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Diode Clipping Circuit On A Tone Pot


Justin G.E.
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I am familiar with the results of adding a diode clipping circuit to a volume pot (ground) for saturation effects, but what are the results of putting a similar circuit onto the ground of a tone pot? As a sidenote, if I had any extras (appropriately rated diodes) lying around I would just test this out and post my findings if in fact those findings were post-worthy, but I don't have any here.

I'm sure that anyone with advice knows what a clipping circuit is, and I would post a visual scheme of what setup I am referencing, but I the sim programs I have only export to strange formats unreadable on the web.

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Diodes are (as a bad analogy) a discrete electronic version of a one-way valve. They will conduct one way and not the other. That said, they only START conducting at around 0.7v on average, dependant on the material the diode of constructed from. Silicon diodes (1N4148 for example) conduct forward at around 0.7v bang on. Germanium diodes (0A48?) are "fuzzy" around the threshold, and have a softer forward conductive threshold.

Basically, if you GROUND the forward end of a diode (anode) to earth as you explain, then as soon as your guitar signal exceeds the forward conductance threshold (avalanche point) then the signal is dumped to earth, read: no signal. A sine wave like that would look something like this:

diode_saturation_dump_to_earth.jpg

This will give you a horrible pseudo-crossover distortion which is unmusical and unrelated harmonically to your original signal. The usual way of using diodes to clip a signal is to cap it at the diode threshold voltage, not just to conduct it to earth. I may be slightly wrong here in that the diode may actually maintain the 0.7v on the cathode side but I'm sure it conducts everything to earth here.

The one thing I haven't pointed out is that a guitar signal is AC with reference to ground (0v) so you would need to add two opposing diodes to conduct one the positive and negative going excursions for conductance at 0.7v and -0.7v

Edited by Prostheta
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This image/schematic IS NOT MINE, it is from jag-stang.com, created by Yun-Jinn Pequesso. This is merely to make my question easier to visualize.

schematic_saturation.GIF

The diode clipping circuit is connected to the volume pot, for use as a saturation effect. So my original question was this: What effects will this same clipping circuit have on my sound if instead I connect it to the Tone Pot?

Btw, Prostheta, thanks for the info and the lengthy response. Does your response still hold true for the above schematic?

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A tone pot is a tap which dumps high frequencies to ground, so using a diode clipper on that type of circuit seems a little odd. Distortion/clipping adds mostly odd order harmonics as you're turning the signal into something more resembling a square wave. A tone control does the opposite by removing higher frequencies so in theory it would be pointless. I stand to be corrected on this as ever of course.

I think the circuit you've displayed would dump signals higher than 0.7v or lower than -0.7v to ground. The tone pot just reduces the high frequency components from the pickups which the saturation pot would "add back in" as odd order harmonics. A better circuit would be to add a 9v battery running a small op-amp circuit with the diodes in the amplification feedback (negative feedback?) loop of the op-amp. That way you can increase the "saturation" effect greatly. Just your standard fuzzbox type of distortion.

For reference, I would recommend using germanium diodes instead of silicon diodes as the forward voltage before conductance is "fuzzy" typically between 0.6v and 0.7v and gives you a smoother sound.

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Prostheta, you've missed one important aspect of diode clipping - it only dumps that portion of the AC signal that exceeds the diode's forward voltage to ground, so it limits the signal voltage at that level, and "clips" off the "tops" of the waveform. It's more like a pressure regulator than a valve, bleeding off only what exceeds the desired pressure. HTH

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That's what I was trying to remember TBH - whether the voltage at the cathode remains at 0.7v or not. Now you mention it, of course it would else you'd end up with some crazy non-linear circuits!!

Apologies for going off half cocked there....!!! I've not seen distortion done at a "passive" level like that. I was more used to making fuzz boxes using a complemented pair of diodes in the negative feedback loop of an op-amp or transistor circuit. It seems I need to brush up on my fundamentals.

whistling.gif

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Doesn't matter if it is passive or active. The voltage across the diodes (assuming a clipping pair arrangement) will remain at the forward voltage level. After all, the diode doesn't know what kind of circuit it is in.

If you look at the waveform you presented - the resultant frequency due to the diode shorting everything to ground would be double the input frequency - if it actually worked that way. Octave fuzzes normally have a full wave rectifier configuration to provide that double the input frequency generation.

IIRC... germanium devices usually have about a 0.2 volt forward voltage. The 0.6-7 volt range for silicon seems a bit high for passive pickups to reach on anything but the initial peaks and that's considering a high output pickup.

I normally figure 0.050 - 0.1 volts the average output level for passive pickups. Jack Darr liked the 50 millivolt figure, but I think the average output has increased since his time.

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I normally figure 0.050 - 0.1 volts the average output level for passive pickups. Jack Darr liked the 50 millivolt figure, but I think the average output has increased since his time.

That's interesting - I use 250mV P-P input as an informal design center, figure 500mV P-P for "clean" boosters, and try to make sure I've got at least 1 volt P-P transients handled unclipped for pristine stuff like compressors and EQs - not always easy, especially with a 9 volt supply. I doubt that the 50mV figure was typical for humbuckers even in Darr's heyday, but I suspect that since tube amps handle transients a lot more gracefully, it wasn't much of a problem until 9 volt stompboxes (and cheap, under-designed solid state amps) became prevalent, and any discrepancies could be largely ignored.

Jack Orman has some interesting figures on his Pickup Signals page, indicating that "moderately hot" humbuckers give an initial attack of over 1 volt P-P. I'd be really interested to know if anybody else has done any serious measurments, and what their results have been.

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I don't interpret that schematic as having a saturation/tone control - the volume control has a saturation circuit on it, not the tone control. Read this AMZ article for more on diode saturation control topolgies: http://www.muzique.com/lab/sat.htm

I think a saturation control is fine, but I wouldn't put it on a passive circuit like a guitar. A stompbox, yes, but not a guitar - I don't think it will sound very good. That's just my opinion, of course. :D

Edited by Paul Marossy
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I normally figure 0.050 - 0.1 volts the average output level for passive pickups. Jack Darr liked the 50 millivolt figure, but I think the average output has increased since his time.

That's interesting - I use 250mV P-P input as an informal design center, figure 500mV P-P for "clean" boosters, and try to make sure I've got at least 1 volt P-P transients handled unclipped for pristine stuff like compressors and EQs - not always easy, especially with a 9 volt supply. I doubt that the 50mV figure was typical for humbuckers even in Darr's heyday, but I suspect that since tube amps handle transients a lot more gracefully, it wasn't much of a problem until 9 volt stompboxes (and cheap, under-designed solid state amps) became prevalent, and any discrepancies could be largely ignored.

Jack Orman has some interesting figures on his Pickup Signals page, indicating that "moderately hot" humbuckers give an initial attack of over 1 volt P-P. I'd be really interested to know if anybody else has done any serious measurments, and what their results have been.

Sorry... I use RMS and not peak-to-peak. Since .1 volt RMS roughly equals .282 p-p I think we're talking the same thing at least for "design center" considerations. Sure, I love headroom and agree about stompboxes... and the need to allow for extra voltage input.

I don't know how Darr got his figures - humbucker, single coil or whatever pickup. But he did recommend a 50 milivolt signal (which I took to be RMS, only because that's the way I think... and I don't remember if he specified) as an average guitar signal level when substituting a signal generator for the input over a guitar so as not to introduce unintentional distortion.

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Since .1 volt RMS roughly equals .282 p-p

http://www.cebik.com/edu/edu4.html

RMS of .282 peak is .199, so roughly .2 but not .1

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Since .1 volt RMS roughly equals .282 p-p

http://www.cebik.com/edu/edu4.html

RMS of .282 peak is .199, so roughly .2 but not .1

You are of course absolute correct in what you are saying. However it appears that you missed a rather significant point: the discussion was about "Peak To Peak" and RMS - not "Peak". Peak to peak sometimes just P-P or even PP, for a sine wave is twice the peak.

So as an example, in order to pass a 10 volt peak sine wave signal without clipping, the circuit has to be able to pass 20 volts peak to peak. The RMS value will be 7.07 volts.

I "call" your website, but I don't care to "raise" :D

RMS AND PEAK TO PEAK VOLTAGES

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Now what will really confuse everyone is the 120v coming out of the wall socket is actually 340v p-p.

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