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I'm trying to find the transference function H(s) of the next op-amp circuit:

enter image description here

I get the next following equations:

When I put the equations together, this is an absurd. How do I solve this?

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  • \$\begingroup\$ What is absurd ? Algebra ? \$\endgroup\$
    – efox29
    May 16, 2016 at 5:13
  • \$\begingroup\$ Where's the input signal? If it's at the V+ terminal then your V+ equation is wrong. \$\endgroup\$
    – Chu
    May 16, 2016 at 7:26
  • \$\begingroup\$ Yes - the circuit (with the shown input node) makes absolutely no sense. \$\endgroup\$
    – LvW
    May 16, 2016 at 8:12

1 Answer 1

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I know it is a late answer but I'll give it a shot in case someone else needs it.

The issue in this configuration, if you analyze the circuit, is that you are essentially forgoing the positive feedback path. The reason is that by fixing your input signal to the \$V^+\$ input, the positive feedback path (the one with \$C\$ and \$R_2\$) does not have a say in what you are inputting into \$V^+\$.

Feedback networks, as their name indicates, feed a fraction of the output to the input, but in this circuit configuration, there is a constant source established. To better illustrate what I am saying, here is what's happening at \$V^+\$:

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see, the voltage going into \$V^+\$ is \$V_{in}\$, which doesn't depend on the feedback network. But even without thinking too much about what I just explained, you can start your analysis with:

$$ V_{out}=A(V^+-V^-)$$ When there are positive and negative feedback networks all in the same OPAmp circuit, it's not always safe to assume that \$V^+=V^-\$ since that only applies when the net feedback is negative. Anyways, here is the analysis: $$ V^+=V_{in}$$ $$ V^-=\frac{R_3}{R_3+R_4}V_{out}$$

Now plugging this into the first equation:

$$ V_{out}=A\bigg(V_{in}-\frac{R_3}{R_3+R_4}V_{out}\bigg)$$

After some algebra you can obtain your transfer function:

$$ \frac{V_{out}}{V_{in}}=\frac{R_3+R_4}{R_3+\frac{R_3+R_4}{A}}$$

Recall that \$A\$ is really large so that your transfer function now becomes:

$$ \frac{V_{out}}{V_{in}}\approx\frac{R_3+R_4}{R_3}$$

As you can see, it is independent of the positive feedback path (no \$R_2,C,or R_1\$).

To further demonstrate, here is a simulation I built in LTSpice

enter image description here

As you can see I have \$R_3=1k\Omega\$ and \$R_4=10k\Omega\$ so the voltage gain, \$\frac{V_{out}}{V_{in}}\$, should be around 11. Here is the output voltage now for sine input of with 1V of amplitude:

enter image description here

The green plot is the output and the blue one is the input.

Now, I have \$R_3=1k\Omega\$ and \$R_4=20k\Omega\$ so the voltage gain, \$\frac{V_{out}}{V_{in}}\$, should be around 21. For the same input (1V sine wave), here is the output:

enter image description here

You can play around with different values for the positive feedback network (\$R_2\$,\$C\$,or \$R_1\$) and you will find that this, ideally, has no effect on the gain of the circuit. I hope this helps!

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