## Power capability question

This forum is for specialized infomation important to the construction and safe operation of the high voltage electrical supplies and related circuitry needed for fusor operation.
JoeBallantyne
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Real name: Joe Ballantyne
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### Re: Power capability question

Dan, why don't you just get a rotary phase converter? Even if you don't get one that can match the power output capacity of your power supply, you could get a 5 or 10 HP phase converter, and limit the current you pull from the supply to contol the max load it places on the converter.

Then you don't have to worry about how to derate 1 phase versus 3, and can just limit the current so that at max voltage you draw less power than what the phase converter can supply.

Joe.

Dan Knapp
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Real name: Dan Knapp

### Re: Power capability question

Joe, I could easily put together a three phase converter from a three phase motor, but I don’t need the full design power of the supply. I’m just trying to determine how much power is possible operating in single phase mode. I keep finding the figure of 57% (or 1.73; 1/1.73 = 0.57), but I’ve yet to find and explanation of how the figure was determined. I’m still searching.

JoeBallantyne
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Real name: Joe Ballantyne
Location: Redmond, WA

### Re: Power capability question

Never really took a power systems class since even 30+ years ago when I was in college they had already deprecated that from the curriculum.

The 1.73 is a rounded version of the square root of 3.

208 3 phase power is 120 volts in each line with a phase separation of 120 degrees between the lines. It is called 208 because that is 120 mulitplied by the square root of 3.

When you calculate the power of a 3 phase 208 line with 30 amps of current you are supposed to multiply 208*30*1.73. Which gives you 10.8KVA. Which is the same as 120*30*3.

This was a link I found: https://www.raritan.com/landing/three-p ... -explained

Effectively the 1.73 is the square root of 3 applied to the RMS voltage and separately to the RMS current that flows in a single line of a balanced 3 phase circuit, so that when you calculate the power delivered in the circuit it is 3 times the power delivered by a single line. Since 1.73*1.73=2.9929

I suspect that is where this factor 1.73 you keep seeing comes from.

Joe.

JoeBallantyne
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### Re: Power capability question

It is also possible that when you calculate the RMS (real mean squared) voltage for 2 nominally 120V RMS lines phase shifted by 120 degrees that you get 208V as your output.

Question is what does the graph of cos(x)-cos(x-120) look like. If you integrate the square of that over one cycle what do you get.

After 30 plus years the math is rusty and so I will leave it as a basic exercise for the reader. (I always DESPISED it when professors did stuff like that.)

Joe.

JoeBallantyne
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Real name: Joe Ballantyne
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### Re: Power capability question

For your scenario where you are running 240V RMS through a transformer to get 208V RMS, the total power you are delivering is just the 208V times whatever RMS current you are supplying.

Note that the phase is different because your lines are 180 degrees out of phase instead of 120.

I suspect if you push your supply hard (ie: try to pull out as much power as the nominal amount of power you can supply on the input), it will NOT stay in spec for ripple, or regulation, because it is expecting to be delivered a more steady/continuous supply of power than it is getting.

If you need your supply to perform up to its specifications, you should build that rotary phase converter, and feed it 3 phase.

If that is not really important in your application, then I suspect if you never pull more than 1/2 the nominal power you can supply the input (say 208V * 20A = 4160VA) it will likely work reasonably well. So never pull more than 2KVA on the output if you are on a 20A 240V circuit.

My 2 cents. (Which are very possibly worth absolutely nothing, no FED intervention required at all.)

Joe.

Rich Feldman
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### Re: Power capability question

I was about to suggest derating to 2/3 of nominal power, based on average current in the input bridge rectifier.
Two of six diodes are idle when you run on single phase, so the high side diodes (for example) each conduct 1/2 instead of 1/3 of the DC bus current.

Under that condition, I bet the ripple current in DC bus capacitors will be greater, and true-RMS current in the 4 diodes that are working.

Could simulate it with a free SPICE program if not confident in one's analytic answer.
Even get fancy and simulate front end power-factor-correction circuit.
All models are wrong; some models are useful. -- George Box

John Futter
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### Re: Power capability question

That is why I said use the 240 volts natively multiplied by root of 2 gives 339 volt dc buss
208 x root of 3 phase gives 359 volts DC buss.
In fact you might need a boost transformer to bring it up a little ( 20 volts) but I reckon that it will go well until the DC buss caps complain about the high ripple current

check to see if it works if it locks out on under voltage you know what to do

i seriously doubt it will lock out on over voltage

Rich Feldman
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### Re: Power capability question

There's no place for sqrt(3) in conversion from sinusoid RMS to peak voltage.

Sqrt(3) is the ratio between 208 and 120 volts.
208 is a nominal phase to phase voltage for three-phase in 120V land.
Bridge rectifier will charge bus capacitors to 208 x sqrt(2) = 294 volts. In six pulses per cycle, or 2 pulses if one wire is not connected.

p.s. the next higher voltage for 3-phase here is 277 V phase-to-neutral, 480 V phase-to-phase.
AFAIK, 3-phase motor nameplates etc would say 480 V, while light fixtures for the same high-bay room would say 277 V.
All models are wrong; some models are useful. -- George Box

John Futter
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### Re: Power capability question

okay I thought 3 phase rectifiers did around approx 2.5 times the rms over single phase root 2 times the rms
correcting my mistake above
Where: VS is equal to (VL(PEAK) ÷ √3) and where VL(PEAK) is the maximum line-to-line voltage (VL*1.414).

VDC = VS x (cube root of 3/pi) = 1.65 x Vs

208 x 1.4142/1.44 = VS = 204.3

VS x 1.65 = 337 VDC

240 volts single phase gives 240 x 1.4142 = 339 volts

close enough for rock and roll

Rich Feldman
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Real name: Rich Feldman
Location: Santa Clara County, CA, USA

### Re: Power capability question

I just learned a new trick in LTspice, while preparing this demo.
Can interactively set differential voltage probes between, say, phase 1 and phase 2.
3phase_rect_sch.png (9.8 KiB) Viewed 573 times
The three "sine" voltage sources each put out +/- 170V (for 120 V RMS).
Chart shows their voltages (with respect to neutral) in green, blue, and red. Phase 3 node is called n001 because my name tag missed the wire.
Three bigger sinusoids represent the instantaneous phase to phase voltages, which are +/- 294 V ( 208 V RMS ).
At the top we see DC bus voltage, the voltage on n_VP with respect to ground. It peaks at 294 volts minus 2 diode drops.

Just for fun, I reran the sim after disconnecting phase 3. That's analogous to running on 208 volt single phase.
Three differential voltage probes remain on the diode junction that used to be phase 3, and now floats around, weakly tied by reverse bias behavior of the diode model.
Colors might be different. We could add charting of phase currents, diode currents, and capacitor currents to get insight mechanically.
All models are wrong; some models are useful. -- George Box