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Re: Variable Clipping Threshold

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11/25/1999 12:09 PM

Re: Variable Clipping Threshold

OK I think I got your way of looking at it.

The (-) pin is forced to be at the same potential as the input (+), so the current at the resistor to ground is proportional to the input, and the current from the output to pin(-) must be equal to this, so the current at the loop is proportional to the input. The output voltage will be the input voltage plus the voltage developed by this current through the loop. In an inverting amplifier the output voltage is only dependant on the current on the loop, not on the input voltage. If you've got only linear resistors that's the same thing, but with diodes on the loop it makes a difference.

If you consider the (+) input to be ground and the ground at the loop to be signal, it's like an inverting amplifier, but you also have to reconsider where the OUTPUT will be referenced to.

Anyway if you want to look at the variclip as an inverting amplifier here it goes. The current through the loop is given by the input and by the resistor to ground. This current must be equal to the current through the feedback resistor plus the current through the diodes. But the current that the opamp puts out is greater than this because the lower part of the "threshold" pot shunts some current to ground. The output voltage depends on the current through the diodes and on the current through the upper part of the threshold pot. It's not just like adding series resistance to the diodes, you're "stealing" some current to ground and that's what effectively increases the apparent clipping threshold.

Of course you do have an increase in series resistance at the same time, so you have a mix of both effects. If you add another opamp so that you can steal current from the diodes while still driving them with a low impedance then you've got an "ideal" variclip. I'll draw an schematic and post it.

11/25/1999 10:34 PM
Jack	I have modified the schematic of the Variclip to show how to use another opamp to buffer the threshold control and drive the diodes with a constant low Z source. http://www.muzique.com/misc/variclip.gif regards, Jack

11/26/1999 11:51 AM

Jack,

The output of Ic1a is biased at Vr and the pot goes to ground. There's Vr VDC on the pot and the pot position will change the bias of IC1b.

The threshold pot should be grounded to "Vr" and you don't need the cap from the second opamp to the diodes because the output of IC1b will also be at 4.5V.

OR move the output cap before the pot and the other cap to the pot wiper - then there's no DC on the pot. You'll need to add a resistor from Ic1b pin(+) to Vr.

11/26/1999 2:24 PM

>>The output of Ic1a is biased at Vr and the pot goes to ground. There's Vr VDC on the pot and the pot position will change the bias of IC1b.

yes, I realized that when I drew it up. It will work fine in this implementation using the output to bias the input of the second opamp... one thing you have to consider is that sometimes when we post things we don't always reveal the very best version! Gotta save some goodies for my new e-zine.

best regards, Jack

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