Referring to the attached image file, I am trying to figure out what the output of the second diagram will be. It looks apparent that the reverse (feedback) resistance will be substantially lower in the second case giving a much lower output voltage.
Please, what is the mathematical relation of the input/output voltages?
This circuit displays the log function when you apply a narrow input amplitude, between 0.3 and 0.6 V.
Or, between -0.3 and -0.6V (for the circuit which has 2 diodes).
Since I was not sure about the output waveform, I used a simulator to check. This circuit is a clipping circuit, which limits the output voltage to 0.6V amplitude.
I think the first mathematical expression is wrong because of the polarity effect. I have forgotten what to do with uni-polarity signals in maths.
Frank