Let's assume we have three-phase diode (not SCR, ignore those in the picture) rectifier like in the picture:
Assuming we have no source inductance and other ideal conditions, diodes from 1 to 6 are conducting at following angles (of course we need to add some load):
Diode 1: 30º < X < 150º
Diode 4: 210º < X < 330º
Diode 3: 150º < X < 270º
Diode 6: 330º < X < 450º (= -90º)
Diode 5: 270º < X < 390º (= -30º)
Diode 2: 90º < X < 210º
As we can see, each of the diodes conduct exactly 120º per cycle. Also, conduction of diodes 1,3,5 and 4,6,2 does not overlap.
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. Subsequently, the conduction angles start overlapping. For example, diode 1 conducts from 20º to 160º and diode 3 conducts from 140º to 280º. Diodes 1 and 3 are overlapping for 20º.
My question is – how is the angle of conduction and overlapping angle affected? I know that it changes with source inductance and also current load on DC side, but I would like to know exactly how. There are lots of pages discussing firing angle of the SCRs in this case, but this is not what I am interested in (I only have plain diodes).
Best Answer
simulate this circuit – 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.
To play with the simulator:
I set it up to display from 100 to 150 ms so that the traces would be stable after startup.
Try setting L1, 2 and 3 to zero, 20 and 50 mH to see the effect.
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.
Does this answer help?
From the comments (1):
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):
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°).