Gallery of Lights
Lamps => Modern => Topic started by: Vince on November 24, 2010, 09:26:12 PM
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Maybe you are one of those who never completely understood the principle of power factor. Well, now that I recently learned it at school, I can explain it!
What is power factor?
To explain what is exactly the power factor, we must first know three basic terms, real power, apparent power and reactive power.
The real power is measured in Watts. You work with watts all the time. It's the part of the electrical power that does the job.
Apparent power is measured in VoltAmps (VA) and represents the whole power consumption. In the diagram that follows, it's P(max), or peak power X 0.707. Don't confuse it with the average power in the diagram! The average power is P(max) X 0.636. We won't need average power in this article LOL.
Reactive power is measured in Reactive VoltAmps (VAr) and represents the part of apparent power that isn't really used, but is still consumed in the electrical circuit. That may sound weird, "How can power not be used, but still consumed?". You'll see it in just a minute ;D
Let's figure everything out in this diagram of a sine wave AC current:
(http://upload.wikimedia.org/wikipedia/commons/thumb/d/db/Power_factor_1.svg/685px-Power_factor_1.svg.png)
The real power is:
Volts(eff) X Amps(eff)
Each of the efficient values can be calculated by multiplying their peak value by 0.707.
As long as there are not any inductive or capacitive load involved (only resistive loads), this principle applies. If there's an inductive (i.e. ballast) or capacitive (i.e. a cheap fluorescent night light with capacitive ballast) load, a phenomenon will occur, the phase angle of current or voltage won't equal 0 degrees anymore!
Let's see the diagram again:
(http://upload.wikimedia.org/wikipedia/commons/thumb/d/db/Power_factor_1.svg/685px-Power_factor_1.svg.png)
This diagram represents what a purely resistive charge would consume. The current (in green) and voltage (in red) sine waves both get to their peak at the same time .
Now, when power is applied to an inductor, the latter will create a 90 degrees phase angle, the voltage will lead the current. Because an inductor opposes the changes of current, it lags the latter. In my sine wave diagram it would give this:
(http://upload.wikimedia.org/wikipedia/commons/thumb/b/b9/Power_factor_0.7.svg/669px-Power_factor_0.7.svg.png)
See the current sine wave? It cycles at the same frequency, but goes to zero LATER than the voltage. Look at 360 degrees (at the very right), at the peak of the voltage sine wave. You'll notice the current sine wave isn't at its maximum value. So, for a maximum voltage, you don't have the full current, which gives you a lower real power, but still draws as much current over a complete AC cycle!
This, my friends, is the effect of a low power factor! :D
To make the phenomenon even more obvious, let's push it to the furthest and create a 90 degrees phase angle!
(http://upload.wikimedia.org/wikipedia/commons/thumb/2/23/Power_factor_0.svg/619px-Power_factor_0.svg.png)
In this diagram, no real power is consumed. Yup, no power! Can you find out by yourself why? If so, don't read any further! ;D Just continue when you find!
[...]
Correction time! XD If you found that the value of the current sine wave was 0 at the voltage peak, you are correct! Because any Volts X 0 Amps always gives 0 Watt.
But ... how about capacitors?
Capacitors will create the same phenomenon, except for one thing, it leads the current , and not the voltage! That is exactly why ballasts are combined with a capacitor to correct the power factor. Like (-) (-) = (+), the capacitor cancels the effect of the ballast on power factor, by LEADING BACK the current.
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Voila! That's it! I hope you now fully understand the principle of power factor. If you don't, feel free to contact me, by PM here, or via email. Just go on my profile page here (http://www.galleryoflights.org/mb/index.php?action=profile;u=6), you have everything needed to contact me. I'll gladly give you further explanations.
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Nicely Done Vince. ;)
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Oh, yeah. That is what I just learned from Circuit II class that I am currently attending. Now I know what the power company use both resistive power and reactive power to calculate the total apparent power. That is why we add capacitor to compensate the lagging power of the inductive ballast.
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This was about pure harmonic voltage/currents (taking phase shift only in determining the power factor), but the real fun start with non-sinusoidal waveforms.
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Yeah, like square, triangle or sawtooth waves. But we didn't see those at school so far LOL.
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I myself have a USB Digital Sampling Oscilloscope I got for free from a very nice deaf electrical engineer!
It makes those kinds of similar graphs on my computer screen when testing on electricity...
GREAT JOB Vince! Proud of ya man! You the man!
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Thanks for writing this informative article Vince, nicely done!
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I've been searching for the capacitor values to correct a normal-power-factor (NPF) HPS ballast to make the circuit high-power-factor (HPF). Usually small wallpacks and floodlights have NPF ballasts (single tap 120v). Often, a correcting capacitor is suggested on the circuit label of the ballast which is connected across the input.
These are the values I have, for all single 120 volt tap without capacitors:
35w HPS = 14 ?F (also 13 ?F shown)
50w HPS = 20 ?F (some suggest 22 ?F)
70w HPS = 28 ?F
100w HPS = 35 ?F (seen 40 ?F in some)
150w HPS = 55 ?F (also 52 ?F is suggetsed)
?F is micro farads, and 240 volt capacitor is wise.
Please make any corrections. Add capacitance values for mercury ballasts (that would be great) as well.
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For home use, there is no reason to worry about the power factor, so no reason to complicate the fixtures unless they were build as HPF from the factory. To really utilize the advantages of the HPF fixtures at home (higher efficiency,...), the power factor correction would have to be incorporated deeper in the ballast design than just a correcting capacitor parallel to the mains terminals.
The HPF is necessary with large installations, where the wiring is loaded to it's limit by just the light fixtures. Then the HPF could make the wiring cost 30..50% less than with NPF ballasts (the same wire gauge could serve twice as much HPF fixtures, so about half of the wiring is needed). But with home use even the rather high power lighting (e.g. 175W yard blaster) load the wiring to 25..30% (3..4A for a NPF ballast on 120V), so with HPF ballast there would be the same wiring, so costing the same.
Don't forget, the utility is billing you for the real consumed energy (so real power), the power factor have very minor influence (it should have none at all, but the meters are not ideal; but nobody know, which direction will help you, as it is pure random error...)
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For home use, there is no reason to worry about the power factor, so no reason to complicate the fixtures unless they were build as HPF from the factory.
I agree. But some people have multiple fixtures on for display or for hydroponics. The lower current benefits the typical 15Amp house wiring.
To really utilize the advantages of the HPF fixtures at home (higher efficiency,...), the power factor correction would have to be incorporated deeper in the ballast design than just a correcting capacitor parallel to the mains terminals.
But I thought that a working ballast (like a cheap one found in wallpacks & floodlights) gets less hot with the suggested capacitor, thereby increasing its life (make a cheap ballast into a better one). Don't be afraid to ellaborate on your deeper in ballast comment.
The HPF is necessary with large installations, where the wiring is loaded to it's limit by just the light fixtures. Then the HPF could make the wiring cost 30..50% less than with NPF ballasts (the same wire gauge could serve twice as much HPF fixtures, so about half of the wiring is needed). But with home use even the rather high power lighting (e.g. 175W yard blaster) load the wiring to 25..30% (3..4A for a NPF ballast on 120V), so with HPF ballast there would be the same wiring, so costing the same.
Don't forget, the utility is billing you for the real consumed energy (so real power), the power factor have very minor influence (it should have none at all, but the meters are not ideal; but nobody know, which direction will help you, as it is pure random error...)
Yes, yes, yes I know WATT consumption does NOT change. And from what I've seen, large installations NEVER use NPF ballasts for the reason you mention. Plus it's the reason 208v/240v/277v/347v/480v taps are incorporated in a good ballast (reduce current in the wires)
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The low wattage street lights here have NPF ballasts and the older low wattage MV fixtures did too. You'd think the electric company would install HPF street lights to reduce the burden on the grid but the NPF ones are the cheapest so that's why they use them... :-\
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Ha. Cheap, cheap, cheap, cheeeeap! Is what the chick said to the hen having her head chopped!
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The low wattage street lights here have NPF ballasts and the older low wattage MV fixtures did too. You'd think the electric company would install HPF street lights to reduce the burden on the grid but the NPF ones are the cheapest so that's why they use them... :-\
If the existing wiring have a margin, they don't have to bother with HPF as well.
Other reason for not using PFC capacitors inside the streetlights is with multiphase installation with fixtures connected in "Y", when the Neutral get loose, the PF compensated fixtures may form low impedance resonators with the other ballasts, creating a lot of consequence damages. Therefore with such systems the capacitors are common for larger group of fixtures and connected to "D".
Other reason to not use capacitors inside the fixtures are the group dimming systems, where the required compensation capacitance changes with the dimming level, so it is usually moved to the dimmer box.
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To really utilize the advantages of the HPF fixtures at home (higher efficiency,...), the power factor correction would have to be incorporated deeper in the ballast design than just a correcting capacitor parallel to the mains terminals.
But I thought that a working ballast (like a cheap one found in wallpacks & floodlights) gets less hot with the suggested capacitor, thereby increasing its life (make a cheap ballast into a better one). Don't be afraid to ellaborate on your deeper in ballast comment.
No, because what the ballast always see on it's input is just plain 120VAC, regardless if you connect the capacitor there or not. So with the same lamp, the primary current will be the same as well.
Where the current become lower, are the wires from the CB box down to your fixture.
Something else is not using the specified capacitor on a (mainly multi-tap) HPF ballast. There the capacitor is usually connected across a winding with the highest voltage. That mean, the complete primary is there to handle the reactive power, so the related current becomes lower. The 120V section then handle just the real power. But when you disconnect the capacitor, the short 120V section will have to handle the complete apparent power, so becomes overloaded. So even the ballast may seem to be working, the missing capacitor causes it's overheating.
But these ballasts I have categorized as the "power factor correction incorporated deeper in the ballast design" and these are always installed with their specified capacitor from the factory (a cheepeese fixture maker use cheaper 120V-only NPF ballast).
Because of this trick, the multi-tap ballasts become actually cheaper as HPF than they would be as NPF, because as NPF, the 120V section would have to be made with thicker wire, what would mean more complicated manufacture (swapping wire gauges), larger ballast body and generally more material to pay, so an additional capacitor with the HPF format is just cheaper solution...
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Ok good stuff. That's why some really good HPF ballasts have such thin windings..so easy to damage if the wrong tap voltage is used (dogh)!