Help with a couple of questions please.

gcbryan

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I'm trying to understand led spec sheets. Using a Cree MC-E it says voltage is 3.7 and max current is 700 mA. It said min lm at 300mA is 370.

Since there are 4 leds in a MC-E are they talking about 370lm x 4 or is this for the entire chip and therefore does this mean the max lumen is only achieved at 700 mA?

I'm also looking at the specs for a flashlight where it says voltage input is 6-16 volts and that it can use 18650 rechargeable or CR123a. It takes 4 CR123's and 2 18650 batteries.

The CR123's are 3.6 volt and the 18650's are 3.7 but there are only 2 of them.
Is this 3.6 x 4 =14.4 V and for the 18650's only 3.7 x 2 = 7.4 V ? How does that work?

I'm just trying to figure out how the numbers add up. In looking at certain flashlight specs with known led chips I'd like to know if it makes sense or not.

Thanks for any help you can provide. I've read a basic faq sheet somewhere on here about leds/bins/vf, etc but these questions weren't answered.
 

buickid

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I'm no LED expert, but I believe 370lm is achieved at 300mA across all four dies in series. Someone else will be able to answer this better. As far as the voltage input, the light probably has a buck converter, which takes a high voltage and lowers it. The range of the buck converter is 6-16 volts, so it can handle a V input of 6-16 volts, and converts it to some lower voltage.
Primary (non rechargable) CR123a are 3.0V each. Rechargable Li-Ion 18650 are 4.20V fully charged, ~3.7V under load. I don't recommend you try 4 x RCR123a (rechargable 123a), as that will be 16.8V. 4x CR123a will be okay, as 3.0V x 4 = 12V, within the range of the buck converter.
 

gcbryan

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OK, thanks, so the buck converter explains the voltage thing. I guess this is so several different types of batteries can be used. That makes sense.

Regarding the MC-E and lumens...if 370 lm is the rating at 350 mA and the flashlight is regulated at 1500 ma what would be lumens be? The Cree spec sheet only mentions minimum lumens at 350 mA.

Regarding the 18650 battery. On DX DealExtreme they have 18650 batteries listed at 3.7V.

I'm no LED expert, but I believe 370lm is achieved at 300mA across all four dies in series. Someone else will be able to answer this better. As far as the voltage input, the light probably has a buck converter, which takes a high voltage and lowers it. The range of the buck converter is 6-16 volts, so it can handle a V input of 6-16 volts, and converts it to some lower voltage.
Primary (non rechargable) CR123a are 3.0V each. Rechargable Li-Ion 18650 are 4.20V fully charged, ~3.7V under load. I don't recommend you try 4 x RCR123a (rechargable 123a), as that will be 16.8V. 4x CR123a will be okay, as 3.0V x 4 = 12V, within the range of the buck converter.
 
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gcbryan

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I would like to understand the spec's as it relates to this light:

Diving Cree MC-E (K-WC) 3-Mode 640-Lumen LED Flashlight Kit (2*18650/4*CR123A/4*16340)

So 640 lumens from 2 18650 batteries and this light is Digital Regulated 1500mA Current.

If the 4 leds are wired in parallel then the voltage requirement would be 3.6 volts I believe and the minimum mA would be 1400 (4*350mA)

If that is how it's done how do you get 670 lumen as the Cree spec says 370 lm at that minimum current?

If it's wired in series the voltage drop would be 3.6 x 4 = 14.4 V so that's not it.

It seems to me that the mA would have to be higher than 1400 for this chip to reach 640 lumen.

What am I missing?

Also, there are some divers locally who do have this light and have tested it out underwater and it is noticeably brighter than a 10 watt HID (Light Cannon) which is rated at 450 lumens. So, it does seem to produce 640 lumens.

I'd just like to understand how.

Any help would be appreciated.
 

Linger

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gcbryan
Using a Cree MC-E it says voltage is 3.7 and max current is 700 mA.
that is per die, each die [email protected]

Since there are 4 leds in a MC-E are they talking about 370lm x 4
No, this is confusing but no (370x4=1340lumen, not cree's rated spec)
chip can be wired differently. Each die addressable or all together. The wiring of the dies in series or in parallel or a combination of these will determine how you want to add the input, either in series thus adding voltage (3.7v+3.7v+3.7v+3.7v) or adding in parralelle thus requiring you to add together the current for each die (700ma + 700ma+700ma+700ma)

I'm also looking at the specs for a flashlight where it says voltage input is 6-16 volts and that it can use 18650 rechargeable or CR123a. It takes 4 CR123's and 2 18650 batteries.

The CR123's
are nominal 3.0v. rcr123's also called 16340's are nominal 3.7v but start at 4.2v 'hot off the charger,' voltage drops with discharge and 3.6v is usually considered empty


the 18650's only 3.7 x 2 = 7.4 V ? How does that work?

so this depends on how the emitter is wired. For example, if the emiter was set 2s2p (two separate groups, each group containing two dies together) you would want 1400ma @7.4v

I'm just trying to figure out how the numbers add up.
and I'm glad you're doing this.
Try batteryuniversity. And read the intro threads (in the stickies at the top of each forum).
 
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Linger

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gcbryan
OK, thanks, so the buck converter explains the voltage thing. I guess this is so several different types of batteries can be used. That makes sense.
Common drivers are buck (higher voltage input is 'bucked down' to a lower voltage for the emitter) and boost (lower voltage current is boosted up to a higher voltage for the emitter). some boards are 'buck/boost' and have a huge operating range (say 0.9v-7.0v): the question is, how efficient are they at different voltages? Some boards have %55 efficiency at the ends of their range

Regarding the MC-E and lumens...if 370 lm is the rating at 350 mA and the flashlight is regulated at 1500 ma what would be lumens be? The Cree spec sheet only mentions minimum lumens at 350 mA.
you're missing some numbers here. Maybe it's easiest to pick a single die emitter first. I totally agree with your methodology, but too many variables is making it harder for you. Take an ssc P4 U2 bin, or cree xpe Q5 as your examples and you'll remove the series/parallel variable

Regarding the 18650 battery. On DX DealExtreme they have 18650 batteries listed at 3.7V.
yes, but re-read what buickid wrote again, it is correct.
 
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Linger

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gcbryan I would like to understand the spec's as it relates to this light:

Diving Cree MC-E (K-WC) 3-Mode 640-Lumen LED Flashlight Kit (2*18650/4*CR123A/4*16340)

So 640 lumens from 2 18650 batteries and this light is Digital Regulated 1500mA Current.

If the 4 leds are wired in parallel then the voltage requirement would be 3.6 volts I believe and the minimum mA would be 1400 (4*350mA)

If that is how it's done how do you get 670 lumen as the Cree spec says 370 lm at that minimum current?
670 / 4 die = 155lumens per die
If it's wired in series the voltage drop would be 3.6 x 4 = 14.4 V so that's not it.
If batteries are end to end (in series) then 2x18650 = 3.7v x 2 = 7.4v (vbatt)
It seems to me that the mA would have to be higher than 1400 for this chip to reach 640 lumen.

What about [email protected]? If emitter is 2s2p (just a guess) then each die is receiving [email protected]
What am I missing?

If mc-e is [email protected] (90lumens per die at 300ma,) if you need 155lumens per die and each die recieve [email protected], that seems OK
Also, there are some divers locally who do have this light and have tested it out underwater and it is noticeably brighter than a 10 watt HID (Light Cannon) which is rated at 450 lumens. So, it does seem to produce 640 lumens.
(4sevens and surefire under-rate their lights, most others overstate, and many china based manufacturors are widely overstated even beyond theoretical possibility (let alone practical losses from reflector, glass, depleted batteries, etc)
I'd just like to understand how.
keep at it
Any help would be appreciated.
 
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