clipping diodes

[boog]

this is just cuz i can't seem to find a straight answer anywhere: in a dirt box like the dist+ theres a pair of clipping diodes at the end of the circuit. if one were to put another pair of diodes after (not in series, but parallel) those first two and the (forward?) voltage required for clipping is higher than the first pair, then they wouldn't clip?

[dougman0988]

I'd have to say you're right.  The signal peaks will be clipped due to the first pair, and if the second pair of diodes' forward voltage is higher than the first, the signal's amplitude will never reach it.  So the second diodes will not conduct.

[mojotron]

1+

There is kind of a grey area if they break over at a voltage that is close - i.e. .69 and .7: If that's the case than the differences in parts may cause the 2nd pair to conduct. Without knowing the parts in question, I think it's safe to say that to above answer is likely the case.

[petemoore]

 You're right. The low clipping threshold does all the current handling, and prevents the higher threshold diodes from ever conducting.
  An SPST switching in lower threshold diode[s across the original clipping diodes = the lower threshold[s of clipping voltage.
  A broken SPST would = original threshold clipping voltage.

[boog]

excellent! at least i figured something out this week. now if only i could determine where that smell is coming from around here.......

[Mark Hammer]

There are a few things being confused here.  Let's untangle them.

Diodes provide clipping because they have a kind of signal-dependant resistance. If the signal is well below the forward voltage, the diode behaves like an open circuit or a brick wall.  Once the signal gets very close to the forward voltage, the diode begins to conduct.  In the case of a configuration like the Tube Screamer, with those diodes in the feedback path, as the output level rises, and the dodes begin copnducting, negative feedback is applied and the signal is abruptly attenuated by,  In the case of the Dist+, once the output approaches the forward voltage of the diodes, they shunt signal to ground to provide a cap or ceiling on the level that we experience/hear/see as "clipping".

You could put a drawer full of diodes of every single type, and a myriad of combinations (1+1, 2+1, 2+2, 3+1, LED, Si, Ge, FRED, Shottky, zener, etc.) in parallel in that position, and if ANY of them provides a path to ground that will conduct at the forward voltage equivalent to the signal output of the op-amp at that moment, THERE WILL BE CLIPPING.

The clipping will not be provided by all, or indeed more than one, of those various diodes, but if there is an acceptable path to ground through at least one of them, the signal will find it.

So, to answer your query, there will still be the same clipping, but the added didoes will not play any role in it because the Ge ones "put their hands up first".

[earthtonesaudio]

Diodes conduct on a nonlinear curve.  Different diodes have differently shaped curves.


Think of it like superimposing a square with side length of 2 over a circle with radius 1.129.  Both have the same area but their outlines overlap.  Similarly two conduction curves of two different diodes may intersect at multiple points.  So while it's true that the one with the lower threshold voltage conducts the majority of the current, this does not tell you which one conducts at voltages below the threshold of both diodes.

[boog]

@mark, that's what i thought would happen, so i'm glad i got that part. but regarding earthtones: you're suggesting that if there are 3 (hypothetical) pair and the first is doing most of the work (lowest voltage), the second perhaps none (highest voltage) then the third (lowish) could still offer some clipping because it's threshold is similar to the first?  i think i understand, just not entirely.  thanks again everyone!

[Toney]

Perhaps this is a good time to add something that I never "got"

Although, I am entirely used to using diode pairs for clipping in this way, I have never understood how the second one achieves anything. That is to say, if there are a pair connected in a simple clipping arrangement from say the output of an op amp to ground, one will be cathode up, the other cathode down.
Surely the diode presenting its cathode to the output with it's a node to ground would never conduct?

[mattthegamer463]

Perhaps this is a good time to add something that I never "got"

Although, I am entirely used to using diode pairs for clipping in this way, I have never understood how the second one achieves anything. That is to say, if there are a pair connected in a simple clipping arrangement from say the output of an op amp to ground, one will be cathode up, the other cathode down.
Surely the diode presenting its cathode to the output with it's a node to ground would never conduct?

(the following is IIRC, dont quote me)

The idea is that the diode does conduct to ground, once the voltage reaches the "voltage" of the diode, it will conduct to ground, clipping the waveform.  Since it is AC we've got here, we also want to clip the negative portion of the wave.  Thats what the second inverted diode does. 

[Toney]

 Makes sense.  

[boog]

fail build #7: didn't connect both diodes to ground. that sounded fairly annoying as i recall.

[MartyMart]

(the following is IIRC, dont quote me)

The idea is that the diode does conduct to ground, once the voltage reaches the "voltage" of the diode, it will conduct to ground, clipping the waveform.  Since it is AC we've got here, we also want to clip the negative portion of the wave.  That's what the second inverted diode does. 

Hence - symetrical OR asymetrical clipping .... for some reason my ears prefer the later, which would
be  2 x diodes one direction and 1 x diode the other, always seems richer in musical harmonics to me, whereas
"symetrical" seems harsh and out of tune with simple chords/ 5th's etc.

MM.

[mikemaddux]

agreed

[merlinb]

Surely the diode presenting its cathode to the output with it's a node to ground would never conduct?

If the diodes are connected to ground then there will also be a coupling cap in the circuit, so the signal voltage becomes AC. You can acheive a similar result without a coupling cap, but only if you connect the diodes to the Vref (4.5V) point, rather than to ground.

[Mark Hammer]

Hence - symetrical OR asymetrical clipping .... for some reason my ears prefer the later, which would
be  2 x diodes one direction and 1 x diode the other, always seems richer in musical harmonics to me, whereas
"symetrical" seems harsh and out of tune with simple chords/ 5th's etc.

But the thing to keep in mind is that the forward voltage of diodes themselves is fixed, and generally occurs in "leaps" that pay little mind to the actual signal amplitude and "sonic goals".

So, yes, any clipping produce by different numbers or types of diodes in each orientation will be "different" for each half cycle.  But how different, and different how?  

E.g., if I use a Ge and a red LED in the opposite direction, that will surely produce a non-symmetrical influence on the waveform since the Ge has a forward voltage this is often as low as 1/7 that of an LED.  Given that what it takes in the way of gain to even begin to nudge the LED into conduction would likely yield serious "square-i-tude" on the Ge side of the signal, why even bother using an LED?

Similarly, it is certainly a simple matter to use a 2+1 complement of the identical type of diode, but the question to consider is whether the gap in forward voltages between the 1-diode side and 2-diode side is what is needed to produce the desired degree of asymmetry, given the intended gain and distortion target.  It may be that the desired degree of asymmetry - for the distortion intensity targeted - is best produced by use of, say, an Si+Ge on one side and an Si+2 Ge on the other, yielding a modest difference.  Alternatively, maybe what a person wants is a pair of Ge but a small resistance in series with one side.  I made a TS808 clone a couple years ago, using the Tonepad layuot, and where the optional extra-for-SD1-enthusiasts diode would go, I wired in a 10k pot to dial in variable asymmetry.  Worked like a charm.

[Toney]

Surely the diode presenting its cathode to the output with it's a node to ground would never conduct?

If the diodes are connected to ground then there will also be a coupling cap in the circuit, so the signal voltage becomes AC. You can acheive a similar result without a coupling cap, but only if you connect the diodes to the Vref (4.5V) point, rather than to ground.

Right, so as the AC signal goes through the negative portion of its cycle, I am assuming that ground is "higher" thus allowing  voltage to flow, with the second diode who's anode is at ground, once the threshold is overcome and so clipping occurs through that part of the cycle, until it is positive again and the other diode takes over.
Weird. I am so used to using these I forgot to ever really think about them...

[MartyMart]

Thanks Mark, interesting stuff as usual - i always seem to pull apart TS-9's/ Dist+'s and add an extra diode on one side !!
I like the pot idea though ..... hmmm

One of the nicest combo's in that circuit was 3 x Ge diodes one side and 5 x on the other - very creamy :-)

MM.

[Mark Hammer]

Well there you go.  3+5 Ge provides a clipping point that is slightly higher than one Si on the 3xGe side, and less than double on the 5xGe side.

Like I say, the ideal gap in forward voltage between each side will ultimately depend on what you're going for, tonally. There is no "ideal" asymmetry or difference in forward voltage that works equally well for all applications.  One of the reasons why I like the variable resistance idea.

[Projectile]

I have to agree with earthtonesaudio. Diodes are often modeled as simple switches that turn on when forward biased voltage exceeds their threshold voltage, but in reality they have a bit of a knee around the threshold voltage. So, a second set of diodes with a slightly higher threshold could definitely still be clipping, but the first set will be doing most of the work.