feedforward clipper

If the overshoot is symmetrical, it's all odd harmonics.

Cheers

Phil Hobbs

Do you even know what feedforward is?

But, this is about two clipping elements in series, in opposing polarity. By construction, any overshoot when device A is active, is mirrored by an undershoot when device B is active. So, there ought to be no odd harmonic 'overshoot' contribution. The symmetry requires true clipping to be odd harmonics, and the effect of an overshoot is even, which makes it easy to detect.

Even harmonics are asymmetric between half cycles. Odd ones aren't. Unless I'm misunderstanding badly, assuming identical devices there will be no asymmetry between half cycles, ergo no even harmonics.

Cheers

Phil Hobbs

You're right. I got confused about how many inversions there were in the second part of the clipping. Instead of making a second harmonic, it'd just phase-shift the fundamental (add some cosine character to a pure sinewave). That's detectable, I suppose, but not very neat.

That's what the author, Douglas Self, calls this particular example. If you are indicating that you have a problem with the terminology, please take it up with him

If I recall correctly, one of the "advantages" of this circuit mentioned in the text is that R4 gets to be shared between opamps...Haha

I'll see if I can find the passage.

On page 702 of the book:

"Since A1 and A3 are never active at the same time, they can share resistor R4, which make for a very elegant circuit."