New (?) Diodes to Ground Question

[Joe Hart]

I can't seem to find a definitive answer to this question.

Let's say we have a "diodes to ground" distortion circuit. We arrange three separate sets of diodes back-to-back in parallel (not series) to ground. So we have two LED's, two Si, and two Ge diodes. I notice a difference in the sound from, say, just two Ge diodes, but why? If the diodes clip at .7, .5, and .3 V respectively, wouldn't the entire signal just clip at .3 V and create the same net result on the signal as a pair of Ge diodes by themselves? Or am I hearing things that aren't there?
-Joe Hart

[GibsonGM]

There are a few answers to this one, Joe.  If you are clipping off at .3V, yet you hear a tonal difference with the other pairs of diodes in there (that's what I am understanding you talking about), it MAY be due to the junction capacitances of the other diodes.   That capacitance also changes with forward bias.   So there is a small amount of filtering going on (low pF capacitances, like a varactor diode would provide).  Some of the higher harmonics in your signal may be 'getting past the clippers', and filtering out, and the amount would depend on how many/what type of diodes follow your clippers.  Sort of loading down the original pair of clipping diodes.
That's my thought!   

[Thomeeque]

Actual (clipping) voltage on the diode depends on actual current going thru it (according to it's I–V characteristics), so maybe at some moments Ge diodes clip "later" than Si diodes? How does whole circuit look?

[JDoyle]

Actual (clipping) voltage on the diode depends on actual current going thru it (according to it's I–V characteristics)

Flip that - actual current through the diode depends on the voltage across it.

It's all in the voltage (same goes for BJTs) - and if the source can then supply the current.

My guess is that even though the 0.3V threshold of the Ge diodes isn't theorhetically enough to turn on the Si diodes, it will be enough to cause the Si diodes to begin conducting. Also, the knee of a Ge diode is nowhere near sharp and if enough current can be supplied, the voltage across them could easily reach into a Si diode's conduction curve.

For the LEDs I'd have to agree with GibsonGM. I don't think they conduct much, if at all, but their capacitance and it's tendency to change, could certainly cause differences in tone.

Regards,

Jay Doyle

[Joe Hart]

How does whole circuit look?

Like a DOD 250 or MXR Distortion+. So how about if drive the IC to clip, toss some LED's in the feedback loop to clip, then add Ge diodes to ground at the output. Will it just sound like a stock Distortion+ because all the clipping that happens earlier in the circuit is "chopped off" by the Ge diodes at the end anyway? Or are there still some "artifacts" from the previous clipping that would change the end result?

I guess what I'm asking is, if you clip off the top 10% of the wave form, THEN clip an additional 10% off that, THEN clip off another 10%, would the end result sound like a circuit that just clips off 30% from the start?

-Joe Hart

[Brymus]

Does having a gain stage inbetween the diodes to ground sets do anything for the clipping?
As in the standard anti-parallel pair to ground > gain stage or two > anti-parallel trio to ground ?
In an attempt to get symetric then asymetric clipping on top of each other.Or to asymetrically clip the symetric waveform?
I have often wondered the exact same thing as Joe.

[soggybag]

What if there was a cap between each diode? Would that make a difference. This AC stuff sometimes throws me. I'm wondering if the voltage after the cap would be at 0v.

[GibsonGM]

Not likely, the cap will send some highs to ground - but not all.  Caps have a reactance (AC resistance) that lessens with higher frequencies. Guitar signals are made up of a mess of frequencies (fundamental plus many harmonics), so some will remain, unless the cap is large enough to completely shunt the signal.  Given that there is SOME resistance in series with the cap ('effective series resistance', since caps aren't perfect), I bet some signal would remain...

[Joe Hart]

I have often wondered the exact same thing as Joe.

Okay, so it's not just me!

Along the lines of this thread, why does a Big Muff Pi have multiple clipping stages? If the last clipping stage was the most severe, wouldn't the last one's clipping be all that "survived"? Or if the first clipping stage was the most severe, wouldn't the signal be squashed enough to go through the rest of the circuit without being clipped? And I know that the signal gets boosted after the first clipping stage in the BMP, but then that makes the last clipping stage the most severe (see my first scenario).

Hmmm...
-Joe Hart

[petemoore]

  The BMP Boosts:Boost-clips:Boost-Clips:TC:Gain
  The second set of diodes sees a clipped, then boosted again signal, boosted enough to exceed the diode threshold.
  If what I read is two sets of BTB diodes, fairly close in thresholds which are paralleled between the signal path and ground [and seeing enough voltage to exceed threshold], and that 'does something'...the something it'd do imestimation is to soften the clipping knee.
  Would be interesting to see if a scope can show this, what'd seem an easy way to soften the clipping curves.

[Joe Hart]

It seems that whatever you do to a signal, if you then clip it enough at the end it would pretty much negate all the clipping that happened before it? But still no one with a definitive answer?
-Joe Hart

[R.G.]

It seems that whatever you do to a signal, if you then clip it enough at the end it would pretty much negate all the clipping that happened before it? But still no one with a definitive answer?
-Joe Hart

The definitive answer:

It depends.

It depends on how big the signal is trying to be versus the clipping process.
It depends on how much of the signal is clipped off.
It depends on the sharpness of the clipping knee and how much of the signal to be clipped resides inside the knee.
Finally it depends on how flat the top of the clipper is.

If the clipper has a dead flat top, then any information above the flat top will be completely lost. If the clipper has a rounded region leading to an ever-flattening curve, then the information above where it starts rounding over is still there, but ever-more squashed together.

Imagine - what happens to a sine wave that peaks at 0.8V when it runs into a 0.7V diode? Diodes actually start rounding over very slightly at perhaps 0.5V, and are pretty flat topped - although not perfectly so - at about 0.7V, so the information in the wave between 0.5V and 0.7V is progressively squashed, until almost total squashing happens above the 0.7V mark.

Compare that to what happens to a 100V peak sine wave into a 0.7V diode clipper. The result is an almost perfectly sharp-cornered square wave. Everything above 0.7V is contained in the flat top. A 10V sine wave into a 0.7V clipper is almost indistinguishable from a 100V sine wave into a 0.7V clipper, but a 1V sine wave is easy to distinguish.

The devil is ...always... in the details.