I am creating this question to ask you about a "strange" behavior I am observing in Matlab about an op-amp circuit. I am reading "Fundamentals of Electric Circuits" the 5th edition, and in the chapter about op-amps I was trying to solve the practice problem 5.1

enter image description here

I have tried to solve it analytically (by hand) and I couldn't get the results the authors give as solutions. Here are my two different attempts, the first one spanning 4 pages is a branch current method and the last (5th) image is the mesh current method: https://i.sstatic.net/c7Qyf.jpg.

To further confirm the results obtained I have modeled the circuit in Matlab in two different ways:

1) with a finite gain op-amp element: !Can't post the link to this image...

2) with a controlled voltage source and input/output resistances equivalent circuit: !Can't post the link to this image...

The first model backs my results. The second model backs the authors results.

I know that the different results (8.99959 vs 9.0004) differ by less than 0.1% and that it probably wouldn't even be worth creating this question, but hey, this is not homework nor am I an Electrical Engineering student, I am just curious about why this is happening.

Well assuming I actually modeled an equivalent circuit, why is Matlab giving me two different results for the output voltage?

How come is my analytical result different from the authors? Did I make any error in my calculations?

  • \$\begingroup\$ 1) imgur.com/tZgkMdc \$\endgroup\$
    – Scout
    Commented Nov 1, 2014 at 1:10
  • \$\begingroup\$ 2) imgur.com/mnYKep4 \$\endgroup\$
    – Scout
    Commented Nov 1, 2014 at 1:11
  • \$\begingroup\$ I'm not sure I totally follow your different models, but it looks to me like what you found is that the non-zero output impedance of the op-amp is the dominant source of gain errors in the circuit. Although it looks like you used a 50 ohm output resistance in that model, and that sounds like too high a value, even for a '741. \$\endgroup\$
    – The Photon
    Commented Nov 1, 2014 at 2:08
  • \$\begingroup\$ I'm sorry, I have said in my initial post that the Example 5.1 is in the previous page and it's an example where the authors use a 741 to build an inverting amplifier just to show how you can model an equivalent circuit for a non-ideal op-amp. The authors just give us the values for Ri, Ro and the open loop gain of the op-amp. I don't know where they got the values, the values may have just came out of the blue, but the authors tell us to use the same values for the Practice Problem 5.1 \$\endgroup\$
    – Scout
    Commented Nov 1, 2014 at 11:17

3 Answers 3


The node voltage equations are not so difficult. But it's really a tedious work to type the equations. I got the same result with you, maybe we are both wrong :).

enter image description here


I've noticed an issue with the linked models:

Your answer


simulate this circuit – Schematic created using CircuitLab

Book answer


simulate this circuit

At first I was very surprised the book got a "reasonable" answer considering it doesn't look like it has negative feedback for an ideal op-amp. However, consider what happens when Vout = 9.0004V (I'm going to ignore Ro and Ri for now, i.e. Ri = \$\infty\$ and Ro = 0):

\begin{equation} V^+ = V_{out} \frac{ 5k\Omega}{45 k\Omega} = 1.000045V \end{equation} Checking this is compatible with the open-loop op-amp model: \begin{equation} V_{out} = G (V^+ - V^-) = 9.0004 V \end{equation}

Out of pure serendipity this happens to be a "valid" solution. While this is a valid solution, it isn't at all stable and I suspect a more "accurate" op-amp model may be able to capture the stability issue more easily. Plugging a more "accurate" op-amp model into LTspice hints this is true: Vout clips to one of the rails (in my case it was the negative rail), as expected for op-amps with no negative feedback.

tl;dr: your answer is right. It remains right even after I plugged a more "accurate" op-amp model into LTspice. I also solved it algebraically using a computer and got the same answer as you (as well as diverger). However, the output is not at all understandable so I'm not going to post it.

  • \$\begingroup\$ I wonder where the "book answer" comes from? Apparently, it has a positive feedback. \$\endgroup\$
    – diverger
    Commented Nov 1, 2014 at 7:33
  • \$\begingroup\$ @diverger Scout posted two links as comments to his question; the second one matches the "book" answer to every digit listed, so presumably the author used this one to find the answer. Of course, some other non-ideal effect might be at play, though I tried input current leakage and offset voltage and they just didn't line up so nicely. \$\endgroup\$ Commented Nov 1, 2014 at 7:35
  • \$\begingroup\$ I was also surprised that even though it should have positive feedback, the solution happens to be very close to the negative feedback solution. \$\endgroup\$ Commented Nov 1, 2014 at 7:39
  • \$\begingroup\$ I checked the 5th picture, the difference between the two method is: the former is node voltage method, and the latter is mesh current method, the structure of the circuit doesn't change. And they also give same result. \$\endgroup\$
    – diverger
    Commented Nov 1, 2014 at 7:39
  • \$\begingroup\$ Yes and no, if you stare carefully at the OP's equations in picture 5 (particularly Eq. 4) you can see the convention assumes V+ is on the side of Rc/Rq, i.e. the sign convention of what I said was the "book's" answer. \$\endgroup\$ Commented Nov 1, 2014 at 7:49

I believe we can see by inspection that the book answer must be in error, since the non-ideal factors that are introduced (finite gain and input resistance and output resistance) will all tend to decrease the gain from the ideal number of 9.0000.

  • \$\begingroup\$ Is this always true for an op-amp? If we consider non-ideal factors then the feedback gain will always be less than the ideal value? That is a real fast way to validate the results obtained if always true. \$\endgroup\$
    – Scout
    Commented Nov 1, 2014 at 11:00
  • \$\begingroup\$ I can't think of any exceptions.. bias current, offset voltage, etc. either have no effect or reduce the gain. \$\endgroup\$ Commented Nov 1, 2014 at 13:40

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