Operational amplifier with reverse diode

Question

On the first picture is a simple schematic. For example, we can get the transfer function simply by equaling "ud+u2" and "Ad*(u1-u2)". The transfer function is: u2 = (Ad*u1)/(1+Ad) - ud/(1+Ad)

On the next picture, that diode is turned in other direction. I am not sure how can I get transfer function. My logical explanation (which is probably wrong) is that the diode does not conduct any electricity, so the current there is zero.

I am asking for the explanation, when and why do we have a voltage on u2 (on the last picture).

Answer 1

The schematic corresponds to a Precision Rectifier. This type of circuit is used as a half-wave rectifier signals whose amplitude is less than the threshold voltage of the diode.
The connection with an operational amplifier, linearizes transfer curve for a diode, making it more ideal. This allows rectify signals from virtually zero.

The diagrams that you have posted, corresponding to half-wave rectifiers. An improved implementation of a precision rectifier is:

^{simulate this circuit – Schematic created using CircuitLab}

For a full-wave recitifier, the assembly is:

^{simulate this circuit}

If this circuit you invert the connection of the diodes, you get the output pulses with positive polarity. In the circuit shown, the polarity of the pulse is negative.

Here, you can view a CI that implements a precision rectifier in this manner.

Answer 2

In both cases, there should be some load resistor shown to compute gain but even with that when the diode is reversed the OP Amp becomes open-loop and saturates, so neither are practical.

The improved design is as follows.