Has anyone tried soft clipping followed by hard clipping to ground?
From some previous tinkering I've learned that for a 1 V amplitude sine wave, soft clipping with silicon diodes will cap the voltage at 1.6 V (1 + 0.6 = 1.6) with some clean character to the sound, while hard clipping with silicon will cap the voltage at 0.6 V with virtually no clean character to it and a compressed sound.
I was just thinking that proper selection of diodes in each stage might give you the best of both worlds. For instance, silicon in the soft clipping loop giving you a 1.6 V signal followed by red LEDs clipping to ground chopping it back down to 1.4 V. More input/higher gain settings would move you from a Tubescreamer sound to a RATT/Dist+ compressed sound.
Is this a standard option? Or a standard fail, and hence never used? Either way I will breadboard it tonight and see what happens.
I would think you would end up clipping whatever you "gained" in the feedback loop would you not? Ergo defeating the purpose?
Now what about 1 silicon in the feedback loop and one to ground?
As psh mentioned, hard clipping tends to "cover up" whatever was gained by soft clipping. In fact, by clipping off everything above some threshold, all signal over the hard clipping threshold is lost. Soft clipping is "soft" because the signal above the start of clipping is more or less squashed down, but still there in a way.
Isn't that the idea behind the Plimsoul?
Schematic here - http://www.maketune.net/files/attach/images/7464/144/584/004/e9825b691169078cabef794982c43a14.PNG (http://www.maketune.net/files/attach/images/7464/144/584/004/e9825b691169078cabef794982c43a14.PNG)
quite a few dod pedals do this, classic tube for one.
QuoteInsert Quote
I would think you would end up clipping whatever you "gained" in the feedback loop would you not? Ergo defeating the purpose?
but would there be any difference in slamming your hard clippers with an unclipped signal and hitting them with a soft clipped signal.
if the hard clipped threshold was just just below the soft clipping, enough so to completely cut it off, lose it. wouldn't there be some difference in how quickly the hard clippers go from not conducting to full conducting/flatline clipping? my thinking is that it would make the transition from not conducting to full conducting... well, softer. :( o.k. heres where i find out i'm wrong :icon_lol:, but better, find out why! :icon_cool:
I remember a project (Wedgie Box? Widgy Box? Wiggy Box? Widgey Box?) in an electronics mag that had hard and soft clipping.There was something in it about how hard and soft clipping interacted when used together.
Okay, I found it. It's called the Widgy Box and was in the June 2006 issue of EPE (Everyday Practical Electronics). Ah never mind, it's hard clipping followed by soft clipping.
Let's not forget this awesome site and the GM arts pedal.
http://www.gmarts.org/index.php?go=217
Quote from: Quackzed on March 24, 2014, 08:28:48 PM
quite a few dod pedals do this, classic tube for one.
QuoteInsert Quote
I would think you would end up clipping whatever you "gained" in the feedback loop would you not? Ergo defeating the purpose?
but would there be any difference in slamming your hard clippers with an unclipped signal and hitting them with a soft clipped signal.
if the hard clipped threshold was just just below the soft clipping, enough so to completely cut it off, lose it. wouldn't there be some difference in how quickly the hard clippers go from not conducting to full conducting/flatline clipping? my thinking is that it would make the transition from not conducting to full conducting... well, softer. :( o.k. heres where i find out i'm wrong :icon_lol:, but better, find out why! :icon_cool:
I think this is part of the ticket. I've heard RG himself talk a number of times about the idea of keeping as much of the signal as possible "in the knee" of the diode conduction, and I think that the TS-style soft clipping arrangement helps with that. It gives us a more predictable signal going into the to-ground clipper. It would take some careful tweaking of gain and selection of diodes, but it might turn out interesting. When you talk "hard clipping" I think a lot time we think about the Rat, which tries to slam a couple thousand volts through between the two diodes, and ends up passing that knee pretty quickly, but if you don't go that far...
And yes, you could also carefully select diodes and set gains in such a way to have a fairly smooth transition from "Tube Screamer" to "Rat" style sounds. Getting the actual controls to work within a useful range could be interesting.
Quote from: aron on March 25, 2014, 04:07:01 AM
Let's not forget this awesome site and the GM arts pedal.
http://www.gmarts.org/index.php?go=217
Why does he list the MXR Dist+ and the Marshall Drivemaster as soft clipping devices? it seems like he's using the type of diode (LED/Germanium) to qualify a design as soft vs. hard clipping, rather than the arrangement.
Or am I completely wrong...
Quote from: electrosonic on March 24, 2014, 07:55:31 PM
Isn't that the idea behind the Plimsoul?
Schematic here - http://www.maketune.net/files/attach/images/7464/144/584/004/e9825b691169078cabef794982c43a14.PNG (http://www.maketune.net/files/attach/images/7464/144/584/004/e9825b691169078cabef794982c43a14.PNG)
Very interesting! I will have to check this out.
Last night I breadboarded a simple non-inverting gain stage with Si diodes followed by red LEDs clipping to ground. One twist is that I put a 10k in series with the Si diodes to try and tune the soft clipping response. The test signal on the scope did indeed show a transition from soft clipping to hard clipping which was tunable with the 10k pot, but in practice the sound wasn't as varied. I will continue with some experiments later this week, maybe even try building the Plimsoul to see what it's all about.
Quote from: YouAre on March 25, 2014, 12:03:00 PM
Quote from: aron on March 25, 2014, 04:07:01 AM
Let's not forget this awesome site and the GM arts pedal.
http://www.gmarts.org/index.php?go=217
Why does he list the MXR Dist+ and the Marshall Drivemaster as soft clipping devices? it seems like he's using the type of diode (LED/Germanium) to qualify a design as soft vs. hard clipping, rather than the arrangement.
Or am I completely wrong...
I was a bit confused by that as well. The capacitors will round off the edges, but I thought that diodes to ground would be considered "hard clipping" regardless of the final shape.
No, it's pretty much about the final shape...
With that in mind, the usually widely different diode currents in feedback loop vs. shunt clipping -schemes result into diode operating at rounder or harder "knees", respectively. However, it really has more to do with diode currents than the actual clipping scheme, like you noticed in practice: control diode current and you can control hardness vs. softness of clipping.
And I agree; the audible difference between hard and soft clipping is in practice almost negligible. I don't know why people focus so much on it. That, of course, in my subjective opininon.
Quote from: teemuk on March 25, 2014, 04:52:09 PM
No, it's pretty much about the final shape...
With that in mind, the usually widely different diode currents in feedback loop vs. shunt clipping -schemes result into diode operating at rounder or harder "knees", respectively. However, it really has more to do with diode currents than the actual clipping scheme, like you noticed in practice: control diode current and you can control hardness vs. softness of clipping.
And I agree; the audible difference between hard and soft clipping is in practice almost negligible. I don't know why people focus so much on it. That, of course, in my subjective opininon.
There's something about this that I don't like about this....
Clippers in the feedback loop affects the gain of the amplifier (what most of us think as "soft" clipping). Once the output signal of amplifier exceeds the forward voltage of the diode, amplifier gain drops from X to unity. This is only affecting the GAIN of the amplifier. So theoretically if you feed the amplifier with an already large signal (larger than the forward voltage of the diode), the gain still drops to unity, putting out a large signal. There is no hard limit. Please correct me if I'm wrong.
Clippers to ground will take any size signal ( "hard") , and clamp them to be fixed at the forward voltage. Using ln914's for example, no matter what signal you hit them with, you're only getting ~.7 volts out. This is a hard limit.
Quote from: YouAre on March 25, 2014, 05:11:23 PM
There's something about this that I don't like about this....
Clippers in the feedback loop affects the gain of the amplifier (what most of us think as "soft" clipping). Once the output signal of amplifier exceeds the forward voltage of the diode, amplifier gain drops from X to unity. This is only affecting the GAIN of the amplifier. So theoretically if you feed the amplifier with an already large signal (larger than the forward voltage of the diode), the gain still drops to unity, putting out a large signal. There is no hard limit. Please correct me if I'm wrong.
Clippers to ground will take any size signal ( "hard") , and clamp them to be fixed at the forward voltage. Using ln914's for example, no matter what signal you hit them with, you're only getting ~.7 volts out. This is a hard limit.
I agree with this in general. There are a number of noticeable practical differences between the two arrangements. Feedback diodes create a sort of crossover distortion which is significantly different from the "hard limit" of diodes to ground (or opamp clipping, etc).
It's worth noting, though, that the diodes themselves don't really cause the limiting, it's actually the limit of the opamp feeding the diodes. With an ideal voltage source the "output" of the diode clipper will continue to rise until they or something else along the way explodes from passing too much current. A completely academic point since real world components do run into real hard limits, but a diode is not a switch, and no part of its conduction curve is either linear or zero slope.
> the gain still drops to unity.... There is no hard limit.
"Hard" is relative.
Some of these things have gain of 100. When gain drops from 100 down to 1, that's pretty hard.
Most of these things also have variable gain. If you turn-down to gain of 3, then the drop to gain of 1 is "not hard".
As I see it.
QuoteOnce the output signal of amplifier exceeds the forward voltage of the diode, amplifier gain drops from X to unity.
This depends entirely on stage's architecture. If the gain stage is non-inverting, diode decreasing its internal impedance close to zero ohms indeed converts the stage to voltage follower with about unity gain. However, if the gain stage is inverting, diode decreasing its internal impedance simply keeps decreasing gain while the diodes shunt to virtual ground at inverting input. Gain is not unity in such case, if it was the clipping could not even work.
QuotePlease correct me if I'm wrong.
You have been corrected.
QuoteClippers to ground will take any size signal ( "hard") , and clamp them to be fixed at the forward voltage.
And this is really no different from diodes in the feedback loop of an inverting amplifier. The only thing different is the magnitude of current flow through the diodes: with diodes in feedback loop the clipping of the stage itself limits current flow through diodes, with shunt clipping diodes you have the full output swing from the gain stage hitting the diodes which results in much higher current flow when diodes are forward biased. Because the diode currents in these two schemes tend to be on very different magnitudes there's a drastic effect on softness or hardness of crossover from full reverse bias to full forward bias. Look it up in diode datasheet. You can easily replicate that effect by limiting the current that flows through those ordinary shunt diodes. For example, there's a big difference whether series resistance between opamp's output and the shunt diode circuit is 100R or if it is 10K. The diode current is what in practice defines the softness or hardness of the clipping, not so much the stage architecture. With right circuit design you can get hard or soft clipping with both methods.
And yes, it is also very much relative: If your maximum signal voltage swing is about 600 millivolts then at waveform peaks the operation is pretty much at the "knee" of the diode's transfer curve and never "above". The clipping will be soft because the "knee" is about 1/5th of the signal swing. If your signal swings close 10 volts then you likely get almost a square wave out and the little "knees" at waveform edges have relatively much lesser effect on overall waveform because the ~100mV "knee" region is only very little percentage of the entire signal.
Also, if you examine clipping just on waveform perspective then you can filter a hard clipped signal heavily and the wave will
look rounder at lower frequencies. Don't fall for this trap. At higher frequencies, where filters have less effect, it would again look like normal hard clipping. In essence, filtering does not really alter softness vs. hardness of clipping. It just removes harmonics, which happens to change the shape of the signal. Same thing with other filters, like for example a generic mid-range notch: pass a signal through such filter and a hard clipped square wave will transform so much that you don't even recognise that it was once a hard clipped square wave.
Quote from: PRR on March 26, 2014, 12:11:54 AM
"Hard" is relative.
Some of these things have gain of 100. When gain drops from 100 down to 1, that's pretty hard.
Most of these things also have variable gain. If you turn-down to gain of 3, then the drop to gain of 1 is "not hard".
As I see it.
True, I guess I'm being pedantic with my semantics. But let's look at 2 cases.
Case 1: Input signal is small, going into non-inverting opamp with feedback clippers. Then what you described applies. With enough gain, we'll get a square wave as we amplify the hell out of the signal and watch the output get squashed.
Case 2: Input signal is big. A portion of the cycle is amplified until we hit that diode forward voltage, and our gain stage becomes a unity gain buffer. But the input signal is still rising, because we a have a big signal. So what happens? I assume the signal keeps rising and then falling as it would in a buffer until we go negative and the process repeats, right? So the output is not getting compressed anymore. This is all assuming that we're not smacking the rails of the op-amp of course.
Same stage, but different input signal levels will yield different output levels.
Now with shunt clippers (just diodes to ground, nothing else), no matter what the input signal size is, we're going to get a maximum output voltage of the diode's forward voltage. This is what I mean by "hard" limit. With the other case I described above, we get compression only up until a certain point.
Quote from: teemuk on March 26, 2014, 10:06:05 AM
This depends entirely on stage's architecture.
My apologies, you're right. I was referring specifically to non-inverting configurations ala the Tubescreamer.
Quote
Gain is not unity in such case, if it was the clipping could not even work.
Please clarify.
Regarding the rest of your response (thank you for taking the time to address this, by the way), I'm assuming that we're not altering the shunt clipping with resistors (ala Jack Orman's warp control) or filtering it in anyway. Yes, this will soften the effects of "hard" clipping the waveform, but I'm analyzing the base cases described above.
What I'm getting at in "soft" vs "hard" is the nature of the attenuation. Yes, diodes in a non-inverting feedback loop can sound like a squarewave, while shunt diodes to ground can barely clip the signal.
I'm thinking that a workable definition (one that I've been using personally at least) is that "Soft" clipping affects the gain of a stage, while "hard" clipping effects output level. Does this make sense?
Quote from: YouAre on March 26, 2014, 12:26:34 PM
My apologies, you're right. I was referring specifically to non-inverting configurations ala the Tubescreamer.
Quote
Yep, and so was everybody else except teemuk. I personally can't think of any popular designs that have diodes in the feedback path of an inverting opamp stage.
Did some more experiments last night. First I checked the voltages coming from my guitar, and was surprised to see that my EMGs were putting out a maximum of about 400 mV from the neck single coil but up to 2 V (!!) from the bridge humbucker. I think these values were limiting the usefulness of my previous arrangement because the voltage was going to be in one range or the other all the time, hence no real change when moving the pots.
Second, I ended up using pairs of Si diodes in series with a 10k pot for both the feedback loop and the shunt to ground. This allowed me to adjust the clipping threshold for each pair.
This time I did notice a few differences when playing around with the 10k pots and the gain pot. With gain up to 100x and beyond, adjusting the feedback pot did virtually nothing whereas adjusting the shunt pot increased the volume while reducing the amount of compression. This was as expected, as one of the effects of the series resistance pot is that it raises the effective voltage drop of the diode pair (i.e. it can make a Si pair behave more like an LED pair in terms of voltage, but with continuous variability). At high gain, even an LED in the feedback loop would be maxed out.
At lower gain settings, the effects were more noticeable although still fairly modest. With the shunt pot dialed up, the feedback pot became a volume control although this time I did hear a difference in the quality of the distortion, it reminded me of a low-speed chainsaw without all the fuzzy buzz. I've heard this before when using both Ge diodes and LEDs as clippers, and as cool as the description sounds I'm not really a fan of it. Interesting though how series resistance with the diodes can change the character of the sound. It probably has to do with the amount of current and subsequent "knee-region" being used as mentioned above by R.G. and others, and Mark Hammer in an old thread, and less to do with the diode material.
One thing I did not get was the smooth transition from stinging overdrive to over-the-top distortion without twisting any knobs like I was hoping for, although in retrospect that was a stupid expectation as the single notes will always be quieter than the chords. I did however find a few settings that made the transition fairly well when moving from the neck to the bridge.
Let the experiments continue!
One note about feedback clipping, it doesn't seem like the diodes ever turn the opamp into a unity gain buffer - if it did, the output would never exceed the input. Going back to the basic opamp equation:
Vout = G(Vnon - Vinv)
Let's assume that the output exceeds the diode clipping threshold "D", and use a simple model of the diode as a switch with a constant forward voltage drop, also D. Then the inverting input will be held at the output voltage minus the voltage drop:
Vinv = Vout - D
Putting the equations together:
Vout = G(Vnon - Vout + D)
(1+G)Vout = G(Vnon + D)
Vout = (G/(1+G))(Vnon + D) ~= Vnon + D
(*this approximation is valid since the open loop gain G is so high, somewhere around 100,000)
This sum seems to be in line with what I've measured for Tubescreamer style gain stages (1.6 V output with a 1 V input), and also with this analysis that asserts the output is a mix of clean and clipped signal (http://www.bteaudio.com/articles/TSS/TSS.html).
The voltage gain is:
A = Vout/Vin = (Vnon + D)/Vnon
Which means there is always some gain being applied even when the diodes are fully conducting, even though it's not a simple multiple of the input.
"dynamic forward resistance" ?
Quote from: Lurco on March 27, 2014, 02:36:32 AM
"dynamic forward resistance" ?
Are you talking about the feedback diodes, or is this a strategy for making a smooth transition from Tubescreamer to Dist+?
QuoteGain is not unity in such case, if it was the clipping could not even work.
QuotePlease clarify.
Ok, this example naturally applies to inverting scheme: We have the said inverting gain stage configured to unity gain. There is also a silicon diode pair in feedback loop, which will limit the output to about 600mV peak values. If we input 100 mVpeak signal we get 100 mVpeak signal out. If we input a higher signal, say 5 Vpeak, we get an output limited to about 600 mV. After forward voltage has been exceeded the dynamic resistance of the diode has decreased and the gain of the stage has decreased as well. This causes the clipping of the signal: dynamic resistance of the diode in feedback loop decreases the gain. If the stage gain was "stuck" on unity the output would follow the input instead and no clipping could happen.
It really isn't all too different in the non-inverting scheme: Dynamic resistance of the diodes keeps decreasing similarly but in non-inverting scheme 100% feedback results to unity gain, unlike in non-inverting scheme where gain can keep decreasing below "1".
Don't forget, the reason a standard non-inverting opamp provides gain is that the feedback and ground resistors form a voltage divider. This lowers the voltage at V_ below Vout, and the opamp increases the gain to compensate. In the non-inverting buffer configuration, there is no voltage divider so the voltage at V_ is equal to Vout, which itself is equal to V+, and only unity gain is applied.
If you start with the standard non-inverting gain formula:
A = 1 + Rf/Rg
then you would also expect unity gain if you placed a forward biased diode in the feedback loop, since its resistance decreases dramatically once it starts to conduct (implying that Rf(total) = Rf||R(diode)-->0). However, this equation does not apply when using a non-linear feedback element like a diode.
The gain actually depends on the difference between V_ and V+. When there is a forward biased diode in the feedback loop, V_ will always be less than Vout by the forward voltage drop of the diode. Working out the gain equation under these assumptions gives you:
Vout = Vin + V(diode)
A = Vout/Vin = (Vin + V(diode))/Vin
Of course this equation only applies when the diode is conducting.
Quote from: PBE6 on March 27, 2014, 11:53:30 AM
Don't forget, the reason a standard non-inverting opamp provides gain is that the feedback and ground resistors form a voltage divider. This lowers the voltage at V_ below Vout, and the opamp increases the gain to compensate. In the non-inverting buffer configuration, there is no voltage divider so the voltage at V_ is equal to Vout, which itself is equal to V+, and only unity gain is applied.
If you start with the standard non-inverting gain formula:
A = 1 + Rf/Rg
then you would also expect unity gain if you placed a forward biased diode in the feedback loop, since its resistance decreases dramatically once it starts to conduct (implying that Rf(total) = Rf||R(diode)-->0). However, this equation does not apply when using a non-linear feedback element like a diode.
The gain actually depends on the difference between V_ and V+. When there is a forward biased diode in the feedback loop, V_ will always be less than Vout by the forward voltage drop of the diode. Working out the gain equation under these assumptions gives you:
Vout = Vin + V(diode)
A = Vout/Vin = (Vin + V(diode))/Vin
Of course this equation only applies when the diode is conducting.
It may work out about like this in practice, but the reason is that R(diode) isn't actual 0 when "forward biased". The resistance of the diode is better thought of as being exactly enough to divide the output voltage down to its forward drop.
You know, just about any circuit that uses a 9V supply, and sticks an op-amp gain stage ahead of some sort of clipping components going to ground, is really a "double clipper". The Distortion+ is the poster child for that, but other pedals/circuits do it too. The limitations on voltage swing of most op-amps make such that you can only expect to bring the signal up to maybe +/-3V before the chip says "Nuh-unh, can't do that". Figure out how much gain needs to be applied to the average guitar signal to bring it to that amplitude and it ain't all that much.
So the chip clips, and then hands the signal off to the diodes, which clip yet again. Myself, I have no idea how to classify what happens in the op-amp itself.
Quote from: ashcat_lt on March 27, 2014, 12:13:09 PM
Quote from: PBE6 on March 27, 2014, 11:53:30 AM
Don't forget, the reason a standard non-inverting opamp provides gain is that the feedback and ground resistors form a voltage divider. This lowers the voltage at V_ below Vout, and the opamp increases the gain to compensate. In the non-inverting buffer configuration, there is no voltage divider so the voltage at V_ is equal to Vout, which itself is equal to V+, and only unity gain is applied.
If you start with the standard non-inverting gain formula:
A = 1 + Rf/Rg
then you would also expect unity gain if you placed a forward biased diode in the feedback loop, since its resistance decreases dramatically once it starts to conduct (implying that Rf(total) = Rf||R(diode)-->0). However, this equation does not apply when using a non-linear feedback element like a diode.
The gain actually depends on the difference between V_ and V+. When there is a forward biased diode in the feedback loop, V_ will always be less than Vout by the forward voltage drop of the diode. Working out the gain equation under these assumptions gives you:
Vout = Vin + V(diode)
A = Vout/Vin = (Vin + V(diode))/Vin
Of course this equation only applies when the diode is conducting.
It may work out about like this in practice, but the reason is that R(diode) isn't actual 0 when "forward biased". The resistance of the diode is better thought of as being exactly enough to divide the output voltage down to its forward drop.
This sounds reasonable and most likely true. I just wanted to point out that the non-inverting clipping stage doesn't ever provide unity gain as it caused me quite a bit of consternation when I was trying to develop a rough model for the Tubescreamer gain stage output.
Yeah, no, like I said, in practice your thing is pretty close to truth. If you look at that equation from the right angle (not "a" right angle...) you might notice that it's basically saying that the output is about the same as if you used a hard clipper and the mixed that with a unity gain clean version of the signal. That is, a Rat with a clean blend sounds a whole lot like a TubeScreamer. In my attempts at modeling these things for JSFX plugins I found this was the best way to get the TS thing happening.
Quote from: Mark Hammer on March 27, 2014, 01:00:23 PM
Myself, I have no idea how to classify what happens in the op-amp itself.
I would call this hard clipping. No matter what, you're not getting above a certain threshold.
I'll stop harping on semantics now.
my flying spaghetti monster uses soft clipping in the feedback of the opamp, and a hard clipper on the arse end. sounds great.
i think it depends on how many stages you got to work with... if you have enough, i think soft clipping sounds great when it's re-clipped later.. very rich fuzzy distortion rich in overtones.
With feedback loop clipping, the output is limited to the input plus the voltage drop of the diodes. This means that:
1) The signal maintains some of the dynamic range of the input. For example, with .6v diodes in each direction, 1v p-p in is limited to 2.2 p-p out, 2v p-p in is limited to 3.2v p-p out.
2) With a large enough input signal the opamp will still hard clip. For example, if the opamp can only put out 6v, and you feed it 5v, it will try to put out 6.2v with the diodes above but instead be hard clipped at 6v. You could combine soft and hard clipping just by putting a booster before a feedback clipping stage.
With shunt clipping, the output is limited to the forward voltage of the diodes per side, period.
One could theoretically arrange things so that a signal would be soft clipped in the feedback loop but retain enough dynamic range to only be hard clipped by the shunt clippers in louder moments. The key to this would be using shunt clippers with a higher voltage drop than the feedback clippers, or else dividing down the signal between the two. For the example above, using shunt clippers of 1.5v per side would do it. A 1v signal would be limited to 2.2v in the soft clip stage and not clip hard, whereas the 2v signal would come out of the soft clip stage at 3.2v and be hard clipped to 3v in the shunt stage.
Of course, there are other factors to play with, like the relationship between current and forward voltage of diodes as well as what happens when you exceed the gain-bandwidth product of an opamp. Those issues would seem to take a little more math than I'm prepared for.
Keppy, I think we've pretty much covered all of that so far, but thanks for chiming in! ;D
Something I've been trying to say for this entire thread - something I only just learned myself recently - is that real world diodes are not switches which go from infinite resistance to zero resistance at a given voltage. The "forward voltage drop" actually just identifies the voltage that causes some specified amount of current to flow through them, and is in fact not any kind of hard limit. The V-I curve of a diode is, in fact, an exponential curve all the way up and all the way down. It just happens that below that "knee region" the slope is extremely steep - the effective resistance gets really big really fast - and on the other side it flattens out so that it takes bigger and bigger changes to get any appreciable change. But if you zoom in on any part of an accurate enough representation of the curve, it will be "curvy" all the way until it explodes. In practice, they work almost like that switch except right around what we call the forward voltage where the bend in the curve is noticeable.
That gain-bandwidth thing is actually pretty easy to sort out, in that it is essentially just a lowpass filter. If you know how much gain you're asking from the opamp, and you divide that into the open-loop bandwidth, you get the cutoff frequency. It's interesting to note that a Rat wide open has it's GBP filter rolling off at the bottom end of the guitar frequencies. I was used to looking at graphs of the Rat frequency response without this figured in, and it looks about like a bandpass centered around 750 (IIRC), very much like what you see on the Tech of TS page. But it don't sound like that. The GBP filter skews the whole thing to actually slope downward all the way across.
And if we're going to add that into the discussion, then we also have to add in the idea of slew-rate limiting. It seems to be somehow connected to the GBP thing, but it is not exactly the same phenomenon. Slew rate limiting is like distortion and filtering at the same time. A real Rat has a lot of this.
A typical TS stage has much less gain than Rat does, so the GBP filter will roll off higher, and you'll have less slew-rate distortion in the audible spectrum even given the same opamp, and of course any part of the signal that has pushed it past the diodes into the "not exactly unity gain" section will be affected much less by these as well as the other filter action in the gain section.
Quote from: ashcat_lt on March 28, 2014, 11:02:05 PM
Keppy, I think we've pretty much covered all of that so far, but thanks for chiming in! ;D
The background info was definitely covered, but I don't think the conclusion was, at least not totally. The part I wanted to draw attention to was the progressive clipping from soft to hard IF the shunt diodes are sufficiently larger that the feedback diodes, which was the OP's question. Every design I've seen that uses both (except the Plimsoul linked earlier in the thread) seems to use the same diodes in both locations, or uses larger diodes in the feedback loop. Sorry if it looked like I didn't read the thread (I did), but I thought a summary would help my post make more sense.
Haven't gone back to try this in a while, but I was just noodling around on Excel and made some pretty pictures I thought I would share.
This is what soft clipping followed by hard clipping looks like (theoretically):
Soft: Ge
Hard: Si
(http://tapatalk.imageshack.com/v2/14/10/30/74b42b576dbc322230c6e1f867490a4a.jpg)
Soft: Si
Hard: LED
(http://tapatalk.imageshack.com/v2/14/10/30/085ea7310020c97293749306737efb86.jpg)
Soft: Si
Hard: LED+Si
(http://tapatalk.imageshack.com/v2/14/10/30/a8c7b8c269f52123cf3fec2229a9adfb.jpg)
(In all of the above, the blue dashed line is the input signal, the green dashed line is after soft clipping and the red solid line is after soft & hard clipping)
Will have to find out what these sound like and take some actual readings tonight.
Check out the Madbean Yellowshark for this. It is an excellent sounding overdrive.
I've come to the conclusion that harmonics aka hardness of the clipped edges of a signal deliver harmonics we can perceive
as hard or soft clipping and can be manipulated with filters somewhat, but dynamic response is best achieved by multiple gain stages.
-> high gain: put more effort into the filtering afterwards,
-> low to medium: step by step clipping.
A new one to me: watch out for the source/emitter behaviour (Vds or Vce) as there it seems where the magic happens (mushy saturation), especially with following stages diving them into saturation.
So if we've sort of established that soft clipping followed by hard clipping might be a bit inaudible and just end up sounding like hard clipping, what about having 2 or more clipping stages in parallel instead of series. So for example for a three stage clipper.. Actually I just wrote up an explanation of what I mean but a quick and dirty schematic would be more informative..
(http://oi62.tinypic.com/30hv37k.jpg)
something like that with 5 or 6 stages
Don't know, just a random thought, has probably been tried and "just sounds like any other clipping"
Looks interesting, I there may be some issues with the mixing though:
http://sound.westhost.com/articles/audio-mixing.htm
I think an inverting opamp mixer like the one in Figure 4 of that article would work better.
I imagine the resultant mix would favour the 3-diode clipping just because the signal would usually be much stronger. I think that would mean more of the original signal would get through, maybe end up sounding more like soft clipping that way? But with more sizzle. Interesting, would love to hear how it sounds!
EDIT: Possibly more like hard clipping actually, just did a quick calculation and the top of the mixed waveform is still flat.
You usually put a resistor (~~1K) between the opamp and the diodes. Otherwise the diodes must clamp the full ~~30mA which the opamp can supply. Power supply demand goes way up. The clipping is quite abrupt.
Then I'd mix with ~~5K resistors. The passive mixer will work OK.
Quote from: PBE6 on October 30, 2014, 01:51:38 PM
Haven't gone back to try this in a while, but I was just noodling around on Excel and made some pretty pictures I thought I would share.
This is what soft clipping followed by hard clipping looks like (theoretically):
[ommitted pics]
(In all of the above, the blue dashed line is the input signal, the green dashed line is after soft clipping and the red solid line is after soft & hard clipping)
Will have to find out what these sound like and take some actual readings tonight.
Those waveforms don't look 'right'.
Whats the math behind it (formula)?
electrip
Quote from: PRR on October 31, 2014, 04:58:48 PM
You usually put a resistor (~~1K) between the opamp and the diodes. Otherwise the diodes must clamp the full ~~30mA which the opamp can supply. Power supply demand goes way up. The clipping is quite abrupt.
Then I'd mix with ~~5K resistors. The passive mixer will work OK.
Sorry yeah, both of those things... Forgot resistor after opamps and didn't change the resistor values from default... Quick n dirty ;)
Anyway my thoughts are that it'd probably sound just like any other clipping with diodes, but if someone has a spare half hour to breadboard it, go ahead! Obviously you had, say 4 stages, you would want the input signal to reach a peak amplitude of 2.8V at least (4x0.7V) so maybe have an adjustable gain stage that allows you to adjust the input signal level to get the maximum effect of the clipping diodes
Quote from: electrip on October 31, 2014, 05:47:51 PM
Quote from: PBE6 on October 30, 2014, 01:51:38 PM
Haven't gone back to try this in a while, but I was just noodling around on Excel and made some pretty pictures I thought I would share.
This is what soft clipping followed by hard clipping looks like (theoretically):
[ommitted pics]
(In all of the above, the blue dashed line is the input signal, the green dashed line is after soft clipping and the red solid line is after soft & hard clipping)
Will have to find out what these sound like and take some actual readings tonight.
Those waveforms don't look 'right'.
Whats the math behind it (formula)?
electrip
Original signal is just a sine wave.
Soft clipping is:
• v(clip) = v(in) {v(in) < v(diode)}
• v(clip) = v(in) + v(diode) {v(in) > v(diode)}
Hard clipping is:
• v(clip) = v(in) {v(in) < v(diode)}
• v(clip) = v(diode) {v(in) > v(diode)}
Quotesomething like that with 5 or 6 stages
Build it. :icon_wink:
Anyway, the depicted circuit is simply going to "average" the clipping: The clipping threshold is always less than that of the "longest" diode string and more than that of the "shortest". The outcome is practically equivalent to plain diode clipping but with different forward voltage. If all clipping devices have the same characteristic curve the tranfer function of the circuit won't even change drastically.
As is, IMO, it's a complex circuit to gain practically nothing. An approach where you average between a hard characteristic curve and soft characteristic curve (instead of different Vf's) might make a little more sense but even then the result is just the average of them: Not quite soft clipping but not quite hard either. (I think you folks have an answer there how the waveform should end up looking like). In practice you can usually achieve the same characteristic by easier means as well, like simply varying series resistance of the clipping diode strings.
But it's a wortwhile experience to build it and discover that despite all complexity it isn't really performing in a groundbreakingly different manner from any other diode clipping scheme.
An interesting thing, BTW, happens to waveforms when you sum LP, BP, and HP clipped signals in similar parallel arrangement: The clipped waveforms do not have the traditional "tops clipped off" -look but start to resemble more the shape of Mesoamerican pyramids.
Quote from: PBE6 on November 01, 2014, 03:29:40 AM
Original signal is just a sine wave.
Soft clipping is:
• v(clip) = v(in) {v(in) < v(diode)}
• v(clip) = v(in) + v(diode) {v(in) > v(diode)}
That ain't right dude. It will almost work if you change the first one to be v(clip) = gain*v(in){gain*v(in)<v(diode)} and the second condition to {gain*v(in)...} In this type of circuit, if there's no gain, then there's no clipping at all, period. There shouldn't be that big infinite slope step where the diodes "turn on".
Quote
Hard clipping is:
• v(clip) = v(in) {v(in) < v(diode)}
• v(clip) = v(diode) {v(in) > v(diode)}
Both of these, of course, are for ideal diodes that actually act like switches. Real diodes don't, but I guess it's close enough for the basic thing you're trying to show.
If you want to get a lot closer, use the hyperbolic tangent (scaled for the Vf of the diode) on the gained up input signal for the hard clipping, and then add the original (unity gain) signal to that for the soft clipping.
The gain in my simulation is 1. The signal strength (2 V) happens to be above the diode threshold, so there is clipping. I could adjust the equations to require a gain input instead of signal strength, but it accomplishes exactly the same thing so why bother? A practical distortion pedal would include a gain control as you say, as real world guitar signals are often on the order of 0.1 V and would not clip unless boosted.
With regard to the infinite slope step, you're right that would not happen in real life. However, if you measure the output of a Tubescreamer-style clipping stage it's really not that far off (visually speaking):
(http://tapatalk.imageshack.com/v2/14/11/01/18a3b6c8696553cc25bd616ffd9397f9.jpg)
http://www.bteaudio.com/articles/TSS/TSS.html
In answer your second comment, yes my simulation uses ideal diodes. When I have some time it would be fun to implement a better diode model to get a more realistic curve (especially if I ever figure out how to do Fourier analysis on the Excel output to estimate the harmonic content of the clipped signal), but ideal diodes are fine for showing the general behavior of signals clipped soft then hard. I will give the hyperbolic tangent method a try.
Nope. The only way the TS-style soft clipper works is if there is actually gain when the diodes are "off". Otherwise, it's just plain unity gain. Fine, it goes over the Vf of the diode...and it's still unity gain, so no change.
The steep slope you see on the bottom of that new image is the effect of the gain on the input signal. In your pictures, there is a section around 0 which is a lesser slope (equal to that of the original curve) then a sudden infinite step up when it hits the Vf of the diode. It is completely different. Your thing for soft clipping is nowhere near correct. Sorry.
Ah! Just ran through a simple calculation, yes you're right it doesn't work without a voltage divider in the feedback loop.
I'll try a diode different model when I get a chance.
Like I've said a couple times, "soft clipping" is almost exactly the same thing as adding the "hard clipped" signal (after gain) to the unity gain original signal. Definitely close enough for what you're trying to do.
^ This. And if you think about it, push-pull circuit's clipping distortion is also largely combined of clean signal (from one half) summing up with clipped signal (from the other half).
I finally got a chance to redo my soft clipping model and do some experiments on the breadboard with the scope and the guitar. Here are a few musings - pretty pictures to follow! :)
Based on the previous discussion, I decided to try a more realistic diode model and ended up choosing the Shockley diode equation:
http://en.m.wikipedia.org/wiki/Diode#Shockley_diode_equation
This model has a theoretical basis and gave good results, but required me to use a brute force numerical solution to find vD. However, now that I have my Excel sheet set up to do that it will be straightforward to try other diode models (like the hyperbolic tangent model mentioned previously in this thread) and see how they compare.
In a Tubescreamer-style gain stage with diodes in the feedback loop, the current flowing out of the ground resistor is equal to the current flowing through the diode plus the current flowing through the feedback resistor (omitting the capacitors for simplicity):
iG = iD + iF
v-/Rg = iD + (vout - v-)/Rf
The voltage (vout - v-) is simply the voltage across the diode vD. Also, the opamp will apply gain to keep both inputs equal, so v- = v+ = vin. Subbing these into the above equation gives:
vin/Rg = iD + vD/Rf
Tubescreamer gain is controlled by varying Rf. High values of Rf force more current through the diode while low values of Rf steal current away from it. In the extremes, Rf = 0 bypasses the diode completely creating a unity gain buffer while Rf = infinity forces all the current through the diode giving maximum distortion. Interestingly enough, this second case can be achieved by removing Rf from the circuit altogether! This result surprised me, but was verified on a bench circuit. It seems that as long as current can flow out of the feedback loop through Rg, the diode will conduct (one caveat - if there is no capacitor in the grounding path, Rg must be connected to Vref instead of ground or the circuit becomes a buffer).
One other interesting thing about this equation is that iG = vin/Rg sets the total current for the feedback loop. Decreasing the absolute value of Rg will increase the amount of current flowing through the diode, pushing it further up the conduction curve. This will result in slightly more volume and distortion due to the small rise in vD, but will also affect the quality of distortion by changing the shape of the diode conduction knee. This effect is fairly subtle, but it is interesting to hear the difference between two circuits with the same gain ratio settings (1 + Rf/Rg) but vastly different values of Rg.
A fun experiment to try is to remove Rf and use a pot for Rg (connecting it to Vref if there's no ground cap). Removing Rf keeps the distortion maxed out, allowing you to audition different values for Rg while keeping the gain fairly constant.
(At very low Rg settings the waveform changes from the familiar bullet-shaped Tubescreamer output to a type of slanted square wave which seems to be the opamp clipping..but at Rg = 0 even weirder things start to happen! The output becomes gated, keeping silent as long as the input is below a certain level and then lurching to life when the input crosses the threshold. Weird and broken sounding, but cool in its own unique way.)
Back to the equation. Subbing in the Shockley diode model gives:
vin = Rg*{Is*(exp(vD/(n*vT))-1) + vD/Rf}
Where:
Is = 1 x 10^(-12) A
n ~= 1.2 or 1.3
vT = 0.026 V
(These values gave good results for Si diodes)
Solving this equation for vD is non-trivial, but it does succumb to numerical methods. Using the fact that (vout - v-) = vD as above, we have:
vout = v- + vD
Again the opamp keeps both inputs the same, so v- = v+ = vin. Therefore we have:
vout = vin + vD.
So, knowing the input vin we can solve for vD and then add them together to find vout.
(An alternative method would be to calculate the diode resistance, use that to calculate Rf||Rd, then use that to find the gain, but this method didn't agree with scope observations except at high gain settings. This could be because of precision limitations with my calculation method, or because Rd = vD/iD is the incorrect formula for diode resistance in this case, or because this method doesn't deal with diode voltage drop properly - but in any event the equation above is much simpler to work with and gives good results.)
As noted previously in this thread, this equation shows that a Tubescreamer mixes the clean signal and the diode distortion together in a 1-to-1 ratio. A very similar effect can be achieved (in a more complicated, but perhaps more flexible way) by blending the output of a clipping-diodes-to-ground-style pedal with the original clean tone. I made a simple prototype based on this concept using individual volume controls for the clean and dirty signals, and the results were encouraging but I think its primary use would be to dial in a bit of clean bass in a bass distortion effect.
Here are some of my results.
Input 0.25 V
Rf 20k
Rg 10k
Gain 3
(http://tapatalk.imageshack.com/v2/14/11/22/e3a01e72e31853c6fa995b103d248d50.jpg)
(http://tapatalk.imageshack.com/v2/14/11/22/ea600e707ea563f66dacce962e0b08be.jpg)
Input 0.25 V
Rf 1M
Rg 10k
Gain 101
(http://tapatalk.imageshack.com/v2/14/11/22/583b39e3b5a3a9a3b777e688783bdb5f.jpg)
(http://tapatalk.imageshack.com/v2/14/11/22/45e211c129765539cc7ccde94eb4403c.jpg)
Input 1 V
Rf 20k
Rg 10k
Gain 3
(http://tapatalk.imageshack.com/v2/14/11/22/7e87fd10532b3f4359b2b67230f5ec43.jpg)
(http://tapatalk.imageshack.com/v2/14/11/22/51226a011ef3598d9478d74cccaac160.jpg)
As you can see, there is better agreement for higher input signals and high gain settings. Adjusting n for each input signal amplitude (1 V vs 0.25 V) seems to help. At this point I'm not sure if the issue is error in the general reasoning, calculation resolution (vD is only calculated to the nearest 10mV), or inherent limitations of the Shockley equation for small currents, but I'm happy with the performance for larger signals.
Oops! Just found a glitch in my spreadsheet, the input signal was accidentally referenced to "n" instead of "amplitude". Much better numerical agreement now all around.
(http://tapatalk.imageshack.com/v2/14/11/22/ed46c013a6afc1ea225f6628e11ccc05.jpg)
(http://tapatalk.imageshack.com/v2/14/11/22/1c9689a55e5cdbd3e065404e07186514.jpg)
I've been was messing around with my spreadsheet again - I added a hard clipping simulation (diodes to ground), upgraded my approximation method from brute force to a binary search, and added a crude harmonic content calculation. Here are some further musings..
In a Dist+ style circuit, the signal is first amplified by a non-inverting opamp (or similar) gain stage before passing through a limiting resistor and finally being clipped by a pair of diodes to ground. This circuit can be modified by adding a series variable resistor between the diodes and ground to create a "saturation" control.
The current through the limiting resistor is equal to the current flowing through the diodes-ground resistor combination plus the current flowing through the load. Summing these currents, we have:
(vin - vout)/Rin = iD + vout/RL
This equation shows that the smaller the input resistor, the more current will be available to flow through the diodes (even down to 0 ohms which maxes out the opamp current, although possibly at the expense of stability). Also, the higher the load resistance the more current will be forced to flow through the diodes. Both of these effects will push the diode current further along the conduction curve, modestly altering the harmonic content of the distortion and providing slightly more output voltage. (As will be shown below, the effect of the saturation control is much greater.)
The current flowing through the diode is simply iD, which gives rise to a diode voltage vD. The current through the ground resistor is equal to the current through the diode, which gives rise to a voltage drop of iD*Rg across the ground resistor. The output voltage is then:
vout = vD + iD*Rg
Although it's not immediately apparent by looking at this equation, the addition of resistance to the diode path limits the amount of current flowing through it. The diode current is always iD because that's how it's defined, but in order to have the same current flow through the diode/resistor combination as just the diode alone, the voltage must increase. When Rg = 0, the above simplifies to vout = vD. As Rg increases, less current flows and the diode has a smaller and smaller effect, causing vout to approach vin (provided RL is large enough not to form a significant voltage divider with Rin - which actually isn't the case with the MXR Dist+!). The effect of the saturation control is therefore very similar to a blend control. Contrast this with the Tubescreamer, where the original signal is always present with a gain of 1x and the gain control adds more and more distortion to it.
Subbing the above expression for vout into the first equation and solving for vin, we have:
vin = iD*(Rin + Rg + Rin*Rg/RL) + vD*(1 + Rin/RL)
This form makes it easier to input into Excel, especially when using the Shockley diode equation:
iD = 1E-12*(exp(vD/(n*0.026))-1)
For this circuit, n = 1 seemed to give the best results, in contrast to the soft clipping simulation where n = 1.25 gave the best results.
Using the above equation, I was able to find vD for a given input vin using a binary search. This gave much better results than the previous method. Using the values of vD I was able to calculate iD*Rg, and adding them together gave vout. Here are some results:
Rin = 10k, Rg = 0
(http://tapatalk.imageshack.com/v2/14/12/04/aadc4421a11e422a9688f45c36489060.jpg)
(http://tapatalk.imageshack.com/v2/14/12/04/9ae35a4be63fab81ce1cd3d0c355f5dd.jpg)
Rin = 10k, Rg = 10k
(http://tapatalk.imageshack.com/v2/14/12/04/8690d25b3f972808119a39a09fbd64a2.jpg)
(http://tapatalk.imageshack.com/v2/14/12/04/1d2fa25fb27748867a801d2e9ab66c21.jpg)
Rin = 1k, Rg = 0
(http://tapatalk.imageshack.com/v2/14/12/04/4c02144bc5872afdbf3d7f05f8f9fc91.jpg)
(http://tapatalk.imageshack.com/v2/14/12/04/1021b5930e447241bfcde02efbd19da2.jpg)
Rin = 1k, Rg = 10k
(http://tapatalk.imageshack.com/v2/14/12/04/1cb2f3db8b75f041a9e00d515c729132.jpg)
(http://tapatalk.imageshack.com/v2/14/12/04/4e52d661c13fe72a2e77996b8d51d110.jpg)
I was quite happy with the agreement between the measured values and the predicted values - and even happier about the lack of spreadsheet goofs! :D
I also implemented a crude harmonic content calculation, which wasn't perfect but actually performed much better than expected. This involved setting up several columns of sine waves with integer multiples of the fundamental frequency and variable amplitudes. These were summed, and the absolute difference between the summed waveform and the predicted vout data points were calculated. These differences were summed into a total difference, and then I used Excel's Solver plug-in to minimize the total difference by varying the amplitudes. The result was a measure of the harmonic content of the distorted signal.
The results were encouraging. Given a single sinusoid input, hard clipping apparently produces odd harmonics only. A sample result is shown below for Rin = 10k, Rg = 10k (with noise hovering somewhere between -70 to -80 dB):
Measured
1 -4.3 dB
2 noise floor (NF)
3 -27.2 dB
4 NF
5 -36.9 dB
6 NF
7 -49.4 dB
8 NF
9 -60 dB
Predicted
1 -2.3 dB
2 NF
3 -24.6 dB
4 NF
5 -34.9 dB
6 NF
7 -49.5 dB
8 NF
9 NF
The model predicted values roughly 2-3 dB too high for many lower harmonics, and got a bit worse with higher harmonics, but this was still much closer than I expected. It also correctly predicted odd harmonics only, which was another pleasant surprise.
Some things I noticed were that increasing Rg increased the level overall and shifted the spectrum toward the lower harmonics (which makes sense, since it reduces clipping distortion by reducing the effect of the diode) and that lowering Rin increased the distortion and level overall (which also makes sense, since there is more current flowing through the diode), but also that lowering Rin shifted the spectrum slightly toward the lower harmonics (which makes less sense, since more current is available to flow through the diode - although it could be that this change moves the current to a different part of the conduction knee).
What use is any of this? Well, it suggests that you can make subtle alterations to the sound of your hard clipping circuit by varying Rin (say, with a 10k-100k trimpot) and you can make much larger alterations by adding a saturation control.
One suggestion would be to use a large Rin value to increase the higher harmonic content and use a similarly large Rg pot to reduce that higher harmonic content to taste. Another would be to use a small value of Rin (say 1k) to increase volume and decrease higher harmonic content, and use a similarly small Rg to blend and refine the sound even further.
The effect of changing Rin will be subtle, but it's worth trying if you have extra trimpots lying around. Adding a saturation control is a fun mod to make and may breathe new life into your hard clipping pedal.
Quote from: ashcat_lt on March 26, 2014, 12:39:33 PM
I personally can't think of any popular designs that have diodes in the feedback path of an inverting opamp stage.
Marshall BluesBreaker / AnalogMan King of Tone :-)
Interesting. This type of clipping circuit gives the following current balance:
vin/Rin = -vout*(1/Rf + iD/vD)
And since (v-) - vout = -vout = vD, we simply have vout = -vD.
This type of clipping is exactly the same as standard hard clipping*, except that the signal is inverted. Lowering Rin will provide slightly more diode current, while inserting a blocking resistor in series with the diodes will reduce diode current and act like a blend control.
I wonder what the advantage of using an inverting clipping stage is?
EDIT: *Wrong again! :( See below for a more thoughtful answer...
Oops! I'm wrong, inverting clipping shares similarities with hard clipping but it's not the same. Did I mention..oops?
The current balance using a blocking resistor is as follows:
vin/Rin = (vD + iD*Rb)*(1/Rf + 1/(Rb + vD/iD)
When Rb = 0, this simplifies to the expression in the previous post. So far so good. What I didn't realize is that while hard clipping takes a signal and squishes the top without any additional gain, the inverting feedback configuration does apply gain. While the amplitude of a hard clipped signal must be less than or equal the input amplitude, the amplitude of this circuit's output is only bound by the voltage across the clipping path. This means that for small signals, the output can exceed the input. In practical terms this means the output can get much closer to being a square wave (albeit with a slightly rounded top) than hard clipping can:
(http://tapatalk.imageshack.com/v2/14/12/05/9cb7f9c018bff9d5e00ac74b0d6bce76.jpg)
(I inverted the input graph just to make it look nicer, in reality the input and output will have opposite polarities.)
That's all well and good. What I think is really interesting is that as the series resistance with the diodes is increased, the waveform changes from a gnarly oblong shape to a Tubescreamer shape to a pure gain shape!
(http://tapatalk.imageshack.com/v2/14/12/05/c5b4e0edeb9742834078f691c3380475.jpg)
(http://tapatalk.imageshack.com/v2/14/12/05/c5549ebc9ec7a64b1b8ee827fc102e7a.jpg)
(http://tapatalk.imageshack.com/v2/14/12/05/239cb3254a4349a3fca552fad2a8d3bb.jpg)
(http://tapatalk.imageshack.com/v2/14/12/05/72b4bdcc26f667a2457b9dfbd4106355.jpg)
Fascinating - while the ground resistor in the hard clipping lets you move from hard clipping to clean, the blocking resistor in this circuit lets you move from fuzz to Tubescreamer overdrive to simple boost! Adjusting the gain also allows you to move from fuzz to hard clipping - 4 sounds in 1!! I'm definitely going to have to build one of these :D
It's sounding to me like the difference between hard and soft clipping might be that soft clipping is actually log-conversion.
I vaguely recall seeing designs for speech-compression by multiple-diode-breakover log-conversion in old (pre-90's) issues of QST (or was it Ham Radio or 73 or CQ), back when I was first getting into ham radio. I did a quick search but turned up nothing on-topic in the first few screens. Pity; from what little I recall, the stages might have looked like Big Muff Pi stages.
I did glance at the log-conversion pages in Walter Jung's IC Op-Amp Cookbook, and there are some things to experiment with in there, using a transistor in an op-amp feedback loop to log-convert a unipolar voltage, which implies that symmetrical-about-ground conversion using a pair of transistors might be useful. I put this out for those with more time-to-play than I have at present. Maybe one of you even has the relevant QST.
single stage: http://www.diystompboxes.com/smfforum/index.php?topic=106760.msg966905#msg966905