Passive clipping circuits - with Schottky diodes

[JohnH]

I was asked to make a post about using Schottky diodes to make passive clipping circuits. These diodes have very low ‘forward’ voltages, so they are able to achieve some tone-altering clipping action at the low voltages that are generated by guitar pickups. Also, at even lower voltages, they are quite leaky which means that they switch on softly and affect the tone right through the decay phase.

Commercially, a similar device called the ‘Black Ice’ can by bought from StewMac, and here is an earlier thread about it
Black Ice

Here is how they wire them in, which also applies to home-built versions described below:
Black Ice – installation

Don’t expect too much from them!, but I tried a few arrangements of my own with schottky diodes in some guitars.

Ideally, one, two or three diodes are installed with a pot, in exactly the same way as a tone control, substituting the tone cap for the diodes. Here are some arrangements, with the Schottky circuits before the volume pot:



This is a clip that I made a few years ago, with one diode and about 20k in series with it, on a Strat, direct to pc:

Schottky test1

Here is one that I have just done, using an LP. This has a back to back pair and no series resistance, miced from an amp:

Schottky test2

The first part of this clip is clean, then the diodes come in, but with a no-gain JFET buffer before the diodes. Then the same thing with no buffer. You can hear how the diodes cut into the signal of the un-buffered pickups.

To try these, just wire them across your output outside of the guitar, to see if you think there is mileage in them. I think they are interesting, and they are possibly useful but best after an active output IMO.
YMMV!

Regards

John

[newey]

John-

Thanks for putting that up! 1+!

At some point, I think we sort of reneged on the rule about not asking questions in the Schematics board, so long as they are pertinent to the schematics at hand. So, at the risk of violating my own wholly-unworkable rule, let me ask a few:

What type of Schottky diodes did you use? Did you experiment with different ones?

On your 2-and 3-diode diagrams, you show diodes in parallel wired in "antiparallel", yet the 2 wired in series in the 3- diode plan are in regular series. Would there be any point in wiring those in "antiseries", and is there a reason for the choices you made in doing it the way you did?

How about a 3-way switch to switch between two different diodes, choosing both in the center position?

[JohnH]

newey - I only tried with one type, which was 1N5819, just because that was what was available. Others may have different characteristics. Germanium diodes can also be put into the mix, thiough standard silicon are too insensitive. A switched arrrangement could certainly be used ti try differnt combinations, such as in flaterics recent TS diode switching mod.

cheers
John

[sumgai]

newey,

Sometimes a reference topic like this can benefit from Q&A, so I don't think we're doing a bad thing here....

Schottky diodes still work in the most basic mode we know for a diode - they conduct in only one direction. Thus, two of them in parallel will yield no benefit - there's still only conduction in one direction, and the voltage drop doesn't change. (This configuration would be useful only in high-power situations.) Two in anti-parallel will each conduct on one-half of the full cycle, so both halves of the waveform get through, albeit with one diode's worth of voltage drop on each half. This sounds different than modifying only one-half of the waveform.

Two in series will conduct in the same direction, but with twice the voltage drop. Again, a noticble effect on the tone. However, two in anti-series will never conduct at all - each half cycle is blocked by one or the other diode.

And finally, using multiples, either symmetrically or asymmetrically, will introduce further voltage drops, but being "soft", they tend to build up their effect on tone gradually, and not harshly.

HTH




sumgai

[newey]

One more question, then. (I raised this in a PM to John, asking him to post this here, since the search function no longer seemed to retrieve his original post):

Could one deal with the drop-off in output with the diodes by attenuating the clean signal in a roughly equal amount, using an appropriate resistor? Thus, the output wouldn't change much flipping the diodes on or off. Could the approximate value required be calculated?

[ashcatlt]

I haven't done a whole lot of experimenting with this type of thing, but my understanding is that the one diode, and 2 + 1 diodes will give a more asymmetrical clipping, which should encourage more even order harmonics, and be a bit more tube-like. More like warmth, or overdrive, than harsh distortion.

Clipping the waveform reduces overall dynamic range by reducing the ratio between peak and RMS (average, or "continuous") output. A resistive volume adjustment (voltage divider) will reduce both peak and RMS voltages together, but retains the overall dynamic range. I guess it depends which voltage you're talking about. Most of us, when we're talking about volume, are talking about RMS.

This same basic question has been asked in stompboxes forums: Is it possible to gang the gain and volume pots on an OD/dist box so that output volume remains the same as we increase the overall distortion.

The answer is, it's likely doable, but is not so much simple to calculate, and probably better dialed in by ear.

[JohnH]

Feb 12, 2010 23:58:38 GMT -5 newey said:

Could one deal with the drop-off in output with the diodes by attenuating the clean signal in a roughly equal amount, using an appropriate resistor? Thus, the output wouldn't change much flipping the diodes on or off. Could the approximate value required be calculated?

Yes you could, but I think it's a shame to have to cut down not only the volume, but also the clear tone that you get from a pickup at full volume. I think these passive cliipers are best used with a significant resistor in series, so they dont cut down volume much.

I basically got there, then I realized that these things are so much better with a bit of active help, and that given that, there are better gain and drive sounds to be found, and that these are best put in a pedal - so I have stopped building distortion circuits into guitars and just keep them clean!

But the best purely passive sounds Ive found this way are from low impedance settings, such a parallel coil wiring of not very hot humbuckers.

However, just the other day I came across a maker down here who builds passive stomp boxes - check this out:

Cave Passive Effects

Listen to the 'Mini Miff' and '60's breakup' , which I think probably work somewhat like we are discussing here, and sound a bit similar to me.

John

[ashcatlt]

I was reviewing this thread today for BAY's thing with the strap button hole, and had a thought.

Above, JohnH wrote something to the effect that this would be wired like a tone control, only with the diodes in place of the cap.

Now, I don't know about Shottky's, but one thing about diode clipping in general is that they often create some pretty harsh and nasty high frequency harmonics. In a purely passive configuration, these might be reduced by cable capacitance and pickup inductance and all the rest of that stuff, but if this is tacked on after an active stage somewhere (like in BAY's acoustic/electric) this won't help quite so much.

So I'm thinking that we might leave a cap there in parallel with the diodes. Wire it exactly like a tone control, only with diodes in parallel with the cap. The cap value would require some experimentation, but it could help quite a deal in getting a more usable overdrive sound.

Edit - Since nobody's replied, I'll add a second thought here:

A No-Load pot, or one of those switched pots might be warranted for this application, to take the clipping section completely out of the circuit when not wanted.

[newey]

So I'm thinking that we might leave a cap there in parallel with the diodes. Wire it exactly like a tone control, only with diodes in parallel with the cap. The cap value would require some experimentation, but it could help quite a deal in getting a more usable overdrive sound.

Better yet, switch the diodes in/out, in parallel with the cap, leaving a regular tone control with the cap when the diodes are switched out of circuit. Whether the cap value that works best with the diodes would be a value of interest for the tone control is a question, of course- some compromise value might be needed.

JohnH's point about this being perhaps better in a stompbox, outboard of the guitar, is sensible for most real-world situations. I was thinking of my on-again, off-again travel guitar- it might be nice to have an onboard distortion so no need to cart a pedal around and no batteries to leak when the thing sits in the case for months on end.

Might be just the thing for a bit of grit in the hotel room without insane amp volumes in play. I think some further experimentation is in order. JohnH's does sound a bit harsh, but I think he's barely scratched the surface of what might be doable.

Of course, John's with the buffer does sound quite a bit better, but again, if you're going to go active, it sort of defeats my purpose FWIW. I could live with the reduced output if the distortion could be made "smoother" (whatever that means . . . )

[Yew]

Let me see if i got this correct

Diodes only let current flow in one direction, Right

When a current goes through a diode, all the frequancies above volume a, get reduced to volume B. (this is compression?)

Using a single diode, the tone gets harsh as the clipping will cut out a lot of the frequancies, however using teo in parallel, the amount of clipping done by each diode is reduced (as the voltage through each is halved,

If you could put a capacitor in parallel with the diode pair, you could have clipping on the lower frequancies, but maintain reasonably clean treble frequancies

if you had a harsh treble, you could put the capacitor in series with the diode pair, giving you a more compressed to end (smoother and whatnot) that might givw the effect of a reasonably clean rhythm tone, and a more compressed rhythm tone, (but... this would make the tone fatter right


ive gotta go, ill finish this later

[BlackAngusYoung]

Sept 1, 2010 8:08:01 GMT -5 Yew said:

Let me see if i got this correct...

I hope you do because your explanation was more understandable to me.
And reading it probably means that when I read the other explanations again to get all the details, I'll understand them better. So, thanks for the notes.

What you're describing as a hopeful outcome sounds pretty good if that's how it works.

[Yew]

If im right, I think i can see many new switches on my guitar very soon

[ashcatlt]

Nope, sorry. Yew’ve gone of the rails a little bit here.

It is true that a diode will only conduct in one direction – when one end is more positive than the other – and then only when the voltage potential between the two ends is above a certain threshold, which we call it’s Forward Voltage Drop.

In this discussion, we’re basically talking about putting diodes across the jack, between the tip and the sleeve – what we often call “hot” and “ground”. But that ain’t exactly true. Sometimes the Tip is more positive than the Sleeve, but sometimes the Sleeve is more positive than the Tip. It goes back and forth at a fairly rapid rate.

If we place a diode across the jack in such a way that it conducts when the Tip is more positive than the Sleeve, it will clip off the waveform when this condition is true. Let’s call that the “top” of the wave. The “top” gets flattened out, while the “bottom” goes through without any alteration. This is what we call Asymmetrical clipping. It tends to accentuate the even-order harmonics, and is often said to sound fairly natural and pleasant.

But sometimes (quite often actually) we want more. So we put a second diode in parallel with the first, pointing the other direction. This one clips the “bottom” of the wave, when the Sleeve is more positive than the Tip. This creates a far more symmetrical output wave, accentuates both even- and odd-order harmonics. It’s not necessarily harsher, but is not as natural sounding, coming across more synthetic and dissonant.

So far we haven’t really talked about frequency, but I did talk about harmonics. These harmonics can be thought of as copies of the input signal, which are shifted upward by some whole number of octaves.

Let’s just say that the highest frequency we’ve got going into the clipper is the highest frequency we want to hear from a guitar. Anything higher starts to sound harsh and nasty and fizzy. Well, now the clipper makes copies of the input which are an octave, or two, or more higher than that highest frequency that we want to hear. That’s what makes the diode clipping sound nasty. To un-nastify this output signal, we need to cut some of those higher harmonics back out.

A capacitor only passes signals which have a frequency above some threshold. Well, it’s not really a brick wall, and this is an over-simplification, but let’s go with it.

If we put a cap in parallel with our diodes – between the Tip and sleeve of the Jack, remember? - the frequencies above that cutoff frequency will end up completing their circuit without having to go all the way through the amplifier. They won’t get amplified, and we won’t hear them. This is the exact same arrangement as our standard Treble Cut Tone Control.

If, on the other hand, we put a cap in series with the diodes, so that it goes Tip>diodes>cap>Sleeve, then only the frequencies which can pass the cap will have any chance of “opening” the diode. Lower frequencies will ignore the entire thing, and won’t get clipped. Only the highest frequencies will get distorted, and we’ll get those octave+ copies of those highest frequencies on the output. This is not usually what we want to hear. It is similar to the Aphex Aural Exciter process, but that’s a different deal altogether.

I hope that helps.

[sumgai]

Hold on there, Bobalooie!

Sept 1, 2010 10:50:56 GMT -5 ashcatlt said:

If we place a diode across the jack in such a way that it conducts when the Tip is more positive than the Sleeve, it will clip off the waveform when this condition is true. Let’s call that the “top” of the wave. The “top” gets flattened out, while the “bottom” goes through without any alteration.

That's the exact opposite of what happens.

Forward voltage drop, which you correctly described (and I didn't quote) is the condition where current does not flow until the voltage difference exceeds a set amount. (For germanium junctions, that's usually in the neighborhood of 0.35vDC, for silicon junctions it's usually about 0.65vDC.) Accordingly, no part of the incoming waveform appears at the other end of the diode until this level is exceeded, and all of the incoming waveform that exceeds this value will appear at the diode's other end.

More to the point, the waveform is not clipped (the top lopped off), it is instead forced to rise very suddenly to the nominal value of the incoming voltage. Also, be aware that this condition holds true on the descending edge of the waveform - as the voltage drops towards zero, the waveform is suddenly cut off - it no longer passes through the diode.

Both "losses" contribute to a distorted waveform, to be sure - but it is not clipped. In fact, the peak voltage of the output waveform remains exactly the same as that of the incoming signal. However, the total time the signal is conducting is reduced, therefore one's ear will perceive that the signal is weaker. That's because the total "on" time is slightly less, and even a moderately sensitive ear can detect that kind of difference. (This leads to the definition of RMS, which is germane, but let's not go there just now, eh?)

Now keep in mind that most pictorial diagrams show only one half of the waveform as being absent - the remaining half (the conducting half) is usually shown as being complete from zero to peak and back to zero. Not true. In point of fact, no diode conducts for exactly 180° of signal - there will always be some additional signal missing. The amount of that missing signal (from the conducting half) depends on the total input voltage level. (And to a lesser extent, the frequency.) Many texts (and nearly all websites that deal with this topic) assume some high level of voltage, 10 vDC being common. In these cases, an additional few degrees might amount to no more than 5 or 10, for a total of 185 or 190° of missing (non-conducted) signal.

But us guitarists? Hey, we're dealing with no more than a volt to start with! At that point, we're really in the soup - 1 volt isn't a helluva lot more than a silicon diode's forward voltage drop conduction point - 0.65vDC. Hence we move "back" to germanium diodes, for this very reason. Which tends to give us more signal for the ear to hear.

So, what's this look like, anyways? Good question. Google failed me, miserably. So I've drawn my own diagram, which suffices, I'm sure.

Normal signal, one cycle thereof, prior to being fed into a diode:



Signal that's been through a diode. Note that to make things more obvious, I've exaggerated the time before the wave "kicks in" or "drops out".


Like I said, all of this depends on both the voltage level and the frequency of the signal. And also of course, as luck would have it, our guitars are not very well known for having pure, single frequency notes, eh? IOW, there are lots of harmonics in there, and they all play a part in how something gets through the diode. Again, this is all for illustration purposes, I didn't "screen capture" a real signal.

........................

Now, why all the foregoing blather, when "everyone just knows" that diodes clip? That's not a bad question. Not a good one either, but not bad.

It's because you never see diodes acting directly on the signal - they are always used in a negatve feedback loop. Whazzat?   It's where some of the output signal is fed back to the input, in exactly 180° out of phase. The net effect is, the opposing phase signal will tend to cancel out some of the incoming signal. However......

What if we were to monkey around with that feedback loop's signal, what about that, eh? Now you're cookin' with gas there, bunky! ;D

Indeed, if we partially obliterate some of the feedback signal, then in turn, the incoming signal won't get combined with an out-of-phase signal, which then lets the incoming signal go through at full blast. Now, for the $64 dollar question..... what would happen if we put a diode in the feedback loop, such that it acts like my second diagram above. What would be the net effect on the incoming signal, eh?

Hint: a certain four-letter word should come to mind....


So ends today's lesson in Electronics 101. We return you now to your favorite Forum, already in progress.......


;D





sumgai

[ashcatlt]

Okay, what you've shown in the diagram is what's going through the diode, but not what's going to get to the amp. These diodes are across the jack, remember, not in series with either of the signal wires. The diode is shorting the jack while conducting, flattening the top of the waveform and leaving the remainder (the bits closer to 0) more or less untouched.

Also, you quite often do find diodes outside a feedback loop. Some OD/dist/fuzz boxes do put the diodes in the feedback loop, but many actually just stick them from "hot" to "ground" after the amp stage, just like what's described above, and it works the same way except that it's got a bigger voltage signal to work with.

It is very rare to see series diodes outside of a feedback loop, though, becuase we generally find that hacking out the "middle" of the waveform (crossover distortion) is not as pleasing as when we clip off the tops, for whatever reason.

[sumgai]

ash,

Ah, the old "diodes in parallel with the amp" trick - I see!

In essence, you're correct, when the diode conducts, any signal that exceeds the forward voltage drop should go to ground. This would indeed be clipping. And it would also be the prime source of so much square wave action that it'd be incredible if it were tonally acceptable.

More to the point, the greater the voltage presented to the diode, the greater the percentage of total clipped signal. IOW, in this kind of circuit, we don't want too much voltage at the diode input, we do want to have something get through to the amp.   The math would work out something like this:

Assume 5v present at the diode input. (Not a bad figure for a preamp circuit, or the output of some stompboxes.) A silicon diode will conduct 4.35 volts worth of signal, leaving a mere 0.65 volts worth of signal to go out the jack. That's a whopping 80+% loss, no? Now if we just use our pickups themselves, and they generate only about 1 volt, then the signal that passes out of the jack will be the same, 0.65v, but that's about 65% of the total signal, yes?

Test time: Which would you rather have, a signal that's almost fully present, or a signal that's been clipped to the bone, and beyond? That said, it's still a matter of time as well, to reach the clipping point, so what the ear hears from that 80+% lost signal will be really harsh - ringing and square waves out the gazoo.

Yes, this can all be made to work, no matter what we cobble together. But despite my training (for which I paid good money), my proclivities still tend towards the KISS principle. 'Nuff said.


~!~!~!~!~!~

You said that diode are sometimes used outside of the feedback loop, and that's true, but also as you stated, the net effect is a crossover notch. Sometimes this is musically acceptable, sometimes not - depends on what order harmonic is most affected. However, this is often ameliorated by using an assymetrical diode setup. The notch might still be there, but it will be "swamped" by the remaining signal that isn't clipped as severely. Or so it appears to my ears, back when I used to breadboard this stuff all the time. (Still got those boards too, but hardly ever use 'em now. Too many other irons in the fire (and not a horse in sight that needs a good shoeing!).)

We're both on the same side of the fence here, but I think my explanations and diagrams will be of value to others as they roam around in The NutzHouse..... I hope.

HTH



sumgai

[roadtonever]

Thanks for the clips JohnH. These are some of the nicest overdrive tones I've heard. Considering there has been no pre-eq applied the fat crunchiness, pleasant bite and lack of harshness is mind blowing.

[Yew]

Assume 5v present at the diode input. (Not a bad figure for a preamp circuit, or the output of some stompboxes.) A silicon diode will conduct 4.35 volts worth of signal, leaving a mere 0.65 volts worth of signal to go out the jack. That's a whopping 80+% loss, no? Now if we just use our pickups themselves, and they generate only about 1 volt, then the signal that passes out of the jack will be the same, 0.65v, but that's about 65% of the total signal, yes?

so if you used a volume pot before and another after the diode (not sure if you could do this)

you would essensially get a distortion control, then a normal master volume?

[ashcatlt]

Yep.

Of course, JohnH has already provided a "distortion", which might actually work better in passive application.

[JohnH]

I found the only way to approach a good tone is to get as large a signal as possible before the diodes. And standard silicon diodes sound particularly bad in a passive circuit - they stay mostly clean and just cr@p out on the high peaks.

Anyone who wants to try this is encouraged to obtain a couple each of Schottky, silicon and germanium diodes - and hear for yourself

Germanium and Schottky diodes have two characteristics that make them better in a passive circuit than a standard silicon diode.

One is a lower cut off voltage - actually its not as simple as an on/off conduction above or below a single voltage, all diodes have a logarithmic relationship between voltage drop and current. But a silicon diode drops about 0.65V at 1mA current, germanium about 0.4V and Schottky about 0.15V. We are working with much less current, so lower voltages apply.

The second thing, I think is more important:
When you plot out the voltage/current graphs for these diodes, once the voltage drop across the silicon diode gets below say, about 0.3V the current gets extremely low. Effectively, the 'resistance' become very high, ie so high that it has little effect on the guitar signal. That explains why a silicon diode across the guitar output sounds clean, with cr@p on the peaks.

By contrast, the effective 'resistance' of germanium and Schottky diodes (ie the ratio of voltage/current) remains within a range that has a significant tonal effect on the guitar, all the way down to zero signal. I recall plotting this out when I looked at this before and I think I got about 75k at very small signals. Given that that is a highly non-linear effect based on signal strength, it shows how the Schottky can keep messing with the waveform, all the way down through the decay phase. That results in a gradual, smoother effect.

cheers

John

[ashcatlt]

JohnH just explained why it is that we want to clip these things so hard.

sumgai's 1V peak-to-peak is only really going to be the loudest part of the signal - the initial pick attack. The sustain/decay phase of the note will be significantly lower than this. We'd like to have some harmonic distortion during that portion of the note event as well. If all that's getting clipped off is the loudest part, it just ends up sounding like there's something broken somewhere.

We had a request in the Effects section for some info re: how and why various effects circuits work. OD/distortion is pretty simple in general, but a pretty broad topic. I have started a topic over there, to continue some of the discussion re: active clipping circuits.

[sbgodofmetal]

can someone post a simple diagram for wiring the diodes to a tone pot and is it possible to get a passive heavy distortion this way

[JohnH]

Mar 26, 2011 21:01:36 GMT -5 sbgodofmetal said:

can someone post a simple diagram for wiring the diodes to a tone pot and is it possible to get a passive heavy distortion this way

All you do is wire them exactly like a tone control. The simplest form of this idea has just one schotky diode. Just wire it exactly where the tone cap would be, and omit the cap. It does not, in this case, make any difference which way round the diode is. In the schematic diagrams in the first post, there are more variations, eg, two diodes in parallel pointing in opposite directions etc.

Its an interesting experiment, but please dont get too hopeful that this will deliver a killer tone (more likely, a 'tone killer' IMO!). Much more power and grunt can be obtained by even a simple active circuit.

A good way to test is just wire it up in series with a pot, and clip it across hot and ground outside of the guitar - you should hear a drop in signal level and a the distortion with it. Then you can decide for yourself.

cheers
John

[sbgodofmetal]

thanks johnh i can't read schem's yet but i'm slowly learning how which reminds me of this [Old chinese proverb: i hear and i forget, i see and i remembr, i do and i understand] if you think about it this old saying can be applied alot on this website

[allmektig]

A friend of mine asked me if I could make a varitone-like circuit for his guitar, and maybe if I could "throw in some other stuff".
Since I had a 1p12t rotary switch lying around, I added 4 different passive clipping arrangements as well.

1 is a single bat41 schottky diode.
2 is 2 bat41 schottky diodes, wired in anti parralel.
3 and 4 is the same, just using 1n34a germanium diodes instead.




Sounds like:
Bridge pickup
soundcloud.com/a-melodyc-shivering-form/passive-clipping-in-a-guitar/

Neck pickup:
soundcloud.com/a-melodyc-shivering-form/passive_clipping2

My guitar playing skills does leave a lot to be desired...


Edit: added another soundclip and sorted descriptions.

[JohnH]

That's good test on these things. To my surprise, I liked the last one best, the two germaniums. Do you think they sound useful?
J

[allmektig]

Jan 19, 2013 18:09:58 GMT -5 JohnH said:

That's good test on these things. To my surprise, I liked the last one best, the two germaniums. Do you think they sound useful?
J

I think they might be usefull, since they do add something to the tonal palette of the guitar. But if you're going to use any distortionpedals or high gain amps, I'm not sure if there's much point to it.

It seems I got a bit confused about wich pickup is wich when I uploaded the tracks. I'll have to check that...
Edit: Nope, it was right.