What is the bode plot of an inverting op amp if you replace the resistors with caps? What if you now add a capacitive load to the output?


I ran a quick simulation and the bode plot looks as follows. It looks as though there are two poles but I am still confused about how to analytically derive the transfer function and analytically calculate the poles. Would it just be -C1/Cf where the frequency terms (jw) cancel out? enter image description here

Update 2:

It seems some clarification is needed on this question. A non-inverting op amp is shown below. In this common configuration, I want to analyze the circuit if you were to replace the resistors with capacitors. I ran a quick simulation as seen in my first update. I now understand that the gain of the circuit is -C1/Cf. However I am confused as to how to hand calculate poles for the bode plot of this circuit. Any help is much appreciated.

enter image description here

  • \$\begingroup\$ Comments / bunfight have been moved to chat here. Not enough relevant technical content. \$\endgroup\$
    – Russell McMahon
    Sep 22 '20 at 11:04
  • \$\begingroup\$ Please carefully read @frr 's answer which explains why this circuit cannot actually work. You're asking about the behavior in response to a load, which would only be an issue for a real op-amp. \$\endgroup\$ Sep 22 '20 at 14:45

I understood the question as being a standard text book inverting opamp, with R1 and R2 replaced by capacitors C1 and C2.

The resistors become complex (imaginary) impedances Z1=1/jwC1 forward and Z2=1/jwC2 feedback. The TF is the ratio -C2/C1, as already mentioned by the OP.

(I do suggest the OP update the diagram to show the intended circuit.)

I assume this is homework, and thus the OpAmp is ideal. The TF is then a constant and independent of f.

There are many tutorials, in writing and on youtube, to derive the TF of such an arrangement. It's not complicated. One example is https://masteringelectronicsdesign.com/how-to-derive-the-inverting-amplifier-transfer-function/

Here is an example to plot the TF, from http://sim.okawa-denshi.jp/en/opampkeisan.htm

Note the capacitance is in the uF range, and the resistors are near zero (calculator will not accept R=0)

enter image description here

and the plots are flat well into the GHz range but not beyond that, due to the non-zero R.

enter image description here

In reality the TF will not be constant for f, as the capacitors and the OpAmps internal resistances will interact to form low- or high-pass filters.

Also, in practice, the capacitors and inductance of the op-amp leads will interact to form resonant circuits.

So this flatness only persists up to a certain frequency, and then it becomes a battle of the parasitics.

  • \$\begingroup\$ Thank you! This helps clarify a few things. I was unable to find a youtube tutorial that deals with the capacitors instead of the resistors. Is there something you may be able to share? Currently I am looking at the internals looking something like what is described here in Figure 1 (using 2 poles) ecircuitcenter.com/OpModels/2nd_and_3rd_Poles/… . I am just super unclear on how to calculate the poles. \$\endgroup\$
    – user263489
    Sep 22 '20 at 5:41
  • 2
    \$\begingroup\$ I found no example exactly like your question with 2 Cs, and I think there is none because it's practically non-sensical. But look at the ones that have a general Z in the fwd and fbk. Then just fill in your Z=1/jwC. I'll add a link to the answer. \$\endgroup\$
    – P2000
    Sep 22 '20 at 5:44
  • \$\begingroup\$ This answer is a little bit dangerous, as it only vaguely hints that there are practical issues but fails to explicitly make it clear enough that the analysis is meaningless when the circuit lacks a DC path. It's not enough to consider how the circuit responds to change, it's also necessary to consider how it gets its initial conditions. \$\endgroup\$ Sep 22 '20 at 14:43

In reality: as you walk the frequency axis from DC towards higher frequencies, AC gain will remain at -1 until some frequency, where the op-amp's output drive capability will get overloaded by the capacitive load. That's neglecting the impedance of your signal source (should be as low as possible for the experiment to be valid).

Working into capacitive load is generally a troublesome scenario for amplifiers. The capacitance amounts to a short circuit at high frequencies.

Another practical aspect: without DC-biasing the (-) input, the amp will quickly DC-drift away to one of the rails. Which one, that depends on the inherent disbalance of the differential input (a real-world property that has to do with manufacturing tolerances / randomness in the process). If you want to mitigate this particular effect, add a high-ohm resistance parallel between output and -in . Say 1-10 Megaohms. It will affect the transfer function of the circuit very little for any meaningful AC, but will nail the gain to 0 for DC.

  • \$\begingroup\$ Indeed, any conclusion which ignores the lack of a DC path is meaningless. \$\endgroup\$ Sep 22 '20 at 14:41

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