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I've been learning more about the characteristics of op amps and I found in a non-inverting amplifier simulation that the gain of an inverting op amp falls off as frequency increases. This behavior is expected and is due to the -3db bandwidth capability of the specific op amp. However, I also noticed that when you vary the gain-setting resistors (but not their ratio), the half-power point for the circuit varies as well. As the value of the gain-setting resistors increase, the frequency at which the half-power point occurs also increases. I was wondering what parameter/characteristic of the op amp causes this and whether some op amps are more immune than others. Here is my simulation:

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In the simulation, I've set the gain to 2 for each step. I've set Rg to start at 10k and step to 100k in increments of 10k. So running the simulation results in 10 bode plots, each of which has a different half-power point due to the gain resistors. I realize the difference in gain for each half-power point is pretty negligible for most applications but I'd still like to know what is causing it. Can anyone tell me this?

Here is the datasheet of the op amp I'm using in the simulation.

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    \$\begingroup\$ The size of R3 determines how much current the output may apply to the input. The smaller R3 is, the more current, the faster the loop. \$\endgroup\$
    – Janka
    Mar 8, 2020 at 3:58
  • \$\begingroup\$ Replot using 10 kHz and 20 kHz. Plot so that the 3 dB points can be adequately seen in the graph \$\endgroup\$
    – Andy aka
    Mar 8, 2020 at 9:53
  • \$\begingroup\$ I found some relevant answer of your question about inverting amplifier or operational amplifier \$\endgroup\$
    – Khan life
    Mar 8, 2020 at 13:29
  • \$\begingroup\$ You really should follow Andy aka`s recommendation and show us the simulation results over a much larger frequency range (at least up to the 3dB points). \$\endgroup\$
    – LvW
    Apr 24, 2020 at 8:48

2 Answers 2

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The resistors are acting as the op amp's load in the absence of any other load.

When there is a larger load (smaller value resistors) the op amp's output stage transistors must be driven harder to supply the extra current and still keep the output voltage at the voltage that it should be. This means that extra current is required to flow back and forth between the op amp's differential input stage and its second stage (output stage driver). The only way to get this extra current is if feedback causes a bigger difference voltage between the inputs.

Now, the difference voltage between the inputs is caused by the output lagging behind the input signal and there must be extra output phase lag to create a bigger input difference signal which causes a larger current out of the input stage which drives the output transistors harder to supply the load and maintain output voltage accuracy.

Like any single pole (low pass filter) phase lag is unavoidably linked to output amplitude and therefore the extra output phase lag causes an inevitable reduction in output amplitude and a resulting reduction in bandwidth.

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Depending on the model you use for the opamp, the amplifier's input impedance (and parasitic capacitance) may look like some additional element in parallel with R2, thus increasing the gain.

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