What you have initially described is a 2nd order low pass filter then you've made a bit of a complication of things because the bandwidth is \$\omega_n\$ i.e. \$\omega_n\$ **is** the 3dB point when \$\zeta\$ is not causing the filter to peak i.e. has a value of \$\frac{1}{\sqrt2}\$: -

OK you may be trying to derive a general expression but that doesn't help much visualize the problem so for now I'm assuming \$\zeta\$ = \$\frac{1}{\sqrt2}\$.

Anyway, moving on and ignoring your expression for \$\omega_0\$, the bandwidth of the low-pass filter is from dc to \$\omega_n\$. Then you've modified the 2nd order low-pass filter expression with a 1st order high pass filter (\$s +z\$).

When s is very low, the low pass filter is unaffected other than having a "gain factor" of z. If z is unity then the low pass filter response is unaffected until jw approaches z in magnitude - this then marks a lower 3dB point and the net response climbs with increasing frequency until the original \$\omega_n\$ is reached then, because the low pass filter is a second order, the response starts to fall again.

What you have proposed is a band pass filter with finite gain at dc.

I'm not sure about the bandwidth

If the high pass filter (\$s +z\$) comes into play at significantly lower frequencies than \$\omega_n\$ then the net bandwidth reduces from \$\omega_n\$ to \$\omega_n - z\$.

That's how I see it anyway.

I would expect the gain-bandwidth product to be a constant, but it's worth comparing the two on the bench before designing it in (I suggest measuring both the gain-of-10 -3dB point and the gain-of-0.1 -3dB point to have a point of comparison with the datasheet limits).

Note that the slew rate is typically a fairly good 5V/us, but you'll run into that at high amplitudes.

## Best Answer

Here is the open-loop bandwidth of a certain op-amp shown in red: -

The blue line is when certain closed-loop components are applied to the op-amp.

Bandwidth is normally measured at the 3dB point of the frequency response and in the case of an op-amp (open-loop) this will be at 24Hz in the diagram.

If closed loop components were present, the gain would be reduced to (say) 20dB (blue line) but the bandwidth would increase to 1MHz.

The above example is for simple resistors "closing" the loop with negative feedback and the resulting bandwidth (3dB point) is always greater.

However, if an op-amp filter circuit was required that cut-off frequencies above 10Hz, the filter would have a bandwidth of 10Hz. In this example the closed-loop bandwidth is less than the open-loop bandwidth.

Does this help?