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how do I use a 532nm 5mW laser module ?

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Hi Timelord, hi guys, hi gals,
I'm new to this forum, so be kind with me !! I was actually in another laser forum but it became sooooo boring... everybody got banned (not me), etc. I've got some 532nm 5mW laser module of atlasnova and I never knew how to use it because I want to power it to infinity without using my precious batteries. Anyway, I need to know what resistor I must use to reduce the currrent from 5V to 3.2V because I only have a 70 ohm (+/-10%) resistor and a 15 ohm resistor of the same tolerance. I also want it to connect it to my PSP recharger which is supposed to give 5V in continuous current (which fried my UV LED :-[)
Since I can't post links, the item is still on ebay. Also I would like to know if I can use it continuously for like 6 hours or 10 minutes.
Help would be appreciated !
 

cdanjo

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Hello,
 I don't know what specs your laser diode runs at, so this is a simple lecture in using Ohm's law (V=IR).  If you put a current 'I' through a resistor 'R' you will measure a voltage 'V' across that resistor.   For series circuits, the current is the same for all components in any circuit where all the components are strung together in series.  All the "sink" voltages must add up to all the "source" voltages.  Since you have a 5v source in series with a resistor in series with the diode, the voltage at the diode will be lower than the source voltage, and that voltage difference will appear across the resistor.  As an example, using your values:
 Vsource = 5V
 Vlaser = 3.2v
 Thus, Vresistor = 5-3.2 = 1.8v.
Now the question is how much current do you want in your laser diode?  Well, your 70-ohm resistor will give you:
 I = Vresistor /R = 1.8v/70ohms = ~26mA
26mA is not going to hurt your laser diode, so I would start with that.  If you want to be even more careful, you can put the 70-ohm and 15-ohm resistors in series so that total resistance is 70+15=85 ohms.  Then I = Vresistor / R = 1.8 / 85 = 21mA, but the diode might not even start!

If you decide you want 40mA, then you calculate the needed resistor this way:
 R = Vresistor /I =  1.8v / 0.040A = 45 ohms (a 47-ohm resistor from Radio Shack would be fine).
Or for 55mA,
 R = 1.8/0.055 = 33 ohms
Now if you only use the 15-ohm resistor, watch out!
 I = 1.8/15 = 120mA ~~ Probably would fry the laser diode!!

There's one thing I've been assuming here that isn't exactly true in reality: That the laser diode voltage is ALWAYS 3.2v.  It's not, but it changes a little bit with the current in the laser, but not much.  Once a laser or LED turns on, there is a very steep Current/Voltage slope, that is, any additional voltage will cause a huge increase in current.  That's why your UV LED burned out, because there was nothing there to limit the current!  Assuming the UV LED was supposed to operate at 4.0v, then the current in the LED was I=(5v-4v)/0 ohms = infinity amps!  (In reality, maybe 0.5-2A... way too much!!).  

Hope this helps and good luck!
 
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Thanks for your message cdanjo ! On my PSP recharger it is marked 5V --- 2A but the LED was connected before with 2x3V button cells batteries so amperage can kill diodes right ? So I guess I could attach the two resistors "en serie" for more security ? Thanks for your replies !
 




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