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Old 06-16-2015, 11:30 AM #1
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Default Hi, sorry for my noob question

Hi everyone. I'm thinking in make a tiny red laser from a 10440 flashlight host, LPC-826 diode and linear driver, but I'm thinking that the voltage may fall short. Anyone knows the voltage drop of a LM317? I have a few and I'm considering use how driver


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Old 06-16-2015, 11:34 AM #2
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Default Re: Hi, sorry for my noob question

It is around 3v
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Old 06-16-2015, 11:41 AM #3
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Default Re: Hi, sorry for my noob question

If I power the lm317 with 3.7V I will get only 0.7v??
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Old 06-16-2015, 03:03 PM #4
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Default Re: Hi, sorry for my noob question

Around there. To power red, you will need around 5.5v
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Old 06-16-2015, 03:30 PM #5
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Default Re: Hi, sorry for my noob question

thanks for the info, I will use a host of two batteries

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Old 06-16-2015, 05:28 PM #6
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Default Re: Hi, sorry for my noob question

If you run the LM317 in constant current you have the drop of the current set resistor at 1.25 Volts plus the 2-3V drop over the regulator plus the laser diodes voltage drop .
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Old 11-04-2015, 10:09 AM #7
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Default Re: Hi, sorry for my noob question

It may above 3v ,
you can upload this picture to google and it will return you the similar product and then you can click for more info.
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Old 11-05-2015, 12:55 AM #8
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Default Re: Hi, sorry for my noob question

There are low drop out regulators that work similarly to the LM317, but you will still lose the 1.25 volts across the sense resistor regardless. You will end up needing 2 batteries in series with this approach.

Running a linear driver to power a red off a single cell is quite a challenge. It can be done, but probably not reliably with commonly available components. You'd either need to work with a low-vgs-on mosfet or a dedicated chip like the amc 3175. The latter has a fixed current of 350 mA or so, which might be a bit much for something small.
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