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Go Back   Laser Pointer Forums - Discuss Laser Pointers > Lasers > Drivers, Test-load, Power Supply, & Batteries



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Old 01-27-2011, 02:42 PM #1
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Default battery and regulator supply question

Using the LM317 regulator circuit-

Is a 9v battery supply to much voltage to the diode?

I see that the AA size 3.7v lithium ion batteries are a popular choice. What is the actual number on these batteries? If i were to say go to radio shack to get them?

If using the lithium batteries as an option, are these good for both low and higher power lasers?

When finding the appropriate power level for a diode on a test circuit, the current that is measured across the resistor on the test bed is the actual mW of the laser? For instance, if i am measuring. 05mA across the resistor then i can assume the laser diodes output is 5mW?

And lastly, at what diode power level is the LM317 circuit capable of supplying? 200mW? 400mW?

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Old 01-27-2011, 02:58 PM #2
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Default Re: battery and regulator supply question

From my experience, 9 volt batteries both supply too much voltage while at the same time not supplying enough current. As to the lithium-ion batteries, the most common choice is an 18650 battery, which is actually a big wider and taller than a AA battery. I've never seen them at a radio shack, so you would probably have to find it online. An 18650 will be good for pretty much everything but the highest powered handheld lasers.

For measuring actual power output of the diode, you cannot simply use the current. If you want to be exact, you will have to use a laser power meter, but if you don't have one of those, you can get close by looking at a power/current graph for the diode that you have.

Lastly, the LM317 circuit is capable of supplying up to 1.5 amps to the diode, which will usually be enough for most applications. If for some reason that is not enough, you can replace the LM317 with an LM350, which is good up to 3 amps.
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Old 01-27-2011, 02:59 PM #3
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Default Re: battery and regulator supply question

This REALLY depends on what diode you're driving.
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Old 01-27-2011, 03:24 PM #4
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Default Re: battery and regulator supply question

Wow that is really good information. The current project is a diode from a 20x dvd burner, and my future one will be for a 1 watt 405nM.
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Old 01-27-2011, 03:53 PM #5
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Default Re: battery and regulator supply question

I forgot to ask.....the acceptable voltage for laser diodes is 7-8 volts then? (3.7v x 2 = 7.4 volts) i guess the main question here is whether or not a 9v battery (with lm317 circuit) is gonna fry my diodes?
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Old 01-27-2011, 05:18 PM #6
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Default Re: battery and regulator supply question

Red Diodes take ~3v. 405/445 Diodes take ~4.5v. A 9v will not fry your diodes, but might just have enough current capacity to run your 20x diode, but the battery will not last long. You'll need a much higher current output for a 405/445 than a standard 9v can supply, so that you will want to go with some sort of lithium or rechargeable.
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Old 01-27-2011, 05:28 PM #7
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Default Re: battery and regulator supply question

Ok i think i am starting to catch on to it. Since i am using a lm317 circuit i will have a 1.25v drop, a red Ld will use 3v, so i would need a min of 4.5 volt input to power up the circuit and laser? If that is the case would 3 higher capacity AA rechargables work for 605nM and 4 high capacity AA for 405nM? I am aware of the 3.7v lithiums......i assume you would use two of those in either set up?
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Old 01-27-2011, 06:21 PM #8
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Default Re: battery and regulator supply question

Two rechargeable 3.7v (4.2v fully charged) will work for 650nm/405nm/445nm. The extra voltage will be converted to heat by the LM317, so You'll need to heat sink it some way.
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