Tube screamer "clean mix"?

[Fancy Lime]

Hi there!

I have come across the claim, that "the Tube Screamer mixes/blendes some clean signal into the clipped signal" (paraphrasing) many times. I always dismissed it as a misunderstanding, since there is nothing in the TS circuit that I would think can be described as blending or mixing of a clipped and an unclipped signal path. But now Rob has made me aware of this post of his:

https://www.diystompboxes.com/smfforum/index.php?topic=121964.msg1148986#msg1148986

And when someone like Rob says it, I'm at least listening. So what do people mean by that (including you, Rob, if you are reading this)? The soft clipping waveform produced by diodes in the negative feedback loop of a non-inverting opamp stage can of course be thought of as a linear combination of a hard clipped signal with gain x (depending on the gain setting) and an unclipped signal with gain 1. But that is just a different description of what we usually call soft clipping. So is it that what people mean? Or do they mean the fact that bass notes can sound fairly clean while the overtones are quite distorted, due to the high-mid boost of the TS gain stage?

Thanks,
Andy

[antonis]

The soft clipping waveform produced by diodes in the negative feedback loop of a non-inverting opamp stage can of course be thought of as a linear combination of a hard clipped signal with gain x (depending on the gain setting) and an unclipped signal with gain 1.

I'm not sure if i get you right, Andy but with such a configuration (diode pair in the NFB loop) lower amplitude signals are  more amplified than higher amplitude ones..
(e.g. a 10mV signal can exhibit the full x60 gain of the amplifier where a 100mV can't..)

Or, maybe, you refer on some analysis from different point of view (e.g. Fourier series waveform synthesis..)

[ElectricDruid]

I'm interested to see what people say to this too. I've never really believed that "mixes clean signal in" thing, but I've never bothered studying it closely enough to know for sure. As you say, it maybe sounds a bit like it because of the softer clipping.

[anotherjim]

Any non-inverting op-amp always gives x1 clean signal + whatever the feedback gives. It's all in the gain formula init!
(Rf/Rin)+1. That +1 is a pristine copy of whatever is on the non-inverting input.

[EBK]

Any non-inverting op-amp always gives x1 clean signal + whatever the feedback gives. It's all in the gain formula init!
(Rf/Rin)+1. That +1 is a pristine copy of whatever is on the non-inverting input.

This is correct.  There is another thread somewhere where I sneered at this clean signal thing repeatedly until I did the circuit analysis myself and saw what people were saying.  It's not a blend/mix as we might typically think about it, but if you turn the gain all the way down, you get the clean out (mostly--there is some fixed resistance in the feedback loop that prevents you from completely zeroing the non-clean), and if you turn up the gain, you get the same baseline clean signal with the clipped signal riding on top.  Basically, if you could short out the feedback loop, you'd have a unity gain clean signal.

[Fancy Lime]

The soft clipping waveform produced by diodes in the negative feedback loop of a non-inverting opamp stage can of course be thought of as a linear combination of a hard clipped signal with gain x (depending on the gain setting) and an unclipped signal with gain 1.

I'm not sure if i get you right, Andy but with such a configuration (diode pair in the NFB loop) lower amplitude signals are  more amplified than higher amplitude ones..
(e.g. a 10mV signal can exhibit the full x60 gain of the amplifier where a 100mV can't..)

Or, maybe, you refer on some analysis from different point of view (e.g. Fourier series waveform synthesis..)

Hi Antonis,

sorry, my bad. I phrased that a bit confusingly. What I mean is: The *total gain* of a signal (picture in you head a sine wave of +/- 0.1V amplitude around 0VDC) in a tube screamer type gain stage can be though of as a simple addition of two simple wave forms (let's assume that the gain pot is set to a gain of 10 = 20dbV):

1) The original signal times the gain factor until the signal swing reaches the diode forward voltage (let's say 0.7V and assume the diode has an infinitely hard knee). This is a sine wave cut hard at +/- 0.7V, exactly as if the diodes were placed to ground after the gain stage. At high gains it approaches the form of a square wave.

2) The original unamplified input signal. This sort of sits on top of the other wave, and makes the difference between hard and soft clipping arrangements.

It may be easier to explain with the gain formula for non inverting opamps:
V(out) / V(in) = 1 + ( Rf / Rg )
Where Rf is the feedback resistance and Rg the resistance to ground from the inverting input of the opamp.
V(out) - V(in) is the voltage across the feedback loop, the resistance of which is Rf. Therefore, Rf becomes zero, when V(out) - V(in) > 0.7V. That means that above this threshold, the gain equation becomes: V(out) / V(in) = 1 + ( 0 / Rg ) = 1
To get back to the earlier attempt at describing it: ( Rf / Rg ) is "wave 1)" and the 1 is "wave 2)" in the above text.
Does that make sense now?

Nevertheless, that is a rather theoretical way of looking at it. Just a deconstruction of the gain formula. Not the same thing as actually blending those two signals. Then again, the same analysis of a hard clipped amplified signal that is actually blended with a non-clipped, non-amplified original signal, would yield the same result. So I guess "clean mixed in" is a valid way of looking at this. I just find it an odd angle and a bit counter intuitive as it relates to the rest of our common terminology. If we place a variable resistor in series with the diodes, we would call it a "softness" control, or something, but not a "blend" control, would we? Yet, from the point of view discussed above, those are the same thing.

Cheers,
Andy

EDIT: and just as I was about to post it, I see that Jim and Eric beat me to the explanation with the Formula. I need to learn to type faster...

[antonis]

If we place a variable resistor in series with the diodes, we would call it a "softness" control, or something, but not a "blend" control, would we? Yet, from the point of view discussed above, those are the same thing..

I hereby declare, with the sole responsibility, not to solemnly respond to any of your queries anymore..!!! 

(you bloody-minded overthinker..) 

[nickcordle]

Gratuitous abuser / hacker / builder of NFB-clipping circuits here ...

My mental picture of it has been pretty much what Andy describes.  It's 1+(Zf/Zs) for basic non-inverting gain, but once either diode is conducting, Zf goes to zero, and you have a unity gain buffer for the extremes of the signal swing.  This totally ignores non-ideal diode behavior, but it's the way I've understood this forever, and basically jives with what I think I hear when I switch different diodes in and out.  Higher threshold => more of the signal swing which gets the mid boost, more drive required before the corners become really sharp and the higher harmonics occur.

Caveat to this is with a 9V supply and mid to high output humbuckers.  The typical op amps involved (4558, TL072, anything like that) are really seriously not rail-to-rail.  So on a loud signal with some drive applied, the "clean" parts (i.e. high and low extremes of the waveform during which the diodes are in conduction) are subject to the rail behavior of the op amp.  Again, subjectively ... you can set up extremes that bring this out.  If you build one with 9V supply rails and LEDs in the feedback loop, you'll get a lot of this behavior.  It's nastier but sometimes in a good way.  Use something rail-to-rail like a TLC2272/2262, or a charge pump to get a wider supply spread, and it kinda ... cleans up.  Subjective and hard to describe but it's plainly visible on a scope if you look at an XY plot and leave out some of the filtering to keep the phase shifts from cluttering the picture.

The other notion of clean mix in a TS is in a variant like a Maxon 820 where there's not only bipolar supply but also a second signal path parallel to the NFB drive stage, summing stage after them, and the second side of a dual-gang pot controlling the mix, loosely speaking.  No clipping in the second path, but pretty significant filtering.  A dedicated handful of people who engineer metal tones adore the 820 because it can be a bit more controlled than an 808 in front of a really raging amp like an Engl or 6505 or whatever.

[ElectricDruid]

I'm not totally convinced by this "gain equation" explanation of where the clean signal comes in. The "+1" doesn't say "add clean signal", it says "add x1 gain", which is not really the same thing.

Let's have a look at an alternative formulation of the gain equation. We know that op-amps like to have their two inputs at the same voltage, right? And that they'll make their output whatever it needs to be to make that happen? Ok, good.

So say we've got Rf = 24K, and Rg = 1K

The feedback signal at the negative input is therefore 1/25th of the output. This is a simple voltage divider:

  feedback = Rg / (Rf+Rg)

Since the op-amp likes the two inputs to be equal, the output must be x25 of the input - the gain is x25:

  gain = (Rf+Rg) / Rg

So where's the clean signal now?

This equation is clearly equivalent to the earlier one, since you can break it up into two parts and then simplify:

  (Rf+Rg) / Rg = Rf/Rg + Rg/Rg = Rf/Rg + 1

I'm not saying that anyone is wrong, since like I said, I've never really studied it to my satisfaction, but I'd be interested to hear your views based on this alternate formulation of the gain equation. The "+1" is still in there, obviously, but changing the equation seems to undermine the argument somehow.

[EBK]

The op amp "copies" the voltage at its non-inverting input to its inverting input.  In this case, that is the clean signal. The output voltage is feedback current times the feedback impedance, referenced to the voltage at the inverting input, i.e., plus the clean signal. 

[anotherjim]

The clean signal effectively exists as the DC reference level to my thinking. In the non-inverting amp, the DC reference is only wiggled by the signal. The amp doesn't know if it's reference or signal that it's obeying.

[Fancy Lime]

@Tom,

you are forgetting the diodes. Below the clipping threshold, your formulation is indeed exactly the same as mine and there is no "clean signal added" (for lack of a better word). The "clean signal" aka the "1" in the equation only shows itself above the clipping threshold, where the diodes effectively clamp Rf to 0 (only true if we look at Delta Rf, of course, not for the whole equation). And when that happens, Rf is not a ohmic resistance anymore, which is why all of the usual formulas break down anyway. To illustrate my earlier point in terms of opamp inputs: Above the clipping threshold, the opamp still needs the inverting input to follow the non-inverting. But because the effective Rf is zero for the portion of the signal swing that is above the clipping threshold of the diodes, the output change to make that so is exactly equal to the signal on the non-inverting input (assuming ideal diodes, which aren't a thing).

There's a million different ways to think about wave combinations. So much fun.

@Antonis
Aww, admit it: the overthinking is part of the fun!     It's a strange love but I'll try to learn to stop worrying and love the fuzz. Promise!

I'm wondering: is there any commercial pedal that puts a variable resistor in series with the diodes to "mix in more clean signal"? Or increase the "remnant gain above clipping", if you want to look at it from that angle instead? If I were the kind of person who thought "how can we sell a Tube Screamer to bass players", I'd probably have a hard look at that. I have been doing that sort of thing with a cap instead of a resistor, which indeed makes the most usable bass overdrive I have ever heard. Needs careful tone shaping before and after, though. I have never seen that in a commercial pedal either for that purpose. It is however exactly the Big Muff clipper arrangement, just in an opamp. Which is why reducing the feedback caps in a Big Muff to 22n or something of the kind makes a MIGHTY bass fuzz.

Cheers,
Andy

[ElectricDruid]

The clean signal effectively exists as the DC reference level to my thinking. In the non-inverting amp, the DC reference is only wiggled by the signal. The amp doesn't know if it's reference or signal that it's obeying.

Above the clipping threshold, the opamp still needs the inverting input to follow the non-inverting. But because the effective Rf is zero for the portion of the signal swing that is above the clipping threshold of the diodes, the output change to make that so is exactly equal to the signal on the non-inverting input (assuming ideal diodes, which aren't a thing).

Ok, I'm convinced. Somehow these two points together make the light come on for me. In the inverting amp, the reference is tied to a fixed ground, so the other input becomes a virtual ground. In the non-inverting amp, the situation is *completely* different, and quite a bit more interesting - like Jim says, the reference wiggles. But then there's the diodes themselves, as Andy points out, and if the diodes are on, the output must follow the input directly.

Thanks very much to everyone. I'd always just used these building blocks before now and never really thought deeply about what was going on inside the black box.

Tangentially, I do find it fascinating how many different ways there are to look at these configurations, and the numerous different ways to understand and explain them. It leaves so much room for creativity, even with something so well understood.

[Rob Strand]

I've put this sheet together showing how the clean signal gets in.   Hopefully the conclusions makes sense on their own.

[PRR]

> The "clean signal" aka the "1" in the equation only shows itself above the clipping threshold,

Not "threshold". The +1 happens when the diodes are MAXed-out.

Small signals are gained-up by typical 100:1. Over say 40dB of mid-level, signals are clipped by diodes to a fairly consistent level. But at very high levels the diodes come down to near-short, say <1k. Now the thing IS a Unity Gain Follower. Higher levels WILL copy right through undistorted.

Up to a point. Peak diode-limited level may be 0.6V, and on 9V supply peak follower output may be 3V, so there's 14dB of "clean-ish headroom" after the diode-limiting has given-up.

[Fancy Lime]

@Paul
Yes, our discussion so far assumed infinite headroom of the opamp and an ideal diode that has infinite resistance below and 0 Ohm above the threshold. In other words: an infinitely sharp knee and complete disregard for all the non-idealities of a real diode.

@Rob
Nice graphical summary on the topic! I hope the younglings find it when tinkering with their first screamer. I always found the "soft" and "hard" clipping terminology somewhat counter-intuitive in terms of how they sound. A lot of "hard clippers" sound quite soft; Klon, OCD, Blues Driver... come to mind. On the other hand, you can get very harsh "hard" sounds from a TS topology by simply tweaking some caps and resistors.

Maybe we ought to straighten out our terminology     Yeah, right, like that's gonna happen. But seriously: Is there a sort of compendium of diystompbox terms? That explains briefly and concisely what the words mean (and what they are commonly mistaken to mean) that we use? If not: Would we want that as a (heavily) moderated sticky thread? May not be the best format for such a thing, though... Opinions, anyone?

Cheers,
Andy

[ElectricDruid]

I've put this sheet together showing how the clean signal gets in.   Hopefully the conclusions makes sense on their own.

Not to me, sorry. You jump from currents to voltages and then back again in a way which baffles me. I don't understand why you're doing it or how you got there. What does this mean, for example:

Id ~IR1 = Vin/R1

"Id is roughly equal to IR1, which is equal to Vin over R1"? What's Id and where did it suddenly spring from?

Then you move to:

Vout =Vd+Vin

So now Id turned into Vd? I feel like you missed several steps out that I'm supposed to know. Did you, or am I having a slow weekend?!
And that's just the first example...
I think there's a lot of unexplained assumed knowledge on that sheet. That's ok if your readers have got the same background as you, but it's not so good for teaching someone who hasn't.

[anotherjim]

The components only produce the voltage you want as a result of current passing through? And current in a feedback diode is going to approximately equal Rin current (only difference would be input bias current in the amp and the parallel feedback resistance if any of that was significant). We don't know the Vf of the diode, but this logic lets you work it out with whatever the current ought to be.

[Rob Strand]

Not to me, sorry. You jump from currents to voltages and then back again in a way which baffles me. I don't understand why you're doing it or how you got there.

I guess tried to show the keys steps without cluttering everything up.   But I agree if you haven't analysed these things from first principles before it might appear a bit terse or cryptic.

Here's a page showing all the steps for a non-inverting amp,
https://masteringelectronicsdesign.com/how-to-derive-the-non-inverting-amplifier-transfer-function/

If I wrote out all the steps like that the key points would get lost - so I can't win!

What does this mean, for example:

Id ~IR1 = Vin/R1

"Id is roughly equal to IR1, which is equal to Vin over R1"?

Yes that's it.   It's approximate because when the diodes clip most of the current goes through the diodes and not the feedback resistor.

What's Id and where did it suddenly spring from?

When you have feedback it effectively makes the voltages the +input and -inputs of the opamp equal.  That follows from an ideal opamp having infinite gain.  (That's why I start each section with "Feedback causes...")

The circuit connects Vin to the +input.   That means the -input is at Vin.
So from that Vin appears across R1 and so the current down R1 must be Vin/R1.

If the opamp clips then this no longer holds as feedback can no longer keep the +input and -input at the same voltage.

Then you move to:

Vout =Vd+Vin

So now Id turned into Vd? I feel like you missed several steps out that I'm supposed to know. Did you, or am I having a slow weekend?!

OK, so the current flowing down R1 has to come from somewhere.  It is supplied from the opamp output terminal.  There is no current "lost" going into the opamp inputs as these are high impedance.  In order for the current to get to R1 it passes through the diodes.   So in essence a non-inverting amplifier forces a current (=Vin/R1) down the feedback network regardless what the network is (diodes, resistors).

If we pass a current through the diode there will be a voltage drop across it.  The actual voltage depends on the diode, what every it is we call that Vd.   So the Vd comes from passing a current Id through the diodes.

So the last step is adding up the voltages.   The voltage at the output is made up of two parts:  the voltage across R1, which is Vin,  and the voltage across the diodes, which is Vd.   These are in series so the voltage drops add.   So that addition is where we get,

Vout =Vd+Vin

If you go through these steps for a simple non-inverting gain stages you can show the gain is A=(1+R2/R1)=(R1+R2)/R1.     

Many feedback circuits can be analysed by:  recognizing the +input and the - inputs are at the same voltage, recognizing what sets the currents, seeing what currents must be equal,  then adding up voltage drops (making sure you get the signs right).   It's much easier to do it this way than to memorize maths like in the link I posted above and it works for any circuit.

So for the inverting case,
- The -input is at 0V since the +input is at 0V
- The input current is Vin /R1
- The input current must flow through the feedback network (since the current cannot be supplied or lost at the -input)
- The output voltage is just the voltage drop across the feedback network as the side of the feedback network connecting to the -input is pinned to 0V.

And that's just the first example...
I think there's a lot of unexplained assumed knowledge on that sheet. That's ok if your readers have got the same background as you, but it's not so good for teaching someone who hasn't.

It's a fair statement.   I 100% agree.   Unfortunately when you explain a "why" question you have to start somewhere up the knowledge tree.    No doubt that causes some people grief and other's say "that's obvious".

[ElectricDruid]

Thanks Rob. I'd probably have been happy if you'd just said somewhere on your sheet that Id was the diode current. We assume stuff like that is obvious, but it's not, and the more "beginner" you are, the less obvious it is. If you're introducing a variable in an equation, introduce it ("Hi! This is Id, they're the current through the feedback diodes!"?). If you don't, it just appears.

I admit a bit of faux naivety in my last reply, but I was making a point. I think your point that if you spell everything out then the key points get lost in the detail is also true though, so like you say, perhaps you can't win. Sigh. There must be a way somehow, and I don't think it's just a question of balance between detail and abstraction, but somehow a question of which parts need each.