Best Way to Cut Overall Output?

[Box_Stuffer]

What is the best way to cut the output of a circuit without effecting the EQ before going into my final level pot?

I feel like sometimes op-amp circuits can be too loud and I don't turn the 100k volume pot past about 9 o'clock because it has way more output than I need. Sorry, I probably need to order an Electronics for Dummies off of Amazon or something.

[Dormammu]

Just reduce the resistor in negative loop of opamp.
Post the diagram you are working on.

[Rob Strand]

I think you want this idea:




If you don't care about keeping the same impedance then you can use case 1 with no R2.

[Box_Stuffer]

I think you want this idea:




If you don't care about keeping the same impedance then you can use case 1 with no R2.

Thanks. I was thinking it would be something like a voltage divider to cut the power. I had tried to put a 100k resistor in-line - which cut it down to a manageable level, but also changed the sound.

[Box_Stuffer]

Just reduce the resistor in negative loop of opamp.
Post the diagram you are working on.

I am just starting out and I have mostly just done variations on this very simple schematic by Wampler. I have some others I would like to move on to that are more elaborate.

[Dormammu]

Thanks. I was thinking it would be something like a voltage divider to cut the power. I had tried to put a 100k resistor in-line - which cut it down to a manageable level, but also changed the sound.

Why not just take a lower resistance volume pot?
This will enable smoother and more predictable regulation.

[Box_Stuffer]

What if I use less power? Will it negatively effect the performance if I use less than 9 volts?

[Rob Strand]

What if I use less power? Will it negatively effect the performance if I use less than 9 volts?

In that circuit, the output level is largely set by the clipping voltage of the diodes.  So if you want to reduce the output the attenuation must occur after the opamp.    If you modify the circuit around opamp it will change the sound - that's something you don't want to do.

For that circuit, if you don't want *anything* to change you need to use one of the circuits I gave before.   However in this case you can do a reasonable approximation by using the "case 1" circuit with no R2.   The reason you can get away with it is the 10k tone pot is substantially smaller than the 100k volume pot.  So to reduce the level by a factor of 0.4 you would just add R1=150k in series with the 100k pot.

With the tone set to full cut the 22n capacitor sees 100k in parallel with 10k, or 10k//100k = 9.09k.  The low-pass filter cut-off will be fc = 1/(2*pi*R*C) = 1/(2*pi*9.09k*22n) = 796Hz.   When you add a 150k in series with 100k pot that the filter will see  100k+150k =250k in parallel with 10k, or 10k // 250k = 9.62k.   A little bit higher than 9.09k but not a lot.  The filter cut-off will then be fc = 1/(2*pi*R*C) = 1/(2*pi*9.62k*22n) = 752Hz.  Not very different to 796Hz - that's why you can get away with no "R2" for this case.  Moreover, you haven't lost anything since you can back-off the tone control a very small amount and still get the original 796Hz cut-off.

[Dormammu]

What if I use less power? Will it negatively effect the performance if I use less than 9 volts?

As Rob said, the level of signal is significantly determined by the diodes in the negative loop.
Games with power supply voltage will not give anything useful.
But, I would recommend increasing R46 about 5-10 times to your taste.
The tone pot is built incorrectly, in this form, it always gives a strong cut of high frequencies, it is better to build it like a guitar pot (if you can't afford 5-6 extra parts for a normal tonestack), increasing the resistance to 25-50k and capacitance up to 100 nF.
Better yet, build an engineering-correct tone stack, not this incomprehensible nonsense.
And reduce the volume pot to 25-50k.

[DIY Bass]

Are you using an audio or linear pot for the output volume?  That will have a bearing on how far you need to turn it to attenuate the volume.

[Dormammu]

Are you using an audio or linear pot for the output volume?  That will have a bearing on how far you need to turn it to attenuate the volume.

Yes, there is a strong suspicion that an inverse-log pot is being used, or connected incorrectly.

[FiveseveN]

The tone pot is built incorrectly, in this form, it always gives a strong cut of high frequencies

No. See SWTC sim with the current values.

it is better to build it like a guitar pot

Absolutely not.

I had tried to put a 100k resistor in-line - which cut it down to a manageable level, but also changed the sound

It really shouldn't, as long as it's after the tone pot. Or might it be the work of dr. Fletcher and Munson?

I probably need to order an Electronics for Dummies off of Amazon or something.

Having a good grasp of the fundamentals makes it all much easier. Kirchhoff should cover everything dirtbox-related, you don't even have to touch the 1862 update.

[Rob Strand]

Are you using an audio or linear pot for the output volume?  That will have a bearing on how far you need to turn it to attenuate the volume.

That's probably the real issue.

[Dormammu]

The tone pot is built incorrectly, in this form, it always gives a strong cut of high frequencies

No. See SWTC sim with the current values.

The main word in SWTC — is "stupidity" and not in a good way.
Are you so poor that you can't afford 5-6 extra parts to build a better circuit with greatest tone control capabilities?