Diode Clipping

Started by RamonRivera, December 10, 2019, 08:31:33 AM

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RamonRivera

How to I build this simple circuit? I know it's just two diodes in parallel but what does the +4.5V mean, thanks


thetragichero

+4.5v can also be noted as Vbias. it's 1/2 the supply voltage and created by a voltage divider between V+ and ground
just two diodes ain't gonna sound like anything without components before and after it. at the very least: a transistor or op amp gain stage before it and a capacitor and resistor after it.
whatcha trying to build?

j_flanders

#2
What you're showing is only a (tiny) part of some larger schematic.
Here's an example of a circuit using that clipping section:
Proco Rat:


Yours seems to come from this DS1 section:
https://www.electrosmash.com/images/tech/ds1/boss-ds1-distortion-op-amp-gain-stage.png
Full schematic here:
https://www.electrosmash.com/images/tech/ds1/boss-ds1-distortion-schematic-parts.jpg

In the Proco Rat the diodes are connected directly  to ground while in the DS1 to the 4,5V bias voltage (the mid point between 0 and 9volt).
You can use either but there can be a difference beween the two.
You can read the explanation about the difference here:
https://www.diystompboxes.com/smfforum/index.php?topic=104231.0

RamonRivera

#3
Quote from: thetragichero on December 10, 2019, 08:53:43 AM
+4.5v can also be noted as Vbias. it's 1/2 the supply voltage and created by a voltage divider between V+ and ground
just two diodes ain't gonna sound like anything without components before and after it. at the very least: a transistor or op amp gain stage before it and a capacitor and resistor after it.
whatcha trying to build?

Im trying to build a simple clipper, I usually use it to distort some synths, what I do is go directly out of my sound card to the diodes and back again trough the preamps, this way I'm able to get them soft clipping but I find it to mild so I was trying to figure out how to get them to clip harder (not really sure cause Im a complete noob)by looking at some schematics I found this on the DS1, my understanding is the this voltage is used to bias the diodes to get a harder clipping so im trying to figure out how do I build this.

Mark Hammer

Diodes "clip" by conducting when the signal approaches some threshold dictated by the diodes, and things around them.  When they conduct, the signal cannot be increased in amplitude any further.

Terms like "hard" and "soft" clipping are not well communicated, hence not well understood.  In the case of something like the Tube Screamer, diodes are placed in the feedback loop of an op-amp.  The gain of the op-amp is given by how much negative feedback is fed from the output to the input.  When the diodes begin to conduct, more negative feedback is allowed to pass, unimpeded, such that the gain is momentarily reduced.  This results in a more compressed sound/signal, because the gain is being pulled back in response to signal peaks.

What gets called "hard clipping" is generally of the form that j_flanders shows in the Rat schematic, where diodes go to ground or Vref.  How is it different, if at all?  The action of the diodes is not very different, but because their behaviour has no impact on the gain of the op-amp stage preceding them, there is none of the pseudo-compression taking place.  All the gain of the op-amp is applied to all of the signal all of the time.

In my view, clipping is clipping is clipping.  What makes the difference in the clipping we hear is the proportion of the audio signal that the clipping occurs for.  If clipping occurs for a millisecond or two at the input peaks because of headroom limitations, we think of that as simply a weakness of the circuit because it is momentary and not consistent.  If it occurs for the brunt of the initial attack portion, we tend to refer to that as overdrive, in the sense that the amplifying circuit is overwhelmed by plucked/picked transients, and colours the signal with additional harmonic content until the peak subsides.  If the clipping and additional harmonic content persists for much of the duration of the signal input, almost irrespective of the input amplitude, we tend to call that fuzz.

The intensity and duration of clipping will be largely determined by two parameters: the threshold/forward-voltage of the diodes used, and the gain applied by whatever circuit accompanies them.  I like to refer to their combination as "proximity to clip".  More gain brings the signal ever closer to that threshold for conduction, and a lower threshold means less gain is required to reach it.  Naturally, a hotter input signal both approaches the threshold more readily, and requires less gain to do so.

Since diode conduction also sets a limit on maximum signal output amplitude, there are tradeoffs to be made when it comes to clipping intensity.  If I use diodes with a lower forward voltage (e.g., germanium or schottky), they will certainly clip more easily, yielding more perceived distortion, but they will also do so at the cost of a drastically reduced output level.  If I use diodes with a higher forward voltage (e.g., blue or white LEDs), I will get a much hotter output (roughly 5-6 times the amplitude of germanium), but unless gobs of gain are applied prior to those diodes, clipping will be confined only to initial transient peaks.

You can probably see where this is headed.  If you have enough pre-gain to push diodes with a low forward voltage hard, chances are some form of gain recovery will be needed to bring the clipped signal level up to a desirable amplitude.  If the diodes have a high Vf, then much more gain and a hotter input signal will be needed to extract significant clipping from them.

Some circuits may attempt to clip twice or more, by affixing diodes to an additional gain stage in some manner, such that the already-clipped signal (which has lost some amplitude during the clipping) is brought back up in level again, such that a second set of diodes can also clip.

I hope this is clear.

Whether diode pairs go to ground or to Vref will depend on the circuit and how it is biased and conserves bias voltage across the multiple stages within the circuit.

amz-fx

Here is a simple clipper that you can use:



You can put it in the signal path anywhere you want to clip the signal. The larger the signal that goes into the input, the more the sound will be clipped or distorted (up to a point anyway). The output signal is going to stay stable around 1.4v pk-pk no matter how hard you drive it.

The resistor can be any value from 0 ohms to 10k. It is basically just to protect your driving circuit from damage (in case it has no protection).

regards, Jack

j_flanders

Quote from: RamonRivera on December 10, 2019, 09:04:52 AMmy understanding is the this voltage is used to bias the diodes to get a harder clipping so im trying to figure out how do I build this.
The diodes clip at around 600mV.
To make them clip harder you need to use a signal that's (a lot) bigger than 600mV which is why we need to amplify the tiny guitar signal.
If you supply a 100mV signal nothing will be clipped off.
If you supply a 700mV signal then 100mV will be clipped off. (a bit of clipping)
If you supply a 4,5V signal then 3,9V will be clipped off.(much harder clipping)

The guitar signal is like a sine wave that goes from 0V up to +100mV, then down through 0V, and down to -100mV and then back up gain and so on.
It's too tiny by itself to be clipped by diodes. We need to amplfy it beyond that 600mV threshold of the diodes.
The problem is that we use 9V batteries or 9V adapters. They only provide a positive 9V voltage.
We can only amplify the part where the guitar signal goes from 0V to + 100mV
We cannot amplify where it goes from 0V to -100mV because the most negative voltage the battery gives is 0V (and not -9V)

What we do is create an 'artificial 0V point' by adding 4,5V to the guitar signal. 4,5 being right in the middle between 0V and 9V.
Now the guitar signal goes from 4,5V(0V + 4,5V) up to 4,6V (100mV + 4,5V), down through 4,5V(0V + 4,5V), and down to 4,4V (-100mV + 4,5V) and then back up again and so on.
Now we can amplify the +100mV peak 45 times: to 9V (4,5V + 45 x 100mV)
We can also amplify the -100mV peak 45 times: to 0V (4,5V + 45 x -100mV)

This way we can amplify both postive and negative parts of the guitar signal/sine wave.

Right before the signal exits the pedal we remove that 'artificial' extra 4,5V so the 9V positive peak becomes 4,5V and the 0V becomes -4,5V.

That 4,5V is the bias. Inside the circuit it serves as the 'artificial 0V reference around which the signal moves up and down' and to which you connect the diodes.

If you have a dual power supply (-9V to +9V) instead of a single power supply (+9V) there would be no need for this 'artificial mid point bias' and you would always connect to ground or 0V.

antonis

Quote from: RamonRivera on December 10, 2019, 09:04:52 AM
my understanding is the this voltage is used to bias the diodes to get a harder clipping..

Not actually..

This "voltage" can do nothing, 'cause either is refered on DC source (AC ground) or on what said above about Vbias..
For the second case, there is almost always a big filtering cap (AC ground again).
In case of cap absence, things get worse 'cause there are 2 resistors in parallel (forming the Vbias voltage divider) leading to GND..
The equivalent resistance in set in series between diode pair and GND, so it lifts up clipped signal voltage (by the amount of resistor(s) value times current flowing through diodes..)

A brute way to get harder clipping with back-to-back diode pair is to use diodes of lower forward voltage drop..
(the above, of course will result into lower (unclipped)signal voltage, hence lower Volume..)

edit: There are too many fast typing guys around here.. :icon_redface:
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

amz-fx

#8
Quote from: Mark Hammer on December 10, 2019, 09:37:32 AM
In my view, clipping is clipping is clipping.

Heresy!   :icon_mrgreen:

Quote from: Mark Hammer on December 10, 2019, 09:37:32 AM
In the case of something like the Tube Screamer, diodes are placed in the feedback loop of an op-amp.  The gain of the op-amp is given by how much negative feedback is fed from the output to the input.  When the diodes begin to conduct, more negative feedback is allowed to pass, unimpeded, such that the gain is momentarily reduced.  This results in a more compressed sound/signal, because the gain is being pulled back in response to signal peaks.

What gets called "hard clipping" is generally of the form that j_flanders shows in the Rat schematic, where diodes go to ground or Vref.  How is it different, if at all?  The action of the diodes is not very different, but because their behaviour has no impact on the gain of the op-amp stage preceding them, there is none of the pseudo-compression taking place.  All the gain of the op-amp is applied to all of the signal all of the time.

In the case of something like the Tube Screamer, diodes are placed in the feedback loop of a non-inverting op-amp. There is a significant difference with that compared to diodes to ground.

Diodes to ground just hard clip the signal. With the non-inverting op amp, the gain stage can only be clipped to gain=1, so when you drive the TS clipping stage hard, there is always some of the original input signal riding on top of the clipped signal. This softens the clipping. You can easily see this on a scope if you feed a triangle wave into the non-inverting clipper since the pointed top of the triangle is always present once clipping is engaged.

Best regards, Jack

j_flanders

Maybe you want to keep it simple like Jack's example.
You could google for 'black ice distortion' (basically the same Jack shows)
https://www.google.com/search?q=black+ice+distortion
For example:
https://www.instructables.com/id/Passive-Guitar-Overdrive-Black-Ice/
It usually uses diodes that clip around 300mV instead of 600mV, so you don't need to amplify the incoming signal.

RamonRivera

Quote from: amz-fx on December 10, 2019, 09:45:53 AM
Here is a simple clipper that you can use:



You can put it in the signal path anywhere you want to clip the signal. The larger the signal that goes into the input, the more the sound will be clipped or distorted (up to a point anyway). The output signal is going to stay stable around 1.4v pk-pk no matter how hard you drive it.

The resistor can be any value from 0 ohms to 10k. It is basically just to protect your driving circuit from damage (in case it has no protection).

regards, Jack

This is what I've been using to clip my signal, but with out the caps and resistor, found that adding the resistor introduces noise I don't really know why, are the caps really necessary do the filter the frequency in any way?

RamonRivera

Quote from: Mark Hammer on December 10, 2019, 09:37:32 AM
Diodes "clip" by conducting when the signal approaches some threshold dictated by the diodes, and things around them.  When they conduct, the signal cannot be increased in amplitude any further.

Terms like "hard" and "soft" clipping are not well communicated, hence not well understood.  In the case of something like the Tube Screamer, diodes are placed in the feedback loop of an op-amp.  The gain of the op-amp is given by how much negative feedback is fed from the output to the input.  When the diodes begin to conduct, more negative feedback is allowed to pass, unimpeded, such that the gain is momentarily reduced.  This results in a more compressed sound/signal, because the gain is being pulled back in response to signal peaks.

What gets called "hard clipping" is generally of the form that j_flanders shows in the Rat schematic, where diodes go to ground or Vref.  How is it different, if at all?  The action of the diodes is not very different, but because their behaviour has no impact on the gain of the op-amp stage preceding them, there is none of the pseudo-compression taking place.  All the gain of the op-amp is applied to all of the signal all of the time.

In my view, clipping is clipping is clipping.  What makes the difference in the clipping we hear is the proportion of the audio signal that the clipping occurs for.  If clipping occurs for a millisecond or two at the input peaks because of headroom limitations, we think of that as simply a weakness of the circuit because it is momentary and not consistent.  If it occurs for the brunt of the initial attack portion, we tend to refer to that as overdrive, in the sense that the amplifying circuit is overwhelmed by plucked/picked transients, and colours the signal with additional harmonic content until the peak subsides.  If the clipping and additional harmonic content persists for much of the duration of the signal input, almost irrespective of the input amplitude, we tend to call that fuzz.

The intensity and duration of clipping will be largely determined by two parameters: the threshold/forward-voltage of the diodes used, and the gain applied by whatever circuit accompanies them.  I like to refer to their combination as "proximity to clip".  More gain brings the signal ever closer to that threshold for conduction, and a lower threshold means less gain is required to reach it.  Naturally, a hotter input signal both approaches the threshold more readily, and requires less gain to do so.

Since diode conduction also sets a limit on maximum signal output amplitude, there are tradeoffs to be made when it comes to clipping intensity.  If I use diodes with a lower forward voltage (e.g., germanium or schottky), they will certainly clip more easily, yielding more perceived distortion, but they will also do so at the cost of a drastically reduced output level.  If I use diodes with a higher forward voltage (e.g., blue or white LEDs), I will get a much hotter output (roughly 5-6 times the amplitude of germanium), but unless gobs of gain are applied prior to those diodes, clipping will be confined only to initial transient peaks.

You can probably see where this is headed.  If you have enough pre-gain to push diodes with a low forward voltage hard, chances are some form of gain recovery will be needed to bring the clipped signal level up to a desirable amplitude.  If the diodes have a high Vf, then much more gain and a hotter input signal will be needed to extract significant clipping from them.

Some circuits may attempt to clip twice or more, by affixing diodes to an additional gain stage in some manner, such that the already-clipped signal (which has lost some amplitude during the clipping) is brought back up in level again, such that a second set of diodes can also clip.

I hope this is clear.

Whether diode pairs go to ground or to Vref will depend on the circuit and how it is biased and conserves bias voltage across the multiple stages within the circuit.

Hey mark thanks for taking time to answer, what I understand from you post is that in order to get higher distortion my options would be to some how get more gain by using a amp or to use lower threshold/forward-voltage diodes?

antonis

#12
Implementing both can result into more "squared" waveform edges..

(On behalf of not allowed to answer heretics..)  :icon_lol:

"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

Mark Hammer

Quote from: RamonRivera on December 10, 2019, 10:08:24 AM
Hey mark thanks for taking time to answer, what I understand from you post is that in order to get higher distortion my options would be to some how get more gain by using a amp or to use lower threshold/forward-voltage diodes?

You're welcome.  And yes, you are correct in your inferences.

The thing people tend to forget about clipping circuits for string instruments is that a guitar string is NOT an oscillator that produces a stable signal with fixed spectral content.  Its amplitude and spectral content change rapidly over the first 30-50 milliseconds, and continue to change as the string decays.  Anything intended to yield distortion from a guitar signal MUST always factor that into the equation.

PRR

> adding the resistor introduces noise

What value resistor?

In a synth clipper, the suggested 1k (1k-3k) will not add noise, and the cranked synth can deliver the 1 Volt or so needed to "clip" ordinary diodes. (For guitar we have to buffer and boost-up first, so it gets more complicated.)
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RamonRivera

#15
Quote from: PRR on December 10, 2019, 01:48:52 PM
> adding the resistor introduces noise

What value resistor?

In a synth clipper, the suggested 1k (1k-3k) will not add noise, and the cranked synth can deliver the 1 Volt or so needed to "clip" ordinary diodes. (For guitar we have to buffer and boost-up first, so it gets more complicated.)

I was using a 10K resistor pot but I find that im using the pot at maximum and I get more distortion with out it so that why I took it out of the circuit, I don't know how much volts does my sound card can deliver but the specs are as follows Max output level (+4dBu ref): +20dBu, im guessing 7 Volts? Im not sure but this is the waveform I get from sending a sine wave at -12 DdFS




antonis

Your waveform shouldn't be interpreted clipped by no mean..
Can't read your scope co-ordinates values, sorry.. :icon_redface:
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

RamonRivera

Can someone help me, how could I amplify the signal coming from my audio interface and into the diodes? do I need and Opamp? in which case which one and how could I set this up.

iainpunk

¨Electra distortion¨, its more of an overdrive than a distortion, but sounds good for how simple it is

friendly reminder: all holes are positive and have negative weight, despite not being there.

cheers

RamonRivera

Quote from: iainpunk on December 15, 2019, 08:55:45 PM
¨Electra distortion¨, its more of an overdrive than a distortion, but sounds good for how simple it is


Thanks this seems like a nice start, just a question if I change the 680r to a higher value that would give this circuit more distortion?