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First build:) success! PHR photos soon.

Scottyzedinosaur

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Hey everyone! I build my first laser about two weeks ago..but I haven't had the chance to post it. Here goes!

I have my PHR diode in an aixiz module, with a rkcstr driver I set at 120mA.(if anyone could tell me the milliwatts I'm getting out of it at this current, I'd really appreciate it.) I have a radioshack switch, and a rechargeable 9v. So far, it's going strong, with a very visible beam, about as much as my red. It lights matches instantly, pops balloons, you know, all ze laser tricks. I have it in an altoids can, and on a duty cycle of 30 seconds on, 10 seconds off. Overall, I'm extremely satisfied for a first build:D
I'll have pics up soon, I have major studying to do tonight. But I'll have em up tomorrow! Hopefully...
By the way, the camera I have picks up blu-ray as REALLY blue. But in reality, it's a rich violet.

P.S. pictures were taken before the LED was added.
 

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I'm pretty sure 120mA is pushing it, 100mA should give you arround 60-100mW. (quoted from HTD) But, I'll leave it to the experts to tell you what your diode is pushing out now ;)
 

Scottyzedinosaur

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Okay:) I'd like to see how long it lasts...but, if it dies, it's my first build, and I'll learn from my mistakes:)
 
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Weren't you just asking how to do this like 2 hours ago?? If your driver was set up etc, but now you built the laser 2 weeks ago?!
 

chipdouglas

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120ma is considered fairly safe... also it really is hard to tell how many mw's you are workin with, many factors are involved. and all those factors aren't really charted...

michael
 

Scottyzedinosaur

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I don't remember asking anything...but yes, I did build it two weeks ago. I just put an LED into my circuit, last night. I had an LED sitting on my desk, and thought "hey, I've seen people with indicator lights, and they're cool. So, whynot?" so I put it in:)
 

TTerbo

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hmm there are a few places on the driver that if you put that led it might through out the output of the circuit. like if you are running it at 6v the led might take 1.5v out of your overall output.
someone should confirm this.
cheers
 
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hmm there are a few places on the driver that if you put that led it might through out the output of the circuit. like if you are running it at 6v the led might take 1.5v out of your consumption.
someone should confirm this.
cheers
I thought that too, but the diode takes the mA right? So it'll still get the same power, but it'll just drain the batteries faster...right? That's what I thought atleast.
 

Uberstrobe

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Optimal placement of the LED depends on how you turn the laser on and off. If you use an external switch that controls whether or not the battery is connected to the driver you could just put a resistor on the diode and connect it across the driver's battery inputs. If you have a driver that turns on with an on board switch that applies voltage to an enable pin it is rather likely that the enable pin is tied to B+ or something similar when the switch is closed. It might go through a resistor or something though to drop the voltage though. With a little bit of calculation and resistor replacement you should be able to place the diode+resistor on the enable pin after the switch...

Hope this helps,
Sam

Edit: Personally I would NOT put this LED on the driver output. Think about it.
 
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TTerbo

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you couldnt put the led at the drivers output unless it was putting the right Ma
 

Uberstrobe

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Absoluelty.

I just looked up the schematic for the rckstr microdrive. You will have to connect it across the battery terminals of the driver with a resistor in series. This assumes you have an external switch to connect and disconnect the driver from the batteries.

It will not
TTerbo said:
take 1.5v out of your consumption.
... Whatever that means.
But it will draw current from your batteries.

What voltage supply are you using? What is the forward voltage of your LED?(color) Also, if you know: what is the rated current for the LED? These questions determine what value of resistor you will need to use.

I hope this helps.

Edit: I decided to just put down the derivation of the equation used.
V=IR Rearrange
R=(V/I) Determine how many volts the resistor needs to drop at a given current.
R=((Vsource-Vfdiode)/Idiode) Tadah

+Rep? :p
 
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