A three phase alternator with an external full wave bridge rectifier, the load bank maintains a voltage of 28V DC. What is the AC voltage value for the output of the alternator/ input to the rectifier at no load and at full load. The test results show 25.5V AC phase to phase at load, I was expecting 28V AC. Can someone provide a formula to calculate the AC voltage based on the given DC voltage.

 

For three phase bridge rectifier - on idle: Vdc = 1.41 * Vphase to phase.

 

For three phase bridge rectifier - on idle: Vdc = 1.41 * Vphase to phase.

Absolutely correct. You will have an additional approx 0.7 volt on each diode.
a total of approx 1.4 Volts

 

So basically according to cork_ie it will be 25.5V(phase-phase)+ 1.4V AC (voltage drop on each diode)= 26.9V DC(load voltage). Am I understanding it right it seems as were adding AC volts to get DC values.

donpetru formula is at idle so for my situation it would be V phase to phase= Vdc/1.41=28/1.41= 19.81 V so I would expect a lower voltage at ideal than at load, now this is not taking the diode voltage into account correct?

 

DC is more of an average of AC, so ACpeak/max will be higher than DC.
At the load you will subtract the diodes voltage drop.

 

Calculations:

**** im rippped here it goes

(Vmax X 2 X Y x freq X Rload X C) / (2 * Y x freq x Rload X C +1) = Vdc ('x' are multiplication)

where Y is 2 for a full-wave and 1 for a half-wave

But all you have to do is 1.41 multiplied by V
phase to phase

 

This collection of erroneous material is a concern.

Looking at a 3-phase full-wave bridge rectifier with diode drops neglected will give you a DC output of ...

\(V_{\text{DC}}=\frac{3 \sqrt{2}}{\pi}V_{\text{rms}}=1.35V_{\text{rms}}\)

Where Vrms is the rms line-to-line voltage for the 3-phase supply.

 

This collection of erroneous material is a concern.

Looking at a 3-phase full-wave bridge rectifier with diode drops neglected will give you a DC output of ...

\(V_{\text{DC}}=\frac{3 \sqrt{2}}{\pi}V_{\text{rms}}=1.35V_{\text{rms}}\)

Where Vrms is the rms line-to-line voltage for the 3-phase supply.

VRms = VDC with basic rectifiers

 

VRms = VDC with basic rectifiers

I would suggest you think again.

Go back to the OP's particular question and do some research on 3-phase bridge rectifier analysis. You may be enlightened.

In your earlier post you quoted an equation involving several parameters such as load resistance, load capacitance & frequency - none of which were mentioned or relevant to the original question.

Then you wrote "But all you have to do is 1.41 multiplied by V
phase to phase". You can't have it both ways when you now write "VRms = VDC with basic rectifiers"

 

So basically according to cork_ie it will be 25.5V(phase-phase)+ 1.4V AC (voltage drop on each diode)= 26.9V DC(load voltage). Am I understanding it right it seems as were adding AC volts to get DC values.

donpetru formula is at idle so for my situation it would be V phase to phase= Vdc/1.41=28/1.41= 19.81 V so I would expect a lower voltage at ideal than at load, now this is not taking the diode voltage into account correct?

No I think you misunderstand me.
RMS (root mean square) voltage is the equivalent DC voltage that produces the same power dissipation in a resistive load.It is an average of voltage V's time. In inductive or capacitive loads things will be a little different.
For a pure sine wave, the RMS value is the square root of 2 divided by 2 (approx. 0.707) times the peak voltage.
Thus the peak voltage is 1/.707 = 1.414 times the RMS Voltage
In this case what you need to look at is Peak to Peak voltage which will be 1.414 times the RMS - but when passed through Positive and Negative rectifier diodes there is an approx 0.7V voltage drop in each diode. Therefore the output voltage will be theoretically Peak to Peak less 1.4 Volts. As it is a 3 phase rectifier you can consider it as 3 rectifiers in Parallel you will subtract the 1.4V once only and not three times.
I hope this clears the matter up for you

 

This collection of erroneous material is a concern.

Looking at a 3-phase full-wave bridge rectifier with diode drops neglected will give you a DC output of ...

\(V_{\text{DC}}=\frac{3 \sqrt{2}}{\pi}V_{\text{rms}}=1.35V_{\text{rms}}\)

Where Vrms is the rms line-to-line voltage for the 3-phase supply.

Where is the \( {\pi} \) coming from ? I think you may be confusing two different formulae \(V_{\text{DC}}=\frac{3 \sqrt{3}}{\pi}V_{\text{}}\) (xxxxxxx) is probably what you are thinking of. i.e. formula used in a Fourier expansion which isn't really required here.

 

Where is the \( {\pi} \) coming from ? I think you may be confusing two different formulae \(V_{\text{DC}}=\frac{3 \sqrt{3}}{\pi}V_{\text{}}\) (xxxxxxx) is probably what you are thinking of. i.e. formula used in a Fourier expansion which isn't really required here.

If one considers the typical 3-phase bridge rectifier [see attachment], the rectified line-to-line voltage appears across the bridge output with conduction between the various phase-to-phase pairings occurring in identical 60° increments of the full 360° cycle.

The average DC output voltage can then readily be obtained by evaluating the averaged integration of the sine function over any 60° conduction cycle.

This may be shown mathematically using radian measure for instance as ...

\(V_{\text{DC}}=\frac{1}{(\frac{\pi}{3})} \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} V_m sin(\theta) d \theta=\frac{3}{\pi}V_m\[ -cos(\theta)\]_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} =\frac{3}{\pi}V_m=\frac{3 \sqrt{2}}{\pi}V_{\text{rms}}=1.35 V_{\text{rms}}\)

Where Vm is the peak value of the rms line-to-line voltage Vrms

Note: The diode forward voltage drop is assumed to be zero in this analysis.

As you might see this isn't a Fourier expansion. In any event, I would think the Fourier expansion of the bridge output voltage waveform would have the DC or mean output value as the same as above. After all, doesn't the Fourier analysis allow one to find both the mean value and the harmonic content? The DC or mean value would be what a true DC measuring instrument would indicate when attached to the bridge output terminals.

 

For 25 VRMS, L-L or 14.4 VRMS L-N, the rectifier output will be 25/ 0.817 = 30.6 VDC... CURRENT I RMS = 0.817 x I DC...Alt. Power is 1.05 x DC Power.
These figures assume full converter with no commutation. These calculations take into account the (6) forward diode drops (Vf) of silicon ~0.6 Vdc.

Source of this information: Std. Handbook for Elec. Engrs. ISBN 007020974X

Many years designing power rectifiers have shown this information to be accurate.

Cheers, DPW [Everything has limitations...and I hate limitations.]

 

VRms = VDC with basic rectifiers

How can you say 1.41 x Vrms = VDC + 1.4V? This is only true when Vrms = 1V.

 

For 25 VRMS, L-L or 20.4 V L-N, the rectifier output will be 25/ 0.817 = 30.6 VDC... CURRENT I RMS = 0.817 x I DC...Alt. Power is 1.05 x DC Power.
These figures assume full converter with no commutation.

Source of this information: Std. Handbook for Elec. Engrs. ISBN 007020974X

Many years designing power rectifiers have shown this information to be accurate.

Cheers, DPW [Everything has limitations...and I hate limitations.]

I had a look at Fink & Carrol - Standard Handbook for Elec. Eng. ISBN 07-020973-1. Is that the same ?

Table 12-2b therein summarizes various transformer fed single & poly-phase bridge rectifiers. Your comment that "For 25 VRMS, L-L or 20.4 V L-N, the rectifier output will be 25/ 0.817 = 30.6 VDC" doesn't square with handbook table. It states the DC output voltage for the 3-phase bridge topology is 2.34Es where Es is the bridge input line-to-neutral RMS voltage.

In any case, 25Vrms line-to-line isn't equivalent to 20.4Vrms line-to-neutral. 25Vrms line-to-line would be 14.43V rms line-to-neutral.

So according to the table

VDC=2.34*Es = 2.34*14.43 = 33.8V DC

or using the line-to-line voltage

VD=1.35*E_line-to-line = 1.35*25 = 33.8V DC

Were you including diode voltage drops as contributing to a likely practical outcome of 30.6V DC??

As a further comment: neglecting losses, the AC real input power and the DC load power would be the same. The transformer VA rating needs to be 1.05*DC load power owing to the effect of harmonics in the line currents.

I don't have a lot of confidence in the long departed OP's quoted numbers. I guess we'll never know.

 

I appreciate all the responses, I remembered some things and picked up on your blog. The numbers still do not add up to me, I have an alternator output to the rectifier of 25.5 V L-L AC and a voltage output after the rectifier of 28 V DC under load. Even cosidering the diode losses I cant end up with a formul that can make both these known practical voltages add up. Do I have to take into consideration a ripple coefficent somehow?

 

Reduce the output loading and see what happens...or the AC voltage may be sagging under load...or one phase may be in-adequate or a loose connection may be suspect when currents are high.

Cheers, DPW [Everything has limitations...and I hate limitations.]

 

It's possible that if one diode (for instance) is open circuit, that the measured DC output would fall to around 27 to 28V. Is your interest in this based on a suspicion that something may actually be faulty?

The only way to really check such matters is to look at the circuit waveforms using an oscilloscope.

Are you able to test the individual bridge input line currents to ascertain any significant imbalance between the three phases?

 

Yes, my intrest in this is based on suspicion that something might be faulty and to understand how the values are what they are. The bridge currents are well balanced at 132.6, 133.6 & 134. AC voltage L-L are all 25.5-7 approxametly. There is more than one unit that has been tested and they all show similar results. I assme the formulas should give me a close approximte to real practical readings.
t n k, do not have an oscilloscope. Would a general check on the rectifier diodes with a multimeter be adequate with no voltage?
Duane, if the voltage is sagging under load it would be a quick drop then stabilizes correct?

 

What you are testing is a reasonably big rectifier given those line currents. Since you seem to have the same result for more than one unit the indications are that this as good as you are going to get.
You didn't answer Duane's question about what happens at light loads.
One would need at lot more information about system component technical specifications. What is the alternator? What are the alternator's electrical parameters - including equivalent impedances, output waveform "purity", etc. What are the bridge rectifier published characteristics? How does the diode voltage drop vary with rectifier current?
There are many factors which will make a first order simplified analysis indicative only if you need accurate performance data.