EP
POWER ELECTRONICS / INTERACTIVE LAB
University of Vigo
Laboratory index
DIODES · 3~ · 6P

Three-phase full-wave uncontrolled rectifier

The highest phase feeds the positive output and the lowest phase returns the current. Explore six-pulse voltage, series loads and two clearly defined capacitor-branch models.

6 × DR / RL / RE / RLE / R‖CU–V–W
Learning level

Higher levels add blocks. The arrow in each heading opens or closes its block. Changing level preserves the operating point and your work.

—Vo,dc
—Vo,rms
—Io,dc
—Io,rms

One complete source period; RMS includes DC.

Circuit and current path

D₁/D₂/D₃ connect the highest U/V/W phase to +. D₄/D₅/D₆ return current from − to the lowest phase. The source neutral is not connected to the load. Crossings without a dot do not join phase conductors.

Each sine-wave circle is an AC phase source. The sources share the internal star point N; vU, vV, vW are phase-to-neutral voltages. Only U, V and W connect to the bridge: there is no neutral wire.

Fixed arrows define positive references. Teal paths carry rectifier current; the purple resistor and brown capacitor branches have their own currents. Scroll the diagram at readable size or use Fit diagram.

90°

Steady-state waveforms

Source and output voltages

Load and source currents

io enters the complete load. For R‖C, the dashed purple iR is the resistor current and solid brown iC is the capacitor-branch current: io = iR + iC. Coincident traces are drawn at their true values.

Predict, test, explain

Predict which source phases supply and return current at 90°. Check the circuit and all three source-current traces; explain why no neutral is required.

Reveal the explanation

The upper bank connects the highest phase to +; the lower bank connects − to the lowest phase. One phase carries +io, another −io, and their sum is zero. At 90° the lowest V/W phases are tied, so the drawn state is the right-hand limit.

Prepare R‖C. Compare output-current RMS with resistor-current RMS and locate the largest charging pulse. Predict the effect of reducing RC; confirm that capacitor voltage remains continuous.

Reveal the explanation

The resistor sees iR = vo/R. The rectifier must also charge C, so its RMS current can be much larger. RC limits the charging pulse and dissipates RC iC²; it also slightly changes the voltage seen by R. A very small RC is not zero.

Predict, test, explain