[SOLVED] 741 Op Amp Non-inverting Butterworth: Phase shift questions

[event.horizon]

Hi,

I'm having a hard time understanding what causes the more-than-neg90deg phase shift of a first pole, low pass butterworth filter.

The formula is: -arctan(f/fh) where fh is the high frequency cutoff.

Theoretically, it shouldn't exceed -90deg even at extremely high frequencies but that's not the case. When I built one in the lab and simulate on in Multisim too, my phase shift exceeded -90deg and came close to -180deg. I also find out the the more the closed-loop gain of the op amp, the more the phase shift is off. Why?

Could anyone explain this. I know an op amps has capacitance but it would really be great if anyone can give me more detail about this.

Thanks :-D

 

[LvW]

Hi,

at first, a Butterworth response is at least of 2nd order.
That means that only one typical response for a 1st order lowpass is possible.
You are right in assuming that such a 1st order function shifts the phase by 90 deg at maximum.
However, if an opamp is involved the phase is shifted beyond -90 deg because of the real and frequency dependent amplifier properties.*** This is the reason that an active filter should be designed with a cut-off frequency that is approximately 50...100 times lower than the unity-gain frequency of the opamp. Otherwise, the deviations from the wanted response are not negligable.
*** A good opamp macro model incorporates up to 3 poles (phase shift -270 deg max)