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Post your mA and mW

Groover

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Feb 15, 2010
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Casio A130 diode
Micro BoostDrive At 1005mA
Output is 667mW with aixiz 405nm glass
 



drummerdimitri

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Oct 15, 2007
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I have a noobs' question for you pros: How can I approximate their laser's output by measuring the potential difference across the positive and negative leads of the microboost (13.03V) with a battery voltage of 4.14 V? I don't know how to measure the diode's current draw, but I know that the current draw from the battery when at 4.2 V is around 1.5 A. What can this tell me about the laser's output? BTW i used a A130 diode for this test which is currently damaged and will be replaced by a A140 diode soon...
Thanks!
 

Benm

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I dont think it will be easy to come up with a conversion - the driver isnt 100% efficient, so its hard to tell where the electrical power ends up.
 

qumefox

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The only accurate way to find optical power is to use a LPM plain and simple. To do it by the way your asking, you'd need to know the exact conversion efficiency of both your driver and your laser diode. You can find out what the efficiency of your driver is with a DMM, a good stable bench power supply, and some math, however finding the efficiency of your diode would still require an LPM.. and if you had an LPM, you wouldn't be having to go through all this trouble to begin with.
 

ludwigfan

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i always tought it was dangerous to use 2 different types of lithium batteries in series?
Very dangerous. I wouldn't do it myself. I was thinking the same thing when I read that. He has a 3.7V lithium ion and a 1.5V lithium battery in series. Guess what happens when one battery runs low while the other one still has plenty of charge left?

The low battery gets charged with reverse polarity. Don't believe me draw a diagram of the entire circuit and follow the currents with one charged and one discharged battery.

Back in the Ni-Cd days sometimes a Ni-Cd cell would get reverse polarity. This was the cause. Of course Ni-Cds don't burst into flames unlike Li-Ions sometimes do.
 




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