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Running your lights at a lower intensity doesn't really much electricity.


RichardH
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I was really surprised last night by a test I did. Many people on the board have talked about saving power or being able to have more lights on a LOR channel by lowering the intensity of their lights.

Even though it is possible to run them at a lower intensity so they don't take as much power, you really don't save that much. Make sure you get an amp meter to really see what your lights are taking. Just because you Dim from 100% down to 50% does not mean you are using 50% less power. Here is a test I did. What I did was hook up 300 mini lights up to one controller on a amp meter (Kill-A-Watt brand) and then changed the intensity with LOR.

Here are the results.

100% - 0.94 amps

90% - 0.94 amps

80% - 0.92 amps

70% - 0.89 amps

60% - 0.85 amps

50% - 0.80 amps

40% - 0.74 amps

30% - 0.66 amps

20% - 0.55 amps

10% - 0.47 amps

5% - 0.44 amps

2% - 0.39 amps

1% - 0.00 amps

I was surprised to see that at 2% you could not even see the light but it was still using about 50% of the electricity. At 5% you could barely see them.

So you really are not saving a whole lot of electricity by dropping it from 100% down to 70%. The difference is only .05 amps for 300 minis.

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Richard,

Thanks for doing that test and sharing the results. I was very suprised by your findings. I have had several people tell me to run the lights on LOR at 80% to save on power needs. This is interesting.

I have run a liight set hooked up to the Kill-A-Watt meter that blinked and saw a large difference. They were C-9 lights.

Great test Richard.

Michael B

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That is a surprise. I too was under the impression that running at say 80% intensity would give the same look but save a considerable amount of electricity..... but then again why not start the switch to LED's and throw the ampmeters away. (0.020 amps per 50 count string. The possibilities are mind boggling!

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Ok here is another test with 3 strings of C9 lights plugged into one channel of LOR.

I have the Kill A Watt hooked up to the wall and both plugs of my LOR box plugged into the Kill A Watt. With no lights on, the meter reads .01 (because of the power the board takes to run it).

100% - 2.77

90% - 2.74

80% - 2.68

70% - 2.54

60% - 2.40

50% - 2.24

40% - 1.98

30% - 1.71

20% - 1.39

10% - 1.05

5% - 0.85

2% - 0.70

1% - 0.01

0% - 0.01

I would like to see some results of others if you happen to have a meter.

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I think what you do save if you are using minis, is increased longevity of the light string. The bulbs will definitely last longer. Years ago series sets used 14-15 volt bulbs. My parents and others would cut an older set apart into two and three light segments. They would then splice these short sections into regular string. This resulted in the voltage applied to each bulb to be less; therefore they lasted longer.

Bob

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Has anybody found the tech specs for the Kill-a-Watt to know if it is taking true RMS current readings? If you use an amp meter that is not true RMS, I would expect numbers extremely similar to what is being reported here, and that the power consumed is being over reported. The reason is that non true RMS meters just read the peak current during the cycle, and then multiply by 0.707, which is fully correct for resistive loads with sinusoidal voltage applied. The thing is that with LOR or any other phase angle dimming solution, all the way until half duty cycle, your peak current won't change unless all other things are equal, but the time that current is passing through the lights has been reduced...

- Kevin

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Hi Richard,

I did this same experiment back in July 2006. Here's a link with my posted results....

http://planetchristmas.mywowbb.com/forum25/5894.html

Test: 8 strings of Target 100 ct. mini lights on a mini tree.Line voltage= 117 volts,Full brightness current when "on" = 2.5 amps, "Shimmer" current =1.4 amps

"Dim" Experiment using LOR Hardware utility:

100% = 2.5 amps = 117 volts

90% = 2.4 amps = 110 volts

80% = 2.3 amps = 102 volts

70% = 2.1 amps = 91 volts

60% = 1.9 amps = 79 volts

50% = 1.7 amps = 65 volts

40% = 1.5 amps = 51 volts

30% = 1.2 amps = 36 volts

20% = 1.0 amps = 24 volts

10% = 0.8 amps = 11 volts

Conclusion: A little different than what I expected. At 50% voltage, current = 68% of full scale. I probably wouldn't want to go much below 70 or 80% or the lights get too dim, so that only ends up saving about 0.2-0.4 amps on a total load of 2.5 amps (about 10-15 % current savings).

Randy

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If you plot these values, you'll see that the voltage is decreasing in a linear fashion but the watts are not.

Somebody should try this with some LED's to see how they perform (whether they are linear or not).

OK. I'll do it - but I'll have to hook up a LOT of 'em to get a decent reading!

Edit: If anyone has a good meter handy feel free to try it - I've got toassist my son with a little school project. (The classic "Egg Drop" thing. This might take a while.) :)

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ChuckHutchings wrote:

Edit: If anyone has a good meter handy feel free to try it - I've got toassist my son with a little school project. (The classic "Egg Drop" thing. This might take a while.) :)

Ahhh... the egg drop experiment. I remember mine from last year. I used a bunch of free Tempur-Pedic mattress samples.

It didn't work.

But back on the subject, great test! I'm now smarter.:waycool:

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Randy wrote:

100% = 2.5 amps = 117 volts

50% = 1.7 amps = 65 volts

If the above figures are correct the watts (that's what costs you!) consumed is:

100% watts = V x A = 2.5 ampsx 117 volts= 292.5 watts

50% watts = V x A = 1.7 amps x 65 volts = 110.5 watts

That IS a saving of more than 50% , and about what I would expect.

AMPS is only half the equation.

Think about this - if W = V x A and A = V / R, and the R of the lamps CAN'T change much, how can you expect to vary the "A" as much as you vary the "intensity"???? You have to vary the voltage!

John

A Grade Registered Electrician

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I understand what you are saying now. I was flawed in my thinking about only looking at the amps but I didn't think about how many amps at what voltage. It would be true if the voltage was the same but the voltage drops. So here is a test just looking at the Watts with a 100 Watt Flood.

100% - 100 Watts

90% - 97 Watts

80% - 92 Watts

70% - 84 Watts

60% - 73 Watts

50% - 60 Watts

40% - 47 Watts

30% - 32 Watts

20% - 18 Watts

10% - 8 Watts

5% - 5 Watts

2% - 3 Watts

0% - 0 Watts

So you do save a lot more than I originally thought.

Looks like I lead a lot of people astray with my first post. I would change it so people don't get confused in the future, but since you can't change your post after 4 hours, it will have to live on. :?

Thanks for your help John.

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Ok, So I'm just going to confuse this topic even more... Here are my results with an Animated LightingLC-16, with 300 minis. The number in brackets is the 'on' level converted from a percent to what I set in Animation Director

 VoltsWatts Amps

No Load 120.8 0 0.00

Straight120.1 116 .96

LC16 w/no load 120.8 0 .01


On Channel (KillAWatt hooked up to the channel feeding the lights)


%ONVolts Watts Amps

100 (255)118.4 114 .96

90 (230) 116.7 111 .99

80 (204)113.0 105 .98

70 (179) 106.3 96 .96

60 (153) 97.5 84 .93

50 (128)87.5 71 .90

40 (102) 74.7 56 .82

30 (77) 59.3 38 .75

20 (51) 45.4 27 .71

10(26) 37.8 23 .79


On Main (KillAWatt plugged into outlet, controller plugged into it)


%ONVolts Watts Amps

100 (255) 119.4 115 .98

90 (230) 119.5 113 .96

80 (204) 119.6108 .95

70 (179) 119.7 99 .92

60 (153) 119.5 86 .89

50 (128) 119.5 73 .84

40 (102) 119.5 58 .78

30 (77) 119.4 41 .70

20 (51) 119.2 25 .59

10 (26) 119.4 14 .50

What's interesting is that the wattage is not following the amperage in a linear fashion. That has to be due to power factor decreasing as the amount of dimming increases. In fact, at 10% intensity, the power factor is .23!

(From Wikipedia: An electrical load that operates on alternating current requires apparent power, which consists of real power plus reactive power. Real power is the power actually consumed by the load. Reactive power is repeatedly demanded by the load and returned to the power source, and it is the cyclical effect that occurs when alternating current passes through a load that contains a reactive component. The presence of reactive power causes the real power to be less than the apparent power, and so, the electric load has a power factor of less than 1.)

Here is where I'm getting confused...

I know that we are billed on watts used (kWh), so we should be most interested in the watts column. However, I also know that residential power meters have built in power factor correction.

Assume we run a light at 10% for 1 hour. Are we being billed:

  1. .0014 kWh - Using plain old watts
  2. .0597 kWh - Volts * Amps (IE the meter corrects thepower factor to 1, then measures)
For that matter, what's the load on the circuit:

  1. .5A (as read)
  2. .12A (Watts / Volts)
I'm thinkthe answer is #2 in both cases, but I'm NOT an electrical engineer. Can anyone shed some 'light' on this subject?

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oldcqr

I believe the resistance of a bulb is not linear, but changes as a function of current. If you add another column to your first experiment and for each reading divide voltage by current you will see that bulb resistance goes down as the voltage goes down. As far a power factor goes that may be glitch in the Kill a Watt. A bulb is pure resistance; therefore there should be no reactive component.

I've been out of school too many years:P

Bob

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I wonder if the change in power factor is due to using a triac to dim the channel. Since we are only using part of the wave every cycle, wouldn't that drive power factor down as we use less and less of the cycle (ie, dimming more and more)?

Of course, it could be the KillA Watt as you said. I read something about most meters not reading true rms.

In any event, it's interesting to research all of this! :]

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Richard,

you are also forgeting thatwe don't just run 300 minis,

take a display of 50,000 lights and if you have your lights dimmed to 70% at all times, you will have saved 8.3 amps total on the lights, now with my display of 150,000 lights we are up to almost 25 amps of savings that is huge

if i understand correctly. in a colder climate the lights and wires will not heat up as much, therfore less resistance, more energy savings.

you also get the same advantage on a display where the lights blink more, less time on equals less heat

that would be another great test, to stick some mini's in the freezer then test the amps, and the same for minis at room temp

bob

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The first observation is that there is nothing (of significance)inductive nor capacitivein a LOR and filament light bulb system, thus there can be no magnetizing nor reactive current. Note also that the general definition of power factor is for sinusoidal voltage and current waveforms. The results oftriacdimming are decidedly not sinusoidal, and the little I have seen for power factor for non sinusoidal loads hurts my head!!!However, I can see why a Kill-A-Watt will report a low power factor as the lights are dimmed below 50%. I just don't know if the answer it is giving is correct. Because of the triac, the lights have zero current flowing until the phase angle that the triac fires. At 100% the triac is firing right at the start of the cycle, and peak current is in alignment with the peak voltage.. As an example point, at 25% brightness, the line voltage peak is still half way through a half cycle, but the current peak will be where the triac turns on at 3/4 of the way through the halfcycle, so it sees the current out of phase with voltage. However, zero current is still in phase with zero voltage..

Separately, while industrial users are increasingly being monitored, metered and penalized for low power factor, residential users as a rule are not.My understanding and the first few search results all indicate thatthe spinning disk residential meters do not measure reactive loads, only real loads.What that means with triac dimmers is a very good question.

Also, yes, the filament resistance changes with temperature. Both light output and light color are side effects of temperature. Lower brightness levels should show lower filament resistances, though I don't expect it to be a large difference within the range of visible light output.

- Kevin

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While the above discussion is very interesting, one point was not mentioned - longevity of the lights.

I use dimmers on my static minilights, not to save electrical use per se (I am addressing that by switching over to LEDs in a controlled fashion), but rather to save my bulbs!

Run at about 75 - 80%, the bulbs last waaaaaay longer than the 2 - 3,000 hours they are rated to last.

I have seen that first hand over the last 30 years of having displays....

Greg

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Amazing the amount of discussion this topic has caused . . .

Just a couple of facts you can't get away from :

1) The "cold" resistance of a filament lamp isLOWER than ahot one. (The idea of the lights in the freezer - if the temperature of the filiament could be lowered, the current would go up, but I think the filiament heat would overcome the effect of the freezer - someone want to prove me wrong?)

2) The action of a triac on the mains is likely to screw up most metering methods. I know its off topic, but think about this. If you rectify AC with a diode, but no smoothing , you can measure both AC volts and DC volts. So should you use a AC or DC ampmeter to measure any load current?

Now, a triac is akin to a couple of SCR connected in reverse. And what is a SCR? A Silicon Controlled Rectifier. Just like my diode in the last paragraph, except I can turn it on and off as I like . . .

And something else I can tell you . . . a couple of years ago I imported some of your mini lights, with the idea of converting them to our "normal" 24 volt operation. Only idiots would run lights on our mains voltage of 230 volts. However, when I was working out how I was going to do it I thought I would do them in lots of 10 lamps (just like ours) but when I did this the current consumed was double what was "normal" - so I wired them in lots of 12, halving the current in the process, and reducing the brigtness by less than 5% . . .

Now I know why yourlights draw such humungus amps, and all at a voltage we ring telephone bells with - 115 volts - :] Sort of a parallel to the gas guzzling auto's that your country used to be reknown for! And other excess's I won't go into . . .

Save Planet Earth!

Use LEDS!!!

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OK RichardH and Oldcqr:

I think the way our digital dimmers accomplish power reductions will actually "fool" the killawatt meter.

I have had similar results where AMP loads actually appear to increase with slight declines in voltage, however real world reasons for using DIMMING are because your breaker is heat activated and throws because you are using too close to your 15 amp limit lets say, Killawatt meter shows almost no effect, maybe even an increase in AMP load at 90% over 100%, however the breaker that was getting hot and throwing at 100% now no longer throws anymore at 90% and doesnt get nearly as hot. The temperature reduction on the breaker is actually 10%.

The reason killawatt meter is "fooled" is, as we know, A/C power has a "zero voltage" occurrance 120 times a second. The dimming effect accomplished by both Animated Lighting and LOR is by slightly "widening" the zero voltage's duration, by using the triac to synchronize with this zero cross/zero voltage occurrance and actually keep the voltage "off" by increasing milliseconds longer then the raw A/C power does. There are 255 steps in animated lighting software because the time in which you can divide a full sine wave is in 255 slices, a setting of 254 for example extends the zero voltage period by 1/255th of a 120th of a second sine wave. Amazing what super precision equipment we have in LOR and Animated Lighting products!!

Aside from extending the zero cross to accomplish dimming, your lights and line are delivered raw 100% voltage. Its very complicated but this is why your killawatt meter is fooled and deliver incorrect data. Probably because the killawatt meter is too fast and digital as well but real world results will produce saved power on analog power meters, and also the effect will reduce heat generated which will have the desired effect on saving fuses or reduced breakers thrown.

I was reading on an engineer's web page that LED lights - power meters actually only monitor the positive sine wave, so 1/2 wave LED lights set to use the negative sine wave ONLY can actually totally miss the power meter entirely and not get registered (essentially 100% free light). This is interesting but I wonder about heavily loading up a large LED setup that DOES consume some heavy amps about unbalancing your system with everything loaded on the neg sine wave and no load at all on the positive sine in order to have the "totally free power" bragging rights. Some of this power meter "trickery" is partly responsible for the incredible power savings with LED's.

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Some useful infromation.

From OHM's law we know

The Current (I) = The Voltage (E) divided by The Resistance ®

The Voltage (E) = The Current (I) multiplied by the resistance ®

The Resistance ® = The Voltage (E) divided by The Current (I)

since the resistance is a fixed constant for all incandescent strings, it is safe to say that if we lower the voltage, then the current must go down and since we pay for the watts we use, the lower the current with a fixed voltage, the lower the watts, the less the electric bill.

With LED's, they are a Direct Current DS device. Already to use the lowest possible amount of power, changing any of the variables has a minimal effect. Most of these devices "Chop" off the second half of the sine wave. Just with that, they use half of the power. Now, a led requires about 5 VDC. They have a current requirement into the Micro Amp range. That would be something like 0.003 uA. So even having a few hundred of them, they never reallly accumulate to a large resistance.

Lastly, in the Alternating Current world, calculating the resistance is not actually done the same way as with direct current. We actually use the impedece (Z). Without getting too scientific, they have a "real resistance" and an "imaginary" part.

The best example of this is a transformer. Take a piece of wire and measure the resistance. Connect it to a source and it will blow a fuse. Take that same piece and wrap it around a steel core and the resistance will change andtwo things will happen. You will make an electromagnet and a heater. The reisistance will be something totally new.

This is why iteasy to fool these watt meters. They are by deisgn only capableof seing the actual current of the circuit. They don't actually measure the resistance or the impedence.

Hope that helps

David

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Unibits wrote:

This is why iteasy to fool these watt meters. They are by deisgn only capableof seing the actual current of the circuit. They don't actually measure the resistance or the impedence.

Yes... its digital trickery because technically, at 50% dim, the zero cross is expanded to having the power turned off 50% of the time, however, the 50% of the time its on, you are getting microslices of power at full 120 volts. A resistive or analog dimmer would not expand how long the zero cross lasts however what you think is 50 volts would be "truely" 50 volts, instead of 120 volts that is turned off for a specified period of time 120 times a second.

On the power meter, I am told the meter only records the positive sine way, or in otherwords, it only records 1/2 the wave similar to the way LED's operate. So if you have LED's that operate on the negative sine wave, its power that is not registered by your power meter. In theory, you could rig up 5 million LED lights consuming 150 hard amps, make them all 1/2 wave negative sine wave LED's and pay "nothing" for the power use. In reality, most 1/2 wave lights come for example in 60 light sets, 1/2 the string counting 30 lights use the positive sine wave and the other 30 lights use the negative wave. In reality you are only paying for 30 lights on the electric bill on a 60 light LED set.

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Joseph Ayo wrote:

On the power meter, I am told the meter only records the positive sine way, or in otherwords, it only records 1/2 the wave similar to the way LED's operate. So if you have LED's that operate on the negative sine wave, its power that is not registered by your power meter. In theory, you could rig up 5 million LED lights consuming 150 hard amps, make them all 1/2 wave negative sine wave LED's and pay "nothing" for the power use. In reality, most 1/2 wave lights come for example in 60 light sets, 1/2 the string counting 30 lights use the positive sine wave and the other 30 lights use the negative wave. In reality you are only paying for 30 lights on the electric bill on a 60 light LED set.

Does the meter only record the positive sine way but charges you for a full sine wave? If this was the case, then half on + and half on - would equal each other out. Now if they were all negative then you could run them for free and if they were all positive, you would be charged twice the amount of power you use.

Maybe this is why half run at negative and half run at positive.

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