Question about the opamp gain calculation in a DS1

[nonoxxx]

Hello,

I am actually experimenting with some mods on an DS1 i had forgot for years and the mods I have done sounds greats : asymetrical clipping with two diodes and one led, electro caps upgrade to film, higher value for R17 for bringing low mids , I got some more tweaking to do with the tone stack but it' turning fine .

I will test this evening to get more gain when turning the gain pot clockwise by lowering the value of R13 to 2k I think (i will also change the value of the cap to keep the same frequency).

But when looking at the Electrosmash analysis of the ds1, I am wandering why R11 don't affect the gain calculation (I was thinking that it will have the same role than the 51k resistor in a screamer opamp loop) but the site give this formula :

1 + ( gain pot)/( R13)

Someone can explain?
Thanks

[slacker]

If you look at a Tube Screamer the gain pot is wired as a variable resistor in series with the 51k from the output to the inverting input, there's then a 4k7 resistor and cap from the inverting input to ground. So the maximum gain is set like you said 1+(51k + value of gain pot)/4k7
The gain pot in a DS1 isn't wired like that so the equation isn't the same, on a DS1 there is no resistor and cap from the inverting input to ground so whatever resistance you put between the output and the inverting input the gain is one 1+(100k+gain pot)/infinite resistance = 1 (it won't really be infinite resistance the PCB with have some but it will be massive so you can call it infinite for what we do) Hopefully that explains why the 100k doesn't come into the equation
The way the gain pot actually works is it's a voltage divider on the output that feeds some of the signal back to the inverting input. Opamps try to keep the voltage on their inverting and non inverting inputs the same so if say you have 1 volt going in to the non inverting input and you have a voltage divider that sends half of that back to the inverting input, the opamp will make the output 2 volts so that both the inputs are the same.
The gain calc on the Electrosmash page gets the right answer but the working isn't right. It should be

1/((resistance of the bit of the pot between the wiper and the 4k7 resistor + 4k7)/104.7K)

Hope that makes sense

http://www.electrosmash.com/boss-ds1-analysis link to the electrosmash article

[PRR]

> why R11 don't affect the gain

Because it series-feeds an opamp pin. For practical purpose, the opamp input pin is "infinite" impedance. Say 10Meg. 10Meg/(100K+10Meg) is 0.99X, no voltage-divider action to speak of. (Divider, and gain-set, action happens in the pot as Slacker says.)

I'm not sure why it is there.

I'm not sure what U1a brings to the party. Signal could go to the same pin on U1b.

[nonoxxx]

Thanks a lot for your replies
The more I am learning about pedals mods or designing , the more I realize that I have a lot of things to learn 

Between 2.2k/1uf fot r13/C8 , with my other mods sounds killer 

[riverasampayo]

I really do not understand the Electrosmash analysis and explanation of the distortion/gain structure.  It is not a configuration where there is a gain involved with the 100k pot.  In this case the gain is fixed at unity as the value of R1 is infinite (no resistor from Op Amp neg pin to ground in a Non Inverting config)...so the distortion gain cannot be (1+Vr100K/R13)...even though I still do not understand the comment before on the voltage divider...any further explanation is appreciated

[riverasampayo]

According to my calcs (which unfortunately I still do not know how to upload from my mobile), the max gain is 104.7k/204.7k...nominator VR at max +4k7 and denominator this plus the feedback 100k resistor...I am missing somenthing...

[PRR]

> It is not a configuration where there is a gain involved with the 100k pot.

The pot makes a loss. The opamp gain is the inverse of the loss.

First, the 100K fixed has NO effect while the opamp is in linear gain. (It may control distortion in overload, or it may be an oversight.) It goes into the *infinite* impedance of the opamp, so does nothing.

Next: that drawing is way confusing.

Draw the pot to-ground like a decent voltage divider.

The 100p clearly has no effect except above the audio band.

The 0.47u and 4.7K give a 50Hz pole, so we can "short" the 0.47u and get the right answer for most of the audio band.



Pot wiper "up", unity loss, unity gain.

Pot wiper "down", 22:1 loss, gain about 22.

I guess since this is "backward", they drew the pot "upside down", so it "goes the same way" (to the hand) as the other pots on the drawing.

[antonis]

I'm not sure why it is there.

Neither do I...

It could make sense forming a HPF (R11 + C7) which in turn forms a "constant" LPF for the NFB loop (it "steels" signal from C7//VR1 when VR1 is lower than it's max value..)

Maybe that's why R11 value is the same with VR1 max one..
(both filters -3db f_co is around 16kHz with VR1 at full range)..

Just a guess, of course..

[anotherjim]

R11, and R39 for that matter... oh, and not forgetting D8...
Paranoid designer or anti-copy tell-tales?

There are capacitors connected that could, if power supply voltage fluctuates, momentarily present voltages to the amp inputs that are outside the power pin voltages at some instant in time. Possible still 4.5v on C8 when chip + power down to zero.
Possible D8 is meant to clamp down excessive negative voltages on the input, but I struggle to see a mechanism that could generate much of that unless Q2 collector biases 1 or 2volts lower than 4.5v and drops to 0v before the 4.5 bias supply does.

All that said, all kinds of circuits with all kinds of op-amps can present similar conditions during power up-down cycles, without such protection and don't blow anything up.