Now that weâve explained away the asymmetry between the positive and negative half cycles, I think I understand the asymmetry present within each half cycle.
Your comment gives us a hint.
In the following figure, for simplicity, I replaced the second opamp circuit by just a ground to match the virtual ground that the second opampâs inverting input forms. The output voltage of the whole circuit is proportional to the current through the 100k resistor shown in yellow at the top right of the figure.
The top left trace is the input voltage, the bottom left is the output of the first distortion stage: the voltage to the left of the 10”F capacitor.
The bottom right of the picture shows the current through the capacitor (yellow) with the currents through the diodes superimposed (orange and red).
You can hardly distinguish the three currents because at any time, most of the current through the capacitor goes through one of the two diodes, the very little remaining current goes through the 100k resistor (top right in yellow, note the 25x smaller scale).
The asymmetry of the output voltage comes from the spikiness of the current through the capacitor.
The current through a capacitor is proportional to the derivative of the voltage across it.
So in this case the current is spiky because the voltage out of the first distortion stage has almost vertical edges as it crosses zero and then it flattens out for most of the cycle, so the derivative of the voltage (and thus the current through the capacitor) has a high initial value at the beginning of the half cycle and then a fairly low value for the rest of the half cycle.
At the very beginning of the half cycle, the current through the 100k resistor quickly rises matching the current through the capacitor until the voltage across the resistor reaches the level where the diodes start conducting (~.6V) at which point most of the additional current flows through the diodes and almost no more voltage increase happens on the output stage.
Shortly thereafter the voltage out of the first stage flattens out and the current through the capacitor drops progressively until the end of the half cycle.
Edit: if you want to play with this circuit, it is at http://tinyurl.com/y7aavf3h
While you were preparing that 
, I entertained myself by calculating the âeffective resistanceâ of a generic small-signal diode, as a function of the voltage across the diode. I-V plots are common, but I couldnât find a corresponding R=V/I plot so made one myself by stepping through a simulation:
The red line is the cut-off frequency for the equivalent 10ÎŒF highpass filter (the properties of which keeps changing during the cycle, obviously).
Zooming in:
EDIT: Turns out the default diode I used has a relatively high reverse current and a lower forward voltage than your typical small signal diode (e.g. 1N914/1N4148) so should probably redo this some day to fix the reverse slope and knee voltage, but the general idea still applies.
Hi everybody,
Super interesting topic, thanks for all this knowledge!
Iâm still a newbie, but I am playing around with a design of a distortion module, to try and learn what all parts do.
I sometime encounter this variation in the design:
(this is for a guitar pedal, so I assume the 9V figure, and the high impedance figures are not relevant for synth)
There is a similar thing in this RAT schematic for Eurorack: https://synthrotek.com/Schematics/308_Eurorack_Distortion_Module_Schematic.pdf
What is this, and why can it be removed for simplification? I understand the first 2 resistors, dividing the voltage source by half (at least I think I do ^^). But I donât understand what the 2 capacitors are for (Voltage regulation? A âdecouplingâ thing?). Is this introducing a positive bias (4.5V with the pedal schematic, or 6V with the eurorack schematic)? Or is there a big voltage drop across the 1M resistor (in this case, what is all this part about)?
Also, second question: what are the capacitor+resistance near the input and the output for (in the first picture of this topic, as this is different in the other designs Iâve linked)? Is this a high impedance input, low impedance output thing? Is this a decoupling thing?
Cheers!
The layout is a bit confusing, but your analysis is correct â thatâs a straightforward voltage divider that produces a voltage at VCC/2 (=4.5 V in your case) which is then mixed with the input (via the current-limiting 1M) to give the input AC signal something to swing around without immediately hitting 0 V. Most synth circuits use a dual supply, with â12 and +12 V rails and 0 V in the middle, and the circuits are then designed to swing around 0 V which simplifies a lot of things, but with a single supply you have to shift the signal upwards; instead of â12/0/+12 you use 0/4.5/9. If you remove the bias, the circuit wonât be able to deal with the negative half of the input signal (and unless you use rail-to-rail opamps which the TL08x definitely isnât, youâll run into issues long before that; the TL08x has enough limitations that Iâm not sure itâs a good idea to use it even with a 4.5 V offset with that circuitâs topology, even in a fuzz circuit ).
The caps in the voltage divider part are just there to smooth thing out, and provide an extra energy reservoir. You may not need both of them, especially not with a good & fresh battery, but thereâs a certain better safe than sorry factor here.
The input and output capacitors filter out any DC components in the input signal (which would interfere with the biasing) on the way in, and filter out the bias added in this circuit on the way out.
And yeah, that Eurorack RAT clone is probably built that way to stay close to the original; you could redesign for a dual supply, but why bother when youâre trying to replicate an existing design?
(They could have gone for a 4.5 V bias to stay even closer to the original, but keeping things symmetrical is also helpful. Yet another option would be to add a 78L09 to the board and run the clone bits on 9 V.)
Thank you! Iâll use a TL07x and remove the bias part. Cheers!
If youâd like to start off on something a little easier thereâs always âThe Great Destroyerâ:
The 4049 means you donât have to worry about rail-to-rail voltage and the layout in this strip board is pretty easy to follow. Iâve made this in fact (but I called it a âWave Badgerâ and used USB to power it.)