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In the following circuit, for a Twin-T notch filter, the author is using an additional inverting op-amp for the T-filter "feedback", that is connected to a potentiometer (pot).

enter image description here
Image credit: here

Why are they using this extra opamp instead of just the pot or fixed voltage divider circuit?


UPDATE:

I see there are some good info from the answers of this question:

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  • \$\begingroup\$ For a start it simplifies the analysis as the bottom of \$2C\$ and \$R/2\$ is a low impedance point otherwise you have to consider the pot total resistance, not just its set point. \$\endgroup\$ – Warren Hill Apr 17 at 15:24
  • \$\begingroup\$ Try simulating something like this in LTSpice (perhaps using a pair of resistors you adjust rather than a pot), and then try simulating it without the opamp. \$\endgroup\$ – esilk Apr 17 at 15:35
  • \$\begingroup\$ Because it's easier to add an op-amp then derive the equation for the pot and the notch filter without impedance seperation \$\endgroup\$ – Voltage Spike Apr 17 at 15:37
  • \$\begingroup\$ Simulating this, is a good idea, especially now when I understand the reason. \$\endgroup\$ – not2qubit Apr 17 at 15:38
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    \$\begingroup\$ Somes to prevent loading on previous stage high impedance, sometimes to achieve high impedance ratios, sometimes to draw uA loads for battery operation, can you think of any more? and 1st OA is redundant and 2nd OA output could be used depending on desire for constant or near constant gain \$\endgroup\$ – Sunnyskyguy EE75 Apr 19 at 17:15
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As the pot is varied, the effective resistance 'seen' by the bottom of the twin T circuit will vary if the feedback amplifier is not present, changing the frequency response of the circuit.

Even for a fixed voltage divider, the analysis would be complicated by the effective resistance; using an amplifier means it does not have to be considered.

The amount of variation if the amplifier were not present will depend on the ratio of the effective feedback resistance to the resistors used in the main twin T section. A relatively small feedback resistance will cause a small variation, but to get the notch back to normal would require varying all the resistors to maintain the necessary ratios.

The use of a buffer amplifier means that the driving impedance from the feedback path will remain very low (and consistent) for all potentiometer settings.

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  • \$\begingroup\$ Any idea in what way a fixed voltage divider will affect \$f_c\$? I am asking this because I have a circuit that does just that, and I noticed that the \$f_c\$ was off a few Hz, from what was intended. \$\endgroup\$ – not2qubit Apr 17 at 15:40
  • \$\begingroup\$ I guess you'll have to try analising the circuit to have the full understanding, but basically the opamp serves as a buffer, acting like a voltage supply. I guess if you chose a 1k potentiometer and a higher resistance for R, the potentiometer will act as a load and Fc will vary as you sweep the pot. \$\endgroup\$ – Filipe Nicoli Apr 18 at 20:05
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This is a variable Q notch design with a bit of gain. (example below 1.2dB) Here you can adjust the BW(-3dB) from 240 Hz (Q<<1) down to 0.5 Hz at unity gain for fc=60 Hz with a Q=120.

This is what a variable Q notch looks like:

enter image description here

The variable -3dB BW or Q can be increased significantly with positive feedback from the OpAmp to the return current, so it is negative feedback. This reduces the passband Bandwidth (BWp) while at the same time reducing the bandstop, BWs.

enter image description here

1st consider tolerances of the passive filter.

There are trade-offs between depth of Q and BW.

The result is that at one frequency, the signal on the HPF side cancels out only with perfectly matched parts on the LPF side. For example, if only 1 parts value changes 1%, the notch rises to -30dB and shifts the notch frequency slightly.

But worse, the attenuation starts long before and after the notch frequency, when the impedance of each cap matches the resistance as a voltage divider.

So while you might have a notch of -40 dB, the actual passband can be very wide. Using -3dB half-power as a reference, this filter has its passband between 4 - 253 Hz. Thus it appears to have a low Q passband but a deep notch.

2nd consider the positive feedback and non-inverting gain effects of the OpAmp and pot.

With a gain of Av = 2 = 6 dB, the filter response is almost the same as the passive filter except with gain. As the gain is reduced, the -3dB passband BW is also reduced giving an apparent increase in Q or filter shape factor.

\$Q= f_c/Δf_{(-3dB)}\$

To optimize this filter, you must choose realistic pass BWp(-3dB) and stop BWs(-30dB) for a given center frequency. Expecting more requires more stages or tweaking of R.

Also note that:

  • I only needed 1 OpAmp!
  • To reduce OpAmp drive current, change all R's to and all C's from μF to nF. (This has no other effects, other than raising the input impedance.)
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  • \$\begingroup\$ I don't know why people down-voted your answer, but if they are anything like me, I guess they didn't understand a word of what you were saying. I guess in this case some more pictures and added explanations would help many EE beginners. In addition the diagram feel very backwards, with rotated opamps, and outputs and grounds in the middle. BW of what?(for me BW is some measure between 2 pts on graph...where are they? The only thing I got is that perhaps the explanation of why people use such high values for R, often in the mega-Ohms, is to reduce the current draw for the opamp? \$\endgroup\$ – not2qubit Apr 18 at 11:41
  • \$\begingroup\$ I guess the prerequisite to understand my answers is have a minimal understand of ; what is BW? and how is it measured ( -3dB down) and what is Q Shape factor? Q= fc/Δf(-3dB), but I do appreciate your feedback. @not2qubit and WHat is negative feedback? \$\endgroup\$ – Sunnyskyguy EE75 Apr 18 at 13:56
  • \$\begingroup\$ Here is the best bet to understand what was said above. \$\endgroup\$ – not2qubit Apr 18 at 14:02
  • \$\begingroup\$ That's great. The important take from my answer is the 1st & last statement which no one else realized. The pot near unity gain is a high variable Q with nearly constant gain, unlike all other variable Q designs. And you only need 1 Op Amp. I thought I invented a similar design for my thesis which was embodied a variable Q, but with a variable f BPF so that bed-ridden folks could whistle any tones, to remote control devices like a touch-tone phone. But like most of us, I was probably not the first to discover anything like mine. where one could adjust Q and keep peak gain nearly constant. \$\endgroup\$ – Sunnyskyguy EE75 Apr 18 at 14:09
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    \$\begingroup\$ I've updated your simulation with bigger R and lower C values here. It show pretty good response of -30dB @ 61.4 Hz with -3dB at ~65.6 Hz with an 11kΩ pot set at 381Ω. \$\endgroup\$ – not2qubit Apr 20 at 11:30

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