Electronic – Help understanding circuit with diodes in it

Look at this schematic:

^{simulate this circuit – Schematic created using CircuitLab}

Now, V over R1 is given by: V(R1) = I(R1) * R1.
And V over R2 is given by: V(R2) = I(R2) * R2.

Can you see that there is only one loop for the current? That means I(R1) needs to be equal to I(R2), or I(R2) === I(R1).

Thus the equations become:

V(R1) = I(R1) * R1
V(R2) = I(R1) * R2

Since also given that R1 === R2, you can also substitute those.

V(R1) = I(R1) * R1
V(R2) = I(R1) * R1

And presto, the equation for V(R1) and V(R2) are now exactly the same:

V(R1) = I(R1) * R1 = V(R2) --> V(R1) = V(R2)

Thus the voltages need to be the same, so the point at V2 will be exactly half the voltage at V1.

Let's let you try the same trick for 4 resistors, see if you got it.

The OpAmp is used in an inverting configuration, i.e. when you apply a positive Vi, the output Vo will be negative. So, the current flows through R3 and the upper path (R1, D1), while current through the lower path is blocked by D2.
So, for a positive input, Vo+ will be negative and Vo- will be zero.
The exact value on Vo+ is determined by the ratio of R3 and R1, as the output terminal Vo will be a value which cancels out the voltage at the negative input terminal.

To examine this a bit more, you can do a simulation, even here. I've drawn your circuit using the circuit designer of EE. Click on simulate this circuit and do a DC sweep for Vi.

^{simulate this circuit – Schematic created using CircuitLab}

The following picture shows the result, blue is Vo+, orange Vo-. I've also added Vo (brown) just to see the effect of the diodes. (Click the image to enlarge)