Agrrrr...!!!

3 serially vanished post..  

Correct me if I get this wrong, but it seems to me that the idea is that the parasitic capacitance to ground between the pot and resistor makes for an un- and variable-balanced bridged-Tee filter. This has phase shifts and such that could, in some circumstances, add phase change to the feedback and might cause oscillation at some frequency.

Yeah, maybe. I'd have to do some brain-hurting equation work or analog simm-ing to get a clearer idea of what the chances really are.

The pot is set up as a variable resistor. It will have higher parasitic capacitance than a fixed resistor, although both have some. If the pot is all the way down, near-zero resistance, it can be modeled as the wiper series resistance to the end terminal, and that fed to the opamp inverting input, or, alternately, if the pot is mirrored, to the junction of the resistor and pot. All the terminals, of course, have parastic capacitance, so the devil-in-the-details question is how much, and what that capacitance does as the pot varies. If we ignore the changing internal cross-capacitance of the pot body itself, we can resolve the pot plus resistor plus stray capacitances down to a lowpass path of one or two time constants. The feedback cap is a highpass path, so the bridged-tee comes into play.

Yeah, the lowpass path has variable phase shift depending on pot position and on how big the stray caps are, but the feedback cap seems to me that it would swamp any resistor-path oddities way before the opamp's unity gain point is reached. I can see that there would be some possibility that with a DC-to-daylight opamp, you could get issues; have to calculate/simulate that one.

But that being said, I've done it both ways without giving it a thought and never had issues.

Of course, as I'm fond of saying, that is just an anecdote, and an anecdote, even if true, is only one point of data out of a near-infinite data space.

Quote from: antonis on October 25, 2024, 05:09:59 AMLet's try a more friendly schematic..
(some items deleted, one added and others changed order and/or value..)



P.S.
After R4 deleted, C3 polarity orientation doesn't matter a lot..
(IC1's pins 7 and 2 should have no DC voltage difference..)

I like where this is going! So let's say we roll with James tone stack. Where/how would I place this in this schematic? I went down the rabbit hole yesterday learning about how changing cap/resistor values shift frequency cutoffs + the equation and such. I love learning about this stuff.

And please excuse my ignorance!

Quote from: R.G. on October 25, 2024, 11:58:31 AMBut that being said, I've done it both ways without giving it a thought and never had issues.

Yeaaapp..

But from good designer practice viewpoint, Ron is right..
(most of commercially availiable pedals have that "stopper/unity gain" resistor facing inverting input)

Quote from: antonis on October 25, 2024, 02:53:46 PM

Quote from: R.G. on October 25, 2024, 11:58:31 AMBut that being said, I've done it both ways without giving it a thought and never had issues.

Yeaaapp..

But from good designer practice viewpoint, Ron is right..
(most of commercially availiable pedals have that "stopper/unity gain" resistor facing inverting input)

Should I swap them for good measure?

Quote from: antonis on October 25, 2024, 02:53:46 PM

Quote from: R.G. on October 25, 2024, 11:58:31 AMBut that being said, I've done it both ways without giving it a thought and never had issues.

Yeaaapp..

But from good designer practice viewpoint, Ron is right..
(most of commercially availiable pedals have that "stopper/unity gain" resistor facing inverting input)

So basically one of those "99.9% of the time there is no difference, but there is that 0.1% and you don't want it to bite you" things.

I have done something similar in a design of mine (pot facing the inverting input, but it was in a non-inverting configuration - I suppose there is little difference though). It is through-hole, and the PCB is at home already, so I guess the solution to this is to install a 1R resistor before and solder the actual resistor to the pot. Or just assemble it as designed and make a change if necessary.

Quote from: ElectricDruid on October 25, 2024, 08:51:07 AMSorry Ron, Antonis, I'm not understanding you. Why does swapping over VR1 and R4 change anything at all? It's a series connection of two resistors - the order is unimportant. Please explain

Quote from: fryingpan on October 25, 2024, 08:57:53 AMI don't understand. The inverting input is (virtual) ground, OK. The resistor has a capacitance (small), the pot has a capacitance (larger). The resistor should dampen any oscillations anyway, "isolating" the output from the input, no matter where it is placed.

What ElectricDruid said

The wiring to the pot and the physical size of the pot increase the capacitance to ground around the pot.  When the pot is located at the opamp input that will add a pole to the feedback network.  If the pole frequency is low enough (set by the capacitance and the feedback resistors) it causes phase-shift in the feedback loop and the amplifier can become unstable.

One solution, which is a good practice in general, is to add a small capacitor across the feedback resistor (which are on the schematic).  That avoids the extra phase-shift in the feedback loop.   While people in pedal game often think of adding a feedback capacitance as a low pass filter, it's actually doing more than that.

It's a well known thing,
https://www.ti.com/lit/an/slyt087/slyt087.pdf?ts=1729813753874

Quote from: ToneRangerAudio on October 25, 2024, 12:25:11 PMSo let's say we roll with James tone stack. Where/how would I place this in this schematic?

Here you are..
(Tonestack as posted by m4268588 - post#14)

Quote from: Rob Strand on October 25, 2024, 04:15:14 PM

Quote from: ElectricDruid on October 25, 2024, 08:51:07 AMSorry Ron, Antonis, I'm not understanding you. Why does swapping over VR1 and R4 change anything at all? It's a series connection of two resistors - the order is unimportant. Please explain

Quote from: fryingpan on October 25, 2024, 08:57:53 AMI don't understand. The inverting input is (virtual) ground, OK. The resistor has a capacitance (small), the pot has a capacitance (larger). The resistor should dampen any oscillations anyway, "isolating" the output from the input, no matter where it is placed.

What ElectricDruid said

The wiring to the pot and the physical size of the pot increase the capacitance to ground around the pot.  When the pot is located at the opamp input that will add a pole to the feedback network.  If the pole frequency is low enough (set by the capacitance and the feedback resistors) it causes phase-shift in the feedback loop and the amplifier can become unstable.

One solution, which is a good practice in general, is to add a small capacitor across the feedback resistor (which are on the schematic).  That avoids the extra phase-shift in the feedback loop.   While people in pedal game often think of adding a feedback capacitance as a low pass filter, it's actually doing more than that.

It's a well known thing,
https://www.ti.com/lit/an/slyt087/slyt087.pdf?ts=1729813753874

So, if you have a parallel capacitor, the order of pot and resistor is not important because the cap swamps the pot's capacitance? Good, I always do that. (Also because I know it makes things more stable).

Quote from: fryingpan on October 25, 2024, 04:49:30 PMSo, if you have a parallel capacitor, the order of pot and resistor is not important because the cap swamps the pot's capacitance? Good, I always do that. (Also because I know it makes things more stable).

It definitely side-steps the issue in most cases.

Notice the capacitances of a pot are independent of the pot value.    That means there is a minimum feedback capacitance which fixes the problem.

If you have a very large pot value there will be a maximum capacitance that avoids HF roll-off. For example suppose we want full audio bandwidth with no more than 0.1dB attenuation at 20kHz.  If the pot value is 1M then the feedback capacitance across the pot can't be more than 1.2pF.  That small feedback capacitance might not be enough to compensate for the stray pot capacitance (and wiring).

From a design perspective you would be better-off using a lower value pot, like 100k or 50k.   That lets you use a larger feedback cap to achieve the 0.1dB rolloff at 20kHz.  The larger feedback cap value will be more likely to compensate for the stray pot capacitance.

If you burrow deeper into the problem the capacitance of a pot has a distributed capacitance along the track, some lumped capacitance at the connection points, and a lumped capacitance at the wiper due to the mechanics of wiper.   The lumped capacitance at the wiper mean the stability issue changes depending on the position of the pot.   Pot wiring also needs to be added.  Twisting the pot wires could actually add capacitance between the ends of the pot and add to the feedback capacitance.

It's can get pretty hairy trying to analyse such fine details.   That why it's better to design out the possible issue with lower pot values.   I suspect if you actually did get a case which needed such fine details to understand most hobbyists would dump the project for another day or maybe just put in a larger feedback cap and put up with the extra roll-off.

FWIW, there was a thread where the builder had this very issue,
https://www.diystompboxes.com/smfforum/index.php?topic=119654.0

Pot value; 100k
Feedback cap which fixed it: 22pF     ; probably not the minimum.

Good. My design has a 25k pot and a 30p capacitor. (It will be 33p in the final build). And, I've just noticed, it's actually before the resistor, not after. Yay me!

Quote from: antonis on October 25, 2024, 02:53:46 PM(most of commercially availiable pedals have that "stopper/unity gain" resistor facing inverting input)

Well put. You do NOT want "long" leads going to virtual grounds. Our diagrams so far do not account for wire length. But it is easy to put an opamp one place and its pot way over where more convenient. And not see any problem. It may not be a problem with a few kOhm in the loop. A variable R allows the user to tune for trouble, unexpected and unexplainable.

I have a virtual model I may post later. But I thought the shunt cap on the pot was suppose to eliminate trouble? (For some practical value of stray capacitance.)

Quote from: antonis on October 25, 2024, 02:53:46 PMBut from good designer practice viewpoint, Ron is right..
(most of commercially availiable pedals have that "stopper/unity gain" resistor facing inverting input)

I guess that this is semantically equivalent to "a fair number of commercially available pedals have the pot attached to the inverting input".

English. What are you gonna do?   

I think the main point of the argument is the feedback cap always helps and helps a lot as it bypasses the problem.    Moving the pot connection from one-end to the other helps but it doesn't completely remove the problem.   If you have the choice it's better to put the pot at the opamp output end as a means of minimizing the damage.

There's also the option of which end the pot wiper ends up.  You could argue further that when the pot is maxed out (maximum resistance) it would be better that the wiper ends up towards the the output of the opamp.

There's a lot of possible wiring variations when you consider the terminals and series resistor placement. You can see how some builds have more issues than others, while the schematics look like they are identical electrically.   The feedback cap make things less dependent on the specifics.

That's why I often suggest adding a small cap to Baxandall tone controls.
Another example of oscillation:
https://www.diystompboxes.com/smfforum/index.php?topic=130663.0
https://www.diystompboxes.com/smfforum/index.php?topic=130663.msg1267508#msg1267508

Quote from: PRR on October 25, 2024, 09:55:58 PM

OUCH..!!!

To be honest, I must say that I have never seen schematics of "hifi" or studio-grade preamps with variable gain. They are basically all fixed gain with an attenuator (usually they are multistage, with a good amount of gain in the first stage but calculated as not to overload it with typical signals - 20dB, perhaps? - an attenuator then a second stage with an additional amount of gain). I have always inferred that this is due to noise and stability considerations (and also the practical impossibility of setting a fixed passband with variable impedance in the feedback).

Here is a method of giving a fixed gain to the BAX.

Since it only shows the principle, it does not take stability into consideration. It may be better to change the topology to place the capacitor on the wiper and provide a stopper resistor instead.

Version 4
SHEET 1 800 600
WIRE 896 -592 720 -592
WIRE 896 -576 896 -592
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WIRE 272 -192 208 -192
WIRE 592 -192 592 -208
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WIRE 720 -192 656 -192
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WIRE 656 -176 656 -192
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WIRE 400 -112 384 -112
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WIRE 896 -112 880 -112
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WIRE 384 -96 384 -112
WIRE 880 -96 880 -112
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WIRE 832 -80 656 -80
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WIRE 832 -32 816 -32
WIRE 32 0 32 -16
WIRE 320 0 320 -32
WIRE 320 0 32 0
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IOPIN 400 -112 Out
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FLAG 208 -448 H12
FLAG 160 -512 H13
FLAG 272 -288 L11
FLAG 208 -192 L12
FLAG 144 -288 L13
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FLAG 768 -192 0
FLAG 704 -512 H21
FLAG 656 -448 H22
FLAG 608 -512 H23
FLAG 720 -288 L21
FLAG 656 -192 L22
FLAG 592 -288 L23
SYMBOL signal 32 -112 R0
WINDOW 123 2 120 Left 2
WINDOW 39 2 104 Left 2
SYMATTR SpiceLine Rser=0
SYMATTR InstName V1
SYMATTR Value ""
SYMBOL e2 384 -112 R0
SYMATTR InstName E1
SYMATTR Value 100k
SYMBOL cap 80 -496 R270
WINDOW 0 32 46 VLeft 2
WINDOW 3 0 18 VRight 2
SYMATTR InstName C11
SYMATTR Value 5n6
SYMBOL cap 272 -496 R270
WINDOW 0 32 46 VLeft 2
WINDOW 3 0 18 VRight 2
SYMATTR InstName C12
SYMATTR Value 5n6
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SYMATTR InstName R11
SYMATTR Value 10k
SYMBOL res 32 -272 R270
WINDOW 0 32 62 VLeft 2
WINDOW 3 0 34 VRight 2
SYMATTR InstName R12
SYMATTR Value 8k2
SYMBOL res 272 -272 R270
WINDOW 0 32 62 VLeft 2
WINDOW 3 0 34 VRight 2
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SYMATTR Value 8k2
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SYMBOL cap 256 -272 R0
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SYMATTR Value 33n
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SYMATTR InstName R14
SYMATTR Value 33k
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SYMATTR InstName E2
SYMATTR Value 100k
SYMBOL cap 528 -496 R270
WINDOW 0 32 46 VLeft 2
WINDOW 3 0 18 VRight 2
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SYMATTR Value 5n6
SYMBOL cap 736 -496 R270
WINDOW 0 32 46 VLeft 2
WINDOW 3 0 18 VRight 2
SYMATTR InstName C22_1
SYMATTR Value {5n6/3.6}
SYMBOL cap 704 -592 R0
SYMATTR InstName C22_2
SYMATTR Value {5n6/3.6*(3.6-1)}
SYMBOL res 640 -448 R0
SYMATTR InstName R21
SYMATTR Value 10k
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WINDOW 0 32 62 VLeft 2
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SYMATTR Value 8k2
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SYMATTR InstName R23_1
SYMATTR Value {8k2*3.6}
SYMBOL res 752 -288 R0
SYMATTR InstName R23_2
SYMATTR Value {8k2*3.6/(3.6-1)}
SYMBOL cap 576 -272 R0
SYMATTR InstName C23
SYMATTR Value 33n
SYMBOL cap 704 -272 R0
WINDOW 0 6 24 Right 2
WINDOW 3 6 40 Right 2
SYMATTR InstName C24
SYMATTR Value 33n
SYMBOL res 640 -192 R0
SYMATTR InstName R24
SYMATTR Value 33k
TEXT 0 56 Left 2 !.AC Oct 10 10Hz 100kHz\n \n.Param Hi=0.5 Lo=0.5\n.STEP Param Hi 0.0 1.1 0.1\n+ Param Lo 0.0 1.0 0.1
TEXT 272 56 Left 1 !XU_H_1 H11 H12 H13 Tone_Pot Rt=50K set={Hi}\nXU_L_1 L11 L12 L13 Tone_Pot Rt=50K set={Lo}\nXU_H_2 H21 H22 H23 Tone_Pot Rt=50K set={Hi}\nXU_L_2 L21 L22 L23 Tone_Pot Rt=50K set={Lo}\n \n.SUBCKT Tone_Pot 1 2 3\n.Param P_T 0.000001\nR1 3 2 { Rt * (1-P_T-max(min(Set,1),0)*(1-P_T*2))}\nR2 2 1 { Rt * (max(min(Set,1),0)*(1-P_T*2)+P_T)}\n.ENDS
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Quote from: Rob Strand on October 25, 2024, 05:53:13 PM

Quote from: fryingpan on October 25, 2024, 04:49:30 PMSo, if you have a parallel capacitor, the order of pot and resistor is not important because the cap swamps the pot's capacitance? Good, I always do that. (Also because I know it makes things more stable).

It definitely side-steps the issue in most cases.

Notice the capacitances of a pot are independent of the pot value.    That means there is a minimum feedback capacitance which fixes the problem.

If you have a very large pot value there will be a maximum capacitance that avoids HF roll-off. For example suppose we want full audio bandwidth with no more than 0.1dB attenuation at 20kHz.  If the pot value is 1M then the feedback capacitance across the pot can't be more than 1.2pF.  That small feedback capacitance might not be enough to compensate for the stray pot capacitance (and wiring).

From a design perspective you would be better-off using a lower value pot, like 100k or 50k.   That lets you use a larger feedback cap to achieve the 0.1dB rolloff at 20kHz.  The larger feedback cap value will be more likely to compensate for the stray pot capacitance.

If you burrow deeper into the problem the capacitance of a pot has a distributed capacitance along the track, some lumped capacitance at the connection points, and a lumped capacitance at the wiper due to the mechanics of wiper.   The lumped capacitance at the wiper mean the stability issue changes depending on the position of the pot.   Pot wiring also needs to be added.  Twisting the pot wires could actually add capacitance between the ends of the pot and add to the feedback capacitance.

It's can get pretty hairy trying to analyse such fine details.   That why it's better to design out the possible issue with lower pot values.   I suspect if you actually did get a case which needed such fine details to understand most hobbyists would dump the project for another day or maybe just put in a larger feedback cap and put up with the extra roll-off.

This is a bit more complicated than just saying you want full audio bandwidth.  If you are making a device for general purpose use, whether it is used alone or with a lot of other stompboxes, you may want full bandwidth and maybe less attenuation at 20 KHz than 0.1 db because they are all going to have some attenuation and if you have a number of tham in series, the attenuation at 20 KHz may be more than 0.1 db.  If you are using this device alone, you may want to cut a lot of guitar energy above 5 KHz.  When Phil Collins put out the "No Jacket Requied" album, everything above 5 KHz was gone.  You don't usually need high frequencies from a guitar.