If a standard (in Europe and a large part of the world except North America and Japan) three-phase 400V AC (three lines having 230V RMS voltage if measured to neutral each) mains supply is rectified with a standard 6-diode rectifier like this:

Three-phase, full-wave bridge rectifier circuit

What DC voltage value will come out of the rectifier? How to calculate it having the RMS AC source voltage given?

Are there any other ways to wire diodes to get a different voltage (without using any transformers or anything else then just diodes), what are they and what DC voltage will come out then?

  • \$\begingroup\$ Is this homework? If not, what is it for? \$\endgroup\$ – tyblu Apr 2 '11 at 19:24
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    \$\begingroup\$ @tyblu: No, not homework, I am just thinking of building a three-phase SMPS some time in the future, and want to know what will be my DC input voltage for the SMPS. \$\endgroup\$ – ria Apr 2 '11 at 19:39

If you measure across the load shown in your figure, the peak voltage will be ~565V; the DC voltage will depend on your load and filtering as others have noted.

If you measure from the + of the load in your figure to the neutral of your AC supply, the peak voltage will be ~325V. If you hook up a load like that, you aren't actually using a full-wave rectifier.

The simplest way to get 565V is to start from 400V and apply the standard \$\sqrt{2}\$ scaling from \$V_{rms}\$ to \$V_{p-p}\$. However, starting from 400V is skipping part of the calculation. The more thorough way to derive 565V is to calculate it as:

$$(325 \text{V}) * \max_{\theta} \left\{ \sin (\theta + \frac{2 \pi}{3}) - \sin (\theta) \right\}$$

The expression is maximized when \$\theta\$ is \$\frac{5 \pi}{3}\$, and the maximum value is \$325 \sqrt{3} = 563\$.

There is a detailed analysis including some java applets here.

  • \$\begingroup\$ Passing by, the java applet link is broken. \$\endgroup\$ – Mister Mystère Nov 2 '15 at 17:13

This configuration is commonly known as a star, or WYE configuration. Its easier to see if you break it down into two halves. Phase to neutral is 230 vrms. Three phases each connected to a diode anode and all three of the cathodes tied together. If measured from neutral to the cathode connections, you would expect to see 230 * 1.414 = 325 vdc. This represents the "peak" voltage of the waveform. Now do the same with the other half of the bridge, which will create a negative voltage of equal value in respect to the neutral. The pulses interweave with each other affectively filing in the gaps of the positive pulses, resulting in 6 pulses creating a smoother dc voltage. The voltage unfiltered would be slightly less than 325 volts. If a filter were added such as a capacitor, the voltage would average close to the calculated value minus the "ripple" value which is always present in a filter.

CAUTION:These voltages are lethal and the proper precautions need to be taken to prevent injury or death! The explanation is for illustration purposes only. In real practices this circuit would be built with an isolation transformer and circuit protection, such as fuses.

  • \$\begingroup\$ @Steve: are you sure it will be "slightly less then 650V DC?" Now I found this graph which suggests that three-phase AC rectified with a full-wave rectifier will have voltage slightly lower then peak voltage of each phase, so that would be 325V. Or maybe I don't understand that plot correctly? \$\endgroup\$ – ria Apr 2 '11 at 20:07
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    \$\begingroup\$ @Steve that depends entirely on what your load current is and how much hold up cap you have on there \$\endgroup\$ – BullBoyShoes Apr 2 '11 at 20:29
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    \$\begingroup\$ @miernik: You are correct. The voltage from neutral to + is +325vdc, and the potential from neutral to - is -325vdc. However they are not additive as I first said, because they interweave into one positive (with respect to negative) wave. Thanks for seeing that! Please see my edit. \$\endgroup\$ – SteveR Apr 2 '11 at 21:34
  • \$\begingroup\$ @Steve: it seems we where both wrong, see the other answer which appeared. \$\endgroup\$ – ria Apr 2 '11 at 23:23
  • \$\begingroup\$ link \$\endgroup\$ – SteveR Apr 3 '11 at 0:53

I think Steve and Andy have it pretty well explained but it really helps me to look at the voltage waveforms and see how exactly they add up. Note that the time between peaks ~5.5ms which is a direct result of the three peaks, one from each phase, being offset by 120 degrees and being added together.

Three waveforms are plotted: V(v+) is the voltage from node V+ to ground. V(v-) is the voltage from node V- to ground. V(v+,v-) is the voltage across the load resistors. schematic simulation

Also, you can right click and view image to see larger versions which are much more legible.


Three phase AC through a rectifier produces this waveform:

enter image description here

The "DC voltage" out has two possible meanings: average, and RMS. RMS is how much heating energy a load in this configuration will see.

The output waveform is a sine wave between 60 and 120 degrees, repeated. Take the RMS of a sine wave between those two angles and we get the RMS of the entire wave. RMS is root-mean-square: take the square root of the mean of the square of the sine wave.

\$V_{peak} \sqrt{\frac{\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}}{sin^2\Theta}}{\frac{\pi}{3}}} \$

\$V_{peak} \sqrt{\frac{\frac{\Theta}{2} - \frac{sin2\Theta}{4}\big]_{\frac{\pi}{3}}^{\frac{2\pi}{3}}}{\frac{\pi}{3}}} \$

\$V_{peak} \sqrt{\frac{\frac{\pi}{3} - \frac{\pi}{6} - \frac{sin\frac{4\pi}{3}}{4} + \frac{sin\frac{2\pi}{3}}{4}}{\frac{\pi}{3}}} \$

\$V_{peak} \sqrt{\frac{\frac{\pi}{3} - \frac{\pi}{6} - \frac{sin\frac{4\pi}{3}}{4} + \frac{sin\frac{2\pi}{3}}{4}}{\frac{\pi}{3}}} \$

\$V_{peak} \sqrt{\frac{\frac{\pi}{6} + \frac{\sqrt{3}}{4}}{\frac{\pi}{3}}} \$

\$V_{peak} \sqrt{\frac{1}{2} + \frac{3\sqrt3}{4\pi}} \$

\$.95577 V_{peak}\$

The average is slightly simpler to compute:

\$V_{peak} \frac{\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} sin\Theta}{\frac{\pi}{3}}\$

\$V_{peak} \frac{-cos\Theta\Big]_{\frac{\pi}{3}}^{\frac{2\pi}{3}}}{\frac{\pi}{3}}\$

\$V_{peak} \frac{cos\frac{\pi}{3} - cos\frac{2\pi}{3}}{\frac{\pi}{3}}\$

\$V_{peak} \frac{2 cos\frac{\pi}{3}}{\frac{\pi}{3}}\$

\$V_{peak} \frac{1}{\frac{\pi}{3}}\$

\$V_{peak} \frac{3}{\pi}\$

\$.955V_{peak} \$

And the peak voltage is, of course, the RMS of the input times the square root of 2.


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