hard vs soft clipping.

[aziltz]

Is the difference in the type of diode or how its setup?

http://users.chariot.net.au/~gmarts/ampovdrv.htm

this site seems to state that germanium diodes or led's in a shunt to ground, or ANY diode type in the feedback loop produce soft clipping, while hard clipping is silicon diodes in a shunt to ground.

[GibsonGM]

ANY diodes in shunt to ground will give hard clipping once the threshold (Vf) has been exceeded.  ANY diodes in in the feedback loop will give a softer clipping.  So the 'type' is determined by where they are, and the 'harshness' by what kind of diode, so to speak.

Think of it more WHERE the diodes are.  If they're in the feedback loop, they'll affect gain and have sort of a 'delayed reaction' (for lack of a better term - they will affect the gain loop), and have more of a rounded shoulder when clipping.

At the output, they are like binary "ON/OFF" devices, giving sharper, more square waves.

The type of diode will influence WHEN the clipping occurs (you have to exceed Vf to conduct, when the diode turns on...).  The lower the Vf, the harder and faster the diode will clip.  So, Ge will clip faster than Si, and LEDs will clip even later.  There are many topics to be searched about the tones they deliver due to the Vf  ;o) 

[Caferacernoc]

"At the output, they are like binary "ON/OFF" devices, giving sharper, more square waves."

I have to disagree a bit with that. In a diode to ground scenario I think Germaniums still sound softer than say, led's. They conduct sooner because of the lower voltage threshold, but it's a softer, mushier, distortion.You can also put a resistor in series with the diodes to soften the "knee" where they conduct. It also depends on whether or not there is a resistor right before the clipping diodes. Looking at a DOD 250 or MXR Dist+ type schematic there is a 10k resistor in between the output coupling cap and the clipping diodes. If that resistor is lowered in value or removed the clipping "sharpness" goes down. Like in a Electra style distortion like the Trotsky Drive or COT50.

MXR style hard clipper:
http://www.schematicheaven.com/effects/dod250_overdrive.pdf

Electra, Trotsky, COT50 style softer clipper:
http://3.bp.blogspot.com/_y4AYtND8Hz8/R72W3KU_osI/AAAAAAAAAH4/pwKc0P3ST5c/s1600-h/cot50briggsmods.gif

This article explains some of the difference:
http://www.headwize.com/projects/limiter_prj.htm

But I agree, all things being equal, diodes in the feedback loop clip softer than diode to ground.

[R.G.]

Imagine you have a pair of diodes back to back, fed by a signal source through a resistor, and going to ground.

The diodes have a no-conduction region up to Vdiode, when they start conducting, and a knee, a small region of voltage (Vknee) where the diode's incremental resistance goes from effectively infinity to a couple of ohms. How do you express the sharpness of clipping?

1) it depends on the size of the signal; signals less than Vdiode don't clip at all
2) it depends on the relative sizes of Vdiode and Vknee; signals less than Vdiode +Vknee never get fully limited, so never fully "flat top". The corners of the clipped waveform are rounded, not sharply squared; sharp squares require much higher harmonics to make them sharp. In germanium, Vdiode and Vknee are closer in size than in silicon, hence the reputation for germanium being softer
3) it depends on the size of the limiting resistor in front of the diodes; with high resistances, the available currents are limited , and so the voltage at the diodes limits more sharply. With lower drive resistances, the diodes are more easily driven further into the knee and it's a bit softer
4) it depends on whether there is some additional resistance in series with the diodes or not. If you put additional resistance in series with the diodes, it effectively adds back in some of the signal undistorted, and smooths out the sharper clipping. This came from Kevin O'Connor's TUT, I posted to the forums, and Jack Orman used that as a takeoff for his "Warp Control" clippers by effectively using a different resistor for each diode.
5) This is the one that's not widely recognized. It depends on how fast the signal transits through the knee, which is most easily seen by imagining that a sine wave is applied to the clipping circuit. A sine wave's sides are quite steep near 0V, but the sides get steadily more shallowly sloped as you get toward the top, where it's flat. If the signal tries to zip through the entire knee region in a portion of the wave's period that's small compared to the period, the corner looks sharp for almost all corners. Put another way, if the applied signal would go to Vtop in the absence of the clipper, then the ratio of Vtop to Vdiode+Vknee determines how little of the waveform's period is spent in the knee. If this is a lot of the period, then the wave is softly clipped. If it's a little, the wave is hard clipped.

[Mark Hammer]

2) it depends on the relative sizes of Vdiode and Vknee; signals less than Vdiode +Vknee never get fully limited, so never fully "flat top". The corners of the clipped waveform are rounded, not sharply squared; sharp squares require much higher harmonics to make them sharp. In germanium, Vdiode and Vknee are closer in size than in silicon, hence the reputation for germanium being softer
3) it depends on the size of the limiting resistor in front of the diodes; with high resistances, the available currents are limited , and so the voltage at the diodes limits more sharply. With lower drive resistances, the diodes are more easily driven further into the knee and it's a bit softer
5) This is the one that's not widely recognized. It depends on how fast the signal transits through the knee, which is most easily seen by imagining that a sine wave is applied to the clipping circuit. A sine wave's sides are quite steep near 0V, but the sides get steadily more shallowly sloped as you get toward the top, where it's flat. If the signal tries to zip through the entire knee region in a portion of the wave's period that's small compared to the period, the corner looks sharp for almost all corners. Put another way, if the applied signal would go to Vtop in the absence of the clipper, then the ratio of Vtop to Vdiode+Vknee determines how little of the waveform's period is spent in the knee. If this is a lot of the period, then the wave is softly clipped. If it's a little, the wave is hard clipped.

Not to cut you off at the knees or anything , but the "knee" concept is one the merits further consideration.  Especially since it is a concept which is inextricably linked with time, and because of that, with frequency.

The key question, for me, is whether the time it takes to transition from no conduction to complete unobstructive conduction is one which has some impact on the audible spectrum.  If the difference between knee 1 and knee 2 is, say, 30 nanoseconds, then I would think you simply aren't going to hear it.  The question that arises out of THAT musing is what the actual critical difference in knee time/duration would have to be in order to be able to hear it.

On another matter, point #3 needs some further clarification.  I have seen a number of circuits over the years, where small-value resistors were used to produce "soft clipping".  For example, in this circuit - http://hammer.ampage.org/files/PocketRockit.PDF - you can see a pair of 68R resistors used with a diode pair to ground to do that very thing.

At one level, or rather in some circumstances, the use of a smaller resistance leading up to the diode ought to mean, intuitively, that the clipping is "harder" simply because a smaller resistance will have an impact on the overall output of the circuit.  The MXR Distortion+ and DOD250 both have a 10k resistor prior to the diode pair, but because the resistor provides some attenuation in conjunction with the resistance that follows it (the pot), the actual level passing by the diodes is reduced.  If one drops that 10k resistor to 2k2 or lower, you can bet the pedal will seem to be "harder".

But that is a misinterpretation.  What "softer" means, is not that the level will be lower, nor that the clipping, when it takes places, will somehow be gentler.  Rather, what soft means is that the tendency for the diode to conduct is spread over a broader range of signal levels.  It goes from on/off to something like off/sorta-on/on.  The difference is between punctate and graded, rather than hard/soft.

Silicon diodes, if I understand the underpinnings here, were historically construed as an improvement over germanium diodes, simply because they could retain the on/off characteristic over a wider range of circumstances.  In contrast, germaniums could play the in-between zone more, which is not a virtue for switching purposes (the primary target application of diodes) regardless of what it offers for fuzzboxes.

So, I'm wondering.  Say one had a succession of cascaded Ge diode combinations to ground, each with something similar to the PocketRockit's 68R/68R pair.  Imagine three 68R resistors in series, and after each is another small value resistor and some diodes going to ground.  The first set is a 3+3 back to back sextet, the second a 2+2 quartet, and the last a 1+1 pair.  Whatever gets by the first set is still robust enough to be altered by the second, and the same thing goes for the 2nd and 3rd.  Because the series resistances are small, there is still plenty of current for the "extended knee" thing to occur.  Because they're Ge diodes, the absolute amplitude of signal needed to reach the critical voltage is still pretty modest and easy to achieve with a 9v battery.

Do we hear something like that as generally softer because of the application of 3 separate knees, or as harder because it's been clipped thrice?

[R.G.]

Not to cut you off at the knees or anything , but the "knee" concept is one the merits further consideration.  Especially since it is a concept which is inextricably linked with time, and because of that, with frequency.

Actually, although it seems to be, it's not time related. It's percent-of-period related. That particular bit of subject change took me about three days to figure out. You know what actually started me off on this? Some guy who shall remain nameless said that the only waveform that looks like itself when clipped is a square wave.  That's true, but much deeper in meaning than he may have thought. 

Leaving human ear response aside for a moment, and looking at produced harmonics, the degree of hardness or softness of clipping seems to be related to what harmonics are produced and their relative amounts. I think we agree that a waveform composed mainly of low order (sixth and lower perhaps) harmonics is softly clipped, while a wave form with large amounts of harmonics seventh and above will sound harsher.

If so, can we also agree that the waveform itself, independent of the actual frequency, is the determinant of what harmonics had to be used to build that particular shape up, Fourier-style? And if we accept that, then the determiner of harmonic structure is the actual wave shape itself. Normalizing all aspects of a wave shape to the wave period produces a harmonic string that is constant. We can then do the Fourier analysis on a 1Hz wave of the correct shape, get the relative sizes of the harmonics, and then convert that to the same-shaped wave at any frequency by merely multiplying the frequencies of the fundamental and all harmonics by the ratio of the desired frequency to the 1Hz prototype. Which is another long winded way of saying wave shape and harmonic spectrum have a 1:1 mapping in both directions. Spectrum is merely the frequency domain representation of the time domain waveform.

The key question, for me, is whether the time it takes to transition from no conduction to complete unobstructive conduction is one which has some impact on the audible spectrum.  If the difference between knee 1 and knee 2 is, say, 30 nanoseconds, then I would think you simply aren't going to hear it.  The question that arises out of THAT musing is what the actual critical difference in knee time/duration would have to be in order to be able to hear it.

That is true. And it's another way of saying "how fast did that waveform shoot through the knee?".  In effect, it amounts to impressing the human hearing response onto the spectrum (frequency domain) obtained from waveform-> spectrum mapping.

The time difference between knee1 and knee2 depends on waveshape, right? A 1Hz sine impressed on two different knees will have a much longer difference IN TIME between the two than a 1kHz sine impressed on the same two knees. It's a thousand times longer difference for the 1Hz wave than for the 1kHz, right? And you're so very close to the idea that hit me. Think of a sine wave. The slope of the sine wave depends not on what time it is, but what phase it is - what fraction of the sine's period independent of frequency. The slope of a sine wave is Vpeak*COS(p) where p is the phase angle. The slope goes from 1X at p=zero to zero at p=90 degrees. So the amount of time the waveform spends in any knee is not fixed, it's directly related to the frequency. High frequencies whip through the knees faster on an absolute time basis. But the fraction of the period it spends in a knee is fixed by the waveform, knee size relative to waveform and period.

More simply illustrated, take a sine wave of 0.1V peak through some "source resistance" resistor R. We have a pair of clipping diodes which start conducting at 0.5V and smoothly increase conduction to 1Siemen (i.e. 1A/V) at 0.7V, the actual change being an exponential function of the applied voltage - in other words, silicon.  Where does the 0.1V signal clip? It doesn't, and the time in the knee is zero. If we turn up the gain, the peak starts getting clipped at 0.5V. It flattens to a degree determinied on and instant by instant basis by the source impedance and the diode impedance. As we turn up the gain more, the sine wave top intrudes more and more into the knee, and gets flatter and flatter, but still has some curve. Above 0.7V, we'll say that it's hard clipped, flat on top.

The time the sine wave spends in the knee is dependent on the amplitude and the frequency. But the fraction of its period it spends in the knee is dependent only on amplitude, not time at all. And the waveshape is constant for all frequencies of the applied sine wave as long as they spend the same fraction of time inside the knee. The fraction of the period spent traversing the knee from no conduction to flat topped always shows as a curve, the result of the sine function impressed on the exponential of the knee.

For a sine with enough peak level ahead of the source impedance to cause the clipped Vpeak=0.7V, the entire top of the waveform is curved. That amounts to saying "only lower order harmonics are produced." As we increase the sine wave even more, there is a larger and larger part of the resultant wave which is at 0.7V, and a smaller and smaller fraction of time spent traversing the knee. If we tried to put a 100V sine wave into this setup, the output at the diodes would be clipped at 0.7V, OK, but the fraction of time spent traversing the knee would be miniscule, only milli-degrees of the period. And that would make for a very sharp, sudden transition between no clipping at all, and full clipping. And that would make for many more high order harmonics and much harsher sound.

I agree with you - two knees with only 30nS of time difference spent in them would not be possible to hear, if only because the only possible differences there are to frequencies of about 2x30nS in the Fourier structure - out at about 167MHz. But two knees with 30 degrees difference of the periods the signal spends moving through the knees is going to be huge - if the frequencies involved produce harmonics we can hear. If we can't hear the harmonics, we can't hear the difference, obviously.

At one level, or rather in some circumstances, the use of a smaller resistance leading up to the diode ought to mean, intuitively, that the clipping is "harder" simply because a smaller resistance will have an impact on the overall output of the circuit.  The MXR Distortion+ and DOD250 both have a 10k resistor prior to the diode pair, but because the resistor provides some attenuation in conjunction with the resistance that follows it (the pot), the actual level passing by the diodes is reduced.  If one drops that 10k resistor to 2k2 or lower, you can bet the pedal will seem to be "harder".

But that is a misinterpretation.  What "softer" means, is not that the level will be lower, nor that the clipping, when it takes places, will somehow be gentler.  Rather, what soft means is that the tendency for the diode to conduct is spread over a broader range of signal levels.  It goes from on/off to something like off/sorta-on/on.  The difference is between punctate and graded, rather than hard/soft.

Yes - what I would call a knee with a large radius of curvature and one with a small radius of curvature. I can see these graphs and illustrations in my head. Why aren't you watching them??   

So, I'm wondering.  Say one had a succession of cascaded Ge diode combinations to ground, each with something similar to the PocketRockit's 68R/68R pair.  Imagine three 68R resistors in series, and after each is another small value resistor and some diodes going to ground.  The first set is a 3+3 back to back sextet, the second a 2+2 quartet, and the last a 1+1 pair.  Whatever gets by the first set is still robust enough to be altered by the second, and the same thing goes for the 2nd and 3rd.  Because the series resistances are small, there is still plenty of current for the "extended knee" thing to occur.  Because they're Ge diodes, the absolute amplitude of signal needed to reach the critical voltage is still pretty modest and easy to achieve with a 9v battery.

Do we hear something like that as generally softer because of the application of 3 separate knees, or as harder because it's been clipped thrice?

Shush, Mark. That's a secret. 

There is another metric of clipping sharpness - the ratio of the pre-knee conduction voltage to the in-knee conduction voltage.  If the significantly-curved region of a germanium diode is 0.1V and the pre-conduction voltage is 0.2V (bearing in mind that in fact it's all one exponential curve, not two, and that we're picking out abstractions and approximations here), then the knee is half the size of the pre-conduction region. That will sound softer for a sine wave of Vpeak proportionately the same size as proportionately-larger sine impressed on a silicon diode where the knee is 0.2V  compared to 0.5V preconduction. 

EEs have been building waveshapers for decades that use diodes connected to various voltages and in series with various resistors to shape triangle waves into other things. Using diodes and resistors to set the thresholds and amounts is how that started.

The waveshape is the harmonic spectrum. How we get it doesn't matter, only means to an end. But understanding the means is a biggie. I explain all this in great detail in this book I'm writing... 

[Mark Hammer]

I explain all this in great detail in this book I'm writing... 

[R.G.]

You put those red faces right back in your briefcase, young man! Or I'll show you some of my own.

[frank_p]

I don't know the pertinence of my intervention but this talking reminds me of those two things:  (If any would like to elaborate or explain, this was floating in my head for a while, and I did not studied it yet):

The limiter in the Dogzilla (there are some graphs, always helpfull...):
http://www.dogstar.dantimax.dk/tubestuf/dzart-6.htm

Diode ladders in Pritchard Amps (sectioning the response curve with linear segments with diode ladders to have a relatively rounded response curve) (Patent image document: page two):
http://patimg1.uspto.gov/.piw?docid=US005734725&PageNum=2&IDKey=592B766537CE&HomeUrl=http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO1%2526Sect2=HITOFF%2526d=PALL%2526p=1%2526u=%25252Fnetahtml%25252FPTO%25252Fsrchnum.htm%2526r=1%2526f=G%2526l=50%2526s1=5,734,725.PN.%2526OS=PN/5,734,725%2526RS=PN/5,734,725

For viewing in explorer:
http://www.alternatiff.com/

Hope I am not too lost... 

[GibsonGM]

Right, Cafer, Ge's can sound a bit 'smoother', and will conduct much sooner (why they tend to be found in 'metal pedals' - they will always be conducting).  An LED in the same spot, as a comparison, is going to be 'grittier' due to its Vf and realm of conduction.  I was merely giving a qualitative description (the realm I tend to stay in most of the time; my fiancee forces me to be more about 'how does it sound' than the mathematical interpretations of the electronic devices I build!).  There is always more to the story, ha ha.  I think R.G. and Mark Hammer have nailed down the 'other side' of it quite thoroughly!!

Just stating a general rule of thumb which applies to the most common distortions/OD's you'll see out there (TS, Dist+ etc).  Smoothing caps can also do a lot.

[R.G.]

The limiter in the Dogzilla (there are some graphs, always helpfull...):
Diode ladders in Pritchard Amps (sectioning the response curve with linear segments with diode ladders to have a relatively rounded response curve) (Patent image document: page two):
...
Hope I am not too lost... 

Nope, not lost at all. Those are two ways of approaching the waveshaping I mentioned. I'm once again amazed that the patent office seems to award patents for rewriting old EE text books, at least in part, but it does.  Diode and transistor waveshapers are/were pretty common items until table lookup and direct synthesis happened shortly after uCs got cheap enough to displace them. I guess maybe Pritchard got that under the theory of a new combination of older, known elements.

In any case, the idea both places is to use combinations of diodes, transistors, and resistors to synthesize a clipping characteristic that's mostly knee and much less pre-conduction "Vdiode". They both fall in with this way of thinking.

[WGTP]

I love it when you guys talk dirty (clipping).  Here are a variety of clippers I assembled and posted a while back.  Various attempts at circuits for addressing the clipping issue (are we still trying to be tube like or just sound cool?).  I also posted a distortion with 4 clipping stages because IMHO, Once is Not Enough.  With multiple clipping stages you get a mushy compression that I really like.

http://www.aronnelson.com/gallery/main.php/v/WGTP/Diagram.GIF.html?g2_imageViewsIndex=1

[R.G.]

Yeah, I'm a fan of multiple distortion stages, going back to pre-web days when we could only talk dirty on the uusenet groups.

[Mark Hammer]

So I'm "cleansing my mental palate" at work and thinking about things other than staffing transactions for a moment, when it occurs to me:  "Is this current-vs-diode-knee thing the basis for op-amp preferences in the Tube Screamer?"

I know Jack Orman had suggested at one time that the properties of different op-amps in the classis TS form-factor could be mimicked to some extent by simply sticking a small-value resistor (<1k) in series with the diode pair.  So maybe the amount of current the op-amp is delivering is the basis of some qualitative aspects of the clipping produced.  All of this, of course, depends on diode knees being perceptible, but assuming they are, maybe there is something to the op-amp choice/debate, based on the diode knee produced by different op-amps and their respective ability to pump out current.
 

[R.G.]

Mark, this will sound unrelated, but it's not. Will you remind me to send you one of my new DVDs when I get them in?

Pester me in about a month.

[Mark Hammer]

I will take it on faith, and will pester you within an inch of your life....in about a month.

[brett]

Hi

then the ratio of Vtop to Vdiode+Vknee determines how little of the waveform's period is spent in the knee

Isn't this the reason why Vdiode and Vknee are important? As RG noted way up the top, if the signal is below Vdiode, the signal isn't clipped at all, and if the signal is a sine wave with Vpeak of 100x Vdiode, the signal is clipped to almost a rectangular wave.

As I've said before, the difference between diodes is in Vdiode.  Vknee is almost identical for all of the diodes I've measured (1N34a, 1N60, 1N400X, 1N4148, LEDs, Schottkys, transistor junctions, etc).  Therefore, the ratio of Vdiode to Vknee determines the "hardness" of clipping.  LEDs (high Vdiode, ratio of about 10) clip harder than Ge (low Vdiode , ratio of about 1).  Putting the diode in the feedback loop of an op amp magnifies Vknee and minimises Vdiode (because the knee is "detected" so effectively by the op amp, which reduces gain dramatically as the voltage rises into the knee).
cheers

[earthtonesaudio]

Seems like the voltage on the "other" side of the diode makes a big difference.  If capacitively coupled, I suppose it can be any low-impedance reference, like ground.  But if you're direct-coupled, the diodes should go to the same voltage reference as the signal (1/2 supply for example) like in the OCD.

I might be wrong but I would think that if you have an op-amp supplied by a 9V and with a reference voltage of 4.5V, then if the diodes are DC coupled from the output to ground, one will never turn on, and the other will never turn off.

[R.G.]

Seems like the voltage on the "other" side of the diode makes a big difference.  If capacitively coupled, I suppose it can be any low-impedance reference, like ground.  But if you're direct-coupled, the diodes should go to the same voltage reference as the signal (1/2 supply for example) like in the OCD.

I might be wrong but I would think that if you have an op-amp supplied by a 9V and with a reference voltage of 4.5V, then if the diodes are DC coupled from the output to ground, one will never turn on, and the other will never turn off.

It does make a difference, in that any DC voltage on the "ground" side of the diodes merely adds to or subtracts from the turn on voltage. The size of the knee, and the forward conduction voltage are unaffected, only the absolute voltage where the diode starts to conduct.

For instance. If I have a pair of silicon diodes with Vdiode = 0.45 and Vknee = 0.25, and they are connected to ground at one end, they start conducting at +/-0.45V. If I connect them to +1V, the start conducting at +1.45, +0.55V. If I connect them to -2V, they start conducting at -2.45 and -1.55V.

It does not change the conduction process at all, excepting for offsetting the conduction/limiting to a different DC voltage. You can even put the tails of the diodes at different DC voltages, and they will conduct when the voltage difference across each diode is the proper polarity and direction. This is common in older waveshaping circuits.

However, since the name of the game is to make the clipping softer in most cases, or to at least manipulate it, you get best results if you don't move them around to force them to conduct at odd voltages.

The one exception is hidden in that last "You can even..." of mine. You can even move the tails of the diodes individually to bias voltages that make them conduct more easily. That has the effect of subtracting a fixed voltage from the Vdiode, and leaving the knee curve unaffected. You can make it all knee this way if you're careful with your design. Of course, this requires that you also feed it a smaller signal if you don't want massive clipping, because lowering the cutin voltage for clipping makes any overdrive that gets above the knee's final voltage be more harshly clipped. I've messed with this abit. It does not respond all that well to the oft-repeated mantra of the headbanging guitarist crying for "more gain, more gain".