Electronics Repair (sci.electronics.repair) Discussion of repairing electronic equipment. Topics include requests for assistance, where to obtain servicing information and parts, techniques for diagnosis and repair, and annecdotes about success, failures and problems.

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Default 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.
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Steve Kraus wrote:

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.

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)...



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Default Calculating Power Factor


"Steve Kraus"


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."



** 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


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Default Calculating Power Factor

For god's sake monitor the neutral current ( using a current clamp
probe)


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.
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Default Calculating Power Factor

Jamie wrote:
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)...



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.


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"Steve Kraus"

For god's sake monitor the neutral current ( using a current clamp
probe)


There is no neutral connection so no neutral current.


** Then monitor one of the phases and the voltage of that same phase.


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?



** The common connection is LIVE in your case - IDIOT !!

In effect that's what I've done
except I'm taking advantage of the small voltage drop that already exists.



** 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




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Steve Kraus wrote:

Jamie wrote:

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)...




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.

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..





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Phil Allison wrote:

** The common connection is LIVE in your case - IDIOT !!


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.
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"Steve Kraus is a Krackpot "

Phil Allison wrote:

** The common connection is LIVE in your case - IDIOT !!


I'm aware of that and dealt with it carefully and appropriately.



** Like hell you did !!!!

Only a COMPLETE IDIOT does what you did.



But by all means please continue with your name calling.



** Listen here pal -

you are one ****ing arrogant, pig ignorant, snipping, trolling pile of vile
autistic dung.

And them are all your GOOD points.


.... Phil


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Default Calculating Power Factor

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


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.


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On 11/14/2010 9:02 PM, Phil Allison wrote:
Only a COMPLETE IDIOT does what you did.


Along with several idiots at Xerox.

While i was working on a switching power supply, we had reason
to measure the input (live) AC line. Tectronix to the rescue.
Apparently they knew there were a few insane people that would
want to float a scope for live AC line measurements and they had
a special isolation box that the scope plugged into for just that
occasion.

Of course, some fool removed it while we were at lunch.

Ya know, it's amazing the fwipzzzt noise the ground lead on an
expensive scope probe makes when it goes up in flames when you
turn the power on.

Jeff

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"Steve Kraus = IMBECILE "


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


** No one tiny bit - if you have a clue about PSUs.


*not counting reactive power issues



** Already told ya - there IS no reactive power issue.


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


** Then you already know the VA figure.

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

No if or buts.


..... Phil




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"Jeffrey Angus"
Phil Allison wrote:

Only a COMPLETE IDIOT does what you did.


Along with several idiots at Xerox.

While i was working on a switching power supply, we had reason
to measure the input (live) AC line. Tectronix to the rescue.
Apparently they knew there were a few insane people that would
want to float a scope for live AC line measurements and they had
a special isolation box that the scope plugged into for just that
occasion.

Of course, some fool removed it while we were at lunch.

Ya know, it's amazing the fwipzzzt noise the ground lead on an
expensive scope probe makes when it goes up in flames when you
turn the power on.


** ROTFL


...... Phil


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On Sun, 14 Nov 2010 21:22:33 -0600, Jeffrey Angus
wrote:

On 11/14/2010 9:02 PM, Phil Allison wrote:
Only a COMPLETE IDIOT does what you did.


Along with several idiots at Xerox.

While i was working on a switching power supply, we had reason
to measure the input (live) AC line. Tectronix to the rescue.
Apparently they knew there were a few insane people that would
want to float a scope for live AC line measurements and they had
a special isolation box that the scope plugged into for just that
occasion.

Of course, some fool removed it while we were at lunch.

Ya know, it's amazing the fwipzzzt noise the ground lead on an
expensive scope probe makes when it goes up in flames when you
turn the power on.

Jeff


Nicely done. I usually had a sign hanging on the test mess during
lunch. Something like "Touch anything and you'll die". However,
ty-wrapping the power cords to both the instrument and the power strip
worked somewhat better. I also placed mouse traps in strategic
locations until my boss demanded that I stop (after one got him).

So, why didn't you use two probes, with the scope set to differential?
To get the phase, you would need a 4 trace scope, but those are common
enough. I believe that differential is also the recommended method as
it keeps the scope case at ground level and takes care of the common
mode noise on the line. The isolation xformer method puts one side of
the power line on the case of the instrument, which is not pleasant if
you touch both the instrument case and a nearby ground.

Remember.... you have but one life to give to your profession.

Incidentally, I always kept my old broken scope probes. The
intermittent probes were especially prized. I would destroy one about
every 2 months. I would mark them with a tiny red nail polish dot,
and leave them connected in a conspicuous location, while I hid my
working scope probes. When they would predictably disappear, I would
just smile. I also had a #2 Phillips screwdriver with a self
stripping plastic handle. I think it was borrowed (stolen) at least 3
times, and then returned. I still have it somewhere.



--
Jeff Liebermann
150 Felker St #D
http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
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On Sun, 14 Nov 2010 18:44:27 -0600, Steve Kraus
wrote:

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.


If all you want are the results, the common Kill-A-Watt power meter
has a PF measurement feature. $25-$40 everywhere.
http://www.p3international.com/products/special/P4400/P4400-CE.html
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.
http://www.pge.com/includes/docs/pdfs/mybusiness/customerservice/energystatus/powerquality/nonsinusoidal_power.pdf

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


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


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"Jeff Liebermann"


With a switcher, you're going to have a non-sinusodial current
waveform.


** Same for a transformer based PSU too.



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.



** PF = true power / VA

No phase angles in sight ......



..... Phil



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On Mon, 15 Nov 2010 16:24:55 +1100, "Phil Allison"
wrote:

With a switcher, you're going to have a non-sinusodial current
waveform.


** Same for a transformer based PSU too.


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.

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.


** PF = true power / VA
No phase angles in sight ......


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
150 Felker St #D
http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
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Jeff Liebermann wrote:
If all you want are the results, the common Kill-A-Watt power meter
has a PF measurement feature. $25-$40 everywhere.
http://www.p3international.com/products/special/P4400/P4400-CE.html
I have 4 of these (various models).


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.
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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.

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Jamie t wrote:
Steve Kraus wrote:

Jamie wrote:

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)...




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.

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.


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.


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"Cydrome Leader"


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.



** 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


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Phil Allison wrote:
"Cydrome Leader"

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.



** 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


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