How small signal rectification circuit with opamp works ?

Hello, could you explain how D1 works in this circuit,
If Up > 0 (Up or U1:A(+IP), U+ at opamp) DONE! I got it, I just ignore the left of the original circuit.
That I don't understand is what happens when Up < 0.

At first, we do nothing then Up = Un (U- at opamp) = 0, it's ok !
In the class they say: if Up < 0 then Up < Un so U1:A(OP) < 0 so current goes through D1 then Un=0 I just don't get it.
In my mind, Up=Un then Un must not be zero ! But I verifiy that I'm wrong when I draw this circuit in Proteus.

Could you explain it for me or guide me to some references.
Thank you!

When the input is positive the output goes positive but the diode blocks any feedback so the voltage across R1, R2 and R3 stays at 0V. When the input goes slightly negative the output goes more negative until D1 conducts because of the high open loop op amp gain. Then the negative feedback causes the voltage at the (-) input to essentially equal the input voltage at the (+) input, thus the voltage across R1 and R2 in series with R3 is equal to the input voltage. (Not sure what the purpose of R1, R2, and R3 is.) :?:

Thus the circuit acts as a precision (near ideal) negative half-wave rectifier with the output equal across R1 (and R2-R3) equal to the input for negative input voltages, and equal to 0V for positive input voltages.

Note that in real life the 100Ω resistors are too small for that op amp to drive at much more than a half volt maximum due to the op amps limited output current capability.

Although the circuit works as a rectifier, I wonder if you copied the original circuit wrongly.

The thing is not at resistor, maybe it is 10k instead. But the thing is
"Then the negative feedback causes the voltage at the (-) input to essentially equal the input voltage at the (+) input"
I think so but my teacher says that wrong. He is resonable I sure that, that how rectification works, at the positive input it works correctly as a rectification, but don't look at it, I just threw the D2 from opamp output to R3(2). But at the negative input, I ask why Un = 0 ?

- - - Updated - - -

The thing is not at resistor, maybe it is 10k instead. But the thing is
"Then the negative feedback causes the voltage at the (-) input to essentially equal the input voltage at the (+) input"
I think so but my teacher says that wrong. He is resonable I sure that, that how rectification works, at the positive input it works correctly as a rectification, but don't look at it, I just threw the D2 from opamp output to R3(2). But at the negative input, I ask why Un = 0 ?

Sorry but I really don't really understand your explanation or question. Where is Un on the schematic? Please use short sentences with a capital letter at the start and a period at the end.

Are you sure you've drawn that circuit correctly?