Hi Guys , new to the site , need to know the following. I have a 3phase transformer with 420 vac input and 380 vac output to a 3 phase bridge rectifier stack. The DC out put from the stack is 515 vdc. How is the DC value arrived at and what is the formula to calculate this , thanks , Alexander....
3 Phase to DC Calculations
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mivey
Senior Member
The DC voltage can get up to the net maximum voltage (peak input voltage less the drop across diodes, etc.). For a sinusoidal, the peak is sqrt(2) times the rms voltage (ex. 120 * 1.4142 = 169.7 volts)
For a rectifier with a capacitor filter:
Vdc (approx) = (Vmax*2*Y*f*R_load*C)/(2*Y*f*R_load*C +1)
Vripple = Vdc / (Y*f*R_load*C)
where Y is 2 for a full-wave and 1 for a half-wave
and R_load*C >>1/f
The factor for mean DC output voltage is 1.35 Vac. Hence your roughly 515Vdc.
It's (3/PI)*sqrt(2)*Vrms
I wrote out the derivation long hand but my scanner has gone awol.
mivey
Senior Member
So V_max = 380 * sqrt(2) = 537.4
with V_ripple ~ 2*(V_max - V_dc) you have
V_ripple ~ 2*(537.4-515) = 44.8 or 8.7%
Does that seem high to you or did I overlook something?
You seemed to have assumed peak rather than average.
erickench
Senior Member
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I know this. If you use an ammeter, the reading is average value not RMS.
This is the area under the current curve divided by the time giving you average value.
This is the area under the current curve divided by the time giving you average value.
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mivey
Senior Member
I seem to recall we used peak-peak ripple voltage divided by the dc voltage to get the ripple percent. The peak-peak being approximated by 2 times the difference from peak to dc.
Can't find any notes right now but an 8.7% ripple seems high.
mivey
Senior Member
Found notes. There is a peak-peak ripple percent and an rms ripple percent.
To convert (~sawtooth decay):
V_ripple_rms = V_ripple_peak-peak / 2 / sqrt(3)
So now I have V_ripple_rms = 44.8/2/sqrt(3) = 12.93 or 2.5%
Now the question becomes: What is acceptable % for a p-p ripple or an rms ripple? I guess it would depend on the load needs.
mivey
Senior Member
Nevermind, I'm not paying attention. There is no filter capacitor so we have no ripple control. Also, my first response was for single phase with a filter capacitor. Missed my nap time I guess. :roll:
NP.
My hand-written note was based on unsmoothed three-phase.
This gives a ripple of 0.866Vp to Vp at six times supply frequency.
gar
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091202-0820 EST
Besoeker:
I believe you left out the peak - minimum part of the calculation.
A three phase full wave rectifier provides peaks every 60 deg. The intersection of two sine waves with a 60 degree difference occurs at 30 degrees from 90. The midpoint of the 60 degree difference. The sine of (90 - 30) is 0.866. Thus, the peak to peak ripple is Vpeak (1-0.866) or 0.134*Vpeak.
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Besoeker:
I believe you left out the peak - minimum part of the calculation.
A three phase full wave rectifier provides peaks every 60 deg. The intersection of two sine waves with a 60 degree difference occurs at 30 degrees from 90. The midpoint of the 60 degree difference. The sine of (90 - 30) is 0.866. Thus, the peak to peak ripple is Vpeak (1-0.866) or 0.134*Vpeak.
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gar
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091202-0952 EST
Besoeker:
Now I understand your meaning of the words "0.866 Vp to Vp".
I usually interpret "peak to peak" to mean "+peak to -peak".
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Besoeker:
Now I understand your meaning of the words "0.866 Vp to Vp".
I usually interpret "peak to peak" to mean "+peak to -peak".
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erickench
Senior Member
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- Brooklyn, NY
How 'bout cutting a sine wave in half at the 180 degree mark and then take the integral of that rectified pulse and divide the result by one half of the time period of the full sine wave. This will give you the average value which is what the ammeter will read.
gar
Senior Member
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- Ann Arbor, Michigan
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- EE
091202-1445 EST
erickench:
The discussion is about a 3 phase full wave rectifier. Therefore, the integral from 0 to 180 and averaged is not the signal. Integrate from 60 to 120 and average over that time period for the average voltage.
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erickench:
The discussion is about a 3 phase full wave rectifier. Therefore, the integral from 0 to 180 and averaged is not the signal. Integrate from 60 to 120 and average over that time period for the average voltage.
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0.866 Vp is the minimum. Vp is the maximum.
The difference between them is the peak to peak and the same value that you gave. There is no "-peak" as such.
Here's a graph of instantaneous voltage, average voltage and the fundamental component of the ripple voltage.
Supply voltage is 415V ac.
Average Vdc is 560Vdc
Ripple is predominantly six times supply frequency (also shown) plus higher orders (not shown). The sixth is just under 4% oF the mean Vdc and total ripple just over 4%.
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