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Re: Variable Clipping Threshold

11/23/1999 2:27 PM
Mark Hammer	Re: Variable Clipping Threshold So, just s'posin you were building or modifying a BMP, and thought "What the hell. There's room on the chassis, and the 5k pot is just sitting there unused anyways.", would you stick it in the first diode pair stage or the second? My own hunch is that first stage would be more productive since it constrains the additional harmonics from the second stage. There ARE other tastes, though, and that's why I ask. On a side note, I've seen circuits where a series back-to-back diode pair was preceded with individual variable resistors to adjust the rise time of the half cycle going through each diode. I always wondered about tacking that circuit on the back of Craig Anderton's comparator-type "Ultra-Fuzz", as a way of varying waveform from square to ramp (although my preference would be to have one pot that pans between the two diodes, rather than two pots). I also noticed on the schematic of the BOSS Heavy Metal (HM-2) that there is a back-to-back diode pair in series in the signal path, and I'm wondering if that doesn't present the same generic possibilities for varying half-cycle rise time. I have a Korg Waveshaper from their ill-fated modular series in the early 80's, and it has a waveshape control that goes from square to ramp. The circuit is a little dense to reverse engineer, but I'm wondering now if it does the same thing. Finally, I have a circuit from a late 70's issue of Elektor that provides almost complete parametric control of clipping for each half cycle (threshold, onset, etc.). I'm glad you reminded me of it. Given that it is one of the few articles I'm able to locate, I'd be happy to scan it and send it to you.

11/23/1999 2:27 PM

Mark Hammer

Re: Variable Clipping Threshold

So, just s'posin you were building or modifying a BMP, and thought "What the hell. There's room on the chassis, and the 5k pot is just sitting there unused anyways.", would you stick it in the first diode pair stage or the second? My own hunch is that first stage would be more productive since it constrains the additional harmonics from the second stage. There ARE other tastes, though, and that's why I ask.

On a side note, I've seen circuits where a series back-to-back diode pair was preceded with individual variable resistors to adjust the rise time of the half cycle going through each diode. I always wondered about tacking that circuit on the back of Craig Anderton's comparator-type "Ultra-Fuzz", as a way of varying waveform from square to ramp (although my preference would be to have one pot that pans between the two diodes, rather than two pots). I also noticed on the schematic of the BOSS Heavy Metal (HM-2) that there is a back-to-back diode pair in series in the signal path, and I'm wondering if that doesn't present the same generic possibilities for varying half-cycle rise time. I have a Korg Waveshaper from their ill-fated modular series in the early 80's, and it has a waveshape control that goes from square to ramp. The circuit is a little dense to reverse engineer, but I'm wondering now if it does the same thing. Finally, I have a circuit from a late 70's issue of Elektor that provides almost complete parametric control of clipping for each half cycle (threshold, onset, etc.). I'm glad you reminded me of it. Given that it is one of the few articles I'm able to locate, I'd be happy to scan it and send it to you.

11/23/1999 3:46 PM
Jack	>>My own hunch is that first stage would be more productive since it constrains the additional harmonics from the second stage. My first hunch was that it would be better on the second stage since if it was on the first, the heavy clipping of the second stage might erase the fine variations from the previous stage. Bottom line is that it probably should be tested both ways to see which is best. >>I always wondered about tacking that circuit on the back of Craig Anderton's comparator-type "Ultra-Fuzz", as a way of varying waveform from square to ramp yes, good idea. It will produce a somewhat sawtooth type wave. I have a similar filter posted: http://www.muzique.com/amz/ffilter.gif R.G. left me a note on Aron's BBS for a few items to think about on the Threshold Control so I'm going to be working on that along with some other variations. regards, Jack

11/23/1999 3:46 PM

Jack

>>My own hunch is that first stage would be more productive since it constrains the additional harmonics from the second stage.

My first hunch was that it would be better on the second stage since if it was on the first, the heavy clipping of the second stage might erase the fine variations from the previous stage. Bottom line is that it probably should be tested both ways to see which is best.

>>I always wondered about tacking that circuit on the back of Craig Anderton's comparator-type "Ultra-Fuzz", as a way of varying waveform from square to ramp

yes, good idea. It will produce a somewhat sawtooth type wave. I have a similar filter posted: http://www.muzique.com/amz/ffilter.gif

R.G. left me a note on Aron's BBS for a few items to think about on the Threshold Control so I'm going to be working on that along with some other variations.

regards, Jack

11/24/1999 12:02 AM

R.G.

quote:
"No, sorry, in this particular case it does matter whether the diodes are connected to ground are not. If you just move the diodes from the -input to ground, it does not work the same."

Sorry? Whatever for?

No, it does not matter. Think about it, Jack. The diodes and resistor don't know or care where they're connected. All they know is that they're driven by some source voltage through some source impedance. Change the source voltage or impedance, and things change for the diodes. An opamp works by making an output voltage be the proper level to force the correct *current* through the feedback network. The output voltage is a reflection of the voltage that the current causes on the feedback element.

If you simply connect the diodes to ground it sounds different. Of course it does - with the diodes connected in the feedback loop of an opamp, they're effectively driven by a current source by the opamp's feedback action. Tie them to ground, and you've changed from a current source to a voltage source, zero source impedance - quite a different circumstance. Drive them from the output of the opamp through a 10K resistor as in a Distortion Plus, and you have yet another variation; now the source has a 10K impedance, and is about halfway (to the diodes, at least) to being a current source. They sound different again.

The clipping varies continously from source impedances of zero ohms up to infinity (a current source), and the harmonics vary with that. I noted this in my earlier posts here about a year ago - it's the input resistor at the (-) input of the opamp that determines the current in the feedback elements, and shortly later that using a natural current source output like an OTA might produce some very soft clipping as the diodes were driven by a current source, making the voltage the instantaneous log of the applied current.

In any case, the diode voltages are only whatever the current through them forces. Insert a resistor to take up more of the feedback voltage available, and the diodes can only go to the point on their characteristic that the current lets them.

11/24/1999 1:40 AM
Jack	Thanks for contributing your analysis. It helps to get a better understanding of why the outputs that I'm seeing from the series vs. divider are so different. >>with the diodes connected in the feedback loop of an opamp, they're effectively driven by a current source by the opamp's feedback action. Tie them to ground, and you've changed from a current source to a voltage source, zero source impedance - quite a different circumstance. NOW we're on to something really interesting. I think this is an area that could be productive to explore. regards, Jack

11/24/1999 1:40 AM

Jack

Thanks for contributing your analysis. It helps to get a better understanding of why the outputs that I'm seeing from the series vs. divider are so different.

>>with the diodes connected in the feedback loop of an opamp, they're effectively driven by a current source by the opamp's feedback action. Tie them to ground, and you've changed from a current source to a voltage source, zero source impedance - quite a different circumstance.

NOW we're on to something really interesting. I think this is an area that could be productive to explore.

regards, Jack

11/24/1999 3:37 PM
nic	Observations semi-on-topic. When I tie a diode clamp to my signal ala "black Ice" passive distortion I get the clipping I'd expect... HOwever when I tie the clamp after my factory tube screamer in bypass mode I get now distortion... What is going on here? Impedance issues? Is the output buffer stopping this? I expected the buffer to "push" the distortion even more. I wonder what would happen if I put the clamp before the buffer... What I was going for was a blending of the screamer and the distortion + for a bit more edge on call. nic

11/24/1999 3:37 PM

nic

Observations semi-on-topic.

When I tie a diode clamp to my signal ala "black Ice" passive distortion I get the clipping I'd expect... HOwever when I tie the clamp after my factory tube screamer in bypass mode I get now distortion... What is going on here? Impedance issues? Is the output buffer stopping this? I expected the buffer to "push" the distortion even more. I wonder what would happen if I put the clamp before the buffer...

What I was going for was a blending of the screamer and the distortion + for a bit more edge on call.

nic

11/24/1999 5:31 PM

GFR

Re: Variable Clipping Threshold

quote:
"The diodes and resistor don't know or care where they're connected. "

But the opamp cares of what's on its feedback loop.

quote:
"An opamp works by making an output voltage be the proper level to force the correct *current* through the feedback network. The output voltage is a reflection of the voltage that the current causes on the feedback element."

This is for an inverting ampliflier. For a non-inverting amplifier like what's used on the variclip or the TS it should be:

"An opamp works by making an output voltage be the proper level to force the voltage at the inverting pin to be equal to the input. The output voltage is a such to make the attenuated voltage at the feedback elemnt to match the input".

The inverting opamp stage is voltage-current feedback and the non-inverting opamp stage is voltage-voltage feedback.

Imagine the output of the variclip is at the pot wiper. If the output was there it would be almost a "normal" opamp clipper.

 

 
          |
 
----------| 
 
          |  ____||___pota_______potwiper
 
          |  /  | ||          |     
 
     -----|-/   |             |
 
     |    |/    |             |
 
     |          |             |       
 
     +---R2------             |     
 
     |------------>|----------|
 
     |------------|<----------|     
 
     R1                      potb
 
     |                        |
 
    ===                       |
 
     |                        |
 
    gnd                      gnd

I've separated the two sections of the pot to be clearer.

While the diode threshold has not been reached, the output is the same (gain 1+R2/R1) but with a higher output impedance (pota). When a diode conducts, gain is one, output impedance is low.

So AT the pot wiper this is just like a "normal" clipper.

Now at the output of the opamp when the diodes are not conducting the gain is (1 + R2/R1). When the diode threshold (Vbe) is reached the output of the opamp is at a higher voltage given by Vbe*(1+pota/potb).

So at the output of the opamp it's like if the clipping happened at a higher voltage.

If you add another opamp you can lower the threshold.

I don't think this is merely adding series resistance to the diodes and it does matter if the diodes are tied to ground or to the (-) pin - the diodes can't tell the diference, but the opamp can.

I wonder why people don't try diodes in an inverting opamp configuration - you could play with the VxI curve as you note (by forcing a controlled current through the diodes).

11/24/1999 9:58 PM
R.G.	I'm fairly familiar with how opamps work... Let's play topology. You can also consider the signal at the (+) input to be ground, and the ground at the end of the impedance at the (-) input to be signal; the only oddity is that the power supplies move around in this mode, too, but modern opamps have a LOT of power supply rejection, so we'll discount that. If you look at it that way, the "signal" is still at the inverting input, and the output moves around with reference to the moving power supplies. Let's look at it another way. The impedance at the inverting input always determines the current in the feedback element because the inverting input has to maintain charge neutrality - it can't store charge, so the charge into the inverting input node is equal to the charge out. The only place it can go is out the impedance at the input to "ground" (wherever that is...) and at the same time the inverting input must follow the noninverting input voltage - Ohm's law determines the current without reference to the feedback element again. Let's look at it another way. If the element in the feedback path were not driven with a current determined by the input impedances, you couldn't make a differential integrator with a linear ramp - which is clearly possible, that's my standard LFO. It's all in how you look at it.

11/24/1999 9:58 PM

R.G.

I'm fairly familiar with how opamps work...

Let's play topology.

You can also consider the signal at the (+) input to be ground, and the ground at the end of the impedance at the (-) input to be signal; the only oddity is that the power supplies move around in this mode, too, but modern opamps have a LOT of power supply rejection, so we'll discount that. If you look at it that way, the "signal" is still at the inverting input, and the output moves around with reference to the moving power supplies.

Let's look at it another way.

The impedance at the inverting input always determines the current in the feedback element because the inverting input has to maintain charge neutrality - it can't store charge, so the charge into the inverting input node is equal to the charge out. The only place it can go is out the impedance at the input to "ground" (wherever that is...) and at the same time the inverting input must follow the noninverting input voltage - Ohm's law determines the current without reference to the feedback element again.

Let's look at it another way.

If the element in the feedback path were not driven with a current determined by the input impedances, you couldn't make a differential integrator with a linear ramp - which is clearly possible, that's my standard LFO.

It's all in how you look at it.

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