"Peripheral" Components of the Distortion Plus...?

[ACS]

Have been tinkering with the timeless classic that is the Distortion Plus, whilst trying to school myself on feedback loops, opamps etc. Think I've done a decent job so far, and my knowledge is certainly on the improve...  But I have some questions regarding some of the peripheral components and what they do / how they work...

Caution: stupid questions follow!!

Referencing this schematic, just for consistency!



1. The cap+resistor combo in the feedback loop appears to be a high pass filter, rolling off some bass frequency - but it doesn't seem to match my understanding of a high pass configuration, namely signal passes through cap, followed by resistor to ground. Rather, this has this basic configuration, but as a "side chain". What's going on here? Given the the drive pot is also in series with this, would I be correct to assume that the frequency of the rolloff changes as the drive changes?
2. Similar deal with the 0.001uF cap at the output - I'm assuming this is a filter again, but I can't get my head around what it's filtering and how.
3. What's the purpose of the "extra" 1M resistor after the voltage divider (heading to the non-inverting input)? If all we wanted was 4.5V to bias the opamp, surely we could just take this from the junction of the first two 1M resistors?
4. What's the purpose of the 10K resistor after the DC blocking cap? Is this just to attenuate the signal so that the diodes clip at an appropriate level?

Thanks in advance.

[ashcat_lt]

1)  It's kinda confusing because it's actually a low pass filter, but it's in the Bizarro world that we call a negative feedback loop.    Worse yet, it's a shelving filter with that extra resistor in there.  You can figure the cutoff like any other LPF with the 1M feedback resistor in the cap, then mix in the unfiltered signal as though it was attenuated by the voltage divider created by that feedback resistor and gain pot (ignore the cap) and that will tell you the attenuation at any given frequency.  The inverse of that will be the gain at that given frequency.  Technically, the feedback resistor and cap set the "top" cutoff, and all three components determine the "bottom" where it goes all shelfy, but my way is easier for me to get my head around.

2)  It's an LPF, the cutoff is essentially determined by the 10k series resistor and this cap.

3)  If you put a jumper in place of that 1M resistor, then the input is directly connected to that 1uF cap that is trying to keep AC noise out of the bias voltage.  This will cause huge tone suck!  In fact, it will probably roll off most if not all of the guitar frequencies.

4)  Some might call this a "current limiting" resistor, and it does have the effect of limiting the amount of current demanded of the opamp when the diodes are conducting hard.  It does that, but also is the top half of the voltage divider that makes the diode clipping work at all and therefore has an impact on the character of that clipping.

Hope some of this helps.

[ACS]

Thanks for that - 2) makes sense  - as an LPF it has a cutoff of about 15KHz, so that gels with wanting to get rid of extraneous HF.
Description for 3) also makes perfect sense; great explanation.
4) I'm a little fuzzy on - if the 10K resistor is the top half of a voltage divider, where's the bottom half? (but I get the current limiting aspect!)

1) is still something of a blur - my calculations based on the 1M resistor and a 47nF cap put the cutoff frequency at about 3Hz... Any readings you could possibly point me towards?

[duck_arse]

on 4), the 2 diodes are the lower half of the voltage divider, and they want to shunt all the current they can get from the opamp to ground, so the 10k sets a limit on that current.

mark hammer usually explains the shelving and the gaining and the f_o-ing of the gain setting section very well. (it's best I don't try .....)

[ashcat_lt]

4) I'm a little fuzzy on - if the 10K resistor is the top half of a voltage divider, where's the bottom half? (but I get the current limiting aspect!)

Like duckarse said, the bottom of the divider is the diodes.  In order for that to make sense you have to think of them as though they were resistors.  They are, in fact, resistors for which the resistance is a function of the voltage across them.  Of course, since they're the bottom of the divider, the voltage across them is completely dependent on their resistance.  We end up needing iterative calculations or a W function (W stands for "whatever the hell that its" ) to figure it out, but that's how it works.  I think maybe you can see how as the voltage output from the opamp tries to get big in one direction or the other, the voltage across the diodes also tries to get bigger, but then the resistance of the diode which is forward biased gets smaller, which pulls the voltage across it down so that it can't get as big as the opamp is trying to make it.  Et voila, we have clipping!

1) is still something of a blur - my calculations based on the 1M resistor and a 47nF cap put the cutoff frequency at about 3Hz... Any readings you could possibly point me towards?

That's about right.  The actual corner frequency is way down there.  If it was a low pass filter in the normal world, you wouldn't hear much of anything coming through.  But this Bizarro world, remember, where everything is opposite.  So everything above 3Hz gets at least some gain, with more gain for higher frequencies.  In fact, the main function of that capacitor is to block DC, so that it is not affected by the voltage divider with the feedback resistor over the pot + 4.7K (which I completely missed yesterday!) so that there is no gain at DC and the opamp can swing around the bias voltage without being pushed toward one of the rails.  

But the gain doesn't actually head toward infinity for higher frequencies.  At a certain point it shelves off.  Remember how I said that the gain pot basically mixes in a voltage divided version of the unfiltered signal?  Well, the point where it shelves off is the point where that unfiltered signal starts to dominate because the low-pass action has attenuated those frequencies enough that they don't make much difference.  With the gain pot all the way down (actually at full resistance) that's way down there around 6 Hz, so that pretty much everything (except DC) gets about 2 x gain.  With the gain all the way up (minimum resistance) there's a lot less of the unfiltered signal in there, so it doesn't contribute significantly until around 720Hz.  The unfiltered signal at this point is divided down by 0.005 of it's original size, and so the gain for everything above this will be the reciprocal of that = about 500.

Great!  So how do we calculate that shelf frequency?  Use the resistors on either side of the cap in parallel as the R.  


Of course, so far we've ignored Gain Bandwidth Product.  Real world opamps aren't flat.  They can only provide a certain amount of gain for any given frequency, and they have a harder time with higher frequencies so that there's a whole "hidden" low pass which is dependent on what the gain would be if we were to ignore the capacitors...