0
\$\begingroup\$

Let's say I have a 2nd order low-pass filter in a Sallen-Key configuration (Chebyshev), and the cutoff frequency should be the frequency at which the phase shift becomes -90°, how will the GBW (gain bandwidth product) of the op-amp I choose affect the position of the cut-off frequency?

Will choosing an op-amp with a higher GBW shift the cut-off frequency higher?

enter image description here

This was my problem during my laboratory session. I built a 3rd order low pass (Chebyshev, Sallen-key) filter using the filter design tool from TI. During the lab I changed the op-amps with two TL081s, and I had to connect the input signal to the 2nd stage and measure its cutoff frequency. I got a much higher cutoff frequency from the design tool report by about 10x.

Answer : if anyone wants a detailed answer you can find it in this book there is a section called Dynamic Op Amp Limitations-EFFECT OF FINITE GBP ON FILTERS.

Sergio Franco - Design With Operational Amplifiers And Analog Integrated Circuits-McGraw-Hill Series in Electrical and Computer Engineering

\$\endgroup\$
2
  • \$\begingroup\$ Link in question is broken? \$\endgroup\$
    – Antonio51
    Commented Nov 5, 2022 at 11:56
  • \$\begingroup\$ @Antonio51 should work now, i dont know where to upload the files \$\endgroup\$
    – BonPi
    Commented Nov 5, 2022 at 14:18

3 Answers 3

4
\$\begingroup\$

If you choose a gain bandwidth product (GBW) that is high enough, the filter will behave very close to what the equations predict. If the GBW is lowered too much, the cutoff frequency will not match what the equations predict. The cutoff frequency will be lower in frequency.

You can either write out the equations taking in to consideration a non-ideal opamp or use a simulator to observe what happens. Since I'm lazy, I'll use a simulator.

The following is a simulation using LTspice for a Butterworth low-pass filter with the gain set to 20 dB and the cutoff frequency set to 10 kHz. The Analog Devices filter calculator was used to determine values. The opamp in the simulation is the universal opamp (mathematical opamp model) which allows you to set the open loop gain and the GBW. Two GBW values are used, 100 kHz and 10 MHz. The opamp open loop gain is set to 1E6. You'll also notice that the lower GBW plot deviates from a Butterworth filter (slight peaking in the passband).

enter image description here

\$\endgroup\$
11
  • \$\begingroup\$ Thank you for your answer but could you provide with a theoretical reason why the GBW matters and how it affects? The problem was that i had used a op-amp with GBW of 1MHz in the filter design tool from TI , it predicted that my cutoff was 2.3K. Then i used a op-amp with a GBW of 14MHz during my laboratory session and found that my cutoff become 23K , i measured that cutoff as the frequency where i got a 90* phase shift. So as per you using an op-amp with a higher op-amp should not increase the cutoff , but using an op-amp with a lower GBW will decrease the cutoff. \$\endgroup\$
    – BonPi
    Commented Nov 5, 2022 at 9:12
  • \$\begingroup\$ I have edited the question with more details \$\endgroup\$
    – BonPi
    Commented Nov 5, 2022 at 9:26
  • 1
    \$\begingroup\$ @BonPi Given what you just said, it is more likely to be an error in wiring up the filter rather than choice of op-amp. \$\endgroup\$
    – Andy aka
    Commented Nov 5, 2022 at 10:07
  • \$\begingroup\$ @Andyaka Well i don't think it is probable because we first measured the DC gain and it was exactly the one from the filter tool, we also double checked the capacitor values. cutoff frequency seems to be only thing affected so I assumed it had something to do with the frequency response of the op-amp we were using in the lab. \$\endgroup\$
    – BonPi
    Commented Nov 5, 2022 at 10:41
  • 1
    \$\begingroup\$ @BonPi I would guess that the components you are using aren't what you think they are. Measure each part's value and see if they are correct. In your schematic, show the values you are using. The equations for various filters usually assume an ideal opamp, i.e., infinite gain and GBW, which makes the maths much easier. \$\endgroup\$
    – qrk
    Commented Nov 5, 2022 at 17:36
1
\$\begingroup\$

A simple rule of thumb is that the op-amp unity gain bandwidth should be about ten times greater than the cut-off frequency but, your circuit has overall circuit gain defined by \$1+\frac{R_4}{R_3}\$ so, you must reduce your specified unity gain BW to allow for that gain. Given that your Q factor cannot be too high for the filter you want the simple rule of thumb is adequate for most cases.

Of course, I would encourage you to use a simulator to trial different op-amps or even ideal op-amp configurations where they have adjustable GBWP.

\$\endgroup\$
1
  • \$\begingroup\$ Here is a more conservative "rule of thumb" for active lowpass filters: Opamps transit frequency ft>20*Qp*fp (Qp: pole-Q; fp: pole frequency). Moreover, in many (if not in most) applications it is the limited slew rate which determines the selection of a suitable opamp. \$\endgroup\$
    – LvW
    Commented Nov 17, 2022 at 11:44
0
\$\begingroup\$

could you provide with a theoretical reason why the GBW matters and how it affects?

EE&O ... Here is the formula containing the cut-off "frequency" ... or GBW changed.
Schematic @qrk

enter image description here

\$\endgroup\$

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.