Referrer to the following datasheet for the laser diode I'm using: http://www.lumex.com/pdf/OED-LDP65001E.pdf
If I use the following circuit at the link below to power the laser diodes what should I calculate Ipd to? From what I can see that would be 0.2 mA. Also, should/could I string 10 of these circuits and parallel? http://www.repairfaq.org/sam/laserdps.htm#dpslp3 On Sat, Aug 30, 2008 at 9:03 PM, John Griessen <[EMAIL PROTECTED]> wrote: > Mike Jarabek wrote: > > Robert Butts wrote: > >> Woe... > >> > >> I'm using ten of these and parallel. I WAS going to just use a 1 amp > >> 5 vdc power supply with a 2.8 V zener diode to adjust the voltage to > >> 2.2 V. I take it this is too simple. > >> > > I didn't mean to scare you. You can probably get away with putting a > > bunch of simple current limiter circuits on the diodes, like the ones in > > the dollar store pointers. The circuit looks pretty simple, but I never > > traced it out. The main advantage to a current control circuit is that > > it will allow you to control the brightness as the diodes age, > > especially if you use the PIN diode for power feedback. > > > > I have even seen people drive these like an LED with a series ballast > > resistor, but this does not protect the diodes from destruction as they > > age, or protect them from over current. > > > Right. When you care about replacement effort or price of the laser > diodes, or even output, you will need to control the current or i**2*R > power input according to temperature, and have it go down as temp rises. > The datasheet will define this relation. To choose the current level > to run at 25 deg C, you would measure an individual diode's light power > output with a power detector, or use a "safe level" below the max to > account for random variation. The datasheet may or may not tell you > about the variation in output power to current relation. Volts will > vary even more drastically with temperature rise, so if you have to work > from volts, you will need to change it to mostly depending on current > through a resistor and dropping power in those resistors. A current > controller will save you power, but add complexity and circuit cost. > > > John Griessen > > > _______________________________________________ > geda-user mailing list > geda-user@moria.seul.org > http://www.seul.org/cgi-bin/mailman/listinfo/geda-user >
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