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| Forums > Marijuana Growing > Growroom Designs & Equipment > LED Grow Lights > DIY LED 5spectre | ||
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#1 |
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Newbie
Join Date: May 2010
Posts: 6
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Hi,
I'm workin on this DIY of my own LED grow light. Equipment which I prepared is: DC 12V 20A 240W Switch Switching Power Supply Driver For LED Strip 110V/220V AC 20pcs of: 3W 620-630nm(VF: 2,2-2,8V, IF: 1100mA) 20pcs of: 3W 650-660nm(VF: 2,2, IF: 750mA)(reverse voltage 5,0V) 6 pcs of: 3W 6500-7000K white (VF: 3,4-3,6V, IF: 700mA) 15pcs of: 3W 470nm(VF: 3,2-3,6V, IF: 650mA) 5pcs of: 1W 440nm(VF:3,2-3,8V, IF:350mA) 2 pcs of Al heatsinks (237*325*30mm, 3,2kg) Air cooling system: Basically I'm gonna use my Vents TT 125 Fan (280m3/h), with reduced voltage. Each heatsink gonna be conected through pipeline, to fan. I think you can see it clearly on pictures. I realy hope it's gonna have enought power to cool all diodes. On pictures you can see only one heatsink, in box its gonna be second one next to him. I've got problem with electrotechnical stuff, with LED array. On 1. heatsing it'll be: 10+10 red one, 3xWhite, 8x470nm blue + 2x 440nm blue. 2. heatsing: 10+10 red one, 3xWhite, 7x470nm blue + 3x 440nm blue. I'm using this calc: https://led.linear1.org/led.wiz and it told me: the wizard thinks the power dissipated in your resistors is a concern And also it says that 10pcs of 630nm LED's total power dissipated by the array is 41052 mW. and thats weird because 10*3W led's = 41W ? Can any one help me with this? I've need to design these LED arrays... :( And I'm totaly lost. I mean TOTALY ![]() Thx M. |
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#2 |
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Inveterate Tinkerer
![]() ![]() Join Date: Feb 2010
Posts: 6,090
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41052mw = 41.052w. Milliwatts are .001 watts, so if you multiply .001 x 41052 = 41.052.
The resistors are dissipating 2.662 watts and 8.228 watts, both of which are very high wattage for resistors. That circuit would worry me because of the almost inevitable imbalance between the parallel branches. If you are going to run parallel feeds off of one source, they need to be as closely matched as possible, and individually fused with fast-blow fuses. |
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#3 |
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Guest
Posts: n/a
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I like your heatsink design, looks good. I think Knna's rule about heatsink size was 9 square inches for every watt, so be careful about overloading them. What type of LEDs are you using?
I think maybe higher voltage drivers that output a constant current (like 24V@500mA) would be easier. That way you won't need resistors in your setup, as the LEDs draw what they need and the power supply handles the rest. |
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#4 |
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Newbie
Join Date: May 2010
Posts: 6
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Rives: Thx for response, we'll I'm not happy about that either, but I've already got power supply and LED's as well. What can happen if I don't come up with a scheme which will be stable ? I could use more powerful resistors (10W,20W resistors), but I'm not much happy with the fact, that those resistors will consume that much.
trem0lo: well both heatsinks together have 170 square inches... and I'm planning to use aprox 160-180W but it will be cooled by fan, it's impossible to cool that with passive cooling system... and I'm using LED's from china, ebay... |
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#5 | |
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Inveterate Tinkerer
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Posts: 6,090
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#6 |
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Newbie
Join Date: May 2010
Posts: 6
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So basicaly I have to lower as much as possible power dissipating of resistors ? I'm thinking I'm gonna use only 8 of them (2*4led leg)
Edit: So I've made some calc and this is best I can get... :( https://imageshack.us/f/19/slepenice.jpg/ |
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#7 |
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Inveterate Tinkerer
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Posts: 6,090
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Getting the power down on the resistors would be good, but the critical thing is to get each parallel leg of the circuit as well balanced as possible so that there aren't discrepancies in the voltage drop, and thus the power draw.
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1 members found this post helpful. |
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#8 |
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Guest
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Fast blow fuses aren't fast enough to protect a diode, no.
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#9 |
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Inveterate Tinkerer
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In general, it is a race that the LED would frequently lose. However, if the power supply and the fusing are carefully selected, they can definitely be made to work, and the successful protection is also impacted by the timing of the failure. Losing a string in a parallel circuit after the fuse is up to operating temperature would have a better chance of a fuse protecting the LEDs than if the failure happened on power up. Additionally, there are specialty fuses that are fast enough to protect electronic components that are restricted to working on 1/2 of a 60-cycle sine wave (thyristors). They are pretty exotic and very expensive, though.
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Philips 315w CDM Elite (CMH) - Overview & Information My DIY Hybrid LED/PL-L Luminaire The VolksLED - DIY LEDs for the Masses Last edited by rives; 02-02-2012 at 03:18 AM.. |
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#10 |
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Guest
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Sigh. When you say "use a fast blow fuse", these guys are just buying whatever is at Radio Scrap or Frys, which are WORTHLESS for protecting LEDs from overcurrent.
You guys build what you want. If you don't want or need my advice, fine, I'll just shut up, sit back and watch. Good luck. No more comments from me. ![]() What do I know? I have over 20 years experience in electrical engineering and R&D. I make a 6 figure salary with a fortune 500 electronics manufacturer whom I've been with over 15 years. My stuff is on-the-air in over 30 countries. I have a technical Emmy. I don't know shit, so carry on. LOL |
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