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GBW is wrong?

 

I was designing with FilterPro Desktop a second order bandpass filter with a Q=30, f0=8142, multiple feedback.

When the passband gain(A0) is 1, the minimal GBW is 27Mhz, when i set the passbabd gain to 10, the GBW is 270Mhz.

When the gain is 1, the attenuator (R1 & R3 resistor) is R3/(R1+R3)=~0.000475, when the passband-gain is 10, the attenuator is 0.00475.

But the GBW is related to the gain of the amplifier, so why the GBW are not equal? The op-amp gain of both configuration is 2*Q^2 at f0, independent of the value of the attenuator...

Thank you

  • FilterPro calcuates the required GBW as:

    GBW = 100*G*fn*Q

    for each second order section, where G is the section gain. The factor of 100 is used to assure 1% gain accuracy.

    So, when the section gain changes by a factor of ten, so too does the GBW recommendation. This may indeed be a conservative value.

  • The pass band filter without attenuator (http://mysite.du.edu/~etuttle/electron/circ141.gif) has a gain of 2*Q^2 at f0.

    When there's the attenuator, the section gain become R3/(R1+R3) * 2*Q^2, but the op-amp gain is still 2*Q^2.

    When i set the passband gain as 1 at f0, FilterPro Desktop give me the values of R1 & R3 that:

    Souce signal-> Attenuator with gain equal to 1/(2*Q^2) -> op-amp gain (2*Q^2 at f0)

    When i set the passband gain as 10

    Source signal -> Attenuator with gain equal to 10/(2*Q^2) -> op amp gain (2*Q^2 at f0).

    So, the op-amp, in both configuration is still amplifying 2*Q^2 at f0, regardless of the attenuator value and regardless of the gain of the entire section. So, the GBW should be the same for both configuration.

  • I don't disagree with your analysis for the multiple-feedback case. FilterPro just doesn't take that into account, opting to be ultra-conservative. We'll look at the possibility of changing that in future releases. Thanks for your input.