How to calculate approx lumens from LED spec sheet

Please excuse me if this has already been covered, as the search didn't turn up anything on this topic.

I'm trying to learn how to calculate the approximate lumens an LED will have when it's driven at a certain amperage. I understand this won't be exact due to all the different bins, and things like an efficiency curve, but I'm still trying to learn the general ideas of how it's calculated.

For my example, I've purchased a SST-50 drop-in (WJ bin if that matters) with a 2.5 amp regulated driver. I've downloaded the SST-50 spec sheet from Lumius, and it says "over 1250 lumens", and that the current can be from 1 to 5 amps. So I'm assuming 5 amps = 1250 lumens. Therefore my 2.5 amps will be around 625 emitter lumens, correct?

The rough conversion to OTF lumens is to subtract a third of the output (right?), so you're looking at around 417 Lumens OTF?



For a bonus question, if I'm using an AW 18650 2600mah protected cell, how long can it drive the SST-50 at 2.5 amps (assuming the 2600mah cell really is 2600mah, the regulator doesn't sap any power, and heat buildup doesn't kill efficiency, etc)?

I'm really looking for the proper formula. I want to say a few minutes past 1 hour, but want to make sure. If there's any general rules for power loss from the regulator (like there is between emitter and OTF lumens), I would love to know that rule of thumb. Thanks!

You may already know these things are 'binned' (categorised) according to colour tint, forward voltage, light output etc.

The actual brightness depends on the flux bin.

On page 3 of the spec sheet I have, it gives you an output brightness for the various flux bins at 1.75A
http://www.luminus.com/stuff/conten...2_sst_50_w_product_datasheet_illumination.pdf


The amount of additional power a driver takes depends on its efficiency rating. Quite a few drivers give you a graph or list of efficiency ratings, so you can work it out that way. If you have a driver that is 80% efficient, and you output 2.5A, then it will take about 3.1A... roughly.

My SST-50 is W65S, WJ bin (the full bin I was given was W65S-Gj100-WJ-G3).

The chart lists "Minimul Flux (lm) @ 1.75 A is 500" max is 600. Following that, at 2.5 amps it would be 714 to 857 lumens which is much more than following the way I did it in the initial post (and much more than the 600 lumens the builder quoted, and MrGman got in his sphere readings -even when I add the 33% loss between emitter and OTF lumens).

...so something still seems to be amiss with the calculations.




The Driver board seems I believe to be this one from ShiningBeam:
http://www.shiningbeam.com/servlet/the-132/3-dsh-Mode-Regulated-Circuit-Board/Detail

It doesn't list the efficiency, but if we use your 80% = 3.1 amp intake number, what's the formula to calculate a rough runtime? Do I simply divide the 2600mah battery by 3100 (3.1 amps) which equals .8387 hours, or about 50 minutes of runtime?

See this post for a generic recipe. For the SST-50, the flux bins are specified at 1.75A instead of 0.350A. But the idea is the same.

The relative luminous flux at 2.5A is about 1.4. Thus, the estimated emitter lumens is (500 to 600)*1.4 = 700 to 840. To account for various optical losses, folks often knock off about 20% to 1/3 to arrive at an estimate out the front (OTF) lumens value. So if we take the conservative route and subtract off 1/3, we get about 460-560 OTF lumens. If we use the 20% correction factor, we get about 560-670 OTF lumens. This latter estimate seems consistent with nailbender's and MrGman's 600 lumen figure. They are using an integrating sphere, so their number is a true OTF lumens figure.

That Shiningbeam driver is based on the AMC7135 linear regulator chip. It'll draw 2.5A from your 18650 (actually 2.8A nominal). If your cell has 2600mAh capacity, you are looking at about 1h run time, give or take.

Justin Case said:

See this post for a generic recipe. For the SST-50, the flux bins are specified at 1.75A instead of 0.350A. But the idea is the same.

Click to expand...

Here's the charts from the SST-50 tech sheet (on the very bottom is the chart you linked to).



It seems like I'm unable to use that same formula you linked to. That one has "Relative Lumious Flux" on the left listed in "a.U" (whatever that is), and the SST-50 has it in percentage. Is there a way to convert the percentage number to "a.U." so I'm able to follow the formula you kindly linked to?



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**Edit, just saw your second post.

Would the exact runtime follow this formula: 2600mah battery / 2.8amps nominal (2800) = .9286 hours = ~56 minutes?



(the following is just for my records so I have it saved somewhere) Following the low numbers I get 1.375 Lumious Flux (137.5% from that chart) x 500 minimum lumens for that bin = 687.5 lumens - 33% to convert to OTF = 458 lumens OTF Lumens

MrGman got 470-405 OTF lumens depending on how long the SST-50 was running, with averages out to 432 lumens so those numbers are pretty close to each other.

Thanks for all the help everybody! I really appreciate the information as I'm learning everything.

A.U. is probably arbitrary units. It's a relative, or multiplicative, factor, so there are no specific units on it like lumens, watts, furlongs per fortnight, etc. You also don't need the graphs for the Seoul P4 LED. You are interested in the SST-50 LED. Thus, you need to use the relative luminous flux vs forward current graph for that LED.

75% is 0.75. 100% is 1.0. 200% is 2.0. Etc.

1. Get the flux bin data for the LED of interest. You are interested in WJ, which is rated at 500-600 emitter lumens at 1.75A forward current.

2. You want to run the SST-50 at 2.5A. Go to the relative flux vs forward current graph and pick off the relative flux value for a forward current of 2.5A. That looks like about 1.4 (140%) to my eye.

3. Multiply the flux bin rating from step 1 by the relative luminous flux value from step 2. That gives (500 to 600 emitter lumens) * 1.4 = 700 to 840 emitter lumens. Multiply by your favorite correction factor to estimate OTF lumens.

Run time comes from dimensional analysis. Capacity is mAh. Current draw is in A or mA. Divide capacity by current draw to get hours. It is a rough estimate. I wouldn't get carried away with all of those decimal places of meaningless accuracy. 50-60 min seems reasonable.