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I want to create an guitar distortion effect. What I want is a unity gain buffer op-amp that clips the signal. In the drawing below there is a circuit found on this forum here which will clip a very low voltage (~200mV) signal. However it will only clip at a particular voltage. What if you need (as I do) to clip simply the peaks off every wave?

I could provide some sort of rectified average peak of the input signal int 'mclamp' and 'clamp' such that the clamping voltage roughly followed the input signal as it rises and falls. For this I would presumably use some sort of 'ideal' rectifier (because of the 0.7 breakdown voltage of a diode bridge).

First.. is there an IC that will do this that anyone knows of? And secondly.. Is what I'm proposing viable?

enter image description here

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  • \$\begingroup\$ That's a very unusual requirement. Obviously, you can't clip at a percentage of the peak value until you've seen what the actual peak value is, so any circuit that implements this would have to incorporate a time delay of some sort. The easiest way to do this would be in a DSP chip. \$\endgroup\$
    – Dave Tweed
    Jun 17, 2016 at 16:54
  • \$\begingroup\$ @Dave yes there'd have to be a time delay. I was thinking some small capacitor arrangement. I'd like to avoid DSP if possible. I want others to be able to replicate the effect with easy off the shelf components. \$\endgroup\$
    – Richard
    Jun 17, 2016 at 20:23

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Here's a partial solution that you could combine with the previous circuit. This circuit differentiates the input signal in order to locate where the peaks are, and then uses a set of sample/hold circuits to capture the values of those peaks. The outputs of this circuit could be used (with some voltage dividers) to set the clipping thresholds to some fraction of the previous peak's value. Not exactly what you asked for, but probably about as close as you can get with analog hardware.

schematic

simulate this circuit – Schematic created using CircuitLab

simulation results

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  • \$\begingroup\$ wow.. thanks very much for your work here! That's two approaches I have now. In this circuit, any DC offset is passed into the sample hold because of the placement of C1. Is this intended? \$\endgroup\$
    – Richard
    Jun 20, 2016 at 11:15
  • \$\begingroup\$ Yes, of course it is. C1 is the start of the side-chain that is simply determining where the inflection points (high and low peaks) of the signal are. The actual signal path is the upper wire that leads directly to the S/H blocks. If you want to block DC offsets, that's a separate problem. \$\endgroup\$
    – Dave Tweed
    Jun 20, 2016 at 11:21
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What you're proposing isn't really viable the way you have described it. You need a way to trigger the clipping, which is very simple with a voltage threshold. It's not so easy to say, "start clipping just before it reaches any peak" especially when you don't have any digital signal processing happening. Your drawing of the audio wave also doesn't show the situation where the wave looks like it's peaking, but then goes flat and turns upward again. When should the device begin clipping? When should it stop clipping? It's a fairly complex problem.

The basic idea that you're describing could be accomplished in several different ways, but none of them are very simple. Here's a way that wouldn't be terribly complicated:

First, you need to run your audio signal through an envelope detector to save the varying amplitude of the wave. Also route the original audio signal through an audio compressor to make the entire wave as close to the same amplitude as possible. Then apply your voltage threshold based clipping circuit, and reapply the amplitude from the envelope detector.

In reality, that's not much different from what distortion pedals do, except they don't typically restore the original envelope. It might be fun and interesting to see how that sounds.

By the way, you may be able to test all of this out by using a MOD DUO pedal.

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  • \$\begingroup\$ Oooh. Thank you. Why would we need companding, why not just feed the output of the envelope detector into the clamp inputs? \$\endgroup\$
    – Richard
    Jun 17, 2016 at 21:40
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    \$\begingroup\$ Well, for starters, the clamping voltage would be moving, which would mean your flattened peaks wouldn't be flat. Also, the envelope detector is essentially a low-pass circuit, so depending on how responsive you want it to be, there will always be some lag both rising and falling. \$\endgroup\$
    – TimH - Codidact
    Jun 17, 2016 at 22:48

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