You'd need to be a Scot to understand.......The internet of truth.
Power factor
- Thread starter mbrooke
- Start date
- Status
mivey
Senior Member
Well I ain't from 'round there so...You'd need to be a Scot to understand.......
Are you implying that the Scots can't all agree on the pronunciation?
Not really although regional accents abound. An Aberdonian and a Glaswegian don't sound at all alike.
The difference between Chic Murray and say Billy Connoly is very marked.
But lest we stray off topic too far and invoke moderaters' intervention........
Distortion power factor is a real issue these days.
The waveform measured on my incoming supply:
The flattened top tells the story.
mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
Ok, back on topic
So it would seem when both X and C are in series, a type of "resonance" takes place?
https://www.youtube.com/watch?v=BPoDvcqqlDY
So it would seem when both X and C are in series, a type of "resonance" takes place?
https://www.youtube.com/watch?v=BPoDvcqqlDY
That be L and C.Ok, back on topic
So it would seem when both X and C are in series, a type of "resonance" takes place?
https://www.youtube.com/watch?v=BPoDvcqqlDY
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
Loch Ness Monster of course.............
The internet of truth.
Punch drunk. humourous, and somewhat silly arguments abound...
Y'all train drivers amuse me and provide a somewhat pleasant diversion....
Hint: Neither Scots nor Americans speak English actually.
I think you wil find that Robert Louis Balfour Stevenson made a pretty good fist of it.Hint: Neither Scots nor Americans speak English actually.
Samuel Clemens too.
peter d
Senior Member
- Location
- New England
Awesome incite, thank
But how does on go about net power factor? Consider an HID lamp with a reactive ballast. Its essentially a resistor in series with a reactor, yet the power factor can vary all over the place.
I'm getting in on this late and didn't read all 19 pages, but an HID arc is not purely resistive, the current waveform of the arc is highly non-linear.
mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
Bes, it was a riddle. To continue it, such as:
What is common or not between British and Americans?
Hint 2 :For the above statement see Oscar Wilde, Bernard Shaw, and Winston Churchill.
P.S. Both authors you cite are outstanding IMHO.:thumbsup:
mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
I want to keep this thread on track, politely saying And yes the British and Americans are different in different ways.
Anyhow, why is it that ballasts are parallel X and C circuits and not series circuits?
And yes the British and Americans are different in different ways. Anyhow, why is it that ballasts are parallel X and C circuits and not series circuits?
mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
Leaving this here as it has graphs for series and parallel type ballasts vs line frequency on page 20:
http://www.etsii.upct.es/antonio/html_der/papers/Ballasts.pdf
http://www.ecospecifier.com/media/7...stitute - Ballasts for Fluorescent Lights.pdf
http://www.etsii.upct.es/antonio/html_der/papers/Ballasts.pdf
http://www.ecospecifier.com/media/7...stitute - Ballasts for Fluorescent Lights.pdf
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
I want to keep this thread on track, politely saying
Anyhow, why is it that ballasts are parallel X and C circuits and not series circuits?
As you vary X and C to get close to unity power factor the series impedance approaches zero, which is not much help in limiting the current through the lamp.
mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
As you vary X and C to get close to unity power factor the series impedance approaches zero, which is not much help in limiting the current through the lamp.
Perhaps... If you take say a 5 milli henry inconductor and a 4uF capacitor, would these small components really behave like a solid piece of #12 when driver a 400 watt Metal Halide?
Last edited:
mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
As you vary X and C to get close to unity power factor the series impedance approaches zero, which is not much help in limiting the current through the lamp.
Here is what I have in mind (second picture in)
http://www.candlepowerforums.com/vb/showthread.php?260956-Fluorescent-to-LED-night-light-conversion
The ballast in this case us an capacitor and inductor in between the line. Perhaps not unit PF, but close. Yet even though its said this circuit would not provide impedance it still does somehow. It is this series circuit which puzzles me. The parallel aspect I understand from the equations posts (awesome help btw)
mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
I found this but still can't grasp total series impedance:
http://www.allaboutcircuits.com/textbook/alternating-current/chpt-5/series-r-l-and-c/
http://www.allaboutcircuits.com/textbook/alternating-current/chpt-5/series-r-l-and-c/
You can calculate the impedances as shown but you can't add them up arithmetically - as simple numbers.
If you look down the page a bit you will see calculations for ZR, ZL, and ZC.
You will note that impedances are expressed as two components.
If you take ZL, and ZC you will not that they have opposite signs, one positive, one negative. They subtract.
Do that and you end up with two components, R and X. But note the 90deg difference. You can't simply add them.
Imagine two sides of a triangle with a 90 deg angle between them. A shape like a corner if you like. Join the outer sides to get the diagonal. That diagonal represents the total impedance.
Obviously, it is longer than the two component parts. But not simply the arithmenic sum. You need to use pythagorous.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
I do not see an inductor anywhere in the schematic. Nor in the picture.Here is what I have in mind (second picture in)
http://www.candlepowerforums.com/vb/showthread.php?260956-Fluorescent-to-LED-night-light-conversion
The ballast in this case us an capacitor and inductor in between the line. Perhaps not unit PF, but close. Yet even though its said this circuit would not provide impedance it still does somehow. It is this series circuit which puzzles me. The parallel aspect I understand from the equations posts (awesome help btw)
Using a series capacitor to limit current is common for small loads and it does not provide anywhere near unity PF.
mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
This, to the right:
My undeertanding is that that was the original circuit, not the replacement capacitor input filter.
In any case, it is not clear to me just how the original circuit was wired.
- Status