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Calculating Power Factor

I'm trying to figure power factor for a large transformer-type DC power supply. This is pretty much just for my own amusement (so I can figure power draw from current and voltage readings) so there's nothing critical here.

This device runs on 208V 3-phase (as in each leg is 120V from neutral but there's no neutral connection). The 3 transformer primaries are wye connected but there's no other connection to the wye point. It draws about

18A per leg.

I decided to compare current and voltage waveforms using a dual trace scope.

To look at current, rather than put a small resistance shunt in one line I decided to take advantage of the small voltage drop that occurs between the breaker panel and the cutoff switch for the device. So I had one channel of the scope looking at the voltage difference between one leg at the cutoff switch and the "hot" side of a 120V outlet on the same leg.

For voltage I just looked at one leg vs. neutral.

I'll skip over boring details about having to float the common side of the scope in order to do this without creating a short circuit. Anyway, it seemed to work. Emphasis on "seemed."

What I saw looked like voltage vs. current was out of phase by about 90°. Is that what one would reasonably expect going into a transformer? Seems kind of extreme so I wonder if my entire setup was bogus in some way.

Part of the reason for the question is that in retrospect I should have probably been looking at voltage as measured from one input leg to the wye point, not neutral since that's what a transformer primary sees. But I would think the wye would be similar to neutral.

Reply to
Steve Kraus
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what you're seeing in the xformer is normal. the xformer is returning the current phase from the current applied source which comes back out of phase with it, thus, the current you see is lagging behind from which it was created from the voltage you now see going the other way...

Power Factor (F) is the difference between True power (resistive) and Apparent power (reactive)...

Reply to
Jamie

"Steve Kraus"

** Pure insanity.

The scope common ( or ground) is at the same voltage as one phase.

For god's sake monitor the neutral current ( using a current clamp probe) and one phase.

..... Phil

Reply to
Phil Allison

There is no neutral connection so no neutral current.

But assuming you misspoke, what's the difference between sensing current on one of the legs using a current clamp probe and wiring a shunt into that line and looking at voltage across it? In effect that's what I've done except I'm taking advantage of the small voltage drop that already exists.

Reply to
Steve Kraus

My aim here is to estimate real power draw. I've got a switch mode supply putting out about the same output and it draws about 8A per leg. (I don't know anything about PF or reactive power on that one). This transformer type rectifier is drawing 18A per leg. I know from asking around that this particular brand is known to have inefficiently wound transformers. So that accounts for part of it. But there's also the matter of taking into account reactive power & power factor.

BTW, output in both cases is about 62A at about 22V.

Reply to
Steve Kraus

"Steve Kraus"

** Then monitor one of the phases and the voltage of that same phase.
** The common connection is LIVE in your case - IDIOT !!
** You are so dangerously * STUPID * you have no clue as to what PF even is.

Hint:

In the example you are discussing, phase angle barely comes into it.

.... Phil

Reply to
Phil Allison

It's possible it has a saturated core transformer. What this does is causes the input side to exert the rated operating current at all times but this type of design is best at keeping the transformer at a safe operating Q to help prevent unwanted harmonics and oscillations in the circuit that can take place where capacitors are in the coil design. You see this in microwave ovens due to the cap in the hV circuit.. with out the saturated transformer it can hit a resonance that can become destructive.

This isn't a problem if you plan on using the supply to it's fullest, as far as efficiency goes..

I think what you're looking for his is efficiency measurements, not PF.. Just load the supply to its max and measure the input side and compare the wattage to the output side. You will see that more energy is going to be used on the input side..

Reply to
Jamie

I'm aware of that and dealt with it carefully and appropriately. It's not particularly difficult not to contact anything live for the few seconds I was looking at the traces. Scope front? Plastic. Scope knobs? Plastic. It's not rocket science. You've never worked around something live? Well good for you. I'm sure I don't have to point out that the voltage, while certainly dangerous, is less than in other parts of the world.

But by all means please continue with your name calling.

Reply to
Steve Kraus

No, it won't need to go much more than it is now. These are xenon arc projection lamp supplies. We have sufficient light output at 55A but as the bulbs age it's necessary to raise the current to keep the arc from wandering on the electrode tips. If I have to go much over 65A (70A is max on these bulbs) it will be time to replace the bulbs but that time is coming very soon on the basis of hours alone. Fortunately for me the bulbs last years.

I was just kind of curious what the difference in power draw was and realized simple current & voltage measurements that I would take might be meaningless when going into a transformer. I do know the difference is real since the conduit going to the transformer type supply gets slightly warm after many hours but the one going into the switch-mode supply doesn't seem to rise detectably above ambient so I tend to believe the 8A figure on that one.

If I'd known the transformer type of this particular brand was so inefficent I would not have bought it 3 years ago. The original 20 year old equipment (two of that same brand) failed about 6 months apart in 2007. The first I replaced with a new one of the same make though of larger capacity. When the 2nd old one gave up the ghost I decided to go with the switch-mode supply.

Electricity is a tiny part of operating costs but still, just on principle...8A vs. 18A* for the same output? It seems rather amazing.

*not counting reactive power issues

Oh, in case anyone cares, I'm measuring current with a Fluke / LEM LH2015 true RMS AC/DC clamp meter.

Reply to
Steve Kraus

"Steve Kraus = IMBECILE "

** No one tiny bit - if you have a clue about PSUs.
** Already told ya - there IS no reactive power issue.
** Then you already know the VA figure.

But to find true power in this example - you need a wattmeter.

No if or buts.

.... Phil

Reply to
Phil Allison

If all you want are the results, the common Kill-A-Watt power meter has a PF measurement feature. $25-$40 everywhere.

I have 4 of these (various models).

If you want to do it safely with a scope, it really requires a 4 channel scope (or 2 channels with two differential inputs per channel). Put a small value non-inductive resistor in series with the power line. One pair of diffential input probes go across the resistor for the current waveform. The other pair of probes go across the line for the voltage waveform.

With a switcher, you're going to have a non-sinusodial current waveform. Therefore, you can't just use the phase angle as the power factor and will need to measure the Apparent Power and the RMS powers.

There's also quite a bit of work done on power factor correction these days which probably includes some measurement techniques.

--
Jeff Liebermann     jeffl@cruzio.com
150 Felker St #D    http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann     AE6KS    831-336-2558
Reply to
Jeff Liebermann

"Jeff Liebermann"

** Same for a transformer based PSU too.

** PF = true power / VA

No phase angles in sight ......

.... Phil

Reply to
Phil Allison

True. Once you get away from a pure sine wave current waveform, the phase angle method doesn't work. See the PG&E article I cited.

True. I just hate to agree with you, but I'll make an exception this one time. But don't worry... it probably won't happen again.

Handy buzzword translator: RMS power = True power Apparent power = VA power.

--
Jeff Liebermann     jeffl@cruzio.com
150 Felker St #D    http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann     AE6KS    831-336-2558
Reply to
Jeff Liebermann

Thanks Jeff; that looks like a useful gadget. I may have to get one. But I don't think it's applicable for my situtation which is 208V / 3-phase.

Actually there could be a way to use it for 3-phase. Assuming it measured current on just the hot leg one could run one leg at a time through the meter with the neutral side of the meter connected to the power line neutral OR, even better, connecting it to the Y point of the power supply. The meter would only be exposed to 120V. Basically, it would be imagining this as 3 separate single phase loads. Then just sum the results.

But the meter is only good for 15A and my supply is pulling 18 and up.

Reply to
Steve Kraus

I enjoyed this thread, especially the pranks.

Anyway, I too had worked on a few high power (1000 W) ARC lamp supplies. The last had photofeedback unit that controlled intensity. The intensity got lowered from a fixed level from the feedback unit. The unit had a shutter, so the setpoints were remembered when the lamp was off.

This photofeedback unit was designed in ahigh tech fashion. The light was sampled with fiber optics at the last of the optical path. The sensor temperature was controlled via a thermoelectric block.

Reply to
Ron D.

dumb question time.

What does the magnetron behave like in a circuit? I realize it's basically a giant rectifier tube, but I'm not wise on tube characteristics and behavior. I imagine there's some bizarre resistive component, but that's all I can come up with.

Reply to
Cydrome Leader

"Cydrome Leader"

** Suffice to say that the load presented by a typical microwave oven to the AC supply is close to sine wave current and pretty much in phase.

Resonance is definitely involved.

.... Phil

Reply to
Phil Allison

I just measured mine to see if an emergency generator would like it. Measured .95 on a kill a watt.

Reply to
mike

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