The bandwidth of inverting opamp is lesser than non-inverting opamp. If both configurations have equal closed loop gain and also Gain-Bandwidth product (GBW) for opamp is constant for both cases (which it will be since GBW does not depend on external configuration) then :

For non-inverting opamp : GBW = closed loop gain * cutoff frequency

For inverting opamp : GBW = |closed loop gain+1| * cutoff frequency

Here cutoff frequency represents the 3-dB bandwidth of the circuit. So for constant GBW, inverting opamp has lesser cutoff frequency or 3-dB bandwidth.

What is the intuitive explanation for this ? And from where is this +1 coming in inverting configuration ?

Thanks !

  • 4
    \$\begingroup\$ "The bandwidth of inverting opamp is lesser than non-inverting opamp" no, a) there's no such thing as an inverting or non-inverting opamp. I think you mean the circuits that contain an opamp, and b) I don't even think that statement is true! So, please show your schematic for these two, and explain why one would have more bandwidth than the other! \$\endgroup\$ Jun 11, 2023 at 12:56

1 Answer 1


It's not the op amp, but the configuration, because a low open-loop gain has a different effect in the two configurations.

Consider these two cases, each with a nominal gain of 10:



simulate this circuit – Schematic created using CircuitLab



simulate this circuit

Now let's take out the bandwidth part of the equation and assume that your op amp has a gain-bandwidth product of 10 Mhz. Since your input is 100 kHz, the open-loop gain of your op-amp is only 100, far from an ideal op amp.

Now do the math with this open-loop gain. You can see that in the inverting configuration, the closed loop gain is reduced to 9.009, while the non-inverting configuration gain is higher at 9.09. The gain-bandwidth product of the op amp itself is not affected, but the open-loop gain requirements of the different closed-loop configurations causes the apparent difference. Obviously the 3-db point occurs at different frequencies for the same gain.

This is theory. In actual practice, any op amp operating close to its "cutoff frequency" will have other parameters affecting its output, such as slew rate and phase shift. We are assuming an ideal op amp in every other respect except for the open-loop gain. Also, a designer will not choose an op amp with a frequency response too low for the application, so this effect is usually negligible.

  • \$\begingroup\$ The concept of noise gain is useful here. \$\endgroup\$
    – Dan Mills
    Jun 12, 2023 at 6:47
  • \$\begingroup\$ @John Birckhead if the actual gain of non-inverting configuration is higher then does that mean the actual cutoff frequency or bandwidth for this configuration will be lower ? \$\endgroup\$ Jun 12, 2023 at 17:28
  • \$\begingroup\$ In the two examples of ideal op amps except for the GBW, the open-loop gain is the same at 100 kHz. So since the bandwidth is the same and the closed loop gain is higher for non-inverting, the GBW product is also higher. Each of the designs will experience a frequency roll-off at the same rate, but the inverting gain is lower to begin with, so it will reach the 3-dB point sooner (at a slightly lower frequency.) \$\endgroup\$ Jun 13, 2023 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.