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3 phase bridge rectifier

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shosh

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anyone know how to calculate load voltage and rms voltages for a 3 phase full wave bridge rectifier?
View attachment 61489

i have this equation from the book that states Vout is 0.955*Vs
which isn't what i get from the simulation from multisim.

120v Vrms input
so 170 Vpp

I get 162 vavg using the equation, but in multisim i get 142 Vdc (avg) which isnt right.
 
Hi,

When fed with a Y input scheme, the peak voltage is sqrt(3)*line voltage peak, and the average voltage is 3/pi times that. So the average voltage is:
Vavg=3*sqrt(3)*VLinePk/pi
where VlinePk is the line to neutral peak voltage.
 
Last edited:
If each phase individually is 120 vrms, then taking 1.5 volts off for two regular silicon rectifiers loss, then at output:

Vpeak = 168.3 volts
Vmin = 145.6 volts
Vpk-pk ripple = 22.7 volts

Vrms = 160.7 vrms.
Vavg. = 160.6 volts
 
Last edited:
anyone know how to calculate load voltage and rms voltages for a 3 phase full wave bridge rectifier?
View attachment 61489

i have this equation from the book that states Vout is 0.955*Vs
which isn't what i get from the simulation from multisim.

120v Vrms input
so 170 Vpp

I get 162 vavg using the equation, but in multisim i get 142 Vdc (avg) which isnt right.

Hi again both of you,

There is a technical inaccuracy in the drawing which causes some ambiguity on how to interpret this when combined with the 'given equation' (really the factor).

The thing is, the drawing shows a Wye input, not a Delta input. The line voltage vL=Vs of the two are the same, but the line to line voltages (vLL) are not the same. The drawing is of a Wye input not a delta, so Vs must be interpreted as the line voltage vL not the line to line voltage vLL. Had it been drawn as a delta, then the line to line voltage vL would have been equal to 120vrms or 170v peak. The line to line voltage vLL of the Wye configuration is sqrt(3) times the line voltage vL, so the line to line voltage is not 120vrms it is 207.8vrms which means the peak is 294v peak.
The factor for peak to average for a three phase system is 3/pi, so when we multiply this factor times 294 we get 280.7 volts average. Not including diode drops (as many of these problems do) this is the average voltage that would be measured out of the circuit shown in the drawing.

Now if the drawing was drawn as a Delta and not a Wye, then the interpretation is entirely different. The line to line voltage vLL is only 120vrms then, and that means 170v peak. The factor for peak to average for the three phase system is the same, 3/pi, so multiplying this factor times 170 we get 162.3 volts average.

What this means is that the 'equation' is for a delta input, not a wye, so the drawing is incorrect and should be shown with all three voltage sources in series, not with their neutrals connected together.

There is another ambiguity here however, in whether we are looking for the average value of the output or the rms value of the output. The conversion factor for a three phase system peak voltage to rms voltage is:
K=sqrt((4*pi+sqrt(27))/(8*pi))
which is approximately equal to:
K=0.8407
Assuming a Delta input with line to line voltages of 170v peak we get:
Vrms=170*0.8407=142.9 volts rms.

So from this we can see that if we calculate the average voltage we get a different result than if we calculate the rms voltage. This means the book was giving the formula for the average voltage output not the rms voltage output, and from above it was also for a delta input not a wye input.
 
Last edited:
anyone know how to calculate load voltage and rms voltages for a 3 phase full wave bridge rectifier?
View attachment 61489

i have this equation from the book that states Vout is 0.955*Vs
which isn't what i get from the simulation from multisim.

120v Vrms input
so 170 Vpp

I get 162 vavg using the equation, but in multisim i get 142 Vdc (avg) which isnt right.

The peak will be the line-line voltage so an unloaded 3phase rectifier with a capacitor will charge upto

120*sqrt(2)*sqrt(3) = 294V (assuming 120 is the phase voltage).
Now a loaded 3phase rectifier will have an average output voltage of: 3*120*sqrt(2)*sqrt(3)/pi = 120*2.34 = 280Vdc
 
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