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New 9mm 445 diodes

Looks to be the same. Not unexpected considering this is also a multimode diode.

With 445's though divergence doesn't bother me as much... only a huge problem with the mitsu diodes imo.

Thank you for posting :beer:

Now go home and test the 405 :wave:
 
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DTR said:
p1018980.jpg
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Are they really 445 nm beams. Looks yellowish.

Do you have a beam without lenses?
 
lazeristasUVISIR said:
Are they really 445 nm beams. Looks yellowish.

Do you have a beam without lenses?
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Uhhhh... Thats probably because the photos were taken through yellow-tinted goggles. It's not the color that matters anyways, that photo was to reference beam profile.
 
Meh, I'm not sure it worth the extra $$. Not at a retail level anyway. You also have to consider you need a machined barrel to hold the new lens too. Thats even more money. Its a lot of extra money for a few hundred mWs of power.

Great job Angelos, and Jordan. :beer:
 
Eh. I took the dive a few days ago, can't wait till I get it. If mine is as strong as Angelos', I may sell my current 445 and make a new one with that beast >:D
 
I was talking at a retail level.

The added cost doesn't justify asking $500 for a complete build.

At least we have a clue now. I applaud everyone who took the plunge.
 
I suppose that's true. But we have seen lasers above the 2.2W mark selling for upwards of $500.
 
I wonder if you can just ghetto drive these with a resistor and a battery provided the diode is heatsinked. They may be tough enough now.
 
joeyss said:
I wonder if you can just ghetto drive these with a resistor and a battery provided the diode is heatsinked. They may be tough enough now.
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You can drive any diode like that. I do it sometimes when I want to do a quick test on a diode. As long as you have a constant input voltage and know the Vf you can use a resistor just fine, but it isn't a good way to do it in a portable build since the output will vary alot as the batteries drain. We use drivers so the LD current stays constant even as the battery voltage changes.
 
lazeerer said:
:thanks:

Yes but they have Higher Voltage.:yabbmad:

I think they are the Same in performance as the 5.6mm ones we have now.

Yes about the Lens. Its On Par with the G lenses. However The G lens is Much More cheaper.

In Other Words buying a 5.6mm diode and a G1 lens will be Alot Cheaper then Buying one of these 9mm diodes that come with the lens.

I also notices with the projector lens if its not hit dead on center you will get a nasty dot and it will be crooked.

Yes the Lens works with your Mount Perfect.:) Its the Only way it can work with the 9mm package since it has a Shorter FL then the G1/G2 lens IT has to sit as close to the bottom of the lens nut as possible. If you do it the other which i tried first it dint work. However it did work for 5.6mm diodes.:) I think it might be best for you to test the lens out for us.

The lens also feels Glass/Gel Like. Hard to explain.

Glad you like the Testing Jay.:) Hope this held you.:beer:
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It's glass... AR coated as well.

Maybe you got that 'gel' feeling because of the spots of glue on the lenses from the way they were mounted. (I just carefully scraped off the glue)

I did some initial testing and comparing with 405nm and 445nm. Very close power comparisons with the 405-G-1 power as you mentioned.

As you also pointed out, I did notice a very sharp focused, pin point dot with blu-ray.

It is a little more 'close focus' in design vs the 405-G-1, and that is why the focus is more 'touchy'.

It will also mean a little more adjusting depending on distance and so forth.

But it is a nice lens.

For 445, I can see similar 'side splash', which was expected with such a close focus lens.

I did also notice the two long 'wings' that we saw with the other Ca$io lenses. But not near as noticeable.

Seems to be a pretty good lens! :)
 
Why do you need to know the Vf for resistor-driving? Why not just find out what resistor you need for the given Vin and then determine how much current you need to be able to pass by Ohm's Law: R = V/I?
 
Wolfman29 said:
Why do you need to know the Vf for resistor-driving? Why not just find out what resistor you need for the given Vin and then determine how much current you need to be able to pass by Ohm's Law: R = V/I?
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Vf is needed.

Here is some rough math for a 445

Power supply=5v
diode vf = 4.2
5v - 4.2v = 0.8v
now you use V=IR

0.8v = (Say 1 amp) * R
So you would want .8ohm resistor to drive a 445 off of 5V at about 1A
 
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A small change in Vf can have a great impact on the currant. These 9mm diode have different Vf than the smaller 5.6mm cousins. That is why I rely on constant current, to have a stable output.
 
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