Clipping diodes exposed

[blackieNYC]

Merlin - that AP will measure IM distortion too, right?

[merlinb]

Merlin - that AP will measure IM distortion too, right?

Mine doesn't. It's too old and doesn't have the add-on.

[PRR]

I believe the proper re-normalization is in dB.

The log-log chart implies this data but it takes some point-finding to get a sense.

I arbitrarily choose "1%-10%" increment of THD as my range. I do not believe 1% THD is audible in simple (one instrument) music, while 10% THD is pretty clear, and none of the curves go far beyond 10%. (Some should run to 25%, but maybe only for HUGE overvoltage.)

You can lay the edge of a matchbook on the graph to find the slope of the curve from 1% to 10% THD. I use a lighter so I had to use a computer line-draw, squints, and the TI-30Xa.



The 1N4148 rises from 1% to 10% THD in almost 7dB of input rise.

The 1N4006 and 10Vz rise from 1% to 10% THD in about 5dB of input rise.

LEDs about 3dB input rise. (MOSFETs seem similar.)

Hard clipping, about 2dB. (This is known from power-amp graphs. A clean-clipping amp can be rated about 58% [2dB] higher power if you do not mind "10% THD" on the specs.)

[Johan]

Did you use the exact same setup for the zeners as for the other diodes?  If so, you didn't really measured the zeners the way they are typically used when used for clipping.  They should be in series back to back (or face to face) instead of parallel.  Not that i think it would make a big difference since the reverse diode knee will still be used elevated by the forward voltage.  It's just that the plot would be more complete if you did change the setup for the zeners than if you didn't. ..Just a thought. ..Good work by the way. .
J

[Thecomedian]

The longer a transistor is in cutoff, the harder the clipping is. You can fine tune a circuit to push a BJT transistor to work in it's non-linear region, where it's just above cutoff but in that region where it's non-linear. That will produce "soft" clipping where second harmonic distortion is generated but third harmonic from those square tops on waves are not quite there yet.

The longer a diode is in "saturation" (on), the harder the clipping is. If a diode of some regular or LED kind required a 1.5v drop to turn on, and your signal is 1.6v, surely that will sound a lot different in distortion generation than a diode that requires a 0.5 volt drop with a 1.6v signal. The former will be on for much less time.

Maybe I'm missing something here, though..

[merlinb]

If a diode of some regular or LED kind required a 1.5v drop to turn on, and your signal is 1.6v, surely that will sound a lot different in distortion generation than a diode that requires a 0.5 volt drop with a 1.6v signal.

It would. But you would get the same sound from the 1.5V diodes simply by turning the gain up by x3. The different diodes give different amounts of headroom in the same circuit, but don't "sound different".

[psychedelicfish]

A smoother curve at the start of clipping would result in a smoother audible transition from clean to distorted, right? If so, the green LED would seem to transition into overdrive nicer than the red one...

[Thecomedian]

I went ahead and ran some tests because I had an inkling.

Suppose we take a circuit that has a 1V response at 100hz and 500mV at 1khz. A 0.7V clipping diode will affect the bands of frequncies that fall >= 0.7V. Those outside will be unaffected, although second and maybe third harmonics will be generated from the clipping of the lower frequencies. If a 1khz harmonic is generated, how does it interact with the 1khz fundamental that wasn't clipped? Does it even?

But anyway, now we take a gain of 3x onto this circuit, and both 100hz and 1khz clip, but 100hz clips more and produces more harmonics, while 1khz clips less and produces less. Less harsh high end, more middle-high harmonic content.

Now we change in an LED of some color that raises the voltage requirement to fire to 1.6V. Now the clipping of the original 1V signal gained to 3V is going to be less distorted, and the 1khz is not even distorted at all.

Variations of different diodes with different forward voltages will produce varying degrees of clipping for each fundamental, in between 1V to 1.5V.

Frequency response of circuits can play a big role in diode clipping sound. The ratio of each frequencies voltage to each other compared to the diode FV can determine where and how much harmonics are added, and I guess that people have already known this when using EQ's before a distortion circuit, but I think it's a consideration that people who hear slightly different things are hearing a change because of these voltages. Like in the above example, simply gaining a circuit won't produce an identical result with a higher FV diode, because the ratio of the Frequency voltage to the diode FV changes for 1V @ 0.7FV to 3V @ 1.5FV, and the same happens for 1khz frequency as well, so the ratio of voltage for a frequency compared to FV of a diode needs to remain similar in order for the circuit to behave similar.. That ratio does seem to determine how hard something clips and how many harmonics are generated.

[merlinb]

A smoother curve at the start of clipping would result in a smoother audible transition from clean to distorted, right? If so, the green LED would seem to transition into overdrive nicer than the red one...

They both look the same to me... (remember, the graph is made from discrete test points, so you can't see the true smoothness of the curves).

[merlinb]

Frequency response of circuits can play a big role in diode clipping sound. The ratio of each frequencies voltage to each other compared to the diode FV can determine where and how much harmonics are added, and I guess that people have already known this when using EQ's before a distortion circuit, but I think it's a consideration that people who hear slightly different things are hearing a change because of these voltages. Like in the above example, simply gaining a circuit won't produce an identical result with a higher FV diode, because the ratio of the Frequency voltage to the diode FV changes for 1V @ 0.7FV to 3V @ 1.5FV, and the same happens for 1khz frequency as well, so the ratio of voltage for a frequency compared to FV of a diode needs to remain similar in order for the circuit to behave similar.. That ratio does seem to determine how hard something clips and how many harmonics are generated.

But IMD is always directly related to THD, and therefore to the transfer characteristic. Only the shape of the basic characteristic matters, not it's headroom. Accepting that nearly all diodes produce the same shape of transfer characteristic means that whatever diode you plug in, you could ultimately get the same response with a suitable adjustment of gain and frequency response. Your argument is slightly specious anyway, because if your 1kHz signal is large enough to clip, then other frequencies applied simultaneously will in fact be 'riding on top' of the 1kHz signal, so they will, in fact, be clipped too.

The take home message is basically what Seljer said, don't waste your time on diodes, mess with the EQ (and gain). Or to be more specific, in a NEW design, don't waste your time choosing between silicon diodes or LEDs or colours of LED- they're all equivalent. Instead make the choice between no diodes, Ge diodes, silicon diodes/LEDs, or weird stuff like MOSFETs. Obviously if you're tweaking an existing circuit and you don't want to mod lots of things, swapping diodes is at least an easy thing to do.

[DougH]

I think that brings more merit to the notion of: if it ain't sounding good, mess with the EQ, not the diodes

Here, here!!

Yup. Been saying this for years. Use diodes (if you must- haha) to set headroom but use EQ to set the "goodness" of the sound.

Nice work Merlin.

[Quackzed]

nice! it can be a bit confusing at first, separating the turn on voltage of different diodes from the signal size hitting them. the fact that swapping in different diodes does sound different makes it seem like its the knee, when in fact, as shown here, its more the relationship of signal size to threshold. how fast does the signal go though the 'knee' of whatever diodes you use. i'm a believer already, but some proof is nice.
good work!
also i suspect mosfet differences will line right up with diode--series resistance as in 'softness' or 'warp' controls if you get the resistance right.
thats just a susspicion really,as  i have never dug into mosfets that deeply... just seems like a pn or np junction is what it is. all the rest like leakage, bias, current etc... is 'outside' the box and can be manipulated, but the junction is what it is...

[WhiskeyMadeMeDoIt]

What contribution does the reverse bias capacitance play into the sound differences between clipping diodes? Maybe this has some effect on the sound that isn't accounted for with the knee curves?
http://www.hanssummers.com/varicap/varicapled.html

[PRR]

> contribution does the reverse bias capacitance play

Look at the numbers. The highest C in the LED data you cite is 60pFd. In Merlin's circuit this works against 10K resistance. 60pFd+10K is 265 KHz; higher F for the lower Cs, out to 3MHz.

If a guitar (80-8KHz) or hi-fi (20Hz-20KHz) is filtered with a variable 265KHz-3MHz low-pass, I would not expect *ANY* audible difference.

Remember the small diodes are intended for many-MHz operation. If they had large C they would be no good for the purpose. The LEDs don't have a real need for speed but it seems they don't have much C anyway. The 1N4007 is "fat" and has large C, but also low-low R, so again the speed is good (though maybe not as far above the audio band as the others).

[WhiskeyMadeMeDoIt]

Thanks for the information PRR.

[Thecomedian]

Frequency response of circuits can play a big role in diode clipping sound. The ratio of each frequencies voltage to each other compared to the diode FV can determine where and how much harmonics are added, and I guess that people have already known this when using EQ's before a distortion circuit, but I think it's a consideration that people who hear slightly different things are hearing a change because of these voltages. Like in the above example, simply gaining a circuit won't produce an identical result with a higher FV diode, because the ratio of the Frequency voltage to the diode FV changes for 1V @ 0.7FV to 3V @ 1.5FV, and the same happens for 1khz frequency as well, so the ratio of voltage for a frequency compared to FV of a diode needs to remain similar in order for the circuit to behave similar.. That ratio does seem to determine how hard something clips and how many harmonics are generated.

But IMD is always directly related to THD, and therefore to the transfer characteristic. Only the shape of the basic characteristic matters, not it's headroom. Accepting that nearly all diodes produce the same shape of transfer characteristic means that whatever diode you plug in, you could ultimately get the same response with a suitable adjustment of gain and frequency response. Your argument is slightly specious anyway, because if your 1kHz signal is large enough to clip, then other frequencies applied simultaneously will in fact be 'riding on top' of the 1kHz signal, so they will, in fact, be clipped too.

The take home message is basically what Seljer said, don't waste your time on diodes, mess with the EQ (and gain). Or to be more specific, in a NEW design, don't waste your time choosing between silicon diodes or LEDs or colours of LED- they're all equivalent. Instead make the choice between no diodes, Ge diodes, silicon diodes/LEDs, or weird stuff like MOSFETs. Obviously if you're tweaking an existing circuit and you don't want to mod lots of things, swapping diodes is at least an easy thing to do.

I don't know how to test large bandwidths for individual frequency changes in Spice. I thought they might be allowed to clip if the Diode was "on" during the time a larger AC voltage clipped, but I couldnt say for sure that would be true. It does make sense, and that means that the 1khz frequency would clip during the entire time the diode was on but what about when the 100hz frequency dips to 0.3V and the diode turns off? Or does a 100hz frequency at 0.3V and a 1khz frequency at 0.7V add up to 1V during the time they are in phase together?

[Quackzed]

Or does a 100hz frequency at 0.3V and a 1khz frequency at 0.7V add up to 1V during the time they are in phase together?

yes its ONE signal. high frequencies/low frequencies all in there together.

[Thecomedian]

If I had only two sine waves, 100hz and 1Khz, and 100hz was 1V and 10Khz was 0.5V, wouldn't there be periods where the 1khz is at it's most positive, but the voltage could be -0.5V due to the 100hz being most negative at that time? I suppose I'm envisioning something close to what is seen in movie voice recording in a visual graph mode: there are peaks at certain places and narrow bands inbetween. Would that be how voltage is behaving with all of these frequencies overlayed onto each other?

[Seljer]

I think that brings more merit to the notion of: if it ain't sounding good, mess with the EQ, not the diodes

Here, here!!

Yup. Been saying this for years. Use diodes (if you must- haha) to set headroom but use EQ to set the "goodness" of the sound.

Nice work Merlin.

There is a bit of a balance you have comprise on, with how much headroom and clipping you desire, and how much noise you can tolerate from the amplification step.

[bool]

I think that the most important factor is the "slope" of nonlinearity, but not in "plot" sense - rather in the ratio of avg. signal amplitude versus the onset of breakup (in short and simplified) and the slope of clipping after that onset.

Of course the EQ also has the role in changing the avg. signal amplitude (in relative sense), because only the "right" frequencies (when done right) will contribute significantly and in the right proportion to the avg. signal amplitude - and consequently get clipped "in the right way".

So both factors are important, interlinkked and interactive. Especially with basses and humbucker guitars (imho).

My 2 EuroCents.