A three-phase diode rectifier converts a three-phase AC voltage at the input to a DC voltage at the output. To show the working principle of the circuit the source and load inductances (Ls and Ld) are neglected for simplicity. The DC voltage is divided into six segments within one fundamental source period that corresponds to the different line-to-line source voltage combinations (VLL). In each segment there is a minimum and maximum DC voltage:
- Minimum DC voltage: If one line-to-line voltage is zero, then the DC voltage is at a minimum of VDC = VLL · sin(60°).
- Maximum DC voltage: The DC voltage increases up to a maximum of VDC = VLL, where two line-to-line voltages are equal.
In between the minimum and maximum DC voltages lies the average DC voltage that is given by: VDC,av = VLL · 3/pi. The ripple of the DC voltage occurs at 6-times the line frequency. For the six intervals the signs of the phase currents (Ia,Ib,Ic) are given by:
|Phase interval||Sign of phase currents|
|0°<φ<60°||( 0,-1, 1)|
|60°<φ<120°||( 1,-1, 0)|
|120°<φ<180°||( 1, 0,-1)|
|180°<φ<240°||( 0, 1,-1)|
|240°<φ<300°||(-0, 1, 0)|
|300°<φ<360°||(-1, 0, 1)|
Influence of inductors
As with the single-phase diode rectifier, the inclusion of a load (Ld) and source inductance (Ls) leads to a current commutation interval between two diode pairs. The larger the source inductance, the more time it takes for the current to commutate. For example, after phase interval 1 (0°<φ<60°) the current commutates from diode pair D5/D6 to D1/D6. During this interval Vca remains equal to zero as D1 and D5 are both conducting, leading to a decreasing DC voltage. The drop of the DC voltage is proportional to the source inductance, i.e., ΔVout ~ Ls.
- Change the source inductance from 0 μH to 50 μH and observe the increase of the current commutation interval as well as the load voltage drop.
- Check that a larger load inductance reduces the DC voltage ripple.