I am simulating the usual active, first-order high-pass filter on LTSpice, as below:


The Op-Amp I'm using is the following:


This is the simulation result:


What is the explanation for the behavior above 100 kHz?

The gain starts to fall, and the phase also falls...

  • \$\begingroup\$ What op-amp are you using? Only high performance and RF op-amps can handle 100KHZ+ with ease. \$\endgroup\$
    – user105652
    May 14, 2016 at 2:47
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    \$\begingroup\$ That is not distortion, merely high-frequency rolloff. \$\endgroup\$
    – uint128_t
    May 14, 2016 at 6:01
  • 1
    \$\begingroup\$ Yes, there is a limit to the maximum operating frequency of an amplifier, the "single pole" part of the opamp puts a limit of the maximum frequency the amp can handle (much like a real amp) \$\endgroup\$
    – Sam
    May 14, 2016 at 6:13
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    \$\begingroup\$ And - please, realize that "distortion" is not the right term. Distortion is caused by a non-linearity of the transfer characteristic (time domain). \$\endgroup\$
    – LvW
    May 14, 2016 at 8:21
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    \$\begingroup\$ Of course, it is an active filter. There is no buffer at all. Instead, the opamp is wired as a "differentiator with loss" - which is identical to a first order (inverting) highpass. \$\endgroup\$
    – LvW
    May 14, 2016 at 14:34

1 Answer 1


You have an ideal "single pole" operational amplifier.

A single pole amplifier is an amplifier with a first order low-pass characteristic. The pole determines the corner frequency of the pole. For a real opamp it can be found using the gain-bandwidth product, which is the product of open loop gain and the corner-frequency.

Above 100kHz you see the typical characteristic of a low-pass. A gain roll-off of 20dB/decade and a phase shift of -90 degrees.

Since you are using an ideal opamp it should be possible to set the corner (pole) frequency to a higher value, so that it doesn't affect your transfer function.

  • \$\begingroup\$ Great answer. Indeed I was able to configure the op-amp and set that corner frequency to a higher value. Excellent. I learned a lot. \$\endgroup\$
    – Pedro A
    May 14, 2016 at 14:11

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