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I have been having quite a bit of difficulty analyzing this op - amp. Can someone explain how to find the voltage A?

I know there are a couple of typical configurations of the op - amp circuit, including a summer, amplifier, open loop configuration, and differential op amp. My issue arises here because this is an op - amp circuit that is completely different from anything I have seen before.

So far, I have tried to use KCL by doing at the node at the negative terminal by doing (10 - 11) / 110k = (11 - Vout) / 3M, and solving for Vout, which gives me Vout = 38.2727 V. However, this is not the output I am expecting (my software says it should be 11V at A. Mainly, my problem lies in the fact that there is a split branch between the 3M, 70 ohms, and 250 ohms, and I'm not sure what is in parallel with what and how I can simplify this area. I am really confused as to the approach to solve this problem and would appreciate any help. Thank you!

Also, R3 is 3M, or 3,000,000 ohms.

enter image description here

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    \$\begingroup\$ Hello and welcome to the forum. We will gladly help you, but only if you show how you try to do it yourself and where you fail. "Please solve it for me" is not ok (even if you want it with explanation). What exactly it is that you don't understand? Edit your post, add things that you figured out and point out the things that you can't. We will guide you from there. \$\endgroup\$
    – Ilya
    Feb 16, 2022 at 14:41
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    \$\begingroup\$ Sorry, I will fix this mistake and edit my post. \$\endgroup\$
    – Aditya S
    Feb 16, 2022 at 14:43
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    \$\begingroup\$ Edit: Ok, I gave some additional details! \$\endgroup\$
    – Aditya S
    Feb 16, 2022 at 14:50
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    \$\begingroup\$ If that is your simulated circuit then it looks like there are a few bad connections to me. \$\endgroup\$
    – Andy aka
    Feb 16, 2022 at 15:10
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    \$\begingroup\$ What about V3=1000 V ? \$\endgroup\$
    – Antonio51
    Feb 16, 2022 at 15:19

2 Answers 2

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So, you figured out that both op amp inputs are 11V. Good.

We have 1V across 110k R3. Which means 1V/110k = 9.(09)uA flows through it. This current has to come from somewhere. The only place it can come from is via 3M resistor. So we have 9.09uA via 3M resistor. It gives 27.(27V) across 3M. So point A has to be 11V+27.27V = 38.27V.

It seems to me you calculated it correctly and most likely with the same method. Maybe your software has problems with microcurrents or these large numbers? Or maybe I overlooked something as well? Have you simulated this circuit in some online simulator like falstad or circuitlab?

Here is falstad simulation results. At first I didn't change Op Amp's power rail properties and got gargabe. When I set them to 0 and 1000V, I got 38.371V. Given the rounding of periodic current number I'd call it a win:

enter image description here

And yes, I dropped + input resistor because it does nothing for ideal op amp.

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  • \$\begingroup\$ Yup, I have simulated it in circuit lab, and it seems to give me 11 V there as well for point A. \$\endgroup\$
    – Aditya S
    Feb 16, 2022 at 15:01
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    \$\begingroup\$ @AdityaS edit your question to show us your circuitlab schematic. If it's telling you that there's 11V at point A then there's a problem. \$\endgroup\$
    – brhans
    Feb 16, 2022 at 15:27
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    \$\begingroup\$ @Ilya Confirmed for 38.247 V with a "TL071" ideal opamp :-) \$\endgroup\$
    – Antonio51
    Feb 16, 2022 at 15:30
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    \$\begingroup\$ @AdityaS ... if Opamp has "level 3" parameters ... Not "level 1" = ideal. \$\endgroup\$
    – Antonio51
    Feb 16, 2022 at 15:32
  • \$\begingroup\$ Yes, you guys are right. I used a different software and it works now. Thank you for the help! I really appreciate it. \$\endgroup\$
    – Aditya S
    Feb 16, 2022 at 15:44
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Well, assuming an ideal opamp we know that \$\text{V}_+=\text{V}_-\$. In your case we get that \$\text{V}_+=\text{V}_-=\text{V}_1=11\space\text{V}\$. So we can find the current trough \$\text{R}_2\$ by using Ohm's law \$\text{I}_{\text{R}_2}=\text{I}_{\text{R}_3}=\frac{\text{V}_2-\text{V}_-}{\text{R}_2}=\frac{10-11}{110\cdot1000}=-\frac{1}{110000}\space\text{A}\$.

Using Ohm's law we can rewrite:

$$\text{I}_{\text{R}_3}=\frac{\text{V}_--\text{V}_\text{A}}{\text{R}_3}\tag1$$

So, you need to solve:

$$-\frac{1}{110000}=\frac{11-\text{V}_\text{A}}{3000000}\tag2$$

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  • \$\begingroup\$ When I plug this into circuitlab, I don't seem to be getting the Va from your equation. I get 11 V, which doesn't really make sense to me. Do you have any ideas? \$\endgroup\$
    – Aditya S
    Feb 16, 2022 at 15:03
  • \$\begingroup\$ You are wrong. I simulated it using LTspice and got my answer. So I think that you're something wrong. \$\endgroup\$
    – Jan Eerland
    Feb 16, 2022 at 15:06
  • \$\begingroup\$ But yeah, thank you for the information. Let me go try on falstad really quick \$\endgroup\$
    – Aditya S
    Feb 16, 2022 at 15:10
  • \$\begingroup\$ Hello everyone, just letting you know that this is a correct answer! The reason I was skeptical at first was that my simulation software was giving the wrong output. I guess it is not a good idea to use antiquated software! Thanks again for the help, I really appreciate it. \$\endgroup\$
    – Aditya S
    Feb 16, 2022 at 15:44
  • \$\begingroup\$ The software gives "always" the right answer (except in some very very rare particular cases) ... It is up to you choosing the "right" parameters ... \$\endgroup\$
    – Antonio51
    Feb 16, 2022 at 15:56

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