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

Three-phase half-wave controlled rectifier

Three SCRs take turns, with a neutral return. Measure firing delay from natural phase commutation and distinguish phase current from load-resistor current.

3 × SCRR / RL / RE / RLE / R‖CU–V–W + N
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

Each phase-to-neutral source feeds one SCR anode. The three cathodes share +; the load’s − returns to source neutral N. U, V and W are not shorted together. Each gate driver returns to its SCR’s K.

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.

Gate commands and actual conduction

The logical gate traces use 0/1. Filled lower strips mark actual conduction of each SCR, independently of the gate command. All panels share the probe; click or drag within a waveform to move it.

Predict, test, explain

Predict what changing α from 0° to 60° does to the resistive-load mean voltage. Prepare the case, inspect the gate and current windows, then explain the result.

Reveal the explanation

Delayed firing removes part of the resistive conduction interval. The complete source-current waveform, rather than a cosine rule assumed for continuous inductive current, determines this result.

With an opposing DC load, compare a 5° gate window and a wide window at the same delay. Predict whether the source becomes eligible before the gate ends.

Reveal the explanation

A gate pulse can finish before the blocked load can accept current. A longer gate window can then initiate conduction when the source overtakes the load voltage. Once current is established, the gate ending does not interrupt it.

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