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I have a quick question on this example problem on (ideal) OpAmp analysis. I understand the basic cases of negative and positive feedback, and how there is a virtual short between the input pins of the opamp. Things just seem to get a bit confusing once we implement more than one opamp and start connecting inputs and outputs and, welp, I guess I need some help understanding this one:

enter image description here

My steps towards a solution: (Let's call the OpAmp on the left the first OpAmp and the one on the right the second OpAmp)

  1. There is virtual short between ground and the inverting input on the first opamp. There must be a voltage drop across the first resistor with value R equal to V_in...
  2. This means a current flows equal to V_in / R through the first resistor. This current cannot go through the OpAmp so it goes through the two 4R resistors.

This is where things get tricky. How do you find the voltage drops across the two 4R resistors? You can't just assume the current is split evenly between the two resistors.

What would the next step be? Any help would be greatly appreciated!

Thanks,

Matthew

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  • \$\begingroup\$ There is not a 'virtual short' between the inverting and non-inverting inputs. Quite the opposite in fact - there's a very high resistance, consequently a very small current flows into these inputs and the voltage across them is almost zero. \$\endgroup\$ – Chu Feb 26 '18 at 0:57
  • \$\begingroup\$ @Chu 'Virtual short' means they must be at the same voltage but no current flows between them. \$\endgroup\$ – τεκ Feb 26 '18 at 1:29
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I added a couple of notes to your schematic. It's drawn similarly, though I didn't bother with calling out one side of \$V_{O_1}\$ and \$V_{O_2}\$, since the reference is just ground for both of them, anyway.

schematic

simulate this circuit – Schematic created using CircuitLab

You are right about the virtual ground for the first opamp. And yes, all the current through \$R_1\$ in the schematic must leave via \$R_4\$ and \$R_5\$. So:

$$\frac{V_\text{IN}}{\text{R}}+\frac{V_{O_1}}{4\text{R}}+\frac{V_{O_2}}{4\text{R}}=0\:\text{A}$$

But did you notice that the second opamp also has a virtual node that easy to identify? Clearly, the (-) input of the 2nd opamp will have the same voltage as its (+) input. But this is just the output of the 1st opamp, too.

So, now you have a way of seeing how \$V_{O_2}\$ must relate to \$V_{O_1}\$. And this means that you can replace \$V_{O_2}\$ in the above equation with some new function of \$V_{O_1}\$ and then just solve for \$V_{O_1}\$ (which because you know the relationship already, also solves for \$V_{O_2}\$.)

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For this circuit, look at the second opamp first.

  • What is the relationship between \$V_{O1}\$ and \$V_{O2}\$?

  • What does this imply about the Thevenin equivalent of the opamp and the two 4R resistors, considered as a source driving the input of the first opamp, as a function of \$V_{O1}\$?

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