3W light 500 lumens Possible?

[DavyCrockett]

I see a 3W flashlight for sale, about 5" long, that claims 500 lumens.
It uses a 3.7V, Li-ion 18650 battery.

Is this possible? To throw out 500 lumens?

thanks,

Davy

 

[kramer5150]

At 4.2V 3 watts requires .714A.
Theres no way youre getting 500L from .714A at 4.2V

Assuming the cell sags under load from 4.2V to (say) 3.8V, 3 Watts requires .789A. So again, 500L is impossible

If they are using a driver circuit that will further decrease efficiency. Most driver boards are only in the ballpark of ~70-80% efficient. The bad ones are less than that.

they are claiming 166 Lumens/Watt, which is off the chart efficiency. My most efficient light is my E2L-KX2 which is 78 Lumens/Watt. My favorite is a 6P-M60, and even that is only in the ~45 Lumens/Watt range.

 

[DavyCrockett]

So 4.2V and 3W REQUIRES ~0.7 amps. . . that's a given? without stating the amps? (i'm not doubting, just learning)
Thanks Kramer, it seemed too much to me, the 500 L (marketing i guess)
They never mentioned amps.
So sometimes they'll mention mA's. Is that 1/10th of an amp?
thanks again.

Just thought of another possible ROT, Kramer. (rule of thumb) If lights are at best 80% efficient, and lets say you can
get, being conservative, 70 lumens per watt, then can one multiply 70 times W to get Lumens?
e.g. 70 x 4.2 ~ 290 lumens.(so the above mentioned 500 L light is probably closer to 300 L) or my 24W which i believe is 1500 lumens, maybe slightly more, then 70 X 24 ~ 1600 lumens. What do you think?

Seems it works! I think we have a new forumula. :-) 70 X Volts ~ Lumens. (~ means approximately equal to)

I'm going to go look at your light reviews.

 

[mattheww50]

Power = The integral of Volts x Amps, or from Ohm's law the integral of Current^2 x Resistance is also power.

For DC, the integral is simply volts x amps, or amps^2 x resistance. For AC it gets much more complex because the wave form is a sine wave, and volts and current are not necessarily in phase. I.E. it is possible to draw 100 Amps from a 120V AC supply, and actually consume no power if the voltage and current are exactly 90 degrees out of phase.
The integral of sin(theta) x sin (theta+90) over time is zero. This is referred to as 'at quadrature'. This fact has some important applications in communications.

So in this example:
If the wattage is 3, and the voltage is 4.2, the 3/4.2= amps from simple algebra.

However as other have pointed out, there are no 166 Lumen per watt commercial LED products at this time with anywhere near 500 lumen output. There may be some in a few more years, but not today.

 

[kramer5150]

So 4.2V and 3W REQUIRES ~0.7 amps. . . that's a given? without stating the amps? (i'm not doubting, just learning)

Yes, A watt by definition is Volts x Amps.

Thanks Kramer, it seemed too much to me, the 500 L (marketing i guess)
They never mentioned amps.
So sometimes they'll mention mA's. Is that 1/10th of an amp?
thanks again.

You multiply A by 1000 to get mA, 1A=1000mA

My E2L draws .17A... x1000 to get 170mA

Just thought of another possible ROT, Kramer. (rule of thumb) If lights are at best 80% efficient, and lets say you can
get, being conservative, 70 lumens per watt, then can one multiply 70 times W to get Lumens?
e.g. 70 x 4.2 ~ 290 lumens.(so the above mentioned 500 L light is probably closer to 300 L) or my 24W which i believe is 1500 lumens, maybe slightly more, then 70 X 24 ~ 1600 lumens. What do you think?

Seems it works! I think we have a new forumula. :-) 70 X Volts ~ Lumens. (~ means approximately equal to)

I dont think you can really treat it as a rule of thumb, because Lumen/Watt efficiency varies greatly... even from very similar lights.

I'm going to go look at your light reviews.
Sure, feel free and thanks!!

 

[agt]

Theoretically 683 lm light flux has the power of 1W. This is the absolute maximum. Real LEDs have 70-100 lm per watt efficiency. So LED is very efficient heating device with an efficiency factor 90%:grin2:

 

[DavyCrockett]

Thanks AGT, so it seems 70 times watts will give a conservative Lumens.
I'm using it! :poof::paypal:

 

[kramer5150]

Real LEDs have 70-100 lm per watt efficiency. So LED is very efficient heating device with an efficiency factor 90%:grin2:

... but in the real world efficiency drops significantly when reflective, refractive and thermal losses of the host are factored in.

Look carefully at the bigchelis sticky and see for yourself how many lights are actually ~70 Lumens/Watt... not really enough to use that as a baseline or accurate "ROT".

 

[DavyCrockett]

So Kramer, I respect your opinion and knowledge, about 60 times? as a ROT. :twothumbs

 

[kramer5150]

So Kramer, I respect your opinion and knowledge, about 60 times? as a ROT. :twothumbs

Well. there are a lot in that efficiency ballpark, but not all of them. thats all I'm really saying. So you have to be careful, making generalizations.

 

[DavyCrockett]

Alright. I love ROT's though.
If inflation is roaring at 12%, the the rule of 72 says prices will double in
6 years. 72/% = years to double. :popcorn:

Bin Laden is long dead and Lee Harvey what's his name was framed. :sssh:

Watts X 60 = Lumens :candle:

 

[StarHalo]

Flashlight rated in watts = red flag.

And there is no such thing as an energy-to-output conversion, different emitters have different efficiency, different reflectors and lens and resistance add different variables, etc.

 

[KenAnderson]

Check out Mac's SST-50. Driven at 2.8 amps, using a high capacity discharge RCR123. Mac says it puts out 500+ lumens out the front.

 

[LEDninja]

I bought a light once that had 3W silkscreened on it. It turned out they were using old leftover bodies originally made for the 3W Luxeon. So '3W' means nothing.

The 18650 battery can push 2.8A for a short time. I am pretty sure my Elektrolumens EDC-P7 (new production is EDC-MCE) can produce 500 OTF lumens. But it gets too hot to hold in 5 minutes.
As a result most 1*18650 lights limit the amps to 2A. At which point 500 emitter lumens is possible. Not likely but possible. 2A works out to 7.2W nominal, 8.4W max.

-

A lot of sellers on the internet are technically illiterate, just copies (usually incorrect) info from the competitors websites. Even when they copy from the correct manufacturer they often copy the wrong page. So buyer beware! Get the actual brand name and model # and search for reviews.

Most bright flashlights claim '900 lumens'. 500 lumens seem to be an oddball number so I did some Googling.

The 1st '3W 500 lumen' light turned out to have 3 LEDs. So it really is a 3*3W. Each LED only needs 167 lumens so yes 3*3W can give you 500 lumens.

The 2nd light has the title '3W 500 Lumens UltraFire C1 Q5 LED Aluminum Flashlight'.
Scroll down and I see:
Bulbs / mode: Q5 LED bulbs
Luxeon: 240 LM
Size: 5.7"(L) x 1.3"(D)
Battery: 1 x 18650 (not included)
THERE IS A SLIGHT DIFFERENCE BETWEEN 500 Lumens AND 240 LM.
In the early days 5 mm LEDs are famous for burning out. So when Lumileds introduced the rugged Luxeon, a torch with a Luxeon LED is considered high quality. When the 3W Luxeon came out CPF members found out they can go from 30 lumens to 75 lumens by just changing the battery from a CR123A to a RCR123A. The older 1W would smoke. So 3W is much prized. Note how the seller managed to get 3W and Luxeon into the page. Out of date marketing but don't blame an enterprising person from trying to impress the customers.
This seller also lists
3W 200 Lumens UltraFire WF-501D Q3 LED Aluminum Flashlight
3W 200 Lumens UltraFire WF-502B Q3 LED Aluminum Flashlight
3W 200 Lumens UltraFire WF-502B 3.7V Xenon Flashlight
3W CREE LED Focus Flashlight
3W 100 Lumens UltraFire C3 P4 LED Aluminum Flashlight
3W 500 Lumens UltraFire WF-503B 5 Mode Q5 LED Aluminum Flashlight
Looks like the seller has grouped his Cree torches under the 3W house brand.
This seller also lists 1W for the lower powered and 5W for the more powerful SSC-P7 flashlights.

 

[jirik_cz]

Theoretically 683 lm light flux has the power of 1W. This is the absolute maximum. Real LEDs have 70-100 lm per watt efficiency. So LED is very efficient heating device with an efficiency factor 90%:grin2:

This is the theoretical maximum for monochromatic GREEN light with 555nm wavelenght.

Theoretical maximum for white light is somewhere around 300lm/1W, so current white LEDs have efficiency around 30%. :p

 

[kramer5150]

The other thing to keep in mind too is that the lights BC chooses to measure really are what I consider to be THE TOP-tier flashlights in existence.

He (generally) won't just measure any junk-light off the shelf. Most of his personal lights are custom one-off builds from some of the best CPF builders and machinists. These boutique builds are on the order of $hundreds$, take months to machine, build and fine tune, using expensive materials and premium parts + components. (generally speaking)

Every element of these lights is optimized to maximize efficiency... eek out every last lumen from each watt of power consumption.

Setting a rule of thumb using this population as a baseline is probably not the best way to do it.

 

[rickypanecatyl]

The neat thing about LED's is how both the lumens/watt & total watts they can be driven at is going up. It wasn't long ago surefire said 5 watts was the max for an LED. I've got a 30 watt+ LED!

Recently I got a Eagletac TC02 that claims 4 watts and 380 emittor lumens or 300 OTF lumens. At the same time I noticed in target an Inova light that claimed 5.8 watts and 135 lumens. I don't know if Inova measures out the front or emitor lumens. Even if it is OTF the difference is huge!

135/5.8 = 23 lumens per watt, and the Eagletac is putting out 75 lumens per watt. So even amongst LED's that is a 3X variable there!!

I think it would be easier to make up a "ROT" for loss on regular vs deep throw reflectors but the lumens per watt # is exactly what everybody is intentionally trying to see change - and it is changing fast!