Design of 3 phase ac to dc power supply

[Besoeker]

Because the voltage referenced was assumed to be V_LN.

But why assume a neutral why none is mentioned?

 

[gar]

130217-0850 EST

GoldDigger:

I am wrong. Wiki is OK when I found out what V_peak is. The circuit and top equation are the only things that appear together on my screen. I did not see that they had a V_LLpeak also.

.

 

[Besoeker]

Additionally, to calculate from V_LL is just a simple change in the equation:

V_DC = V_AV = 3V_LLpeak/pi = 0.955 ? V_LLpeak


...or...


V_DC = V_AV = 3√2V_LLrms/pi = 1.35 ? V_LLrms

All covered, and more succinctly I think, in the last three lines of my derivation.
Using the RMS voltage (the last two lines of it) is a convenience in that you know that without calculation rather than the peak voltage.

 

[K8MHZ]

Usman...

Are you designing a system employing a "truck" alternator?

Regards, phil Corso

I thought about that, too, but Prestolite / Leece - Neville only makes up to 185 amp 24 volt (28 volt charging) capacity alternators.

Larger current, up to 300 amps, is available in 12/14 volt systems. The idea behind using 24 volt systems in vehicles is to decrease the amps, so there may not be any 200 amp 24 volt alternators to employ.

 

[K8MHZ]

For some reason, vehicle alternators have a funny way of making a full wave bridge.

There are six diodes. Three are called a 'diode trio' and the other three are called the 'rectifier bridge'. The alternators start out as 3 phase AC and can make up to 90 volts or so if not regulated. I have a scope for alternators and we used to use the ripple wave forms to tell us what was wrong with the charging system. Many people think cars have perfect DC, but that is not at all true.

 

[Smart $]

But why assume a neutral why none is mentioned?

I didn't do the assuming... gar did in Post #2

...

Assume the alternator output is a Y topology, ... The voltage from neutral to one leg is a 28 V RMS sine wave. ...

 

[usman umar]

thanks every one for reply..

• Alternator o/p is 3 phase , delta connected , 28 V , 206.4 A , 100 hz , 0.8 pf.
  
• So O/p power of alternator is = 1.73*28*206*0.8 = 8 Kw
  

• i connected 3 phase bridge rectifier . Output to rectifier is Vdc=1.35*28= 37.8 V. (NOTE: To reduce my Dc o/p voltage i will reduce my alternator o/p with the help of AVR))

• output current of rectifier is I=8Kw/37.8 = 211.6 A

now Question No-1: Is the above calculations are true or not?
Question No-2:How to design a filter to reduce ripple voltage upto 20mv? any equation

 

[usman umar]

Usman...

Are you designing a system employing a "truck" alternator?

Regards, phil Corso

yes it is..

 

[gar]

130218-0841 EST

usman umar:

You have provided some clarification.

What is the reason for your questions?

What is your goal?

You did not define AVR, but it probably refers to automatic voltage regulator. The output voltage is controlled by the magnetic field intensity, and probably that is controlled by the excitation current to a wound coil type of field. Thus, you apparently plan to adjust field excitation to obtain a desired DC output voltage.

Ignore PF on a single generator. PF will be whatever it is, and thus determined by the load.

The 1.35 factor is correct for a full-wave bridge rectifier when the input voltage is a sine wave, voltage is RMS line-to-line of the delta source, the load is a pure resistance, and all series inductance is ignored.

Your ripple question depends upon the load, and internal series inductance of the generator. You need to define the load, and whether the source inductance is significant.

.

 

[gar]

130218-0929 EST

usman umar:

After you have a completed total circuit you can determine a power factor. To do this you will need to define what PF is to mean.

I would probably do this:

PF = DC power output / (total volt-ampere input)

Where
DC power output = Vdc * Idc
Total volt-ampere input = 3 * (Vrms line-to-line / 1.732) * Irms of one line

You might measure real power input to the rectifier, but that should not be any different than the DC power output, except for diode losses.

.

 

[Besoeker]

thanks every one for reply..

• Alternator o/p is 3 phase , delta connected , 28 V , 206.4 A , 100 hz , 0.8 pf.
  
• So O/p power of alternator is = 1.73*28*206*0.8 = 8 Kw
  

• i connected 3 phase bridge rectifier . Output to rectifier is Vdc=1.35*28= 37.8 V. (NOTE: To reduce my Dc o/p voltage i will reduce my alternator o/p with the help of AVR))

• output current of rectifier is I=8Kw/37.8 = 211.6 A

now Question No-1: Is the above calculations are true or not?
Question No-2:How to design a filter to reduce ripple voltage upto 20mv? any equation

A few points in no particular order.
I agree with your 8kW calculation. But note that the output power of an alternator is often limited by the power of the prime mover.
If you fit a filter to get to your required 20mV ripple then it is likely to have a substantial choke in the DC output. That being the case the relationship is that Idc = 1.225Iac. From sqrt(3/2).
So, at the full current rating of the alternator, the Idc would be 252A. This is at 28Vdc so maximum DC power is about 7kW.

It's good that you can use the AVR to regulate the alternator output. This will give you lower ripple in the DC to filter than other forms of voltage control would.
On filtering, I'll maybe come back to that in a later post. Sometimes my old age makes me indolent.........

 

[Besoeker]

Now is later. I thought dinner was imminent - it is now, and my wife besides being a Southern Belle, is a superb cook.:thumbsup:

Anyway, about 1mH and 4700uF should do your filtering depending on what kind of load you have. The 100Hz is also a plus.