Commutation overlap angle
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Schematic for simulation.



Figure 2. With inductances set at zero the current supplied by each phase turns on and off smartly at the voltage crossing points. \$I_3 \$ turned off to improve legibility.

C I V I L: 
In a Capacitor I leads V which leads I in an L (inductor).

From the above, we can expect that with source inductance the source current will lag the source voltage.



Figure 3. Simulation plot of Figure 1 with 20 mH inductance in each phase.


Without source inductance we would expect the blue phase, V1, to suddenly supply positive current when the blue voltage exceeds the other two. Instead we notice that the blue current gives a very lazy start as L1 impedes the change in current.
Without source inductance we would expect the blue phase, V1, to suddenly turn off when the orange phase, V2, exceeds V1. It doesn't. Instead we see a lazy turn off as the inductor dissipates its stored energy into the load.
Same as (1) on the negative half-cycle.
Same as (2) on the negative half-cycle.


To play with the simulator:


Hit Edit on my post.
Edit the above schematic.
Run time domain simulation.


I set it up to display from 100 to 150 ms so that the traces would be stable after startup.


Cancel | Cancel when finished so as not to mess up my answer.


Try setting L1, 2 and 3 to zero, 20 and 50 mH to see the effect.


  However, as soon as we look at some more realistic scenario and add source inductance, diodes start conducting more than 120º per cycle because of the commutation process.


This is due to the lazy turn-on and turn-off clearly visible in Figure 3. Compare the zero overlap in Figure 2 with that of 3.


  My question is - how is the angle of conduction and overlapping angle affected?


Does this answer help?



From the comments (1):


  Could you maybe also explain me how does this lazy turn-on/turn-off affect DC voltage and current?


Run the simulation to see. Add in a NODE on the top of the load resistor. Move the GND symbol to the bottom of the resistor so that it will be easier to understand the resultant graph. 



Figure 4. Because, in this example, the load is resistive the load voltage will track the supply current. i.e., It's horrible!



From the comments (2):


  Without source inductance, diode D1 conducts between 30º and 150º, so 120º total. Let's say we add some inductance so it now conducts for 140º total. Does this mean it will now conduct from 20º to 160º (+10º at the beginning and +10º at the end, so symmetrical)?


That sounds correct but notice on Figure 3 that the diodes don't turn "hard-on" anymore as they did in the non-inductive Figure 2. In your example they would fade in from 10° to 30° (centred around 20°) and fade out between 140° and 160° (centred around 150°).