View Full Version : Kichler LED Fixtures - True Wattage

JimLewis

12-22-2010, 09:40 AM

Something I learned yesterday about the Kichler LED light fixtures kind of frustrates me. I am told that when designing systems with their LED fixtures, you have to account for more wattage than the fixture is actually labeled for. For instance, if you're using their 8.5 watt spot light, you need to account account for 12 watts. I was like, "Huh???" The rep. (from my supplier, not from Kichler) explained to me that he didn't understand it perfectly either but from his understanding it was something about how the LED fixtures aren't as efficient. So while they put out 8.5 watts of energy, they use 12 watts of energy to do so. Something about using more amp watts than a normal fixture??? That's when I got really confused. Not sure what amp watts are.

Anyway, WTF? If the fixture consumes 12 watts, then why not just call it a 12 watt fixture? Why label it 8.5 watts but then tell everyone "Oh, sorry. That's just on the output side. You're actually going to need 12 watts to power it." Just call it a 12 watt fixture then, right?

Tomwilllight

12-22-2010, 05:02 PM

As I understand it, there is always some power lost to (or "used by") the driver in an LED lamp. This is very similar to the Ballast Loss issue when calculating loads for HID and Fluorescent lighting. Ballast Loss varies depending on the type of ballast used to power the light source.

The takeaway: unless the LED's specifications list the System Watts, the manufacturer may not be disclosing the actual watts used to power their lamps. They are listing the watts consumed by a portion of the system only - the LED(s) - not the entire system (driver + chip) required to make the lamp work.

Somebody who understands the calculations for Power Factor should step up here and help us out. I'm over my head trying to balance my check book.

Tom

JimLewis

12-22-2010, 05:23 PM

Yah, that's basically how it was explained to me too. But regardless of why, it's all sort of B.S. to me. Just call it at 12w fixture, if that's what it uses.

sprinklerchris

12-22-2010, 08:15 PM

Yah, that's basically how it was explained to me too. But regardless of why, it's all sort of B.S. to me. Just call it at 12w fixture, if that's what it uses.

Most of the other LED manufacturers are rating their fixtures at actual watts consumed. Wonder why the desire to be misleading?

S&MLL

12-22-2010, 09:04 PM

They don’t desire to be misleading. In many of the catalogs it puts a chart right next to the LED section. Part of a installers job is to know the product he is installing. This is something they cover in slideshow they do at open houses. So maybe the problem falls on your distributor. Kichler tells there reps to tell the distributor. Distributor tells you….. And its less then 10 percent. But let me ask you this question. Are you designing a system with such tight tolerances that you cant allow for something so small? 9 watt is really a 12.4 . That is 3.4 watts.

And just a fyi Im pretty sure for the past 2 years they have had them labeled them correctly. 12.4 watts is what its called in every book in my office and on their website.

There are actually 2 ratings on the fixtures, the wattage and the VA (volt amps). In most cases, wattage and VA are the same but when you have the driver or ballast, they are not. You always need to design to VA, not wattage. It's always going to be the higher number and shows your true power comsumption.

INTEGRA Bespoke Lighting

12-22-2010, 10:16 PM

….. And its less then 10 percent. 9 watt is really a 12.4 . That is 3.4 watts.

Check your math there: 3.4 watts discrepancy on a 9 watt rating is 37.7% 3.4 watts discrepancy on a 12.4 watt rating is 27.4 %

Either way you slice it, it has the appearance of being somewhat inefficient. It would be more concise and accurate if the product is labeled and rated for the total power it consumes.

S&MLL

12-23-2010, 01:16 AM

Check your math there: 3.4 watts discrepancy on a 9 watt rating is 37.7% 3.4 watts discrepancy on a 12.4 watt rating is 27.4 %

Either way you slice it, it has the appearance of being somewhat inefficient. It would be more concise and accurate if the product is labeled and rated for the total power it consumes.

You caught me :hammerhead: My math was not to good on that one. But still 3 watts is not gonna kill you.

JimLewis

12-23-2010, 03:01 AM

When you multiply it by 20 or 30 fixtures, that 3 watts can make a difference, yes. It makes a difference on which transformer you purchase or budget for. And it would suck to get that far into a job just to find out that you need to upgrade to a larger transformer. Customer isn't going to be too happy to hear about an upcharge.

If you really want to see what its drawing use your amp meter around one of the wires feeding the light. Find out the exact voltage and use OHM's law to find out the wattage. I(amps) x E(voltage) = P(watts)

When we first started replacing incandescent lamps in the electrical field we performed this test and we were very surprised of the results compared to the marketing information provided. Some of the electronics consumed more wattage than the LED itself.

Any how I hope that helps for the people wishing to perform the experiment themselves.

Ken

jmiller@kichler.com

12-29-2010, 01:58 AM

Hello Jim,

There is a very clear and good explanation as to why we show a VA and Wattage on our LED fixtures. Have you ever seen the formula Volts X Amps = Watts? That is a standard formula for incandescent lighting. What is not present in that formula is that the Volts X Amps also get multiplied by the Power Factor. In incandescent the power factor is one so it never really shows up since it wouldn't change anything.

In LED the Power Factor is less than one. Now when you multiply the Volts X Amps times a number less than one, things change. At Kichler we give you the VA. The total VA is what is being drawn on the transformer becuase it is taking the electronics into account (something incandscent doesn't have).

So one has to total the VA of all their LED fixtures to determine the size of the transformer they need. The reason we still show the wattage is because that is what the home owner will still be charged from their utility company.

Although a 10 light job using our 8.5 watt fixtures may have a total VA of over 100, that tells you the transformer size you need. Again however, the total wattage being used and what the electricity company charges the homeowner is 10 X 8.5 watts or 85 watts of energy use.

That is why we show both numbers. It allows a contractor to be accurate in sizing the transformer, yet allows the contractor to show the homeowner the actual 2/3rds energy savings they will receive in comparison to an equivalent MR16. Not to mention the 40,000 hour life of the LED.

Hope that helps you understand it a little more. If you have any further questions, please don't hesitate to call me direct at 216-573-1005 ext. 6216.

I'd be happy to discuss.

Jack

steveparrott

12-29-2010, 09:24 PM

Thanks Jack, your explanation is very helpful. This is an important concept that all landscape lighting installers need to learn.

To provide further info for those who want to understand more:

When you have a simple circuit with a resistor (such as an incandescent filament), the voltage sine wave moves in perfect sync with the amperage (amps) and power (watts). V x A = W. (see first diagram below.) This is analgous to hitting a punching bag with both hands where your punches are timed exactly to the swing of a bag.

When you replace the filament with a circuit containing various electronic components (like a flourescent ballast or an LED driver), the voltage and amperage move out of sync with the power. V x A = W/Power Factor. (see second diagram.) This is analgous to hitting the punching bag when your timing is off. Some punches hit the bag when it swings towards you, some as the bag swings away - very tiring. In both cases, you're striking with the same speed (V) and the bag weighs the same (W), but with your timing off (low Power Factor) you end up using more force (A).

The utility company only charges for the watts, but they do so with the understanding that your household devices have a power factor of about 0.9. If they find a building with a very low power factor, they may increase the KWhr rates since the higher VA's require them to invest in bigger transformers and wire.

While achieving a high power factor in an LED device is not a big issue, LED engineers always try to achieve 0.9 or above. One reason is because lower power factors put more stress on the LED components themselves.

jmiller@kichler.com

12-30-2010, 08:58 AM

Thanks for the additional comments Steve.

Have a Happy New Year

Jack

Chris J

01-05-2011, 09:09 PM

I've installed many many LED systems now, and I always find that even though I use the VA rating of the Kichler fixtures when designing my systems, the true amp load is always much less than I anticipated. I will continue to use the higher VA rating when designing a system, but as someone already pointed out..... even if it is a 20 or 30 light system, your only talking about 60 watts or so. Leave some room for improvement, and trust your meter.

jmiller@kichler.com

01-10-2011, 11:55 PM

I apologize that this is so confusing to everyone, but it is confusing. LEDS are not your father's lighting. LEDS are not new technology, but they are new to our industry. Not only do they operate quite differently than the lighting we are used to, there is technology that is new to all of us andwe have to get familiar with it. No different when people started learning 12volt landscape lighting if it was new to them.

Our electronic drivers convert the 12V AC current to the specified DC current by the LED chip manufacturer. When electric power is converted from AC to DC there are losses and changes known as the "Power Factor". In the formula W=V X A that we are all familiar with in our 12V AC systems, the power factor is 1. Basically the formula is actually W=(V X A) X PF. But since the power factor is 1, it is not used.

In DC the power factor is less than one and that will change things. We determine the VA for you since we know the Power Factor.

The reason we do not call our 8.5W fixture a 12W fixture is because the wattage is the energy consumed. The fixture is only consuming 8.5W. The homeowner will only be charged for 8.5W from their electric company.

However the load on the transformer is the VA. So one must add up the VA to determine transformer size.

Just like a 20W MR16 - the wattage is not the amount of light, it is the energy consumed.

We are not trying to fool or confuse anyone.

This is a simple explanation that hopefully is easier to comprehend than if an electrical engineer was explaining it.

I apologize that this is so confusing to everyone, but it is confusing. LEDS are not your father's lighting. LEDS are not new technology, but they are new to our industry. Not only do they operate quite differently than the lighting we are used to, there is technology that is new to all of us andwe have to get familiar with it. No different when people started learning 12volt landscape lighting if it was new to them.

Our electronic drivers convert the 12V AC current to the specified DC current by the LED chip manufacturer. When electric power is converted from AC to DC there are losses and changes known as the "Power Factor". In the formula W=V X A that we are all familiar with in our 12V AC systems, the power factor is 1. Basically the formula is actually W=(V X A) X PF. But since the power factor is 1, it is not used.

In DC the power factor is less than one and that will change things. We determine the VA for you since we know the Power Factor.

The reason we do not call our 8.5W fixture a 12W fixture is because the wattage is the energy consumed. The fixture is only consuming 8.5W. The homeowner will only be charged for 8.5W from their electric company.

However the load on the transformer is the VA. So one must add up the VA to determine transformer size.

Just like a 20W MR16 - the wattage is not the amount of light, it is the energy consumed.

We are not trying to fool or confuse anyone.

This is a simple explanation that hopefully is easier to comprehend than if an electrical engineer was explaining it.

Not trying to be negative here but I'm an electrician and I'm not understanding what your saying at all.

Halogen: Add up all wattages plus wire which is all your resistance and this is the total load the transformer will see. Halogen lamps consume the wattages written on the lamp and the lumens are what they are. Place an amp meter around your primary wire on the transformer find the amps, find the voltage and you will have the total watts being consumed by the entire lighting system.

LED Fixtures: If they consume 12watts and that is what the transformer will see than that is what the customer will pay...period. Forget about the power factor formula, stick an amp meter around the wire on the fixture find out the exact voltage and do OHM's law to find the ampacity, wattage etc.

If you have to size the transformer to 12 watts per fixture (even though its advertised at 8.5 watts) and that fixture is drawing 12 Watts you will be charged the 12 watts for usage not the 8.5 watts you stated.

Generally your lighting fixtures are an electrical appliance which consumes energy. Tell me how much energy via total consumed wattage it will use when connected to a power source.

Standards...if every company is going to advertise something different other than what the total energy being consumed is for each fixture than this industry is going to have issues. From my perspective I would like to see on the package and instructions the total amount of watts (energy) it will consume so that I can size my transformer accordingly. The degree of beam spread and lumen output would be nice too.

Just my 2 cents today.

Ken

steveparrott

01-13-2011, 10:51 AM

Not trying to be negative here but I'm an electrician and I'm not understanding what your saying at all.

Halogen: Add up all wattages plus wire which is all your resistance and this is the total load the transformer will see. Halogen lamps consume the wattages written on the lamp and the lumens are what they are. Place an amp meter around your primary wire on the transformer find the amps, find the voltage and you will have the total watts being consumed by the entire lighting system.

LED Fixtures: If they consume 12watts and that is what the transformer will see than that is what the customer will pay...period. Forget about the power factor formula, stick an amp meter around the wire on the fixture find out the exact voltage and do OHM's law to find the ampacity, wattage etc.

If you have to size the transformer to 12 watts per fixture (even though its advertised at 8.5 watts) and that fixture is drawing 12 Watts you will be charged the 12 watts for usage not the 8.5 watts you stated.

Generally your lighting fixtures are an electrical appliance which consumes energy. Tell me how much energy via total consumed wattage it will use when connected to a power source.

Standards...if every company is going to advertise something different other than what the total energy being consumed is for each fixture than this industry is going to have issues. From my perspective I would like to see on the package and instructions the total amount of watts (energy) it will consume so that I can size my transformer accordingly. The degree of beam spread and lumen output would be nice too.

Just my 2 cents today.

Ken

Hi Ken, I agree that alll manufacturers should standardize the way they state electrical specs. But, I think you have your facts mixed up.

As Jack's and my posts here have tried to explain, the energy company only charges for watts; the transformer is sized by volt-amps.

You can't simply measure volts and amps (with LED's) to get Watts. The equation is not V x A = W; it is V x A x 'Power Factor' = W.

Hi Steve,

Ok so I have been reading lots trying to understand the electrical theory with LED's.

A volt-ampere (VA) is the amount of apparent power in an electrical circuit, equal to the product of voltage and current. It is equal to the electrical power measured in watts for Direct Current (DC) circuits. ...

Power factor= True power/Apparent Power

To correct the power factor you will incorporate a capacitor to adjust it to 1 as close as you can. I get all of that.

I am totally not understanding the 8.5 watts to the 12 watt measurements?

Lets just pretend for a minute to help me understand this.

We have just installed a system of 10 LED fixtures, (forget about wire resistance etc lets just focus on the fixtures)...

We have used a Cast Journeyman 300 watt transformer. I have turned the system on and now I will take my amp meter and take the Amp reading on the primary loop. I will then take the primary voltage reading. I will calculate the total primary watts by using OHM's Law which is P= ExI.

I can perform the same on the secondary side of the system. What will I see for total watts on the secondary side? 80.0 Watts or 120 Watts?

Please help me understand?

Thanks

Ken

Ken, when reading the amps on the wire and multiplying by voltage, you're getting volt amps, not watts. In order to get watts, you have to multiply by the power factor, which is 1 in a halogen system so VA = W. So in your exampe, if you have 10 fixtures that are rated at 12VA each, you'll read 10A, or 120VA (assuming 12V) with the meter but if the LEDs themselves have a power factor of .8, then you have to multiply the 120 by .8, and you'll come up with 96W. The 12 rating on the Kichler fixture is the VA rating, which is what the transformer is sized for, but the fixture actually only consumes 8.5 watts, even though it is pulling 12VA from the transformer.

Thankyou BCG,

For what ever reason the way that you have worded that finally sunk into my thick skull...:)

Ken

David Gretzmier

01-15-2011, 01:50 AM

sheesh, this is confusing. but in reality, if we are talking watts consumed, a good old kill-a-watt meter tells all in easy to understand watts. plug in kill-a-watt to wall, plug in transformer. witness one important thing- your trans when on, is pulling watts when running no fixtures are hooked up at all. add a 20 watt halogen fixture, and the watts on the meter do not go up from that baseline number by a linear 20 watts. rarely does the meter go up in a perfect linear fashion on the 2nd 20 watt fixture either. LED's are no different. unless that driver is per100% perfectly efficient, and they are not, That driver may have to work harder from that led to create the identical light from a 9 volt feed or a 15 volt feed. The resistance loss in the wire is still going to vary based on how the guage, how much wire you have out there and load on that wire.

the same fixture affects on a transformer or drivers for LED's may make your kill-a-watt read a few watts up or down just based on ambient temperature. or even the age of the bulbs or led's in place. LED's may affect this less, but I am willing to bet that LED drivers vary more than a few percentage points from 5 degrees to 105 degrees.

David you have that right as it is very confusing. I am still not 100 percent in agreement of what has been stated. My earlier post was only to reflect that I understand what they were saying but I am on the fence post if I agree with what is being said.

OHM's Law, Power Factor, Kirchoff's Law its all electrical theory and no matter what you say you cannot change those one little bit.

Volt-amps simply is wattage. A power factor of less than one or higher than one is ineffecient.

If the industry is selling our power sources (transformers) in watts then why do we want to involve terminology of volt-amps? That is why its confusing. What were the marketing ideas behind this other than to be different from the rest of the group. That is why there are standards so that the whole world can understand it and follow it.

Steve, are you ready to market your 300 what transformers in (120v-3A) Volt-amps?

I guess to solve this mystry I am going to have to obtain one Kichler fixture, hook it up to a transformer and put my amp meter around the wires and test for voltage on the primary and the secondary. My voltage tester and OHM's law will not tell me something new.

Confused and testing.....

KKKKEN

Anybody,

Can you tell me how to prove that the fixtures are only consuming 8.5 watts using an amp meter and OHM's law? You have explained how you are using the 12VA's but not the 8.5 watts.

Thanks.

Ken

ok here is what I found everybody. This helped me understand totally.

Since watts is volts times amps, what is VA? VA (or volt-amps) is also volts times amps, the concept however has been extended to AC power. For DC current

VA = Watts (DC current).

In AC if the volts and amps are in phase (for example a resistive load) then the equation is also

VA=Watts (resistive load)

where V is the RMS voltage and A the RMS amperage.

In AC the volts and amps are not always in phase (meaning that the peak of the voltage curve is does not happen at the peak of the current curve). So in AC, if the volts and amps are not precisely in phase you have to calculate the watts by multiplying the volts times the amps at each moment in time and take the average over time. The ratio between the VA (i.e. rms volts time rms amps) and Watts is called the power factor PF.

VA·PF = Watts (any load, including inductive loads)

In other words, volt-amps x power factor = watts. Similarly, KVA*PF = KW,

Or kilovolt-amps times power factor equals kilowatts.

When you want to know how much the electricity is costing you, you use watts. When you are specifying equipment loads, fuses, and wiring sizes you use the VA, or the rms voltage and rms amperage. This is because VA considers the peak of both current and voltage, without taking into account if they happen at the same time or not

its been a while since I took electrical theory and it has been like learning it all over again. Sorry if I may have confused everybody but thanks for following my own dimise.

Ken

steveparrott

01-16-2011, 12:33 PM

If the industry is selling our power sources (transformers) in watts then why do we want to involve terminology of volt-amps? That is why its confusing. What were the marketing ideas behind this other than to be different from the rest of the group. That is why there are standards so that the whole world can understand it and follow it.

Steve, are you ready to market your 300 what transformers in (120v-3A) Volt-amps?

KKKKEN

Ken, I think part of the confusion comes from the history of the landscape lighting industry. The electrical industry specifies transformer capacity in volt-amps (VA) or kilovolt-amps (kVA). At some point in our industry, transformers came to be specified in watts instead of kVA - perhaps because incandescent lights have a power factor of one, making VA = W. It also made it easier to estimate transformer size required, e.g. 600 lamp watts needs a 600 W transformer.

Given the new LED wave, I will start changing our transformer specs to kVA because specifying them in watts no longer makes sense.

JimLewis

01-16-2011, 07:08 PM

You know, I know there are a handfull of "lighting experts" out there who know all this stuff up and down. You can quote Ohm's law and every other electrical calculation off the top of your head. You understand electricity and math and are able to memorize that stuff because you're doing lighting all the time that you could run circles around most people who do L.V. lighting.

But I think I speak for probably 80% of the industry when I say, "I don't want to have to know all that stuff," just to be able to install outdoor lighting. The very reason that L.V. lighting started to work it's way into the landscaping market - trying to get your average LCO or landscape contractor into L.V. lighting - was this concept that, "We've made this easy! Now you guys can do outdoor lighting too!" And I know a lot of people here in the lighting forum are probably offended by that. Some of you may feel that people who don't fully understand every electrical law, fully understand electricity, have an electrical license, etc. shouldn't be messing around with outdoor lighting at all. But I don't think that's reality. The reality is, you DO have a lot of guys out there installing lighting who aren't electrical geniuses. And they've made outdoor lighting fairly simple. You take a few classes, learn some basics about transformers, wire sizing, wire connections, load ratings, fixture placement, etc. and then you are in business! The whole concept for the last 15 years that I've heard at any class I've attended from FXL, Unique, Vista Pro, Kichler is this same concept..... "This isn't that hard. You guys can pick this up. You just need to understand a few basic concepts and you can be installing lighting too!"

So with that said, if the industry wants to keep this something that is fairly simple for the average LCO or landscaper to get into (without really forking things up) then they need to keep things like this simple. Label your fixtures and transformers so that we can easily understand how many fixtures we can put on a transformer!

I'm not an idiot. I have a college degree. I know math, algebra, calculus. I am a fairly bright guy. But even while I have a little more education than most LCOs or landscape contractors out there, I still don't really want to have to know all that much about it, if I don't have to. I want to know enough to do a good job and make sure the lighting I install lasts a long time and makes my customers happy. But I also want it to be simple enough that I can understand it quickly and more importantly, my workers can understand it quickly. You can show me the calculations all you want and how Watts aren't really watts when you're doing LEDs and how the watts used aren't the watts you're charged for. Whatever. I don't really care too much about that. All I want to know is how many fixtures can I get on a controller? And I want to be able to explain that to my workers easily too. I shouldn't have to tell them, "Well, this fixture is a 8.5w light. But not really. It's actually a 12w light. Well, not really. Long story. But let's pretend - for transformer purposes - that it's a 12w light. Okay?" That just sounds screwy.

If you guys want to start using the term VA instead of watts, fine. I could care less. It just needs to be simple. I think probably a good 80% of contractors who install outdoor lighting would agree. It should be fairly simple and straightforward. And it always has been before LED.

Hey Jim,

Sorry that I kinda hijacked your thread but it was a learning experience for me and I am sure for others and I do apologize.

I am sure that the landscape lighting industry is going to be very confusing and frustrating in the next little while. The reason is "change". Technology is changing our industry and we must learn to adapt to these changes. Some of the people that contribute and share on this site have been designing with light for a long time and they have been through similar changes before. Examples of this are changing from 120 volt lighting to 12 volt lighting. Par Lamps to Halogen Lamps and now Halogen Lamps to LED's.

I am a guy who just loves what I do (in the summer) and that is designing and installing lighting systems. I have not been doing this for a long time but wow what a ton of information and learning I have done. Now I have to do it all over again and so does everybody else. We all have to learn how to paint with LED's and we also have to learn how to engineer the lighitng system all over again especially if you want it to last.

I feel like I have spent more time in my yard than in customers yards this past summer. I have been experimenting with different samples and different manufacturers. Most of what I have played with is no longer available as new versions have already replaced them and that is how fast things are changing.

One thing is for sure and that is we really need standards so everybody is on the same road to success. I am not too sure what body of manufacturers or government developed them for the incandescent and halogen but I hope they step up and help us all with LED's.

KKKen

I just wanted to follow up with this as I think its important to understand. The LED's are consuming 8.5 watts and the total watts consumed combining the electronics and the LED's are 12 VA. Clear as mud.

What a perfect question to ask at a round about table conversation on LED's at the conference. I will bring my meter.

steveparrott

01-21-2011, 01:14 PM

8.5W = energy consumption - 'Real Power' (what the homeowner pays for)

12.0VA = 'Apparent Power' due to the electronics' distorting the current (what the transformer and wire is sized to)

8.5W = energy consumption - 'Real Power' (what the homeowner pays for)

12.0VA = 'Apparent Power' due to the electronics' distorting the current (what the transformer and wire is sized to)

Hello Everybody,

I am sorry to be repeating this all over again but this is obviously bothering me. It looks as though I am going to need an electrical engineer to explain the theory and measurements to me so that I can fully understand this.

As an electrician who fully understands OHM's law and power factor I still do not understand how the utility meter (which uses a magnetic field created by the volts and amps used to turn an aluminum disk or be measured by electronics) is only going to see 8.5 watts (Real Power?) when it consumes 12 VA(Apparent Power)? 8.5 watts of "real power" is not equal to 12 volt-amps"apparent power". We all know this because we have to size the transformer for greater than 8.5 watts because of the distorting current.

How did Kichler measure the 8.5 watts? If someone could explain that or show me how to measure that using my multi meter that would be great. This would also relieve my temple pain :)

I guess we all could just ask Kichler to remove the Watts portion from all the labels so that it would be easier for everybody to size the transformer properly.

Just having some fun trying to learn maybe something new or learning of a mistake that someone else might learn from.

Smiles

Ken

Hello Everybody,

Just for those of you that have been following this particular thread I have completed some testing with the help of some others (thank you) and thought that I would post the results.

Testing LED's for Fun

Test 1

Kichler Transformer : K-15PR300SS

Kichler LED Fixture : 15742 BBR

NO LOAD

Primary Voltage 118.4 v

Primary Ampacity .29 A

Total Watts 34.34 w

Secondary Voltage 11.7 v

LOAD ATTACHED

Primary Voltage 118 v

Primary Ampacity .36 A

Primary Watts 42.5 w

Secondary Voltage 11.6 v

Secondary Voltage .71 A

Secondary Watts 8.24 w

TEST 2

Nightscaping Transformer : T-500-SS

Kichler LED Fixture : 15742 BBR

NO LOAD

Primary Voltage 118.7 v

Primary Ampacity .17 A

Total Watts 20.18 w

Secondary Voltage 12.8 v

LOAD ATTACHED

Primary Voltage 118.5 v

Primary Ampacity .24 A

Total Watts 28,44 w

Secondary Voltage 12.8 v

Secondary Ampacity .67 A

Secondary Watts 8.5 w

Observations

LED Fixture specifications are bang on with the 8.5 watts.

Fixture most certainly does consume the 8.5 watts.

Thoughts:

To provide a power source that can handle a draw of 8.5 watts and only if the power factor was unity than you could theoretically use an 8.5 va transformer but this would mean that it would be at 100% load at all times. By supplying a 12 va transformer the transformer will be running at about 85% load which would be fine for a transformer.

The most surprising observation was the wasted watts that each transformer was consuming once powered up without any loads attached. Remember that the Nightscaping transformer is still a plate designed transformer while the Kichler (MDL) is a new energy efficient torrodial core type transformer. WOW what a difference.

Costs associated with the Kichler transformer : 34.34 watts plugged in using a timer and photo cell will consume 824.16 watts per day. Roughly 30 KW's a month and that is with no load.

You have to add this extra wattage consumed to the wattages of the fixtures and this is what the customer will see being used on their hydro meter. (Just to clarify things Jack)

I am glad that I conducted this test as I learned a lot from doing so. A small example of this is instead of relying on other individuals to explain things to me I have learned that I must test for myself to fully understand and visually see it happening.

I hope that you all have found this interesting.

The only thing that I want to test now is other transformers to see how many watts they consume without a load and then try loading them up to see what happens. Then we will find out the real costs associated with our systems. Everyone will be different.

Ron C. you were correct !!

steveparrott

02-20-2011, 01:10 PM

Hello Everybody,

Just for those of you that have been following this particular thread I have completed some testing with the help of some others (thank you) and thought that I would post the results.

Testing LED's for Fun

Test 1

Kichler Transformer : K-15PR300SS

Kichler LED Fixture : 15742 BBR

NO LOAD

Primary Voltage 118.4 v

Primary Ampacity .29 A

Total Watts 34.34 w

Secondary Voltage 11.7 v

LOAD ATTACHED

Primary Voltage 118 v

Primary Ampacity .36 A

Primary Watts 42.5 w

Secondary Voltage 11.6 v

Secondary Voltage .71 A

Secondary Watts 8.24 w

TEST 2

Nightscaping Transformer : T-500-SS

Kichler LED Fixture : 15742 BBR

NO LOAD

Primary Voltage 118.7 v

Primary Ampacity .17 A

Total Watts 20.18 w

Secondary Voltage 12.8 v

LOAD ATTACHED

Primary Voltage 118.5 v

Primary Ampacity .24 A

Total Watts 28,44 w

Secondary Voltage 12.8 v

Secondary Ampacity .67 A

Secondary Watts 8.5 w

Observations

LED Fixture specifications are bang on with the 8.5 watts.

Fixture most certainly does consume the 8.5 watts.

Thoughts:

To provide a power source that can handle a draw of 8.5 watts and only if the power factor was unity than you could theoretically use an 8.5 va transformer but this would mean that it would be at 100% load at all times. By supplying a 12 va transformer the transformer will be running at about 85% load which would be fine for a transformer.

The most surprising observation was the wasted watts that each transformer was consuming once powered up without any loads attached. Remember that the Nightscaping transformer is still a plate designed transformer while the Kichler (MDL) is a new energy efficient torrodial core type transformer. WOW what a difference.

Costs associated with the Kichler transformer : 34.34 watts plugged in using a timer and photo cell will consume 824.16 watts per day. Roughly 30 KW's a month and that is with no load.

You have to add this extra wattage consumed to the wattages of the fixtures and this is what the customer will see being used on their hydro meter. (Just to clarify things Jack)

I am glad that I conducted this test as I learned a lot from doing so. A small example of this is instead of relying on other individuals to explain things to me I have learned that I must test for myself to fully understand and visually see it happening.

I hope that you all have found this interesting.

The only thing that I want to test now is other transformers to see how many watts they consume without a load and then try loading them up to see what happens. Then we will find out the real costs associated with our systems. Everyone will be different.

Ron C. you were correct !!

Very cool that you're doing these tests. I don't know how you measured secondary amps and watts, but the values seem to be off. Since (V x A)/W = power factor; your results indicate a power factor of 1.0. For these fixtures, you would expect a power factor somewhere between 0.8 and 0.9.

Hi Steve,

I simply measured the secondary voltage and the secondary ampacity with my multi meter and performed OHM's Law to get the wattage. E X I= P. If I know two of the values which are measured by the meter than you can apply the OHM's Law formula to find the unknown value.

I also have pictures documenting the complete test and the procedures on how we measured the values above. Let me know if you want to take a look at them. The only other test I would like to perform and I did not is an OHM test on the fixture leads to find the R value. This would tell me what the resistance of the fixture is.

So the question I have for you is the fixture a resistive load or a reactive load?

Remember that the power factor formula is the following:

Power Factor = True Power/ Apparent Power

Ken

Future Test to be Conducted:

Transformer and an integrated LED fixture

We will be connecting an osciliscope to the secondary and primary to observe what exactly is happening to the AC wave lengths.

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