Differential LPF Cutout Frequency Simulation - Page 1
Forum section: electronics / projects

author: electronx
author_rank: Regular Contributor
timestamp: February 23, 2024, 09:44:45 pm
content: While designing a thermocouple circuit, I saw that a 1nF capacitor was placed at the differential signal input as shown in the figure. This was added for both differential signals, so I can see from the simulation that it is a common mode LPF. We can say that it has a cut-off frequency of 15.87775k, which means approximately 16kHz. However, when I analyzed the Differential lowpass filter, we can say that it is 756 Hz, that is, approximately 0.8khz. However, in this image, the cutoff frequency is written as 1.3khz. What point am I missing?(When I add 1nF and 1nF Caps to the simulation, the cut-off frequency becomes 756.08325 Hz. How was the 1.3KHz cut-off frequency reached?)source analog devices Analog Dialogue 44-10, October (2010)******************************************************************************************************************************************************************************************************************************

author: moffy
author_rank: Super Contributor
timestamp: February 23, 2024, 10:12:17 pm
content: I can only conclude they left out a factor of 2, yours and the calculation I did are exactly 1/2 of what they say. The article:https://www.ti.com/lit/an/slwa053b/slwa053b.pdf?ts=1708725804987&ref_url=https%253A%252F%252Fwww.google.com.au%252Fnicely describes converting between single ended and differential filters.

author: electronx
author_rank: Regular Contributor
timestamp: February 23, 2024, 10:38:23 pm
content: Thanks for this article ,its very informative.According to the documentation you posted, the single ended conversion is done like this. However, I still did not get the same cutoff frequency. I think I missed some points.

author: moffy
author_rank: Super Contributor
timestamp: February 23, 2024, 10:47:07 pm
content: Your single ended conversion is correct except the cap should be 21nF, it looks like they made a mistake, very easily done.

author: Benta
author_rank: Super Contributor
timestamp: February 23, 2024, 10:51:37 pm
content: Interesting.A quick calc gives me 2.653 kHz, which seems to fit with your second sim (the first one is unreadable. PLEASE post zoomed cutouts of the relevant screen areas.

author: Benta
author_rank: Super Contributor
timestamp: February 23, 2024, 11:30:11 pm
content: Are you asking me? Please address the poster or quote the reply.

author: electronx
author_rank: Regular Contributor
timestamp: February 24, 2024, 08:37:52 am
content: Quote from: Benta on February 23, 2024, 10:51:37 pmHow did you manage to get this frequency value? Can you share the simulation result? How can the cutoff frequency be the same in the second simulation? I am confused.

author: electronx
author_rank: Regular Contributor
timestamp: February 24, 2024, 09:07:29 am
content: Quote from: moffy on February 23, 2024, 10:47:07 pm"The definition of the cutoff frequency can vary based on different design methods, leading to different interpretations. When defined at -6 dB, it signifies that the amplitude of a input at the cutoff frequency is halved at the output, resulting in a quarter of the original power. However, if one desires the power to be halved at the cutoff frequency, the definition should be at the -3 dB point.""It is noteworthy that in the 1960s and 1970s, two prominent companies, HP and Tektronix, were known for manufacturing high-quality spectrum analyzers. HP analyzers specified filter widths at the 3 dB points, while Tektronix specified them at the 6 dB points. This distinction became a point of contention between Engineering and Marketing at Tektronix. Despite potential advantages of Tektronix analyzers in practical use, HP analyzers often appeared superior on paper, leading to ongoing debates about the better choice between the two brands."Sourcehttps://dsp.stackexchange.com/questions/84860/half-power-frequencies-is-6db-or-3db/84861#84861The ResultHowever, in this case, it seems that they calculated the common mode filter for -3db and the differential filter for -6db.

author: moffy
author_rank: Super Contributor
timestamp: February 24, 2024, 09:30:43 am
content: Quote from: electronx on February 24, 2024, 09:07:29 amYeah, 1.3k is not two times 756, silly mistakeBut it is confusing to switch between the -3dB def and -6dB for the differential unless they mention it in the article. Good catch though.

author: Benta
author_rank: Super Contributor
timestamp: February 24, 2024, 12:36:30 pm
content: Quote from: electronx on February 24, 2024, 08:37:52 amIgnore it, please, I made a mistake. Cutoff frequency is 758 Hz. @moffy's 21 nF is correct.Concerning -3 and -6 dB points: the cutoff frequency is generally defined as the intersection point of the Bode-plot asymptotes, in this case -3 dB.

author: electronx
author_rank: Regular Contributor
timestamp: February 24, 2024, 08:37:29 pm
content: Quote from: moffy on February 24, 2024, 09:30:43 amIt doesn't matter, the human brain is prone to linear thinking. I do this frequently, too. I think linearly, not logarithmically. Thank you very much for your help. I will ask Analog Devices engineers again and I will share the answer here.

author: electronx
author_rank: Regular Contributor
timestamp: February 24, 2024, 08:38:56 pm
content: Quote from: Benta on February 24, 2024, 12:36:30 pmNo problem, thank you for your time and curiosity.