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I am struggling with the following self-study problem, taken from The Analysis and Design of Linear Circuits (8th edition). enter image description here

There are two things that shock me from this problem.

  • We need to get a higher non inverting gain for the operational amplifier than the inverting gain. For me, this is hard to grasp because if the inverting gain is \$K = -\frac{R_2}{R_1}\$, then the non inverting gain will be \$K = a*(1+\frac{R_2}{R_1})\$, which is always lower than the inverting gain because a is a coefficient that can be added through voltage dividers (and thus \$a<1\$).
  • How can you only have one operational amplifier, yet still get an amplification of -3 after the operational amplifier and before \$v_0\$?

Any guidance is tremendously appreciated. The book does not explain this at all and I am just a very confused student.

Edit:

The transfer function I computed is: $$V_A=5V_1 -V_2+V_0; V_0=-3V_A$$ $$V_0 = -15V_1 + 3V_2 - 3V_0 \Rightarrow V_0 = -3.75V_1 + 0.75V_2$$

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  • \$\begingroup\$ Hint: you can eliminate the -3 gain block at output by adjusting the gains of input blocks. \$\endgroup\$ – nidhin Oct 18 '18 at 15:11
  • \$\begingroup\$ Really? So don't I have to have another physical element at the OpAmp output giving me that gain? Because if I do what you suggest, I wouldn't be representing the exact same block diagram, right? Then I could just use a differential amplifier that verifies the transfer function I wrote in my edit. \$\endgroup\$ – Bee Oct 18 '18 at 15:15
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    \$\begingroup\$ You will be implementing a reduced version of the block diagram. And I think it’s okay as far as the transfer function remain the same and we are not interested in any of the signals other than v1, v2 and v0 \$\endgroup\$ – nidhin Oct 18 '18 at 15:28
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    \$\begingroup\$ Yes, the problem states "...Design a circuit that -realizes- the block diagram...". When it comes to control theory diagrams there are many different ways to say/implement the same thing, and I don't see any particular "use X technique" constraints. \$\endgroup\$ – isdi Oct 18 '18 at 15:37
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It's an exercise in analysis, or algebraic trickery if you will ;-)

Please excuse the handwritten solution. It took several coffees! Your transfer function is correct. The trick is to see that it can correspond to the gain of a differential amp, where the resistors have 4 independent values. (To simplify the working a bit, I use "Vn" for the + and - inputs of the op amp, which are at the same potential of course.)

At the end I do a sanity check by putting some notional voltages of 1V and 2V and seeing that the result makes sense. I think it is OK.

enter image description here enter image description here

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In complex problems, start by generating ANY solution, then understand where you can improve upon your solutions. The human mind strongly benefits from memory aids, as the brain struggles to manage the complexity. Visual documentation allows the eyes to provide a massive increase in recall bandwidth, once the brain has memorized the visual shapes on the sheet of paper.

Try this:

schematic

simulate this circuit – Schematic created using CircuitLab

What is WRONG with this? The full requested transfer function is not achieved, and this (partial) solution uses 4 opamps.

SO we may conclude the INVERTING TOPOLOGY is the WRONG approach.

Since we need some inversion, simply using the non-inverting topology is also the WRONG approach.

How about the common-mode-stripper topology? that has +in and -in

schematic

simulate this circuit

Many years ago, my 2nd engineering manager (an early retiree from NASA) said "Do something. Don't just stand there, confused and not trying anything."

We now have a partial solution using FOUR opamps.

And we may have a useful circuit providing + input and - input.

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