Hi,

i have a 3 phase bridge rectifier with 3 times 230Vrms inputs.
The output I measure on a DC load I connected to it is about 316VDC.

How to calculate this output voltage?

 

See

 

oke so from the picture 560V is also what i expected as output voltage:
2340*sqrt(2)*sqrt(3) = 560V.

But the difference is the null line? How does this change the output voltage?
Sorry i am quite new in power electronics

 

In the previous calculations I used a voltage of 220 V. This is due to the fact that I have been in areas with this voltage most of my life. But now our voltage is officially 230 V, but in reality nothing has changed - as was the average voltage of 220 V, and remains so. Now in the calculation I changed the voltage to 230 V and derived the maximum value of output voltage. I recommend studying LTspice if you want to get serious about electronics.

 

could you share your LTSPICE file?

 

I am sorry.

 

hi,

i am trying to open the drafts but i am missing a lot of symbols.
Could you also share those? Or could someone explain why the peak is at 320V?

 

hi,db,
Vpeak = 1.414 * Vrms ;
220Vrms *1.414= 311Vpk.
E

Look here for Alex's models.
https://forum.allaboutcircuits.com/...nents-models-of-ltspice-free-download.133690/

or here.
https://yadi.sk/d/6nSSUmZiT9oCoQ

 

hi,db,
Vpeak = 1.414 * Vrms ;
220Vrms *1.414= 311Vpk.
E

Look here for Alex's models.
https://forum.allaboutcircuits.com/...nents-models-of-ltspice-free-download.133690/

or here.
https://yadi.sk/d/6nSSUmZiT9oCoQ

Hi,
thanks
No i mean more like the difference between 560V and 320V
I see that the nul line is different between those. but why does that give the difference in voltage?

 

hi.
For 3Phase AC.
230Vrms.
230Vrms *1.732 [ for 3 phase] = 398Vrms
398Vrms *1.414 = 563Vpk.

Have you not studied 3 phase maths.
E

Look here;
https://www.ecmweb.com/basics/article/20897069/calculating-single-and-3phase-parameters

 

hi,db,
Vpeak = 1.414 * Vrms ;
220Vrms *1.414= 311Vpk.
E

Look here for Alex's models.
https://forum.allaboutcircuits.com/...nents-models-of-ltspice-free-download.133690/

or here.
https://yadi.sk/d/6nSSUmZiT9oCoQ

My webpage:
http://bordodynov.ltwiki.org/

 

yes i understand that part but how is that related to the nul line.
I am talking about the picture.

the first circuit has nul line connected and reaches 325V DC max
the second circuit has nul line not connected and reaches 560V DC max

 

hi,
OK, Check with @Bordodynov
I have not download those asc files yet, I am in the middle of another LTS project.
E

 

The top scheme is terrible. I looked at the current through the neutral wire and was horrified. Also the currents through the bottom diodes are huge. I apologize. In this circuit, the phase voltages are applied directly to the diodes, and this will lead to a big bang! I am sorry.

 

i find the spice plots to be confusing for this question. Just take a look at a simple explanation from a textbook such as 3 phase bridge or from 3 phase bridge tutorial
You can see the derivation of how to get the Vdc

Realize that your meter is likely measuring the rms of the 300Hz ripple, not the peak DC.

 

@Bordodynov the 3 phases symbol is not in your library. Could you share it with me?
thanks in advance

 

@Bordodynov the 3 phases symbol is not in your library. Could you share it with me?
thanks in advance

You can find Alex's entire library here:
http://bordodynov.ltwiki.org/

The 16M zip file is the one you want. The symbol (3phases.asy) is in the "Source" subfolder.

 

Hi,

i have a 3 phase bridge rectifier with 3 times 230Vrms inputs.
The output I measure on a DC load I connected to it is about 316VDC.

How to calculate this output voltage?

You should specify if you are measuring the voltage line to line or line to neutral.

1. Plot the three phase output (or look at the simulations others have done). This gives you a feel for what you have to solve for. Keep in mind you will be plotting the absolute value of the sines, for example instead of sin(t) you will plot abs(sin(t)). You can actually do that but include the other three phases also. If you want to you can include the amplitude now or later, which is A=230*sqrt(2).
2. Solve for the crossing points of two successive cross points of the waveforms you see. This gives you two time values t1 and t2. You may have to use a numerical solver to get these values.
3. Integrate one of the waves for t from t1 to t2. Divide the result by (t2-t1). That gives you the average.

Try it yourself it's interesting.
I'll come back later with some results myself.