Transformer and rectifier math

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SceneryDriver

Senior Member
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NJ
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Electrical and Automation Designer
All,
I posted this question over at plcs.net, and didn't really get a good answer so I'm asking here:

My transformer and rectifier math for 3-phase supply is a bit rusty.

I'm looking to design a nominal 70VDC power supply for a large number of servos. They require 75VDC, but with the number of servos I'm looking to run, the factory-supplied power supplies will consume too much room in the control cabinet, and be too heavy.

I'm looking at building a DC supply using:

(3) 1kva 48VAC secondary transformers wired together with a delta secondary (fed from a 208VAC 3-phase source)
(1) 3-phase, 60A bridge rectifier module
(2) 18,000uF filter caps

The transformers are rated at 234VAC primary, 48VAC secondary; if I feed this power supply from 208VAC, that should give me a ~42VAC secondary voltage.

I will rectify the outputs of the transformers, and if my math is right, I should get 42V x 1.65 = ~70VDC. With filtering caps, I should have a fairly solid DC supply. The servos have a 90VDC max rating, so I have plenty of headroom.

Is my math correct to derive my peak DC voltage, using the 1.65 multiplication factor for bridge rectified 3-phase voltage? Nominal DC voltage should be close to peak, with adequate filtering, yes?

My math came from this equation:

power99.gif


I'm also slightly unclear on what my ultimate capacity of the power supply will be; if the transformers' secondaries can each supply 20A, I should have 60A DC available, yes?

I haven't done much power supply design with a three-phase source (any, in fact). A single phase supplied DC supply would be easier, but it won't give me the capacity I need without a monster transformer.


Thanks for the help.

SceneryDriver
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
How are you getting from a 3 phase input to a single phase output?

Also, have you thought about the harmonic currents this will induce into the AC supply?

the factory-supplied power supplies will consume too much room in the control cabinet, and be too heavy.

Before I went too far down the road of building custom supplies, I'd probably ask myself if there is a reason the factory supplied units are larger and heavier than what you have in mind.
 

junkhound

Senior Member
Location
Renton, WA
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EE, power electronics specialty
Scenery - how much is the load?

You can buy off the shelf 5 kW Vienna bridge 3 phase PFC rectifiers a lot cheaper than designing a straight rectifier with big caps.
About the size of 2 packs of playing cards, that should fit, eh.


Post your amperage needs and can probably give you a part number even. Lots of sources, Vicor, Murata, Kepco, Aveox, etc....
 

paulengr

Senior Member
The following PDF contains everything you are looking fir:

https://cds.cern.ch/record/987551/files/p133.pdf

Lots of calculus but they give the solved equations so you can safely ignore the calculus equations.

Second you are missing voltage ripple in your calculations. Vdc=Vp-Idc/4FC where:
Vdc is average voltage out
Idc is in Amps
Vp is the peak AC voltage, RMS x 1.009 for a six pulse rectifier
F is frequency in Hz
C is capacitance in Farads (18,000 / 1,000,000)

This is the formula usually used for capacitor sizing.

Also be aware that these are frequently used as precharge circuits for large drive systems. But one thing missing above is startup. At startup the capacitor appears as a huge short circuit and can easily burn up your rectifier. In a drive they just put a resistor in series. Since the precharge is essentially startup use only a contactor opens at the end of precharge. In your case though since your whole circuit is precharge you'd have to do the opposite...insert a contactor to bypass the resistor on startup.

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synchro

Senior Member
Location
Chicago, IL
Occupation
EE
I don't remember the formula, but I'm pretty sure the DC from a three phase bridge is close to 1.35 x RMS AC

Yes, VDC = 3sqrt(2)/pi x VLine-Line RMS = ~1.35 VLine-Line RMS

About 1.4VDC should also be subtracted from the output to account for voltage drops in a pair of diodes inside the bridge.


What is the most suitable term for the resultant DC voltage from a 6 pulse rectifier, considering the ripple? I hear "average" used most of the time. Why not RMS?

I guess it depends on the requirements of the load that is served by the DC supply.
RMS or Root Mean Square is the square root of the average (aka "mean") of one (or the sum of more than one) quantities that are squared, usually voltages or currents in electrical applications. RMS is more meaningful in AC power applications than the peak voltage of a waveform because power can than be directly calculated as P = V x I =V2/R =I2R for resistive loads when V and I are RMS quantities. Also, heating of conductors is proportional to I2R and therefore to the "mean square" of the AC current waveform even if harmonics or other frequencies are present.

It's more likely that the average value of the DC supply voltage is a more useful quantity than RMS for loads that use DC, unless you wanted to accurately quantify the total power into a resistive load when there's a significant amount of ripple (which would not be a common siituation).
Otherwise, the average DC voltage and peak-peak ripple are more useful quantities. For example, if you had a linear voltage regulator IC fed by a rectifier then the minimum voltage on the waveform due to DC ripple is very important. The voltage drop across the regulator cannot fall below a minimum value or it falls out of regulation. An RMS spec doesn't give you enough information in this case, because very short dips in the DC voltage would have minimal effect on the RMS ripple value.
 

junkhound

Senior Member
Location
Renton, WA
Occupation
EE, power electronics specialty
For those without analysis tools or a math capable scope, here are visual waveforms, 100 Vrms line to neutral. (If you can read it, would not enlarge further)
Note that difference between avg and rms is less than the diode drop variation with temperature.
2.32/1.73 = 1.34 for Vdc to Vrmslinetoline for LOW LOAD, includes about 0.8V drop per diode. diode drop will be higher at higher currents.
Source for this was a 100 Vrms line-neutral 3 phase, full wave 6 diode bridge.
Ripple with no capacitance is about 14% peak to peak
1awf.jpg
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
For those without analysis tools or a math capable scope, here are visual waveforms, 100 Vrms line to neutral. (If you can read it, would not enlarge further)
Note that difference between avg and rms is less than the diode drop variation with temperature.
2.32/1.73 = 1.34 for Vdc to Vrmslinetoline for LOW LOAD, includes about 0.8V drop per diode. diode drop will be higher at higher currents.
Source for this was a 100 Vrms line-neutral 3 phase, full wave 6 diode bridge.
Ripple with no capacitance is about 14% peak to peak

Can you determine the peak diode current? That's one very important parameter for diode rectifier design. Both at stead state, and also when the power is first applied and the caps. are completely discharged.

Although, I still don't know how we are getting from a 3 phase supply to a single phase output, so maybe you don't know what the complete circuit is either.
 

junkhound

Senior Member
Location
Renton, WA
Occupation
EE, power electronics specialty
Can you determine the peak diode current? That's one very important parameter for diode rectifier design. Both at stead state, and also when the power is first applied and the caps. are completely discharged.

Although, I still don't know how we are getting from a 3 phase supply to a single phase output, so maybe you don't know what the complete circuit is either.

here is the common circuit for simple 6 diode 3 phase bridge
You can get EVERY voltage, current, and power dissipation from this type FEA model This is ORCAD PSpice, you can download a free student version off the orcad web site, think this schematic has few enough parts for the free version. there are other free FEA circuit anlaysis programs such a LTSpice - this has generator source impedances from another project being worked, much lower values for connection to the grid. The plot on previous reply has the output cap (C2) set to zero .

1aschdb.jpg
 

paulengr

Senior Member
What is the most suitable term for the resultant DC voltage from a 6 pulse rectifier, considering the ripple? I hear "average" used most of the time. Why not RMS?
RMS has to do with the AC that we don't care about...we want only the DC component.

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
190710-1948 EDT

paulengr:

A true RMS measurement includes both the DC component and any AC component added to it.

Most electronic voltmeters labeled "true RMS" only measure the RMS value of the AC component. An input capacitor strips out the DC component.

An electrodynamometer meter does read true RMS within its frequency range capabilities. And a hot-wire or thermal type also reads true RMS up to moderately high frequencies.

Study the definition of RMS. Look at the calculus equations for calculating RMS.

.
 

paulengr

Senior Member
Agreed RMS includes DC but I didn't say it did not include it, only that it's not the relevant value. Agreed the meters that aren't true RMS don't give correct values just as they don't usually handle nonsinusoidal voltages correctly either.

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Yes, VDC = 3sqrt(2)/pi x VLine-Line RMS = ~1.35 VLine-Line RMS

About 1.4VDC should also be subtracted from the output to account for voltage drops in a pair of diodes inside the bridge.




I guess it depends on the requirements of the load that is served by the DC supply.
RMS or Root Mean Square is the square root of the average (aka "mean") of one (or the sum of more than one) quantities that are squared, usually voltages or currents in electrical applications. RMS is more meaningful in AC power applications than the peak voltage of a waveform because power can than be directly calculated as P = V x I =V2/R =I2R for resistive loads when V and I are RMS quantities. Also, heating of conductors is proportional to I2R and therefore to the "mean square" of the AC current waveform even if harmonics or other frequencies are present.

It's more likely that the average value of the DC supply voltage is a more useful quantity than RMS for loads that use DC, unless you wanted to accurately quantify the total power into a resistive load when there's a significant amount of ripple (which would not be a common siituation).
Otherwise, the average DC voltage and peak-peak ripple are more useful quantities. For example, if you had a linear voltage regulator IC fed by a rectifier then the minimum voltage on the waveform due to DC ripple is very important. The voltage drop across the regulator cannot fall below a minimum value or it falls out of regulation. An RMS spec doesn't give you enough information in this case, because very short dips in the DC voltage would have minimal effect on the RMS ripple value.

Thank you for the thorough explanation!
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
If you apply a voltage source (AC, DC, mixed) to a resistor, current will flow and the resistor will dissipate heat. The average power dissipated will depend on the RMS applied voltage, and this is true for AC, DC, mixed AC and DC.

-Jon
 

paulengr

Senior Member
I'm not trying to be snarky, but that is not a good response. I want my money back ;) That was basically my question: why not apply the RMS principle to DC like we do with AC. Nothing says RMS is for AC only.
In DC servo motors torque is proportional to voltage (less counter EMF which is hopefully a small value) so average torque is proportional to average voltage. If the voltage has ripple the average torque is proportional to average voltage. On pure AC if the frequency was slow enough it would just oscillate back and forth. On an AC motor RMS voltage is what matters, to a point. On an AC power meter on resistive loads V x I = P with RMS V and I just as with DC power.

Usually when taught in school we talk about the fact that in the analog era making a cheap mostly RMS AC meter is simple with just a single diode rectifier circuit but true RMS let alone doing square roots in analog is nontrivial but that's teaching first year students, not second years in the motor class.

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
190711-0846 EDT

paulengr:

In DC servo motors torque is proportional to voltage (less counter EMF which is hopefully a small value) so average torque is proportional to average voltage.
That is an incorrect statement.

Torque is proportional to current for a constant magnetic field in which that current is flowing. It is true that if a servo spindle is at zero RPM that current will be approximately proportional to applied voltage. But it is not true that voltage produces torque, it is current.

Counter EMF is approximately proportional to RPM. Why is that hopefully small? If you are talking about a servo there are many times, and possibly mostly when the servo is rotating.

On a CNC machine mostly doing drilling the X and Y servos are mostly stopped. However, both Z and spindle are moving. But if you are contour milling, then X, Y, and Z are mostly moving, and spindle is at a constant high speed. On most CNC lathes the servos are mostly moving, and the spindle is usually at a high speed.

.



 
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