TS circuits - dry signal leaking through with hot input signal?

[BAARON]

Thanks, I realized my sentence is kind of garbled, what I meant is that a non-inverting amplifier circuit's output will have an exact copy of what it has on the non-inverting input added to whatever kind of gain you are making happen with the feedback loop to the inverting input.

Um, no.  No, that is wrong.  What 1 + R1/R2 means is that the gain will never fall below unity.  Besides, if it were an opamp with infinite headroom and no clipping (hypothetical situation time!) and you added the amplified signal to the un-amplified signal at the output, they just add together anyway and increase the amplitude, blending into one signal that ends up being louder... how loud, you ask?  Why, 1+R1/R2!

Opamps use their output to try and keep their inputs at the same voltage, so with a circuit using feedback to the inverting input, any voltage seen on the noninverting input will be amplified at least by a gain of 1, to try and make it back to the (-) input.  Put something in the way, like back to back diodes, and the output instantly rises to a diodes turn on voltage {editor's note: no. no it doesn't. This is the root of your misunderstanding, I think.}, and once there, will continue to rise if the (+) input is higher than the diode turn on voltage.  That's why small signals (P-P less than the turn on voltages of the diodes) don't show up sounding clean on the TS's clipping stage output, there is a ton of clipped gain from the output trying to produce something that after being shoved through the TS's complex feedback circuit, will look like a unity gain signal to the (-) input.  But a big signal, like a big strummed guitar with atomic humbuckers, bigger than the turn on voltage of the diodes, will make the output of the opamp turn them on and continue to go bigger to make a matching signal on the (-) input as it sees on the (+) input.

Again, not quite: you started off on the right track, but then started getting things backwards.

1. How the non-inverting op-amp circuit works, in regards to R1 and R2
Yes, op-amps like to see the same voltage on their + and - inputs, but they can't do anything to control the voltage level at their INPUTS...  The only place they can directly control is the voltage at the output.  Because there's a feedback loop in the non-inverting circuit that attaches the output to the - input, they can indirectly control the voltage the - input sees by changing the output voltage.

If the + input sees +1v DC, the op-amp will produce an output that would change the voltage on the - input to the same thing (+1v).  If the output and - input were shorted with a wire, then all the op-amp would have to do is produce the exact same signal at the output as it saw on the input to make the - input see the same thing.  However, in the non-inverting op-amp configuration we use, there is  something in the way: R1 and R2.  They form a voltage divider.  If they are both 1k, then any voltage seen in the middle of the divider (which is connected to the - input, by the way) will be reduced by 50%.  Therefore, the output of the opamp must output a signal of +2v DC in order for the - input to see the same +1v that the + input sees (2v x 50% = 1v).  Thus, we have a gain of 2 at the output, because the output voltage is 2x the voltage at the + input.  NOT a signal with unity gain, plus some other signal to satisfy the feedback loop.

In other words, in the non-inverting op-amp circuit, the voltage at the - input only exists because of the voltage at the output of the op-amp, and because the voltage divider of R1+R2 attenuates the signal on its way from the output to the - input, the output signal (Vout) will always have a gain of Vout = Vin*{1+(R1/R2)}

2. How the capacitors factor into it.
Capacitors offer more impedance toward low frequency signals than high frequency signals.  The smaller the capacitor, the more impedance toward low frequency signals.
The 47pF/51pF capacitor in parallel with R1 offers so much impedance toward low-frequency signals that they ignore it completely and travel through R1 instead.  To a very high frequency, though, the impedance of the tiny cap is lower than the impedance of R1, so they pass through the cap instead.  When this happens, the capacitor is essentially part of the voltage divider and the high frequencies are attenuated LESS than if they had passed through the higher-impedance R1.  Thus, the output of the op-amp does not need to amplify the high frequencies as much as the low frequencies in order to make the + and - input match: therefore, high frequencies gradually get rolled off.  (Note that because R1 and this capacitor are in parallel and R1 changes because it's a gain pot, as you turn the gain up the little cap will roll off MORE highs because the resistor impedance is getting more and more difficult to get through as compared to the unchanging capacitor.)
Meanwhile, the 0.047µF cap at the bottom of the voltage divider ensures that the higher the frequency, the easier it is for an AC signal to travel to ground.  At high frequencies, the capacitor is essentially invisible.  At low frequencies (which have a harder time passing through small caps), it adds to the impedance of R2, and lowers the gain (because if you increase R2 in R1/R2, you get a smaller number... i.e., less gain).  So because of the presence of the 0.047µF cap at the bottom of R2, low frequencies do not need to be amplified as much by the op-amp in order to make the two inputs match because they are being attenuated less in the voltage divider.
In short, the cap in parallel with R1 lowers the gain of high frequencies and rolls off highs, and the cap between R2 and ground/Vb lowers the gain of low frequencies and attenuates bass.  The result is smoother ultra-high frequencies, and cleaner bass frequencies.
But at no point will the overall gain actually be less than 1 (unity with the input signal).  (See my very first point in this long post.)

3. How the clipping diodes factor into it.
We have two diodes in parallel with R1.  Diodes have something called a forward voltage: this is the voltage at which they begin to conduct power.  With a silicon diode, this is usually about 0.6v.  When the output voltage is lower than this level (whether it be because your gain is turned down or you're using low-output pickups or you're playing softly or the guitar's volume is turned down or etc etc etc), the diodes don't factor into the circuit at all because they can't conduct power, and things stay CLEAN.  However, when the output voltage EXCEEDS the forward voltage of the diode, the power has two paths it can follow from the output to the - input: R1 (high impedance) or the diode (only a few ohms of impedance, once you've passed the forward voltage).  It chooses the diode because it's the path of least resistance (and power is lazy), and thus any output voltage that would be higher than the forward voltage of the diode ends up being clipped at the diode's forward voltage instead. 
You see, the signal won't fall below the forward voltage, because if it did, then the diode would essentially be gone and the gain would push it back up past that forward voltage again anyway.  It won't go higher than the forward voltage, because the impedance of the diode is so low that R1 is practically open, and when we have our 1+R1/R2 formula, we see that the gain is essentially going to be 1... so it won't be able to produce a higher voltage than the diode's forward voltage.  It can't go higher, and it can't go lower, so it stays at the diode's forward voltage.  The result is that it essentially chops off any part of the wave that would pass the diode's forward voltage, and clipping happens.
(That's a crude way to explain it, but that's more or less what happens.  Read RG's article for a more in-depth explanation.) 

Thus, when the gain is turned up or you're playing hard or you have high output pickups, it gets clipped MORE, because more and more of the amplified signal would be pushed to voltage levels higher than the diodes' forward clipping voltages (and more and more signal being clipped by the diodes).
So: quiet input signals are clipped less.  Loud input signals are clipped more.  That's touch sensitivity.  Hot pickups are clipped more, cool pickups are clipped less.

Conclusion
Hot pickups are clipped more in the tube screamer circuit.  They just are.  Period.  Therefore, if it's not the clipping part of the circuit that's sneaking a clean signal to the output, and true-bypass tube screamers never ever exhibit this problem (because there's no path for the signal to take other than the overdrive circuit), then it must be the electronic bypass system used in the Bad Monkey and TS9, because that's the only other difference, and the only way the clean signal could sneak past and be blended with the output of the overdrive circuit.

Maybe then the most scientific solution to clamping the output is not clipping diodes on the output of the opamp, but clipping diodes (same as the ones in the feedback loop, Si probably) on the non inverting input, to Vref (or through a cap to ground).  That way the opamp's (+) input never sees a signal bigger than the clipped signal the opamp makes with the diodes in the feedback loop.  But the original poster's diodes after the output will work fine too.

This doesn't work, because that's not how non-inverting op-amps work.
If you do this, all you'll be doing is putting clipping diodes on the input of the circuit, which would be like putting clipping diodes across the volume pot on your guitar.  The only time they'd ever conduct/clip is if your guitar's pickups were able to put out a signal with a greater voltage swing than the combined forward voltages of the two diodes.  That isn't going to solve the problem, because it's completely unrelated to the problem (leaky electronic bypass).

[Processaurus]

Yes, #1 and #2 are good and accurate explanations of how the resistors and capacitors make gain and a bandpass filter, taking a look at #3 though, what you have explained is more like what happens with clipping diodes in an inverting opamp configuration, where the diodes are a voltage clamp, and nothing over the diodes forward voltage will ever come out of the opamp's output.  There you can get a gain of less than 1.  In the non-inverting amplifier, you can't (unless there is active amplification in the feedback loop).

Imagine a text book voltage follower (with the old school bipolar supply to keep things simple), the output is connected to the (-) input by a wire.  (-) input sees what the output puts out, which is equal to the voltage on the (+) input.  Gain is 1.  Put a Si diode in there instead of the wire.  A forward conducting diode basically behaves like a wire once it has it's forward voltage met.  The output will be 1 + the forward voltage of the diode (~.7v for Si, 1.4v for red LEDs, .3v for Ge).  This is because on the other side of the diode on the (-) input, it is seeing the output voltage minus the voltage drop of the diode, and it wants that to be equal to the voltage on the (+) input.   Small voltages will be insignificant when added to the turn on voltage. 

Imagine 10mV on the (+) input, for the (-) input to see that, the output puts out .710V, which is the 10mV plus .7v (turn on voltage of the Si diode).  The 10mV is mostly lost in the output, the amplifier is more akin to a comparator. 

But imagine a larger voltage on the (+) input.  Say, 5v.  The (-) input wants to see 5v, so it makes the output put out 5v + .7v (the turn on voltage of the diode).  We just got a much bigger voltage out of the non inverting amplifier than the diode's turn on voltage.  The diode does not act as a clamp.  Same thing happens with the big AC guitar signal.  Some humbuckers are unusually hot, and can put out a couple volts peak to peak with a big strum.  That will pop right out the output. Single coils are more around ~200mV P-P, and wouldn't be noticed compared to the larger signal the output is generating to overcome the diode turn on voltage in the feedback loop.

Don't think it is badly designed bypass in this case, unless the bypass is broken and the bypass signal is always on.  An easy way to test that is to turn the effect volume down and listen if the bypassed guitar is still on.

Cheers, thanks for getting me to reread my opamp booklet.

[BAARON]

Hmm.  After much lying on my bed and thinking through the math (I took a spill today and my knee is out of commission for a while, so I had time to think about it), yes.  You're more right than me on point 3: I definitely misread RG's tube screamer article and had that part somewhat wrong.

The diode does not act as a clamp on the original input signal, but it DOES act as a clamp on the amplified part of the signal.  In other words, whenever the gain (determined by the 1+R1/R2 divider) causes the voltage at the output to be higher than the forward voltage of the diodes, it clips the amplified signal and adds adds the clean signal from the + input to the diode's voltage drop.  http://www.bteaudio.com/articles/TSS/TSS.html has a very nice picture of this happening (the second oscilloscope picture down).

If the problem is thus that the clean signal is too large compared to the diode-clipped amplified signal (hence the clean signal being too audible), wouldn't it make more sense to use LEDs as clipping diodes (with their larger forward voltage) to change the ratio of amplified/clipped-to-clean than to add hard-clipping diodes to ground somewhere in an attempt to chop off the clean part of the waveform (which will have a large effect on the touch-responsiveness of the circuit)?