OpAmp circuit with diodes: conduction conditions
Lets analyse this circuit by Kirschoffs law:

as long as your (perfect) OPAMP is not saturated it tries to regulate the output like like that the diferential input is 0V.

M1: \$U_{R1} + u_{diff} - V_2 = 0 =&gt; U_{R1 }= V_2\$

M2: \$-U_{R1} + U_{D1} - V_1 + U1_{OUT} = 0 =&gt; U1_{OUT} = V_2 - U_{D1} + V_1\$

M3: \$-U_{R1} - U_{D2} - U_{R2} + U1_{OUT} = 0 =&gt; U1_{OUT} = V_2 + U_{D2} + U_{R2}\$ 

Lets say both diodes are ideal and conduct @0.5V
And lets also say the part D1 is not assembled then you get:

\$U1_{OUT} = 2*V_2+0.5V\$ (D2 is always conducting a \$U1_{OUT}\$ always bigger than \$V_2\$).

With this knowledge we add now D1 and get:

\$2*V_2+0.5V = V_2 - U_{D1} + V_1\$

\$V_2 + 0.5V = -U_{D1} + V_1\$

\$V_2 = -2*0.5V + V_1\$ => D1 is conduction between \$V_2=0\$ and \$V_2 = 1V\$

So between \$V_2 = -VDD..1V\$ voltage \$U1_{OUT}\$ is determined by:

\$U1_{OUT} = V_2 + V_1 - U_{D1} = V2 + 1.5V\$

Above \$V_2 = 1V..VSS\$ its determined by:

\$U1_{OUT}=2*V_2+V_{D2}\$