Electrical – phase matching power supply circuit

Adding the capacitor smooths the rectified AC voltage and not having one probably accounts for you measuring different voltages when you applied different loads. Pure half wave, as a waveform looks like this: -

If you add a capacitor the red waveform is transformed into the blue waveform for normal loads. This is the first benefit you get - the dc voltage remains largely at the peak value of the AC waveform and is therefore more predictable when you measure it with your meter.

When you connected DC motors this has a similar effect in that between half wave pulses, the back emf from the motor "free-running" holds the dc voltage and prevents it falling to zero (like in the red waveform).

The voltage you measured was 7.1V off-load and this largely corresponds to what a half-wave rectifier produces. Mathematically it is \$ \frac{V_{ACpeak}}{\Pi}\$ = \$ \frac{16.3\times\sqrt{2}}{\Pi}\$ = 7.34V.

A better option is to use a full bridge rectifier as this fills in the gaps on the negative cycle of the AC waveform: -

This has a better average dc voltage without a capacitor of twice what the half wave circuit gives i.e. about 14.7V. Also the smoothing capacitor can be smaller (when using one) because it gets charged-up to peak voltage twice as often.

There are a whole load of other things I could go into but this document from wiki says most of them and covers the basics above.

Remember that there are voltage harmonics, because of the current harmonics. I mean, that voltage harmonics are "generated" by current harmonics. So, if the three lines impedances are not exactly the same, then the voltage drop generated by the 3rd harmonic current in each line will be different.

Have you tried to measure currents instead of voltages? You may go to phase voltage from line current (analizing the voltage drop)

*If I wasn´t clear I can write it in spanish (Im from Argentina and my eglish isn´t good enough sometimes)