 = 0
Vo = 0

Now add another voltage source; connect it to the Vo node. Let this additional source have a variable voltage of X volts.

You should have no trouble solving the network again. This time the solution would be:

V1 = (X + 12)/3
V2 = (X + 12)/3
V3 = X/2
Vo = X

Notice that if we let X = 0, we get the previous solution, as we expect.

Notice further that the expressions for V2 and V3 are different. But, and here is the important thing to notice, there is a value of X that will make V2 and V3 the same.

It's easy to find that value of X. Just solve V2 = V3, or (X + 12)/3 = X/2.

Well, this is what the opamp does. By means of the negative feedback from the output of the opamp to its - input, it forces the + input and the - input to be the same. And the value of X that makes (X + 12)/3 = X/2 will be the voltage at the output of the opamp in the original circuit when the 4Ω resistor is removed, and the 6 volt battery is left in place.

When your fourth equation is V2 = V3, that's doing the same thing to your network solution that the opamp does--it's making V2 = V3.

10. ### sn00py23 Thread Starter Member

Mar 3, 2010
31
0
but I got 12V for V2, is that possible?

Also, is there any way I can use just Thevenin to reduce this circuit? I realize you are using Nodal to get Thevenin, but my Instructor wants us to use Thevenin only.

Meaning to remove the load and put it in an equivalent Thevenin circuit and solve for open circuit voltage and Rth without using another theorem. Is this possible without using another theorem first?

11. ### Heavydoody Active Member

Jul 31, 2009
140
11
Thevenin's theorem is not a circuit analysis technique. No matter what you do, you will eventually need to use something else to solve the problem, even if its just combining parallel resistors and applying Ohm's law. The purpose of Thevenin's theorem is to make a complex circuit, which isn't going to change, simple, so that other circuits which get their input from it can be toyed with without having to go back and recalculate the unchanged portion again. So in this case, you could change that four ohm resistor with one of a different value and readily determine the new resistor's current. I am no seer, but I would guess that your instructor wants you to find the current through the four ohm resistor by using the Thevenin equivalent only. How you arrive at the Thevenin equivalent should be up to you.

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12. ### The Electrician AAC Fanatic!

Oct 9, 2007
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That is correct. You are wondering how it can be 12V when it looks like the only source of voltage in the circuit is the 6 volt battery, right? Remember that the opamp can put out voltage equal to its supply voltage. The opamp supply voltage is not specified, and since it is taken to be a ideal opamp, it can put out any voltage.

The opamp is configured as a non-inverting amplifier with a gain of 2, so the output voltage should be 2 times the voltage at the + input, or 24 volts.

So your instructor wants you to use Thevenin's theorem to determine an overall Thevenin equivalent? That sounds somewhat circular.

The 6 volt battery and its 16 ohm resistor are inherently a Thevenin subcircuit, but as soon as the opamp is included, things get sufficiently complicated that I don't see a way to just determine another Thevenin subcircuit by inspection.

You can replace the 6 volt battery by a short, but the other source is the opamp output, and you can't just replace it by a short circuit. You have to do a certain amount of algebra on the whole thing.

When your instructor shows you how to do it, would you report back to us and show his method?

By the way, did you get a number for Rth using nodal analysis yet?

13. ### sn00py23 Thread Starter Member

Mar 3, 2010
31
0
I get 1V for Vin, which means I should have 4A across the 4Ohm resistor. If that's right then I know what I was needing to do. If it isn't, well let's not go there.

What did you get?

Whoa, I just read your post because I didn't refresh in time and now the math that I thought was incomplete and didn't make sense because I guessed, is the right answer? Yet, the equation for finding Vin says Vo(Rg/(Rg+Rf) says that Vin is 1V, which then means that if I do a KVL walk from there to ground that I can only get 1V across the 4Ohm resistor, BUT I didn't take into account the balance of the circuit but now knowing V2 is 1V and Vo is 2V, I shouldn't have a problem solving for the balance of the circuit and for all other unknowns to check my work, right?

Can't do it now because I have some running around to do, but will check and get back to you.

Okay, just checked this along with the KVLs and it works. Knowing that Vin is 1V using the calculations for an ideal op amp and knowing Vo allowed me to determine that at Vin which was your V2 that I have 1V, so nothing to solve for. All I would have to do is solve for V1 which is a piece of cake given everything else and having done that I could determine that I have a voltage drop of 1.57V on the 8Ohm resistor, 3.43V on the 16Ohm resistor and .57V on the 32Ohm resistor. Then, checking KVLs, everything works including the point of 2V at Vout to ground. Whew!

Is that what you got?

Last edited: Mar 5, 2010
14. ### sn00py23 Thread Starter Member

Mar 3, 2010
31
0
You're from Northern OH? Where exactly? I'm originally from Cleveland. Actually, I'm not sure that you're correct in your statement that Thevenin isn't an analysis technique when it is used for finding maximum power. I do agree with why it is used but again that goes back to Maximum power. For straight DC stuff, I know my stuff! ;-) but as soon as I begin to apply what I know to stuff I haven't seen or worked with then I can't seem to see the forest for the trees.

Once I found my formulae for gain and saw (by doing) then it became clear to me what I needed to do.

Now, I'm not clear as to why 12V is good unless the Electrician how just emphasized his point of whatever I put in that I will get double out, but I think I have to speak with his further.

Take care and I hope you're enjoying all that snow, you lucky devil you. I miss is LOTS down here, over by the water.

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I think you've gotten ahead of yourself here.

What we were supposed to be doing first was to find the Thevenin equivalent for everything except the 4 ohm resistor.

To do that, we were solving the network with the 4 ohm resistor removed, and I had suggested labels for the 4 nodes of the network.

It's true that the voltage at node V2 given that the 4 ohm resistor is missing is 12 volts. But that's only part of the Thevenin equivalent.

You need to determine Rth, the Thevenin resistance.

Have you made any progress on this front?

In post #13, you gave a number of voltages, which I assume were with the 4 ohm resistor back in place. Those numbers are not correct, and you didn't mention how you used the Thevenin equivalent to get them.

You need to get the correct Thevenin equivalent before you can find the current in the 4 ohm resistor, which is what your original problem statement is asking for.

16. ### sn00py23 Thread Starter Member

Mar 3, 2010
31
0
This circuit will be with me forever.

I went back and took a look at Rth and it didn't jive, so I'm back to square one and a half.

At this point I don't recall how I got the 12V. Am I wrong to think that I have Vin at 1V and Vo at 2V. If that's okay, then I'll sleep better tonight.

Isn't Vin = V2? That's what I was thinking when I found Vin. I'm beginning, after rereading everything you wrote that I have a better understanding of the op amp but not it's effect on the circuit, but I'm going to take a risk and assume that the 32Ohm resistor is feeding back into the circuit through the 8Ohm resistor. Is that correct? If that's true then that would account for the larger drop there. At Node 1 I am getting current (thinking conventional flow) into that node from the 32Ohm and 16Ohm resistor and then feeding back into the 8Ohm to go into Node 2 and then into the op amp? Please tell me that's correct, if for no other reason than to passify me until I get a brain embolism and it puts me out of my misery.

If Rg and Rf aren't really needed but Rf is shorted as everything I see implies then I get 2V to ground then I should be getting 1V there as well as Vin which makes sense, but then my Rth isn't working out.

I'm beginning to wonder why I gave up drinking.......

17. ### The Electrician AAC Fanatic!

Oct 9, 2007
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In post #13, you used the symbols Vin, Rg, and Rf for the first time. They aren't on the original schematic.

It would be a big help if you would post a schematic showing the use of those symbols.

18. ### sn00py23 Thread Starter Member

Mar 3, 2010
31
0
Vin is the + side of the op amp and Rg and Rf are the two 10-Ohm resistors. Sorry for the confusion. Rf is the feedback resistor.

Maybe that's why I'm getting confused.

Also, as I stated earlier, I can't seem to get a good handle on Rth either because every time I calculate I come up with something different.

19. ### Heavydoody Active Member

Jul 31, 2009
140
11
In the first step, I get 12v at V2 (the noninverting input to the opamp), which is what you had earlier. V2=V3, so you have 12v at V3 as well. V0 should be 24v. Of course, none of the other voltages really matter, Vth is the 12v you are getting at V2. What values have you gotten for Rth in the second part?

20. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,282
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When you have a dependent source like you do in this network (the opamp), you aren't going to be able to solve the network by just looking at various resistors and seeing how they are hooked up and where your voltage source is, and then doing what you would with a simple circuit--just using your intuition. You're going to have to solve the network equations.

I've given you the first two and the last equation:

Eq 1: (V1-6)/16 + (V1-V2)/8 + (V1-Vo)/32 = 0
Eq 2: (V2 - V1)/8 = 0
Eq 3: ???????
Eq 4: V2 - V3 = 0

If you'll give a try at deriving equation 3, we can solve for Vth. Then, we can make the necessary small modifications to the equations and get Rth.

So, see what you can do for equation 3.