# 3 phase rectifier

• posted

Hi,

If I need a solid 350VDC output from a 3-phase generator, can someone point me to the equations to use to figure out how much peak to peak voltage I must maintain on the output of the generator to achieve the DC voltage out using a 3-phase rectifier?

Thanks!

• posted

What sort of filter is after the rectifier?

Assuming a standard 3 phase bridge:

The bottom of the ripple is at 3/4ths the peak to peak minus the drop in the diodes. If you need to maintain a voltage without filtering, this is the point you need to worry about.

The peaks of the ripple are at 0.866 of the peak to peak minus the drop of the diodes. If this is a capacitive filtered case, this is the peak of the ripple on the output.

A good estimate for the capacitive filtered ripple amplitude is:

V =3D I / ( 360 * C)

• posted

• posted

1. Do your own math - that's what school is for.

1. Don't top-post

Good Luck! Rich

• posted

This depends on the source at hand. If you have a floating Delta supply, then you are stuck with the phase to phase common practice which in general is a basic DC peaks = RMS * 1.414 minus all the other little crap i'll explain later.

If you decide to use a grounded (Y)/Star type source, then you could use a simple 3 Diode full wave with respect to ground/common.

Since 3 phase xformers are designed with a 120 degree offset in mind, each leg is actually generating more than what you measure from leg to leg. This is needed to make up for the degree offset so that you can achieve required voltage leg to leg how ever, in systems where the (Y)or 1 leg of a delta is ground and thus using a common as the low side. You need to up the actual when converting to single phase/DC.

in which case. Vrms = Vrms * (1/0.8666); for example. in systems that employ 480V 3 phase to common point. Vrms = (480 /2)*1.153 = (480 * 1.153) /2 = 276.72

We'll just round that off to 277.

systems are measured in RMS (mostly), you then calculate your Peak.

Pk = 277*1.414 = 391.68

diodes have a loss of ~ .7 per unit. so, if you use a full bridge in a floating 3 phase, you'll have ~ 1.5 loss which gives you a total of ~390.0

after that is all said and done, you have some ripple from the caps.. I guess if one was to rely on the standard system for better ripple control, one can use that extra voltage to be dropped via a resistor on the reservoir capacitor.

So in the end, it all works out :) P.S. The dropping R must be calculated via the expected load and the ripple error. The R will drive the reservoir cap. Also keep in mind that crappy caps with high ESR can give you some bad ripple.

"

• posted

If nothing is designed yet, may I suggest a 12-Pulse Rectifier (ANSI Circuit 31) configuration? By feeding the rectifiers with two sets of windings, one connected delta while the other is connected wye, the phases will be separated by 60 degree and the ripple voltage will be 1.02% before filtering. Also, the ripple frequency will be twice that of a standard 3-phase bridge rectifier circuit and thus easier to filter out.

See Figure 16.5 of [

], the second figure at [
], and Figure 12.22 of [
].

Also see:

Figure 11.6, _Newnes Electrical Power Engineer's Handbook_ [

]

"modeling the 12 pulse rectifier", _Power Electonics handbook_ [

]

Figure 2.4.4, _Electric Power Transformer Engineering_ [

]

_IEEE guide for harmonic control and reactive compensation of static power converters_ [

]

Needless to say, you should also read the discussion on the nearby pages when looking at the above figures.

If the above is too advanced for you, start here: [

]

I strongly advise that you hire someone who knows what they are doing. 3-phase AC to high-voltage DC is not an area where someone who has to ask basic design questions should be starting out. IHTH.

```--
Guy Macon  Guy Macon
Guy Macon  Guy Macon ```
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Hey, Guy! What's this guy paying you to do his homework for him? ;-p

Cheers! Rich

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