Hi.
I've just read this: "The most common application of the ideal diode is the full wave rectifier. It works by summing one of the ideal diode outputs with the input."
The source page is http://www.uni-bonn.de/~uzs159/diode_tutorial/ (http://www.uni-bonn.de/~uzs159/diode_tutorial/)
Don't you need to add two half waves to get full rectification? Doesn't summing a half-wave rectified signal with a full wave signal just gives another half wave?
Using the ideal diodes shown, would it be a better idea to just feed V1 and V2 into the +ve input of the second (unity gain) op-amp? That would make the "mixer" impedance very high.
thanks for any advice
Just think of it this way: If you add a periodic signal to itself 180 degrees out of phase, the result is 0.
If you split the signal--one side is flipped 180 degrees, the other is divided by two. Then add them. The result is the original signal 180 degrees out of phase at 1/2 the amplitude.
Now take a signal, and it's 180-degree complement for instance. 1/2 wave rectify the 180 degree shifted signal, then add that back to the original. Every half-cycle cancels, and the output overall is a 1/2 wave rectified signal in phase with the input.
Now the final logical step is this: Divide that original signal by 2, then add it with the 180 degree shifted 1/2 wave rectified signal. On the in phase 1/2 cycle, the signal is uneffected. After it dips below 0, then the 180 degree path is rising above one. Since the magnitude of the uneffected is -1/2x, and the 180 degree flipped signal is 1x, we have 1 + (-1/2)=+1/2. The output therefore is a full-wave rectified signal of amplitude 1/2.
I didn't get to read the article in the link because the link was dead. If it's anything like DOD uses in their envelope followers on some compressors, then they're actually leaving the 180 degree flipped signal at unity gain, and adding the noninverted signal at unity + 1 gain to get an equivalent effect.
Hey, that's cool! Very cool.
Thanks.