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How does this power supply work

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FlapJack

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How can this power supply possibly work. It is called a Stacked switching regulator. It is an non isolated topology that uses 208 V 3 phase for an input and an output of 3.3 KW, 330 VDC at 10A.

It's claim to fame is that most of the power passes directly through from the 3 phase rectified line to the output. The regulation is handled by a small inverter run at 25 KHZ with a square wave. The inverter primary is center tapped and supplied with a variable DC voltage controlled by the feedback loop.

The output of the inverter is a single secondary in to a bridge rectifier across a 400 uF 100 V capacitor that is in series with the 208 3 phase rectified line.

The 3 phase line is 208 VAC x 1.732 = 360 VDC - diode drop. For this circuit to work the capacitor has to pass the DC from the 360 VDC and somehow buck 30 VDC doing it.

I can see the inverter output driving the 400 uF capacitor to 30 VDC and it's polarity bucking the 360 VDC but that is assuming that the 360 VDC supply could pass through the capacitor. Capacitors are supposed to block DC!!!

Maybe it is a flux capacitor.

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BTW in the post there are 3 separate pictures labeled page 1, page 2, and page 3, it just looks like one big picture, hover your mouse over the thumbnail to see the file names.
 

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It's very obvious, the switcher operates to fill in the 15% ripple from the rectified 3 phase input. Look a bit harder at the power wiring of the circuit...!

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The buck converter supplies a variable voltage to the push pull switcher to help achieve the ripple in-fill, that way the push-pull can operate near full pwm with no switching losses and reduced diode recovery losses.
 

I do not see it. Why isn't the 100 uF capacitor blocking the 330 vdc line. The capacitor is in series with the output of the 330 VDC supply.
 

Correct, you don't see it, there are essentially two psu's in series, one is the ripply rectified mains, the other is a 10A rated switcher power supply, modulated to fill in the ripple to give smoother DC out, up to ~10A...

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remove the 400uF 100V cap and then follow the power circuit, this may help your understanding
 

Vripple = I load / FC

3 phase rectifier 6 power pulses x 50hz worst case = 300hz

10A / 300hz x 1000uF = 33 volts ripple
10A / 360hz x 1000uF = 28 volts ripple

I am going to try to redraw to power output.

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No matter how i look at it the 100 uF capacitor is in series with the 360 VDC power supply effectively blocking it.

The inverter bridge rectifier cannot pass the 360 vdc around the capacitor because it is the wrong polarity.
 

Oh God, sigh, #1 it is an 400UF cap...! and #2 the bridge is not the wrong way round, have a closer look...!
 

OK, i am seeing this a little differently now.

I redrew it again, this time two capacitors in series with a battery across each one of them to get rid of the clutter. The power supplies are additive, +- to +-.

If that is correct, then the inverter can only add voltage to the circuit. I have to think about that.
 

208ph-ph x Sqrt(2) = 294V peak, thus the other psu must add up to 36V or more to get 330VDC regulated
 

Now i see the diode is ok and thank you for getting nit picky on the 400 uF cap.

The last problem i have is the 360 vdc peaks. A friend of mine in body work used to say "you can't fill a high spot down" and that is what we have, a high peak of about 30 volts we can not get rid of because we can only add voltage.

The ripple calculation says 28 to 33 v and it looks like that is what we need to get rid of to meet out 330 vdc spec. Adding more DC voltage to the 360 vdc peak supply will only make things worse. So i do not see how it is regulating unless they do not mind the ripple voltage and only want to boost up the 360 vdc supply if the 208 vac line sags.

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208ph-ph x Sqrt(2) = 294V peak, thus the other psu must add up to 36V or more to get 330VDC regulated

The schematic clearly shows a 3 phase connection and 3 phase bridge.

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You can not believe how hard it was to find an authoritative source for 3 phase Y to full bridge and DC.

https://en.wikipedia.org/wiki/Rectifier#Three-phase_bridge_rectifier

For a three-phase full-wave diode rectifier, the ideal, no-load average output voltage is

Vdc = 3 x (sq root of 3) x V peak / pi

Vpeak, the peak value of the phase (line to neutral) input voltages.

208v line to neutral = 120 vac x 1.414 = 170 v peak

3 x 1.732 x 170 V peak / pi = 281 vdc

So you were correct, the inverter must add voltage to meet the 330 vdc spec, and this would fill in the ripple.

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Last thought, under the inverter bridge it says 10.7v at 10A. I kept looking all along for the 10.7 volts to be significant but i do not see any relevance. Does anyone else see why this was posted?
 

Sigh, the o/p of the switcher is not flat DC, it is modulated from zero to whatever to fill in the ripple and provide, overall, 330V or so regulated, the 10.7 volts is an error, or typo....

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Also 208 phase to phase (any source, Y or delta) has a simple maximum of 1.414 x rms, the minimum is a little harder to compute, can you work out the valley voltage of the ripple?
 

The valley voltage should be peak_voltage multiplied by sin(120) or 0.866 (I hope I have not goofed up this time)
 

I spent 2 hours searching for information to calculate the valley and even tried putting c_mitra's answer in the search terms, this is clearly classified information.
 

nope, with the aid of a simple sketch and a calculator that does sine functions you can work it out easily...

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or you can run a simple 3 phase ckt on p-spice, LT spice, or look up a year 1 text book on electrical systems
 

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