Inverted Op Amp clipping

[WGTP]

If I use a single diode in the feedback loop and the op amp is inverting it, does that mean each time the signal comes around the loop the diode clips the other side of the wave form from the time before :?:

Also, if I want to put a single diode to ground after the op amp which way to I face it to clip the opposite side of the wave form from the one in the op amp loop :?:

Maybe be a really silly question.

[niftydog]

a diode clips when the signal at it's anode is greater than that at the cathode. (cathode has the band on the body)

Inverting op amp; when the input goes high the output goes low. When the input goes low, it sends the output high.

So, if you have the diode "pointing" towards the output of the op amp, it will clip when the input goes high. (vice versa for when it's pointing towards the input.)

Now, the op amp diode clips when the input is high (read; output is low) so you want the other diode to clip when the output is high (opposites). That is to say that you want the anode to be at a higher voltage than the cathode. I figure you're just connecting the diode to ground - so you connect the cathode to ground and the anode to the op amp output.

clear as mud?

[WGTP]

Thanks, that helps.  With all that inverting and negative feedback stuff going on, I couldn't figure it out.  I am also not sure I can hear a difference between doing it one way or another.  It could be with the op amp clipping on one side of the wave form and the diode clipping the other, that by the time the signal gets to the diode to ground at the end the signal path, it doesn't matter.  Not sure.  Guess I need a scope.  

[puretube]

scope is great: helps (a.o.) with asymmetrical clipping quests...

[WGTP]

Does "inverting" mean amplitude or phase or both???   :?:

[niftydog]

inverting inputs invert the voltage. As a consequence, the phase is therefore shifted by 180 degrees.