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I want to compare the calculated bode plot of an inverting op-amp with the simulated one, on the 1 mHz-200 kHz spectrum.

As expected, theoretically there is no stray capacitance on the circuit above, so the calculated transfer function is -1, unity gain with phase inversion, no matter the frequency, whereas in the simulation, a more acurated aproach to reality (TL072 for example), it's showing a low pass, -0.028 dB @ 200 kHz.

In order to take the calculated plot as a reference, I can't write in my paper that a great deal of similarity between the two plots was observed, because "a great deal" is a personal judgment, what I consider to be very simular may not be similar enough for the so-called doctors of engineering.

I must stablish a standard of what is too much and what is not, or calculate the percentage discrepancy. As I want to take the calculated plot as the reference, what's the percentage difference between 0 dB and any other gain, -0.028 dB in my case?

Please do not tell me to take the simulated plot as the reference, I wanna know how to take 0 dB as a reference in a percentual comparison.

Circuit and Bode Plot

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  • \$\begingroup\$ Cut to the chase: how lesser, percentagewise, is -0.028 dB than 0 dB? \$\endgroup\$
    – Gabriel Santos
    Commented Jul 29, 2022 at 21:15
  • \$\begingroup\$ as voltage gain -0.028dB means 10^(-0.028/20) = 0.9968 That's 0.32% less than 0dB where 0dB means the same as voltage gain =1. Not a slightest idea how significant drop that is where you live. Any electronic component which have its specified quantities within plusminus 0.38% of the specified values should be considered to be premium, substantially more accurate than the commonly used ones. \$\endgroup\$
    – user136077
    Commented Jul 29, 2022 at 21:30
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    \$\begingroup\$ You are basically asking how decibels work? What research have you done already (it's an academic paper, so it should be your own research?) \$\endgroup\$
    – Justme
    Commented Jul 29, 2022 at 21:47
  • \$\begingroup\$ I just changed "unitary" for "unity" I know both are english words, but as Portuguese is my native language, and the Portuguese for "a gain of 1" is "ganho unitário", it sounded better, for my ears, "unitary gain". So, considering "unitary" is more like a political term, as in "a unitary state", can the word "unitary" be used in an electronic context meaning "unity"? \$\endgroup\$
    – Gabriel Santos
    Commented Aug 1, 2022 at 18:11

2 Answers 2

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It is meaningless to compare dBs in percentage terms. Decibels are already relative measurements, so you should just refer to the dB difference (0.028 dB in your case, or about 0.3 % in the amplitude).

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An opamp circuit dropoff is usually mostly due to the limited frequency response of the opamp. Research gain-bandwidth product (GBP). On a datasheet that I found, this opamp has a minimum GBP of 2.5 MHz, which means that for a circuit with a gain of one, the cutoff (-3 dB) will be 2.5 MHz.

At 1 decade from the cutoff (200k vs 2.5M), it is almost flat, but there will be a slight reduction. I calculate that at 200 kHz, the gain will be -0.0277 dB, which rounds off to -0.028 dB, exactly what you are seeing. I calculated with Excel, there are many other ways, including some online tools.

To confirm, you may want to extend your plot out to 10 MHz and see what cutoff the simulation predicts.

So, I recommend that you include GBP in your theoretical calculations, then everything will match.

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