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Direct-Drive 445 Host

DC batteries can spike, but with regular use I don't think they would as much.

On a small sidenote, where are you finding these 5A and 9A drivers? I have a custom one off host for my phlat red LED that I'm going to throw together, but I need a driver.
 





Pontiacg5 said:
DC batteries can spike, but with regular use I don't think they would as much.

On a small sidenote, where are you finding these 5A and 9A drivers? I have a custom one off host for my phlat red LED that I'm going to throw together, but I need a driver.
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It's not a voltage spike. That's not going to happen with a battery. Excessive inrush current would be the problem.
 
LarryDFW said:
Wally;

As long as you have a good heatsink,

the diode will be fine.

I've checked a couple of diodes, and both of them were fine for direct drive.

LarryDFW

P.S. No voltage spikes from a DC battery.
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I may just give it a try tomorrow. I've got three 445nm diodes, and even if I ruin one they're only going to get cheaper to replace, I guess.
 
Not by much until projector prices come down. At the cheapest i've seen a130's, the per diode cost has been $31.something. (don't remember the projector price anymore) Plus something for the time and trouble of extracting them, $35 is the cheapest they'll be.
 
qumefox said:
Not by much until projector prices come down. At the cheapest i've seen a130's, the per diode cost has been $31.something. (don't remember the projector price anymore) Plus something for the time and trouble of extracting them, $35 is the cheapest they'll be.
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Yeah, $35 is a pretty awesome price. They were $50 when I got these and I had to wait on a group buy.
 
I've seen them for $35 in the BST section. Not sure if those threads are still open and taking orders though.
 
I may regret it but I'll try it considering that your battery gives 600ma and people have run their diodes at 1.5-2A. I have a 2 ohm and a 20 ohm resistor that I'll use with it.
That just leaves heat sinking and collimating.
 
qumefox said:
Can you put a 1 ohm resister in series with it and see what current it's really pulling?
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Why bother with a resistor, just read the current draw off the batteries. Since its direct drive you know exactly what the diode is getting.
 
Something's not right because with a 2 ohm resistor in place my diode pulls nearly 780ma of current but it's not as bright as I expected. I suspect that the voltage may be lower than it should be. I know that a 300mw IR diode will quickly burn holes in electrical tape at extremely close range even when uncollimated but with the same resistor and an extra AA battery this one doesn't give enough light to heat anything. Meh I'll figure it out given enough time.
 
psi seeker 34 said:
Something's not right because with a 2 ohm resistor in place my diode pulls nearly 780ma of current but it's not as bright as I expected. I suspect that the voltage may be lower than it should be. I know that a 300mw IR diode will quickly burn holes in electrical tape at extremely close range even when uncollimated but with the same resistor and an extra AA battery this one doesn't give enough light to heat anything. Meh I'll figure it out given enough time.
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How many batteries are you using, and what type are they?
 
psi seeker 34;

Remember that diode voltage matches up well with:

1. A 4.2VDC Lithium-ion battery
(the newest ones I have operate @ 4.35 VDC)
2. Three 1.5 VDC batteries with a small value resistor
(try a 0.47 ohm to test current)

Remember, that you must have a good heat sink for any direct-drive system.

LarryDFW
 
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My resistor is 2 ohm which is the lowest I could find in my pile of electronic scrap.
If I want to go any lower I'll salvage a potentiometer to do the job so I don't have to keep soldering and unsoldering resistors to get the right one.
 
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psi seeker 34 said:
My resistor is 2 ohm which is the lowest I could find in my pile of electronic scrap.
If I want to go any lower I'll salvage a potentiometer to do the job so I don't have to keep soldering and unsoldering resistors to get the right one.
Click to expand...

Or you could use a couple resistors is parallel to get a lower resistance value. The equation for the overall resistance is Rtotal=1/(1/R1 + 1/R2 + 1/R3 + etc...)
You can also use an online resistor calculator instead for the calculation such as this one.
 
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