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I designed a multiple feedback filter using the analogue devices filter wizard. I have never designed this type of filter therefore I followed up by reading about it and manually calculating the components. I used arbitrary values to learn the process. My first filter has Q of 5 and Fc at 50KHz and about 6dB of gain. I also used Ideal op-amp for simplicity. I measured the noise over a wide spectrum of 1K to 1Meg ignoring 0.1Hz to 1KHz as 1/f noise is heavily suppressed as well as I am using an ideal op-amp. enter image description here I have decided to redesign the filter with a higher value for the Q. I designed for Q of 50 and fc at 50KHz and about 6dB of gain. Same as before I used ideal op-amp and the same bandwidth for noise. To my surprise, the filter with higher Q is noisier. What would be the reason for this? I measured the noise produced by R1 as it is 10s higher and the RMS value calculated by LTSpice is about the same. Therefore, the only other major difference is positive feedback. enter image description here Could someone explain how this positive feedback loop works? Also why this positive feedback is not necessary for lower values of Q?

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  • \$\begingroup\$ Can you explain where your 2nd circuit (with positive feedback) came from please? \$\endgroup\$
    – Andy aka
    Commented Mar 31, 2022 at 12:53
  • \$\begingroup\$ @Andyaka I used Analog Filter Wizard and reduced the passband from 10KHz (1st circuit) to 1KHz. \$\endgroup\$
    – Wintermute
    Commented Mar 31, 2022 at 13:29

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To my surprise, the filter with higher Q is noisier. What would be the reason for this?

The noise gain in the 2nd circuit is about 20 dB higher than the first circuit. How do you measure noise-gain: -

enter image description here

I discarded the input source and left it open circuit because the circa 100 Ω resistor was very dominant here. Strictly speaking I should have shorted the input source resistor but, on this occasion it will make very little difference to the result.

enter image description here

The difference in noise gain is about 20 dB hence, the 2nd circuit will handle op-amp noise quite differently compared to the first circuit. And, it is almost entirely down to the use of positive feedback. Here's what the noise gains look like when the feedback is removed in the 2nd circuit: -

enter image description here

Could someone explain how this positive feedback loop works? Also why this positive feedback is not necessary for lower values of Q?

I might be able to but, you need to link to a document that describes that circuit. It's not one I've come across before so I'm a tad out-of-my-depth here but, I'm willing to run through any derivation about this if you can provide a link to a web source that describes it.

Suffice to say (for now) that it is the positive feedback that creates a significantly higher op-amp noise gain.

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  • \$\begingroup\$ Thank you for your answer regarding noise. Regarding Q factor I noticed this post: electronics.stackexchange.com/questions/380518/… and reference to this book: archive.org/details/… \$\endgroup\$
    – Wintermute
    Commented Mar 31, 2022 at 15:08
  • \$\begingroup\$ Page 482 has all the formulas which I used to calculate the Beta (100) of my design. And verify the component values to my satisfaction. Still struggling to see a direct correlation to high Q. Additionally, I noticed a vague explanation of how Beta is selected “The value of Beta could be any positive number but this particular value has been calculated so as to minimize the Q -sensitivity.” [Page 482] I must admit out-of-my-depth here! \$\endgroup\$
    – Wintermute
    Commented Mar 31, 2022 at 15:08
  • \$\begingroup\$ Also, why do I connect the noise source to non-inverting input? Considering that I feed signal to inverting input when the circuit works as a filter? The non-inverting input is my reference voltage to shift the signal. \$\endgroup\$
    – Wintermute
    Commented Mar 31, 2022 at 16:09

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