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This is a typical example of negative feedback in an Op-Amp - enter image description here

For this case, the feedback of the output can be controlled using a resistor divider, and let's call the feedback fraction to be X. We can thus find the closed gain to be -

$$G_{CL} = \frac{G_{OL}}{1 + XG_{OL}}$$ using the open loop gain for the op-amp. As the open loop gain is very high, the equation reduces to 1/X. So, we can get a sizeable gain in this case using a proper resistor divider, but this retains the phase of the input waveform in V_in.

Now here is where I'm facing a problem. If I need to reverse the phase by 180 degrees for the output, maintaining the same magnitude of gain, how should I change the connections in the existing circuit without adding any other component? Using the equation -

$$V_{out} = G_{OL}(V_{non-inv}-V_{inv})$$, I see that I need to ground the non-inverting input and V_inverting should have a voltage of (XV_out + V_in), so that the gain remains same with a change in sign. But I have not been able to implement this, and this configuration distorts my input, though my output voltage is the same as it was with the phase retention gain -

enter image description here

Can someone help me out in this? It would be of great help.

NOTE: It is possible to shift the phase by 180 degrees using a three pole RC filter. I wish to know if there is an alternative to this without adding any new component.

Edit: Here is the actual V_in, and this doesn't get distorted for the Phase retention gain - enter image description here

Here is the output that I get in both cases, the phase reversed or phase retained gains -

enter image description here

Here is the distorted V_in for the phase reversed gain in my case - enter image description here

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    \$\begingroup\$ You need to search for "noninverting amplifier". There is documentation everywhere on this. \$\endgroup\$ Oct 6 '21 at 15:29
  • \$\begingroup\$ I have been searching everywhere and reading up articles. I came to know that I can actually use a cascaded RC filter to shift the phase by 180 degrees, but isn't it possible without using the filter? \$\endgroup\$ Oct 6 '21 at 15:32
  • \$\begingroup\$ What do you expect the output to look like? Show us a sketch of your input sinusoid and your desired output sinusoid. \$\endgroup\$ Oct 6 '21 at 15:38
  • \$\begingroup\$ @ElliotAlderson I've added the simulations. \$\endgroup\$ Oct 6 '21 at 15:51
  • \$\begingroup\$ Why are people still using 741? I used them in 1971, 50 years ago. There are so many better op-amps out there today. \$\endgroup\$ Oct 6 '21 at 17:26
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Let's first remember the inverting amplifier:

op-amp inverting amplifier

(Image source: Electronics Post - Application of Op-Amp as Inverting Amplifier)

What you are trying to do is to build this inverting amplifier with zero-Ohm resistors: Rf=Ri=0R.

Think about what to expect at the output considering the gain formula, Av=-Rf/Ri.

What you need to do is to use equal and non-zero resistances for both resistors, taking the bias currents and source's drive capability into account.

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  • \$\begingroup\$ Ok, I'm trying this out. \$\endgroup\$ Oct 6 '21 at 15:55
  • \$\begingroup\$ Thanks a lot! That worked! I would have to look into the inverting amplifier some more. \$\endgroup\$ Oct 6 '21 at 15:59
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You can't without additional resistors because you short-circuit input with output.

Inverting buffer with op-amps

If you need to switch between inverting and non inverting try this:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I would actually connect a resistor divider in between, I didn't show this in the picture. But then, how do the resistors play a role in inverting the output? \$\endgroup\$ Oct 6 '21 at 15:53
  • \$\begingroup\$ They act as a voltage divider and separate in and out. Op Amp wants to equalise voltage between +in and -in. \$\endgroup\$
    – Jomonger
    Oct 6 '21 at 15:58
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    \$\begingroup\$ Only add a switch between the + input and ground ... And you have the two schematic. :-) And if it is an electronic switch, you have a digital "modulator". \$\endgroup\$
    – Antonio51
    Dec 21 '21 at 20:21

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