Design of 3 phase ac to dc power supply

[usman umar]

i am designing 3 phase power supply. out put from alternator is 28 v , 206.4 A , 100 hz , 0.8 pf..
now i have to use a 3 phase rectifier . DC 28 V is needed at o/p.

what is the output power after rectification . current also? formula??
how to design a capacitor if we want max ripple voltage = 28mv?
how to measure ripples(any cct)?

thanks

 

[gar]

130215-1746 EST

usman umar:

Assume the alternator output is a Y topology, you use half wave rectification, and there are three diodes. The voltage from neutral to one leg is a 28 V RMS sine wave. The average DC voltage is obtained by calculating the average value of a sine wave from -60 to +60 degrees relative to the sine wave peak where Vpeak = 1.414 * 28 = 39.6 V. This average value is 1.17 * 28 = 32.8 V. With sufficient capacitance as the load, then the DC output is 39.6 V.

The time from one ripple peak to the next is 1/(3*100) = 0.00333 seconds. You will need to know the average load current to determine the ripple, and you need to define whether you want RMS ripple voltage, peak-to-peak, or whatever.

Your questions need clarification. As you can see you won't get 28 V DC from a basic rectifier circuit. Phase shift control of SCRs would provide a means.

For some theory see Chapter 1. of "Theory and Application of Industrial Electronics", Cage and Bashe, First Edition, McGraw-Hill, 1951.

.

 

[Besoeker]

i am designing 3 phase power supply. out put from alternator is 28 v , 206.4 A , 100 hz , 0.8 pf..
now i have to use a 3 phase rectifier . DC 28 V is needed at o/p.

what is the output power after rectification . current also? formula??
how to design a capacitor if we want max ripple voltage = 28mv?
how to measure ripples(any cct)?

thanks

Some stuff I did earlier showing the calculation, the rectified waveform, its mean value, and the ripple which is at six times supply frequency.

Hope it's of some help.

 

[GoldDigger]

130215-1746 EST
Assume the alternator output is a Y topology, you use half wave rectification, and there are three diodes.

But as long as the wye, including the neutral, is not grounded, you can also use full wave rectification with a six diode bridge and get less ripple for the same output filter, since you get six voltage pulse per revolution instead of just three.
This is typically done in automotive alternators these days since diodes are relatively cheap.
I think that the formulae that Besoeker gave are for a six diode bridge, but it is hard to be sure at a glance since the graph does not show the input AC waveform.

 

[gar]

130216-1410 EST

GoldDigger:

You are correct. In the book I referenced are equations for full wave and different sources. Also in Reference Data for Radio Engineers are additional combinations and scaling factors.

Besoeker"s calculations were for 3 phase full-wave or 6 phase half-wave.

Whether the load filter is capacitor input or choke input (load current dependent) will determine output average voltage. But none of these provide 28 V DC output from 28 V AC input. A transformer, some sort of chopping, or a series pass regulator will be required to get the 28 V DC.

Also the desired ripple will only be possible after adequate filtering. This he understands from the content of the original post.

I suspect this is a class problem and usman umar should do some study from the information we have presented, and not be afraid to ask more questions.

We need clarification of the original question.

.

 

[Besoeker]

But as long as the wye, including the neutral, is not grounded, you can also use full wave rectification with a six diode bridge.

You can whether or not the neutral is grounded. In UK and most EU countries LV distribution is from the star (wye) secondary of a transformer where the neutral is earthed (grounded).
This works perfectly well with the three-phase six diode bridge arrangement for the front end rectifier arrangement on three-phase input variable frequency drives for example.

I think that the formulae that Besoeker gave are for a six diode bridge, but it is hard to be sure at a glance since the graph does not show the input AC waveform.

The graph shows six pulses over a complete cycle - that's what 3-phase full-wave rectification gives you.

The numbers for the integration may be a little less obvious. Each of the six pulses can be divided into two symmetrical halves hence the integration from pi/3 to pi/2. Six such in a half cycle and half a cycle is pi.
That's where the numbers come from.

 

[Phil Corso]

Usman...

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

Regards, phil Corso

 

[Besoeker]

130216-1410 EST


Besoeker"s calculations were for 3 phase full-wave or 6 phase half-wave.

Yes, it works for hexaphase too with line to neutral. But we are told that it is three-phase.

Whether the load filter is capacitor input or choke input (load current dependent) will determine output average voltage. But none of these provide 28 V DC output from 28 V AC input. A transformer, some sort of chopping, or a series pass regulator will be required to get the 28 V DC.

Agreed.

 

[GoldDigger]

You can whether or not the neutral is grounded. In UK and most EU countries LV distribution is from the star (wye) secondary of a transformer where the neutral is earthed (grounded).
This works perfectly well with the three-phase six diode bridge arrangement for the front end rectifier arrangement on three-phase input variable frequency drives for example.
As long as the DC output does not have to be grounded on one side!
The graph shows six pulses over a complete cycle - that's what 3-phase full-wave rectification gives you.
I guessed that the graph was intended to show only one cycle, but did not want to depend on that assumption.

Thanks for the clarification on the integrals. Other comments inline in blue.

 

[Besoeker]

Thanks for the clarification on the integrals.
Other comments inline in blue

Thank you.
It's often to explain mathematics in words. Suffice it to say that my method gives the same result as your link.

But no assumption required about the one cycle.
The horizontal axis shows 0-360deg for the angle. However, I do accept that the resolution isn't brilliant.

 

[Smart $]

Whether the load filter is capacitor input or choke input (load current dependent) will determine output average voltage. But none of these provide 28 V DC output from 28 V AC input. A transformer, some sort of chopping, or a series pass regulator will be required to get the 28 V DC.

Agreed.

Quite rusty on inverter theory, but I'm just not seeing this using a "line-to-line" bridge. Following formula pulled from wikipedia page GoldDigger linked...

V_DC = V_AV = 3√3V_PEAK/pi = 1.654V_PEAK = 1.654 ? 28 ? √2 = 65.49V (no load, no accounting for commutation)

 

[gar]

130216-1604 EST

Smart $:

In a full wave bridge rectifier, 6 diodes, fed from a delta source with the DC floating, or the delta floating, there are never more than two diodes on at a time and these are fed from the delta line to line voltage. Thus, the peak voltage is Vac RMS line to line * 1.414 .

Draw out the circuit.

This is the standard design in HAAS CNC machines. 230 V RMS line to line produces a DC bus of 325 V. This DC bus floats all over relative to earth with our wild-leg open delta source.

. .

 

[GoldDigger]

Thus, the peak voltage is Vac RMS line to line * 1.414 .

Draw out the circuit.

This is the standard design in HAAS CNC machines. 230 V RMS line to line produces a DC bus of 325 V. This DC bus floats all over relative to earth with our wild-leg open delta source.

. .

The formula used by Smart $ (1.654VPEAK) assumes that Vpeak is the line to neutral voltage (look at the diagram of where the voltage sources are) rather than the line-to-line voltage. Now which one is the 28 volts specified by the OP?

 

[Smart $]

130216-1604 EST

Smart $:

In a full wave bridge rectifier, 6 diodes, fed from a delta source with the DC floating, or the delta floating, there are never more than two diodes on at a time and these are fed from the delta line to line voltage. Thus, the peak voltage is Vac RMS line to line * 1.414 . ...

You're changing the parameters of your initial assumption, which was a wye-configured output of 28V_RMS line to neutral. Thus line-to-line voltage is 28√3 = 48.5V_L-Lrms. Multiply by √2 for 68.6V_L-Lpeak.

As GoldDigger queries, the OP's 28V is line to neutral or line to line...???

 

[gar]

130216-2103 EST

In my first post I made an assumption for a circuit, and the information for that was correct. The factor is 1.17 .

For convenience Besoeker made the assumption of a three phase bridge full-wave rectifier because he already had the calculations and waveforms to describe that circuit. His factor result is 1.35 and corresponds with textbook values and derivations.

From here the discussion has drifted, and I don't want to try to unscramble it.

In Wiki under "Three-phase bridge rectifier" the top equation reduces to 1.169 * V_RMS. where V_RMS is the line-to-line voltage. This is incorrect. The value should be 1.35 as calculated by Besoeker. The derivation of the Wiki equation is not shown.

It is unlikely that usman umar wants or cares about the average DC output without filtering. If he wants a capacitor input filter, then we are only concerned with V_PEAK to determine the DC output voltage.

How you get 28 VDC from 28 VAC is a whole different story that needs clarification on the requirements.

.

 

[Besoeker]

Quite rusty on inverter theory, but I'm just not seeing this using a "line-to-line" bridge. Following formula pulled from wikipedia page GoldDigger linked...

V_DC = V_AV = 3√3V_PEAK/pi = 1.654V_PEAK = 1.654 ? 28 ? √2 = 65.49V (no load, no accounting for commutation)

Well, you can calculate it from V_LN but why would you if you were using this common arrangement of a three-phase bridge:

(Assume diodes rather than SCRs - it was the only pic I had to hand)

 

[Besoeker]

It is unlikely that usman umar wants or cares about the average DC output without filtering. If he wants a capacitor input filter, then we are only concerned with V_PEAK to determine the DC output voltage.

For the level of ripple required, an LC filter in the DC would probably be a better option and be kinder to the AC supply.

How you get 28 VDC from 28 VAC is a whole different story that needs clarification on the requirements.

Maybe the simplest thing would be to adjust the voltage regulator on the alternator to get about 20V.

 

[Smart $]

Well, you can calculate it from V_LN but why would you if you were using this common arrangement of a three-phase bridge: ...

Because the voltage referenced was assumed to be V_LN.

 

[Smart $]

130216-2103 EST

In my first post I made an assumption for a circuit, and the information for that was correct. The factor is 1.17 .

For convenience Besoeker made the assumption of a three phase bridge full-wave rectifier because he already had the calculations and waveforms to describe that circuit. His factor result is 1.35 and corresponds with textbook values and derivations.

From here the discussion has drifted, and I don't want to try to unscramble it.

In Wiki under "Three-phase bridge rectifier" the top equation reduces to 1.169 * V_RMS. where V_RMS is the line-to-line voltage. This is incorrect. The value should be 1.35 as calculated by Besoeker. The derivation of the Wiki equation is not shown.

It is unlikely that usman umar wants or cares about the average DC output without filtering. If he wants a capacitor input filter, then we are only concerned with V_PEAK to determine the DC output voltage.

How you get 28 VDC from 28 VAC is a whole different story that needs clarification on the requirements.

.

Well here's two links to web pages that agree with the equation on wiki'...

http://www.ewh.ieee.org/soc/es/Nov1998/08/CONVERTR.HTM

http://www.eng.uwi.tt/depts/elec/staff/rdefour/ee33d/s3_3pbr.html

It's obvious someone is wrong :happyyes:

 

[Smart $]

Well, you can calculate it from V_LN but why would you if you were using this common arrangement of a three-phase bridge:


(Assume diodes rather than SCRs - it was the only pic I had to hand)

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


(and for SCR-controlled, simply multiply by cos ɑ, where ɑ is the delay angle)


...so...

28V_LNrms ? √3 = 48.5V_LLrms

V_DC = V_AV = 65.5