LED project... input appreciated

[Anonymous]

The LED dissipates about half of that of resistor (132 mW
verse 68 mW ), so that ratio is roughly 1/3 vs 2/3. 1/3 of 9 V is about 3V, which is about what you want for the LED according to the voltage specs. This, assume, that the wizard knows the LED has half as much resistance as the resistor (165 ohm vs 330 of the resistors).

I guess it is safer to drive it with 9 volt, since most LED will have limited life if you drive it above certain volt/amp even if it is within the factory spec. You can get the 12V power supply if you have issue with light intensity, but I can assure you that its life will be cut short when you drive it at that level. The LED cost more than the power supply by several factors, so you may want to be a bit more conservative.

I apologize that I can't be of much help in verifying your values. Sorry.

 

[simico]

Hi thanks again, this is a real help…

…I think I understand, :? but predictably that leads me to more questions… 8O

:?: Is the 132mW a set factor directly/proportionally related to the 330 ohm properties of the resistor?

:?: As such, does the 68mW and 132mW remain constant irrelevant to the voltage value put intro the circuit?

:?: And as such, will the ratio between the to components remain at 1/3 – 2/3?

:?: And as such, to calculate the optimum voltage, is it just a factor of this ratio?

And therefore, if I want say 3.4v…

:?: …Because the LED is a 1/3 ratio to the resistor, does that mean the circuit voltage should theoretically be 10.2v to supply 3.4v to the LED?

:?: …and if that is all correct, 12v would give 4v per LED (0.2v too much) leading to early LED death via excess heat to both the LED and resistor?

If that’s right, then I think I understand! :idea: :!:

Many thanks, Si.

 

[Anonymous]

I think all you need to do is to take a quick look at Ohm's Law. Basically, it says V=I*R, where voltage is V, current (amp) is I , and resistance (ohm) is R. All that is missing is the resistance of the LED at driving voltage. I roughly estimate it to be half of that of the driving resistor, not an exact value.

If the 1:2 ratio is correct, then the resistor will take 66% of the voltage, and the LED 33%. If you put 12 voltage on it, then the voltage drop will be 4 volt for the LED, which is too much.

As an exercise, see if you can use the Ohm's Law to calculate the resistance of the LED given that other two values, that is, driving voltage and current rating. (simple substitution on the formula!). With that value, you can get everything you need to calculate everything.

 

[mr_X]

their light intensity and focus sucks-led's still have quite a ways to go before they'll really be usefull for corals-even in a nano, imo

these guys disagree-
http://solarisled.com/

 

[simico]

Thanks dupaboy1992, that’s been great advice. I going to see what power adaptor best suits, and then wire it all up…( :!: ) It’s never going to be the brightest or cheapest way to light my tank, but when better LEDs come out I can have another go, and I just like the idea of trying something new / different…

…I will test the output with my LUX meter and see what I get…


Hi mr_X, thanks for adding the link. Obviously, the unit is (currently at least) very expensive, 8O

...but what would be really interesting is to know what spec the individual LEDs are that they are using to power the thing… (…I think they are Philips clusters but can’t remember where I read that… - anyone know for sure?) :?:

…also, having Goggled the Solaris I found this out, from…

http://www.reefcentral.com/forums/showt ... did=938675

“The Solaris produces 17k LUX, compared to T5 at 13k, VHO at 9k…

…The MH produces 33k LUX. This is nearly double the output. However, the flaw of MH is unequal intensity distribution unlike fluorescent lamps and the Solaris. In other words, MH lamps offer high intensity levels directly under the lamps, but as light rays travel at angles deeper into the tank, intensity levels (LUX) are greatly reduced.”

…and on…

http://www.nano-reef.com/forums/index.p ... 97985&st=0

…it talks about LEDs with 87.4 to 113.6 lumens output, but this is related to something called K2’s and Lux IIIs, and I am only guessing these are individual LEDs. (!)…

Anyway just opening the debate, comments welcome…

 

[pwe169]

Hi all, I've been reading through this and searching for available LED's at a descent price.
Found:
http://www.besthongkong.com/product_inf ... cts_id=285
at $1.40 they seem very good with an impressive output. Has anyone tried any?

Pete

 

[simico]

Anyone understand what "4 chips" means? :?:

Also, what does the 10000-18000mcd part mean, and why is that so much lower than the final total power output figure? :?:

Iv: 10000-18000mcd @ 80mA, Total power 825,000mcd*

Anyway, if we are actually talking 825,000mcd from one 55 degree 10mm LED, I make that 585.694 lumens! :? - Surely I got something wrong?! :?:

If not, these things are what everyones been waiting for I think...

...and those other LEDs I bought are now a bit rubbish!

...but seriously my friend, those are a great find. Credit to you... 8)

 

[Anonymous]

No idea about 4 chips... my guess it that they use four diodes in the single package?!?!?

mcd means milli-candle. One candle is the brightness of.... you guess it right, one candle.

 

[pwe169]

When I done a bit of research the 825000 just didn't stack up.

In the end I have ordered:-

25 x Collimator Holder PG1N-SE01 (PG1N-SE01) = $9.75
25 x Collimator PG1N-NX45 45 Deg (PG1N-NX45) = $46.50
13 x ProLight 3W Blue LED PG1N-3LBS (BEC0601075) = $58.50
12 x ProLight 3W White LED PG1N-3LWS (BEC0601051) = $61.08

This will be very similar to another well known make of LED hood!!

When I get them and start to play about I'll post some more results

Pete

 

[mr_X]

make 2 of them...one in 72" for my tank :wink: