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ArcticMyst Security by Avery

First laser Volt/Amp question

Joined
Mar 29, 2014
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Hello I'm new to this forum, well this is my first post but I've read a lot on here. I'm very new to lasers in general as well as it was only Awhile ago I became interested in lasers. My first laser (and only laser as of now) is a 445nm 2W m140 laser diode with an X-drive and g-2 Lens I bought from DTR. I use a variable power supply that goes 0-18V and 0-3A which is perfect since the laser said it uses 6-8V 2A Max. When I turn it on and start increasesing voltage and amp it gets up to 5.6-6.5V at around 1.6A however when I turn voltage up the amps have gone down to 1.03 with 9.8V.. I don't know why this happens when I increase voltage I thought they go up simultaneously? Thanks!
 





oahu99

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Feb 23, 2014
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I'm not quite sure I understand your question, but I believe the current draw is dropping because the driver isn't designed for 9.8 volts in. Why not just use the 6 volts you said works?
Not sure if that answered your question or not.
 
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Mar 29, 2014
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Yeah I guess I should have actually asked my questions lol..
1. Why is the increase not proportional?
2. Will a higher voltage alone increase the power of the laser?
3. Is this better or worse for the diode, as it's taking only half the amps while still providing higher voltage?
 
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Mar 29, 2014
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Will that driver survive over 9 Volts?
I think something FAILED here.
HMike

No I don't think so, the laser doesn't seem to have a problem handling it and it seems to be the same brightness. Does the power come from the amps or voltage? (In this specific case I mean)
 
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May 30, 2011
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Wait a second, are you reading the voltage/current off the power supply or a multimeter in line with the diode? I think what is happening is you are reading it off the power supply, in which case it makes sense. If it is taking 1.6A at ~6.2V, this is proportional to 1.03 with 9.8V. In the end the driver will draw different current from your power supply (or batteries) depending on the voltage, and still give out the same constant current as expected, which is why there is no problem with your diode.
 

Things

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The X-Drive is a buck driver, in the simplest (and incorrect) terms, it "converts" the excess voltage into the current needed to drive the laser. So by turning up the voltage, the driver is pulling less current. This is how buck drivers work and generate minimal heat vs a linear driver. More info: Buck converter - Wikipedia, the free encyclopedia
 
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Mar 29, 2014
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Wait a second, are you reading the voltage/current off the power supply or a multimeter in line with the diode? I think what is happening is you are reading it off the power supply, in which case it makes sense. If it is taking 1.6A at ~6.2V, this is proportional to 1.03 with 9.8V. In the end the driver will draw different current from your power supply (or batteries) depending on the voltage, and still give out the same constant current as expected, which is why there is no problem with your diode.

Yes I am reading it from the power supply. So is the driver just going to prevent the diode from drawing too much no matter what? I know this is basic electricity knowledge, but why exactly is this proportional? And how do the proportion of volt to amps effect the total watt output?
 

oahu99

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Feb 23, 2014
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A watt is volts x amps.
Yes since it is a buck driver it puts out more current than it draws. However, it only puts out as many watts as it draws.
Hope that answers your question.
 
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