Hi TI support team,
I have a question on page 16 of the OPA134 datasheet. In the typical application, the low pass cutoff frequency is 30 kHz, but my calculation based on Formula (3) is about 21 kHz. I have simulated this circuit on TINA and the filter's - 3dB frequency is about 30kHz. I am confused about these two results.
Hi Amy,
The fc equation (Eq. 3) that is provided in the OPA134 datasheet is accurate when the 2nd-order active filter section shown has a maximally flat Butterworth response. That filter response is very popular and exhibits a "Q" of 0.71. The filter bandwidth in that particular case is defined by the frequency where the gain is down -3 dB from the flat, passband gain.
However, when the circuit shown in Figure 30 is simulated as you did using TINA Spice, there is gain peaking in the frequency response, with the peak occurring at about 17.5 kHz. That peaking indicates that the Q is higher than 0.71, and the -3dB bandwidth not going to be what is obtained using Eq. 3. It does appear that the datasheet information is for a Butterworth example and some clarification would be helpful in distinguishing how the Figure 30 response differs.
The SLOA049B Application Report that Kai referenced, in Section 7 - Low-Pass Multiple-Feedback (MFB) Architecture, includes the Frequency Scaling Factor (FSF). That factor can be used to account for common filter responses.
If you need to synthesize a particular active filter, having a certain type of response TI offers some filter tools that makes the job easy. There is the older FilterPro program, and a newer beta version Filter Design Tool. You can find them here:
https://webench.ti.com/filter-design-tool/design/93
Here is an example of a 30 kHz, G = 3.53 V/V, Q = 0.71 Butterworth low-pass filter synthesized using FilterPro
If you apply Eq. 3 from the OPA134 datasheet using the component values shown in the FilterPro schematic, you will find that the Fc works out correctly being equal to 30 kHz.
Regards, Thomas
Precision Amplifiers Applications Engineering
