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ArcticMyst Security by Avery

Just a couple general questions about building lasers

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Mar 5, 2012
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So I've built a couple of lasers now. Ive got about 10 of 1.8a using the m140 445 diodes. Also have a 2.2a powered laser using the 9mm 445 diodes.

My main question is how exactly do you measure all the measurements on a laser.

My laser is powered by a 1.8a driver. Uses two 3.7v batteries giving it a total of 7.4v. Now in order to get the Watts of the circuit you would multiple VxA which would be 13.32w.

Maybe im getting it mixed up. Do i have to use the mA off the battery to get the correct calculations?
so if it was a 1200mA 3.7v battery, That would give me 8.88W.

Does that sound about right?

But when I go back to the information when I bought the diode, it says it is a 1.8W 445nw diode. So what is the 1.8w mean exactly? What the output is? What the diode can handle?

Just trying to understand what the information on these diodes mean exactly and how to predict what they can actually handle and what not.

Thanks
 





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My laser is powered by a 1.8a driver. Uses two 3.7v batteries giving it a total of 7.4v. Now in order to get the Watts of the circuit you would multiple VxA which would be 13.32w.

No that is meaningless and has nothing to do with the power the laser is using or its output, the current the driver is providing to the diode will be different than what it's drawing from the batteries, the voltage to the diode will also be different, the driver is constant current and will adjust the voltage as needed, also those batteries will charge to 4.2V but drop immediately under load to about 4V and drop more from there as they are drained.

To determine how much power your laser is using you must measure current from the batteries, not from the driver, measure at the tailcap with your DMM.

Maybe im getting it mixed up. Do i have to use the mA off the battery to get the correct calculations?
so if it was a 1200mA 3.7v battery, That would give me 8.88W.

Does that sound about right?

No it certainly does not. Forget about the mAh printed on the battery, that just gives you an idea how long the battery will last before it needs recharging.

But when I go back to the information when I bought the diode, it says it is a 1.8W 445nw diode. So what is the 1.8w mean exactly? What the output is? What the diode can handle?

Just trying to understand what the information on these diodes mean exactly and how to predict what they can actually handle and what not.

Thanks

The 1.8W is what you can expect the typical output to be, the only way to tell exactly is with an LPM. Your laser will use several times the power of the output of the laser, much power will be dissipated as heat, there is power lost in the driver, in the diode, and in the optics.

Alan
 
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Encap

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So I've built a couple of lasers now. Ive got about 10 of 1.8a using the m140 445 diodes. Also have a 2.2a powered laser using the 9mm 445 diodes.

My main question is how exactly do you measure all the measurements on a laser. My laser is powered by a 1.8a driver.

But when I go back to the information when I bought the diode, it says it is a 1.8W 445nw diode. So what is the 1.8w mean exactly? What the output is? What the diode can handle?

Just trying to understand what the information on these diodes mean exactly and how to predict what they can actually handle and what not.

Thanks

1.8 W of laser light output is just an average of what can be expected from an M140 diode when measured by a Laser Power Meter.

Have a look at DTR's M140 web page---there is a lot of information there and if you scroll down you can see the output of a typical test M140 at different current and voltage settings https://sites.google.com/site/dtrlpf/home/diodes/445nm-h1650-diodes Looks like his example is giving an output of 1931 mW at 1,8 A 4.8 volts . You should be getting an output of 1800 mW at 1.6 A supplied from the driver so at 1.8A you can expect slightly higher output.

The batteries have nothing to do with figuring the lasers output assuming you have enough battery power to feed the driver---it is all about the what they driver is sending to the diode.
 
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APEX1

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Also depends on optics. If your using a g series lens or even the s1 it will give a higher output over the 3element lens. You are assuming diodes are 100% efficient
 
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Differentiate between 'Optical Power Output' ( what you are really interested in) and 'Electrical Power Dissipation' (what should be considered in build design). Ohm's law will give you power dissipation from the current draw on the batt.s As already stated, optical power is related to diode efficiency, which will change from diode to diode (due to manufacturing). It can be roughly calculated (with a bit of work), but should really be metered.
 




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