Op amp filter - too many input/feedback paths for my brain

Thread Starter

to3metalcan

Joined Jul 20, 2014
260
So, I'm looking at the attached, which is the input stage to a small Ampeg bass guitar amplifier that in my opinion sounds pretty good. The stage after it is a Baxendall-style tone control, and then it goes right into the power amp/output stage which is nothing special, so my theory is that this stage must be what makes the amp sound a cut above other small bass amps I've heard. I could be totally wrong about that, but I'd like to figure out what's going on here, either way. At first I thought it was a variation on a Sallen-Key filter, then it kinda looked like an MFB topology, then I started getting a headache and decided to post here.

If I HAD to stick a guess to the wall, it's a HIGH pass filter that has a big Q hump at a low frequency so that it counter-intuitively boosts BASS, but I still don't know exactly why it's executed the way it is, here. Any insight is very welcome!
 

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Jony130

Joined Feb 17, 2009
5,181
Interesting circuit. At the low frequency, we have a Sallen-Key filter but at "high frequency", we have "ordinary" non-inverting amplifier.
For F >> 0.16/(R18 * C14) ≈ 16Hz the gain is (1 + R16/R18) ≈ 11VV. And at F > 0.16/(R19 * C15) ≈ 4kHz gain increases to (1 + R16/(R18||R19) ) ≈ 41 V/V
 

Thread Starter

to3metalcan

Joined Jul 20, 2014
260
Interesting circuit. At the low frequency, we have a Sallen-Key filter but at "high frequency", we have "ordinary" non-inverting amplifier.
For F >> 0.16/(R18 * C14) ≈ 16Hz the gain is (1 + R16/R18) ≈ 11VV. And at F > 0.16/(R19 * C15) ≈ 4kHz gain increases to (1 + R16/(R18||R19) ) ≈ 41 V/V
I agree that's part of what's going on, but I don't think that takes into account that R5 also provides input to the inverting terminal of the op amp. Isn't there going to be some differential action, there, given that both terminals get input? And it seems like from the perspective of the inverting input, the T-network described by R5 and R6 with C14/C15 and R18/R19 is going to do something weird to the response. Since R5 and R6 aren't equal, it's going to attenuate the input (causing a cut/rolloff) and attenuate the feedback (causing a boost) at two different frequencies. Or am I overthinking things?
 

crutschow

Joined Mar 14, 2008
25,398
Here's the LTspice, AC frequency response simulation of the circuit from 20Hz to 20kHz.
The blue trace is for the P1, 2.2kΩ input and the yellow trace is for the J1, 10kΩ input.

It's kind of an usual response, but that's apparently what the designer wanted. ;)

upload_2019-7-3_18-42-30.png
 

Thread Starter

to3metalcan

Joined Jul 20, 2014
260
Here's the LTspice, AC frequency response simulation of the circuit from 20Hz to 20kHz.
The blue trace is for the P1, 2.2kΩ input and the yellow trace is for the J1, 10kΩ input.

It's kind of an usual response, but that's apparently what the designer wanted. ;)

View attachment 180938
Thanks so much! That curve actually makes a lot of sense. It emphasizes the frequencies down near the lowest notes, and also the bright overtones of the strings, and dips in the mid-frequencies, which a lot of players think sound "muddy" in combination with other instruments. I still need to puzzle through a bit of the "why" but that pretty much gives me the "what." Thanks again!
 

crutschow

Joined Mar 14, 2008
25,398
It is a modified Sallen-Key high-pass filter with gain.
The difference is that the gain, as determined by R3 and the parallel impedance of R5/C3 with R6/C4 (component labels in my schematic), increases at high frequency, due to the drop in impedance of the RC combinations.
This causes the gain peak at about 12kHz.
 

crutschow

Joined Mar 14, 2008
25,398
Actually the high frequency gain increase is determined by R6C4, which has a corner frequency of 4kHz.
The R5C3 corner is down at 16Hz, so has no effect on the high frequency gain.

And the high-pass Sallen-Key filter is underdamped, when generates the low frequency peak at 60Hz.
 
Last edited:

Jony130

Joined Feb 17, 2009
5,181
the T-network described by R5 and R6 with C14/C15 and R18/R19 is going to do something weird to the response. Since R5 and R6 aren't equal, it's going to attenuate the input (causing a cut/rolloff) and attenuate the feedback (causing a boost) at two different frequencies. Or am I overthinking things?
At frequency when we can ignore Xc6 and Xc7 we have this equivalent circuit:

22d.PNG

In addition, notice that due to the negative feedback loop around the opamp the voltage at point A will also be equal to Vin.
Therefore there is no voltage drop across R5 hence, no current will flow through R5.

I_R5 = (Vin - V_A)/R5 = (Vin - Vin)/R5 = 0A.

Actually the high frequency gain increase is determined by R6C4, which has a corner frequency of 4kHz.
The R5C3 corner is down at 16Hz, so has no effect on the high-frequency gain.
This is exact what I wrote in post #2 And, the high-frequency gain is R3/(R5||R6)
 

Thread Starter

to3metalcan

Joined Jul 20, 2014
260
At frequency when we can ignore Xc6 and Xc7 we have this equivalent circuit:

View attachment 180979

In addition, notice that due to the negative feedback loop around the opamp the voltage at point A will also be equal to Vin.
Therefore there is no voltage drop across R5 hence, no current will flow through R5.
I think that's what I was missing, the fact that R5 is effectively bootstrapped. I can take it apart in my head and use superposition once I take that into account. Thanks!
 

Thread Starter

to3metalcan

Joined Jul 20, 2014
260
But current does flow through R5 at the low frequency where the Sallen-Key filter is active.
I thought of that, too. The inputs are only going to reject common signal at frequencies both of them can see. Does that simulation show what happens to the phase at those low frequencies? Just curious. Since gain in a Sallen-Key filter is directly tied to Q, is that input path what's damping the filter at cutoff (with this amount of gain in a typical configuration, I'm pretty sure the stage would oscillate!)
 

crutschow

Joined Mar 14, 2008
25,398
Here's the plot with phase:
upload_2019-7-4_22-44-26.png

Below is a plot of the response to a step input:
The output has a large overshoot, showing the expected underdamped response, but no ringing, indicating the circuit should be stable.
upload_2019-7-4_22-48-54.png
 
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