Please explain to me this op amp gain stage

Stompboxes circuits published in magazines, books or on DIY electronics websites.
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sixthfloor
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Post by sixthfloor »

I am stumped as to how this works :

Image

The diodes ensure the clipping. I guess the lower bandwidth is controlled by C2-R1 as in a TS-like gain stage. When the pot is fully clockwise, the output of the op amp should bypass C3 and R2, making the circuit work just as a TS-like gain stage. What I can't wrap my head around is what happens with other pot positions ? Thank you by advance !

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Post by Frank_NH »

I think that basically you will have the resistance from the pot wiper to output lug in parallel with R2-C3. As you turn down the op amp gain (and increase the wiper to output resistance), the signal will retain more treble (corner frequency depends on R2-C3). At least that's how I see it. Depends on the part values obviously and what the output is connected to (probably an inverting gain stage).

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Post by toneman »

check this out--

https://www.electrosmash.com/tube-screamer-analysis

(it was a 1second google)

8)
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Post by plush »

toneman wrote:check this out--

https://www.electrosmash.com/tube-screamer-analysis

(it was a 1second google)

8)
You may want to check his question again to realize your link is irrelevant, same as your flexing over your googling skills.

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Post by sixthfloor »

Frank_NH wrote:I think that basically you will have the resistance from the pot wiper to output lug in parallel with R2-C3. As you turn down the op amp gain (and increase the wiper to output resistance), the signal will retain more treble (corner frequency depends on R2-C3). At least that's how I see it. Depends on the part values obviously and what the output is connected to (probably an inverting gain stage).
Thanks Frank, this does make sense as caps are comparable to frequency dependant resistors. I hadn't thought of it in that light !

In the design I was looking at, R2 = 15k, C3 = 2n2 and P1 = 100k. Connected to an inverting gain stage, good guess :D

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Post by toneman »

true that..doesnt explain the "gain" section.

i did not look that closely...was hoping to encourage a sort of "self-learning".... :scratch:

but, the "screamer topology" has been explained, and re-explained soooo many times.... :roll:

it IS important to note that this is a NON-inverting gain stage.

The resultant harmonics are different for an inverting stage with the same type of gain/diode feedback.

That's what makes this topology one of the most cloned out there.

here's another link WITH non-inverting opamp "gain" explanation:

http://www.geofex.com/article_folders/t ... sxtech.htm
Last edited by toneman on 09 May 2019, 21:36, edited 1 time in total.
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Post by Frank_NH »

sixthfloor wrote:
Frank_NH wrote:I think that basically you will have the resistance from the pot wiper to output lug in parallel with R2-C3. As you turn down the op amp gain (and increase the wiper to output resistance), the signal will retain more treble (corner frequency depends on R2-C3). At least that's how I see it. Depends on the part values obviously and what the output is connected to (probably an inverting gain stage).
Thanks Frank, this does make sense as caps are comparable to frequency dependant resistors. I hadn't thought of it in that light !

In the design I was looking at, R2 = 15k, C3 = 2n2 and P1 = 100k. Connected to an inverting gain stage, good guess :D
The idea of a non-inverting gain stage driving an inverting gain stage, with the gain pot split like your schematic, is found in a lot of Blues Breaker-style overdrives. I like the idea of retaining treble as you reduce the gain, in your case the R2-C3 frequency is about 4.8KHz, so it should add some sparkle to the lower gain tones.

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Post by Manfred »

sixthfloor wrote:
Frank_NH wrote:I think that basically you will have the resistance from the pot wiper to output lug in parallel with R2-C3. As you turn down the op amp gain (and increase the wiper to output resistance), the signal will retain more treble (corner frequency depends on R2-C3). At least that's how I see it. Depends on the part values obviously and what the output is connected to (probably an inverting gain stage).
Thanks Frank, this does make sense as caps are comparable to frequency dependant resistors. I hadn't thought of it in that light !

In the design I was looking at, R2 = 15k, C3 = 2n2 and P1 = 100k. Connected to an inverting gain stage, good guess :D
What the values of C2 and R1 ?

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Post by sixthfloor »

Thanks Toneman, I was aware of the geofex and electrosmash articles, they were some of the first resources I read when I started taking interest in guitar effects. I understand well enough a TS gain stage as it is, as you say, very common :D

Manfred, C2 = 100n and R1 = 1k5 (roughly 1kHz filter). C1 is quite typical, 220p.

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Post by Manfred »

Thanks Sixthfloor.
I have another question, which circuit is connected to the output of the stage.
C2, C3 and the part of P1 forms together with the load resistance a high-pass filter,
it is therefore important to know how the load is.

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Post by sixthfloor »

Manfred, here's the rest of the circuit. It's a design from a russian guy; I encountered this type of gain stage in various of his circuits and wanted to know more about how it worked.

As you can see, the output of the gain stage is connected to the inverting input of another op amp through a cap and a resistor. I'm out of my depth as far as loads are concerned, or their effects on filters :scratch:

Image Image

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Post by Manfred »

Sixthfloor, the voltage between the inverting and non-inverting input of the OP-amp are always in the range of few Millivolts respectively few Microvolts depend on the applied type.
In the case of the non-inverting amplifier the inverting input is virtually connected to the ground although the resistance in between is very high and so it is also known as "Virtual Ground".
The ideal OP-amp have between the inputs no voltage and an infinite resistance.
How it works?
The voltage at the resistor connected to the inverting input generates a current Iin=Uin/Rin because inverting input ist virtually connected to ground,
the current cannot flows into the inverting input now, because the input resitance is theoretically infinite.
The current flows must through the feedback resistor to the OP-amp output, the voltage on the output follows until the condition
Uout = Iin*Rfeedback is met.

Therefore, the connected load for the previous stage is Rin repectively Zin.

And now I have to stop, more in the next post.

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Post by Ben N »

It's a treble bleed circuit, just like in a guitar, except the effect is in inverse proportion to gain rather than volume.

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Post by Manfred »

The following images shows the results of the simulation:
ClipperFilterCurves.jpg
HighCutFrequencyDependOnPotSetting.jpg

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Post by Manfred »

Sorry, the seconde image above is wrongly described.
I wrote high-cut but so it should be called low-cut.
LowCutFrequencyDependOnPotSetting.jpg
I forgot to mention that the I ran the simualtion with 10 Kiloohms load.
That is frequency response between the OP-amp output and the load resistor, thus the out-filter circuit separate.
OutFilter.jpg

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Post by sixthfloor »

Ben N wrote:It's a treble bleed circuit, just like in a guitar, except the effect is in inverse proportion to gain rather than volume.
Thanks Ben, that's a pretty intuitive way to see it :)

Manfred, great explanation and graphs. It is very nice of you taking the time to do this !

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Post by bmxguitarsbmx »

P1 adjusts the gain of both op-amps at the same time. The "Left side" of the Wiper is the Rfb of Op-amp1, the "Right side" of the Wiper is the Rin of Op-amp2

Rfb = Feedback Resistor, Rin = Input resistor as typically used in op-amp gain equations.

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Post by Manfred »

I have named the image above entitled as "ClipperFilterCurves", perhaps it was misleading
it should more properly be entitled as "clipper frequency response for different gain".
Both parts of potentiometer are considered.

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Post by sixthfloor »

bmxguitarsbmx wrote:P1 adjusts the gain of both op-amps at the same time. The "Left side" of the Wiper is the Rfb of Op-amp1, the "Right side" of the Wiper is the Rin of Op-amp2

Rfb = Feedback Resistor, Rin = Input resistor as typically used in op-amp gain equations.
That's a smart trick ! Hence the common use of a non-inverting gain stage driving an inverting gain stage (as Frank mentionned) I guess.

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