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09-16-2014, 02:53 PM #17
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AgelessAnarchy
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Re: Calculating out put on a M140

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09-16-2014, 07:59 PM #18
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KrowBar
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Re: Calculating out put on a M140

Quote:
 I am a electronics engineer
Quote:
 Ok I have a question about DTRs OP on the m140 using a g2. Looking at the pics you can see a wattage increase that's fairly in sink with the mA increase I get that. Bit my question is this. There is also a voltage increase. Is that the diode drawing more voltage as the wats go up or were was the voltage manually increased

No offense Ageless, but you don't exactly sound like an electronics engineer. I've never heard someone familiar with how electricity works refer to something drawing or pulling a voltage. Usually we refer to it as "dropping" voltage. This is simply the potential difference between two points in a circuit. As the diode draws more current, the potential difference across it increases, just like the potential difference between the ends of a wire through which current is flowing (although the relationship is not linear in the case of the diode.) In any event, if you have a device which can supply a specified current while maintaining a variable voltage difference (a constant current source) then it should be able to power your diode in a controlled manner. You can simply simultaneously measure the current draw (should be doing this already, otherwise I'm not sure how you would be specifying the current), and the voltage drop (put a multimeter across your diode) to determine the power draw. If you wish to know the optical power output, then you can use a LPM as mentioned, or make a rough estimate based on power curves reported by others using a similar (same model) diode. The electrical power into the diode might be important for estimating run time or determining which batteries are required, while knowing the output optical power is only helpful if your specific use of the laser requires it.

Quote:
 The m140 should act no different than the class two 2s I've been using the demand around 3.0-3.5 @ around 700mA

Also, (not to say that it can't exist), but I have not heard of any laser diode that has less than 0.05% efficiency. A class 2 laser would output less than 1mw of optical power, but you are saying that yours draw around 2W of electrical power. Somethin just don't seem right there.

09-17-2014, 12:26 AM #19
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AgelessAnarchy
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Re: Calculating out put on a M140

It really doesn't matter what you think, I know the degrees that I have. Besides I just finished up my project this evening so it really doesn't matter anymore. I also never said I had a class 2 at two watts. I said that it drew voltage in the same fashion or close enough. So thanks for everyone that had valid input. Not useless nitpicking like this.

09-17-2014, 02:08 PM #20
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KrowBar
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Re: Calculating out put on a M140

Quote:
 Not useless nitpicking like this
No need to get testy. I'm not picking on you or attacking your background - thus my "no offense" comment. You asked for help understanding something. Your explanation of your background suggested that you would be familiar with certain concepts, yet your questions suggested maybe a refresher was in order. While it may not matter what I think, it does matter what people in general think when you are soliciting their help. If the impression they get of you conflicts with how you are trying to present yourself you are less likely to get the help you want. In this case other responses may have given you the information you were after in spite of this, and that's fine, but I'd hope that you could use the additional info to improve future communication as opposed to lashing out at it as useless nitpicking.

Quote:
 I also never said I had a class 2 at two watts
Quote:
 the class two 2s I've been using the demand around 3.0-3.5 @ around 700mA
I think if you try multiplying 3 volts by 700 mA, you will find that indeed you did say you have a class 2 drawing over 2 watts. As I said, this seems extremely high for a laser that would be putting out less than 1 mW of optical power, so you might want to either double check your measurements, or consider the possibility that the laser is much more powerful than you realized (also acknowledging that it could in fact be a very inefficient diode that converts 99.95% of what you input into heat)

Let's not get off to a bad start here. I like to be of help to those that can use it, be it related to laser building, conceptual understanding, social skills or otherwise. Please don't stomp off all upset just because I might happen to sound condescending from time to time - that would be my own lack of social skills at work.

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