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Thread: Runtime Calculation, is this right?

  1. #1
    Flashaholic* ti-force's Avatar
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    Default Runtime Calculation, is this right?

    I just went to the Kenworster.com runtime calculator and entered some data. I've often wondered this, but now I'm trying to figure it out. Do these numbers look right, or am I doing something wrong? I'm trying to figure out the calculated runtime for a Quark 123-2 R2, using 1x 17670 and 2x RCR123's.

    For the 1x 17670 battery calculation, I entered in the following information:

    Desired LED Current: 882mA
    LED Forward Voltage: 3.4V
    Battery Input Voltage: 3.7V
    Battery Capacity: 1600mAh
    Converter Efficiency: 90%

    When I hit the calculate button, this is the data I get, which seems close enough to me:

    Runtime: 1.78 hrs., which is 1 hr and 46 min.
    Battery Current Draw: 901mA



    Now for the 2x RCR123 runtime calculation. This is what doesn't seem quite right to me, but maybe it is. Here's the data entered into the calculator:

    Desired LED Current: 888mA
    LED Forward Voltage: 3.4V
    Battery Input Voltage: 7.4
    Battery Capacity: 750mAh
    Converter Efficiency: 85%

    Hit the calculate button and this is what I get:

    Runtime: 3.13 hrs., which is 3 hrs. and 7 min.
    Battery Current Draw: .480mA

    I find 3 hours and 7 minutes of runtime very hard to believe. Am I messing up on the voltage or is this correct?
    Thanks!

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    Flashaholic* AnAppleSnail's Avatar
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    Default Re: Runtime Calculation, is this right?

    You've messed up somewhere. You've got 5.92 watt-hours in the 18650 and 5.55 watt-hours in the 2x cr123. Given .882 and .888 mA of LED draw, that's:

    3.7v*1.6Ah = 5.92 Wh
    5.92Wh / (3.4v * .882A) = 1.97 hours ideal
    1.78*.9 efficiency = 1.78 hours

    3.7v*.75Ah*2 batteries = 5.55 Wh
    5.55Wh / (3.4v * .888A) = 1.84 hours ideal
    1.84*.85 efficiency = 1.56 hours

    I can't remember how to calculate battery draw in amps, probably:
    (battery voltage)/(LED voltage * LED current * (1-efficiency))
    My biggest light-hog is my camera.

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    Flashaholic* csshih's Avatar
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    Default Re: Runtime Calculation, is this right?

    also, those battery ratings are usually calculated at .2C.. you'll have a lower cap at a higher drain rate from internal resistance.

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    Default Re: Runtime Calculation, is this right?

    Interesting! What would I put for an SST-50 @ 2.8A?

    Desired LED Current: 2800mah
    LED Forward Voltage: 3...?
    Battery Input Voltage: 6V (2XCR123)
    Battery Capacity: 1500mAh
    Converter Efficiency: ...?

    I'm expecting ~30 minutes.. I've never timed it before when I had it, but just a guess..

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    Flashaholic* ti-force's Avatar
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    Default Re: Runtime Calculation, is this right?

    Quote Originally Posted by AnAppleSnail View Post
    You've messed up somewhere. You've got 5.92 watt-hours in the 18650 and 5.55 watt-hours in the 2x cr123. Given .882 and .888 mA of LED draw, that's:

    3.7v*1.6Ah = 5.92 Wh
    5.92Wh / (3.4v * .882A) = 1.97 hours ideal
    1.78*.9 efficiency = 1.78 hours

    3.7v*.75Ah*2 batteries = 5.55 Wh
    5.55Wh / (3.4v * .888A) = 1.84 hours ideal
    1.84*.85 efficiency = 1.56 hours

    I can't remember how to calculate battery draw in amps, probably:
    (battery voltage)/(LED voltage * LED current * (1-efficiency))
    You've got the voltage for two RCR123's in series at 3.7v, is this correct? I'm not trying to question you, just trying to understand. I measured current at the tail cap for each battery configuration. This is what I get:

    one 17670 = 901mA
    two RCR123's = 480mA

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    Flashaholic* ti-force's Avatar
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    Default Re: Runtime Calculation, is this right?

    Quote Originally Posted by csshih View Post
    also, those battery ratings are usually calculated at .2C.. you'll have a lower cap at a higher drain rate from internal resistance.
    That's good to know, thanks for that. I'm sure this would be different for each battery, but is there a general percentage that could be taken off for a given discharge rate to compensate for this? I'm just trying to get ballpark estimates really, but what do you think?

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    Flashaholic* ti-force's Avatar
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    Default Re: Runtime Calculation, is this right?

    Quote Originally Posted by sfca View Post
    Interesting! What would I put for an SST-50 @ 2.8A?

    Desired LED Current: 2800mah
    LED Forward Voltage: 3...?
    Battery Input Voltage: 6V (2XCR123)
    Battery Capacity: 1500mAh
    Converter Efficiency: ...?

    I'm expecting ~30 minutes.. I've never timed it before when I had it, but just a guess..
    What driver are you using? As for voltage forward, and if SST-50's are lottery like as SST-90's, you will probably have to take the average voltage forward from the datasheet, unless the light is already together, then you can measure the voltage forward, which would be best.
    Last edited by ti-force; 05-05-2010 at 05:53 PM.

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    *Flashaholic* mdocod's Avatar
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    Default Re: Runtime Calculation, is this right?

    Hello Ti-Force,

    Convert to watts, then calculate:

    LED@3.4V X 882mA = 3W
    3W / 0.9 (regulator efficiency) = 3.33W

    17670@3.7V X ~1500mA = ~5.5WH

    2xRCR123@7.4V X 550mA = ~4WH

    (I'm using realistic capacity values considering the expected load here)

    ----

    17670: 5.5WH/3.33W = 1.65H or about 100 minutes.

    2xRCR123: 4WH/3.33W = 1.2H or about 70 minutes.

    ---

    Realistically speaking, the buck driver will have some over-head voltage requirement, and may not operate at full output through the discharge on a single 17670, which will translate to longer run-times with diminishing output. This can not be so easily calculated unless we know the exact Vf at the desired drive level and the over-head requirements of the driver. Also, the driver efficiency will likely vary across the range of voltage input, but for rough calculations 90% is perfectly acceptable IMO.

    -------------------

    Hello sfce,

    For CR123s, we have to de-rate them pretty heavily for the higher expected loads. I think we would have to assume around 2 amp loads at the cells to achieve 2.8A at the emitter. This would de-rate the cells to around 2.35V and maybe 1000mAH. That makes a total of about 4.7WH to work with, and you'll probably be consuming at least 10-12W or more,

    20-30 minutes run-time would be a rough estimate after factoring in regulator efficiency loss.

    Eric

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    Default Re: Runtime Calculation, is this right?

    Quote Originally Posted by ti-force View Post
    You've got the voltage for two RCR123's in series at 3.7v, is this correct? I'm not trying to question you, just trying to understand. I measured current at the tail cap for each battery configuration. This is what I get:

    one 17670 = 901mA
    two RCR123's = 480mA
    Just to answer your question (he's got the numbers right by the way) he's multiplied the 3.7V times two.

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    Flashaholic* ti-force's Avatar
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    Default Re: Runtime Calculation, is this right?

    Quote Originally Posted by gcbryan View Post
    Just to answer your question (he's got the numbers right by the way) he's multiplied the 3.7V times two.
    Oh okay, I see the "*2 batteries" now. I didn't see that before, and I also didn't try to re-calculate his figures, which would've answered my question. He took 3.7 x .75 = 2.775 and then multiplied 2.775 x 2. Same answer, just a different route.
    Last edited by ti-force; 05-05-2010 at 06:52 PM.

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    Flashaholic* ti-force's Avatar
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    Default Re: Runtime Calculation, is this right?

    Quote Originally Posted by mdocod View Post
    Hello Ti-Force,

    Convert to watts, then calculate:

    LED@3.4V X 882mA = 3W
    3W / 0.9 (regulator efficiency) = 3.33W

    17670@3.7V X ~1500mA = ~5.5WH

    2xRCR123@7.4V X 550mA = ~4WH

    (I'm using realistic capacity values considering the expected load here)

    ----

    17670: 5.5WH/3.33W = 1.65H or about 100 minutes.

    2xRCR123: 4WH/3.33W = 1.2H or about 70 minutes.

    ---

    Realistically speaking, the buck driver will have some over-head voltage requirement, and may not operate at full output through the discharge on a single 17670, which will translate to longer run-times with diminishing output. This can not be so easily calculated unless we know the exact Vf at the desired drive level and the over-head requirements of the driver. Also, the driver efficiency will likely vary across the range of voltage input, but for rough calculations 90% is perfectly acceptable IMO.

    Eric
    Eric,

    You're the man!! Thanks for that. I think I'm starting to better understand now.
    Thanks,
    Casey

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    Default Re: Runtime Calculation, is this right?

    Quote Originally Posted by ti-force View Post
    What driver are you using? As for voltage forward, and if SST-50's are lottery like as SST-90's, you will probably have to take the average voltage forward from the datasheet, unless the light is already together, then you can measure the voltage forward, which would be best.
    It's a Nailbender 2.8A module. The LED is the DE bin from photonfanatic @ the marketplace.. Highest flux bin, etc, etc assume that may play a part.
    The color bin is like ~7,000 - 8,000K, bluest color bin. That's pretty much it.

    Man this stuff is complicated! Way more complicated then I'd think.

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    Default Re: Runtime Calculation, is this right?

    I think it's actually pretty easy to figure runtime, if you are OK with a rough estimate, once you just think about what the factors are (as opposed to memorizing formulas).

    What is the battery capable of in Watt Hours?
    For a 18650 you might use 3.7V * 2500mA so 9.25 wH

    What does the emitter require?
    For a XP-G it might be 3.4V and perhaps it's being driven at 1A so 3.4 W

    Just divide your battery capacity 9.25wH by emitter requirement 3.4 W and you get 2.7 hours.

    Just reduce this by the driver efficiently (varies but usually 85-90 %) so 2.7 hours * 85% = 2.3 hours.

    That's how I think of it anyway.

    It's a quick way to determine what a light is likely to do before you buy it and helps you to judge manufacturer claims of runtime and lumen output. Sometimes you can determine that an ad can't be correct. Either the runtime is incorrect or the lumen output is not correct.

    With the above example if a light using a XP-G R5 was advertised as outputting 325 lumens for 4 hours on one (18650) you would know that either it's not being driven at 1A and therefore can't output 325 lumens or it is being driven at 1A and can't have a 4 hour runtime.
    Last edited by gcbryan; 05-06-2010 at 03:45 PM.

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    Flashaholic* old4570's Avatar
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    Default Re: Runtime Calculation, is this right?

    Know your battery ...

    Lets call it 1A with a single 4.2v [ fully charged Li-ion ] 17670 rated 1600mA
    = 4.2W

    Now 2xRCR123A fully charged = 8.4 or 500mA current draw ..

    This is never going to be exact ..

    1A VS 500mA , 1600mA Capacity VS 600mA ...

    So if everything was constant , your talking one and a half hours for the 17670 VS just over one hour for the two RCR123A's ..

    Its different again if your talking primary 2x3v cells as there capacity can be as high as 1400mA for the better cells .

    Dont worry about driver efficiency : as it may vary as well due to the voltage range .

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    Flashaholic* fisk-king's Avatar
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    Question Re: Runtime Calculation, is this right?

    Thanks for this runtime formula but I am having problems with my calculations for an el capitan using a lithium AA. According to Curt, the current draw is 12mAh with a drive of 30mAh at 3Lm. A lithium aa has a capacity of ~2800mAh. After my figuring I came up with ~100.

    1.5V x 2800mAh= 4.2wH

    3.4V(xp-g) x .012Ah = .041W

    4.2/.041 = 102.4 hrs

    Are these calculations correct? Thanks in advance guys.



    Quote Originally Posted by mdocod View Post
    Hello Ti-Force,

    Convert to watts, then calculate:

    LED@3.4V X 882mA = 3W
    3W / 0.9 (regulator efficiency) = 3.33W

    17670@3.7V X ~1500mA = ~5.5WH

    2xRCR123@7.4V X 550mA = ~4WH

    (I'm using realistic capacity values considering the expected load here)

    ----

    17670: 5.5WH/3.33W = 1.65H or about 100 minutes.

    2xRCR123: 4WH/3.33W = 1.2H or about 70 minutes.

    ---

    Realistically speaking, the buck driver will have some over-head voltage requirement, and may not operate at full output through the discharge on a single 17670, which will translate to longer run-times with diminishing output. This can not be so easily calculated unless we know the exact Vf at the desired drive level and the over-head requirements of the driver. Also, the driver efficiency will likely vary across the range of voltage input, but for rough calculations 90% is perfectly acceptable IMO.

    -------------------

    Hello sfce,

    For CR123s, we have to de-rate them pretty heavily for the higher expected loads. I think we would have to assume around 2 amp loads at the cells to achieve 2.8A at the emitter. This would de-rate the cells to around 2.35V and maybe 1000mAH. That makes a total of about 4.7WH to work with, and you'll probably be consuming at least 10-12W or more,

    20-30 minutes run-time would be a rough estimate after factoring in regulator efficiency loss.

    Eric
    "Now they will know why they are afraid of the dark. Now they learn why they fear the night."-Thulsa Doom PAYPAL FEE CALCULATOR; list of lights.

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    *Flashaholic* mdocod's Avatar
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    Default Re: Runtime Calculation, is this right?

    Quote Originally Posted by fisk-king View Post
    Thanks for this runtime formula but I am having problems with my calculations for an el capitan using a lithium AA. According to Curt, the current draw is 12mAh with a drive of 30mAh at 3Lm. A lithium aa has a capacity of ~2800mAh. After my figuring I came up with ~100.

    1.5V x 2800mAh= 4.2wH

    3.4V(xp-g) x .012Ah = .041W

    4.2/.041 = 102.4 hrs

    Are these calculations correct? Thanks in advance guys.

    Hi fist-king,

    Current is not measures in amp-hours. It is measured in amps.

    If the current "draw" from the cell is 12mA from a ~1.5V source, and the current drive is into an LED with a Vf of ~3.4V, then the current across the LED would not be any more than ~5mA.

    It's impossible to have 12mA draw from a 1.5V source to drive a diode at 30mA@3.4V. The only way to do this would be to first invent an over-unity device that actually works. Think about it: 1.5V x 12mA = 18mW. 3.4V x 30mA = 102mW. 18mW in is never going to equal anything more than 18mW out.

    Find out more about the circuit, then we can help figure out some realistic expectations.

    Eric

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