Maxed out Rat (sort of...)

[XAXAU]

Hi guys! I´m researching the Rat and all of the versions of it to build a maxed out pedal. Haven´t built a pedal before, don´t know that much about electronics and it´s been 15 years since I soldered in school but that´s not gonna stop me from trying to build this thing! 

I´ll be using the pedal for hardware synths and sounds coming out of my DAW.

I´ve looked at many kits of many different types of pedals and figured I should build a few different ones for sonic variety but now I feel like since a lot of overdrive/distortion pedals are basically rehashing older designs and are very similar I should make just one big distortion box which can do a lot instead.

So what I was thinking is I use the Rat PCB as a base and modify it to my liking. I love distortion so I looked at all the rat versions and thought why not put every Rat on a toggle to switch in and out? Then I thought it would be cool if I could switch in the distortion in the negative feedback loop as well. A lowpass filter sounded boring so I thought why not drop in a hpf/lpf tilt filter from the Muff but with flat mids instead? And of course a toggle to switch the clipping to full range if I wish to destroy the bass.

Now I´m not sure these mods are going to work. The clipping circuitry is what´s making me scratch my head a bit as I was planning on putting each Rat version on a "ON-OFF-ON" toggle for "SYMMETRIC-NO CLIPPING-ASYMMETRIC." So basically when all clipping toggles are set to "OFF" it´s only the OPAMP that´s being clipped.

The Rat V1 used 1N914 diodes, Rat V2 1N4148 diodes, the Turbo Rat used red leds, the You Dirty Rat used 1N34A germanium diodes and the Fat Rat MosFets (unsure about which kind).

So in my naive mind I see 6x ON-OFF-ON switches on the pedal which I can switch in and out of the circuit as I please to produce a wide range of distortion tones. Even blending multiple switches. But honestly I´m not sure this is going to work. The Rat puts out a lot of gain but can it drive all those clippers? Noob alert here! 

On top of that I´d like to be able to switch the whole clipping network to the negative feedback loop of the opamp but that part I haven´t figured out how to do yet.


1. Will the clipping section work? One at a time for sure but what if I use all the clippers at the same time? Diodes and GE are 1Vf, leds 1,5Vf, the rest I dunno.

2. How do I switch in the clipping network into the feedback loop?

3. Will the Muff tilt eq section work? R1=33k, C1=5.6n, R2=33k, C2=5.6n Pot=25k lin. Will cut Tone 100k, R7 & C8 out.



4. Will the bass boost switch work? The 2 resistor in parallel adds up to 43.361ohms so with the 47ohm I put in there will be a slight level drop I guess. R4&R5=47R, C5=2,2uF & C6=56uF. Cutoffs are 60.5Hz & 1540Hz

I couldn´t upload my amateurish schematic I made but I hope you guys understand what I´m trying to get at.  If not please don´t hesitate to ask.



Cheers guys!

[anotherjim]

As it do already have a low pass. A variable high pass might be better than the BMP tone. You could fit one in the R4-C6 network.

Your clipper network will be ok so long as it's all after that 1k out of the amp. 1k is big enough to protect the amp when the diodes fully conduct.

Whatever your clipper network is, it essentially only has 2 terminals, so a DPDT switch with the clipper network across the middle contacts of both poles will let you throw it between the existing position (R6 & ground) and the feedback loop (amp pins 2 & 6).

[ashcat_lt]

914 = 4148

2 x 914 (or 4148) = LED or close enough for our purposes.

With multiple parallel diodes, the one with the lowest Vf does its thing and the rest can basically be ignored.  I can't see a good way that your idea can give a diode of one sort in one direction and another in the other.  My idea was to have a rotary for each "side" of the waveform, but then I realized that 1+2 is the same as 2+1 and therefore half the possible combinations would be redundant, so I just put the ones that were most interesting to me (1+1, 1+2, 2+2, 1+0, and 0+0) on one DP5T, and then I found out...

The Rat can't actually do assymetric clipping in practice because the clipping section sits between two caps.  Assymetric clipping looks very much like a DC offset, and those caps won't support such a thing.  One or the other (or both) will float toward the center point between the Vfs.  Course, when a Rat is really cranked, it hits the Vf so quickly that the "duty cycle" of the square wave won't be very far off from 50/50 anyway even with the common 1+2 assymetric configs.

Ideal would be to remove those caps.  Then you need one side of the clipper to go to V+ instead of ground.  Then you can almost just pick one symmetrical pair and vary the bias voltage and gain to achieve about the same effect of switching around all those diodes - and a bunch of "in between settings" - using a pot instead of switch(es).  I think probably replace the transistor buffer with an opamp that perhaps includes a bit of makeup gain, especially if you're doing that BMP tone stack.

[XAXAU]

Whatever your clipper network is, it essentially only has 2 terminals, so a DPDT switch with the clipper network across the middle contacts of both poles will let you throw it between the existing position (R6 & ground) and the feedback loop (amp pins 2 & 6).

So from R6 the signal goes to the clipping network > both input lugs on the DPDT > top position goes to ground > bottom position goes to opamp 2&6?

Ideal would be to remove those caps.  Then you need one side of the clipper to go to V+ instead of ground.  Then you can almost just pick one symmetrical pair and vary the bias voltage and gain to achieve about the same effect of switching around all those diodes - and a bunch of "in between settings" - using a pot instead of switch(es).  I think probably replace the transistor buffer with an opamp that perhaps includes a bit of makeup gain, especially if you're doing that BMP tone stack.

Actually I don´t need the BMP tone stack or the Rat lowpass either for that matter. I can shape the output in the DAW so I might just get rid of it. Then again I don´t really need a lot of gain either if the BMP tone stack is bogging down the circuit?

Asymmetrical clipping would be nice though so are you suggesting getting rid of C7 and C9? V+ is the +4.5V right?

I thought different diodes and transistors had a sound of their own and different knee? That´s why I thought of using all the different versions ofrom all the different rats.

So a symmetrical pair of diodes and a pot after one of the diodes?



After som eheavy readin gthe only other "interesting" clipping option is using diode conneted mosfets which has a very soft knee?

[anotherjim]

Clipper on middle lugs. R6/ground top lugs. Pins 2/6 bottom lugs. That's a changeover setup.

[ashcat_lt]

If you actually normalize the curves, most diodes are pretty much indistinguishable.  A 1V signal clipped by a silicon diode looks almost exactly like a 2V signal clipped by two Si diodes in series which looks almost exactly like if it was clipped by an LED.  Ge curves are a little different, but that only really can matter when the signal is in that knee region.  A Rat has a whole lot of gain, so spends most of its time way beyond the knee, so minor differences in the curve just can't really make much change to the sound at all.

When I said V+, I meant +9V, but I was wrong.  They should both go to 4.5V.  It's not exactly non-trivial to vary the signal bias, but...

Edit to add - Now that I think about it, the negative feedback TS-style "soft-clipper" is really a lot like a Rat with a unity clean blend.  If you DC couple through the clipper, add a bias control and a clean mix control, you'd have a whole pile of tweakability.  You wouldn't be able to say that it's a faithful reproduction of any specific model, but nobody will hear enough of a difference to say without looking.

[GGBB]

The Rat can't actually do assymetric clipping in practice because the clipping section sits between two caps.  Assymetric clipping looks very much like a DC offset, and those caps won't support such a thing.  One or the other (or both) will float toward the center point between the Vfs.

Can you explain that more? How does the cap alter the shape of the more clipped side of the waveform so that it once again becomes symmetrical with the other side? I understand your point about offset, but I thought there was a lot more to asymmetrical clipping than just average DC offset. When I think about it, almost all pedals have an output coupling cap, so does that mean that all of them that claim to have asymmetrical clipping actually don't?

[anotherjim]

Signal centres on the average level after removing DC. A symmetrical signal loses any apparent asymmetry. This type of distortion, the op-amp will be mostly clipping, and that tends to remove any asymmetry from the original signal. Chopping a squared wave at different peak levels in the clipper doesn't actually change its shape & the next coupling cap centres it back in the middle.

If you make C7 small so that it is a high pass (maybe 700Hz) -  it will differentiate the squared wave into alternate sawtooth shape pulses. Different clipping levels will give those pulses different width shapes. That asymmetry will survive. But with a fixed filter frequency, you only get it happening in a sweet spot range of barely 1.5 octave. Also, when the signal falls below op-amp clipping, the high pass gives a much weakened signal to the clippers, but it might make an interesting option switch.

[GGBB]

A symmetrical signal loses any apparent asymmetry.

The rest of what you wrote I understood - thanks. But the statement above - huh? If it's symmetrical to begin with, how does it have any asymmetry? What do you mean by "apparent" asymmetry?

[anotherjim]

Yeh, I forgot a detail  I meant a symmetrical with a DC offset - which can appear to be asymmetrical in relation to the centre bias level.

[GGBB]

Okay - so this is what I am getting from this. The cap removes DC bias (obviously). Once the op-amp starts clipping, because the waveform is now square, it can't be made any squarer, so diode clipping does nothing except perhaps alter apparent DC offset, which is later negated by the cap. Right?

I assume therefore that if the op-amp isn't clipping we can have asymmetrical diode clipping. Would I also be correct in assuming that since those op-amp square waves aren't perfectly square in the real world, asymmetrical diodes do have some effect, however tiny and possibly inaudible?

[ashcat_lt]

When the diodes are not conducting, the section between the opamp coupling cap and the cap before the buffer have absuloutely no connection to a DC reference point.  All they have to go by is the average of the AC wiggles.  They will always float to a place where that average is 0.  When they start to clip, if the diode drop on each side is identical, the average is 0 already, but if they are different, then the average starts to skew in that direction.  That forces the coupling caps to charge or discharge to make up the offset.  It ends up centering the wave between the limits, and it ends up clipping symmetrically.  It does depend on those caps charging, though, so it might "start" assymetric, but a sustained note will head toward symmetry, and if you keep playing and don't give them a chance to settle back...

Theoretically, a pulldown resistor tied to ground should solve the whole problem, but in practice it's tough to find a resistor value that makes that reference stiff enough to fight the opamp without messing up other (more important) things like overall level and frequency response. 


Edit to add - There was another question above that I wanted to address.  I said above that a 1V signal clipped by a single Si looks the same as a 2V signal into two Si.  If we cram a 3V signal into a 1+2 assym pair, it'll center itself at around 0.3V toward the double side, and we end up with with a 1.2V signal clipped by a single diode on one side and a 1.8 V signal clipped by two diodes on the other.  One of these is double the diode drop and the other is only half-again.  That must be different, right?

Yes it is, but that difference may not be as much as you may think, and like I said before most of the time this thing doesn't "sit in the knee" long enough to really tell either way.

[GGBB]

That is clear from earlier posts. But it doesn't address true asymmetry between wave shapes on either side of the average DC offset when waves are not perfectly square (which is theoretically never).

[ashcat_lt]

That is clear from earlier posts. But it doesn't address true asymmetry between wave shapes on either side of the average DC offset when waves are not perfectly square (which is theoretically never).

It's always true whether the opamp is clipping or not.  If the diodes are clipping, the wave centers between the diode drops so that they're both clipping by about the same amount.

[PRR]

Define "asymmetric".

Lopsided in voltage, in time, or both?

50:50 time asymmetry through a high-pass will settle out to a symmetric wave same size both sides.

A 10:90 time-asymmetric spike wave will settle with the high side taller.

Speech/music/guitar may be nearly time-symmetric or significantly time asymmetric.

[XAXAU]

I´m way out of my league here but I find it weird that if you clip something asymmetrically the capacitors will make it symmetrical again.

Also, the wave wont go back to what it was before the asymmetrical clipping so it must be distorted in one way or another right? It has to have a whole new sound of it´s own?

[ashcat_lt]

I´m way out of my league here but I find it weird that if you clip something asymmetrically the capacitors will make it symmetrical again.

No.  The capacitors won't let it clip assymetrically for any extended period of time.  If it wants to take 1V off one side and 0.5V off the other, it'll end up taking 0.75 off each instead.

[XAXAU]

I´m way out of my league here but I find it weird that if you clip something asymmetrically the capacitors will make it symmetrical again.

No.  The capacitors won't let it clip assymetrically for any extended period of time.  If it wants to take 1V off one side and 0.5V off the other, it'll end up taking 0.75 off each instead.

Ah ok I see. What if the asymmetrical clipping was done in the feedback loop?

[ashcat_lt]

Ah ok I see. What if the asymmetrical clipping was done in the feedback loop?

That's a good question.  I've never really looked into it.  I tend to think that the opamp will just push to wherever it thinks it should be going based on the output, and have no problem clipping assymetrically this way.  It will try to correct for an actual DC offset between the inputs, but this isn't an actual DC offset, so it should work fine.

[XAXAU]

Ah ok I see. What if the asymmetrical clipping was done in the feedback loop?

That's a good question.  I've never really looked into it.  I tend to think that the opamp will just push to wherever it thinks it should be going based on the output, and have no problem clipping assymetrically this way.  It will try to correct for an actual DC offset between the inputs, but this isn't an actual DC offset, so it should work fine.

So the caps will not allow the formation of asymmetrical distortion but it may allow asymmetrically distorted signals to pass through in this case?