Re: gEDA-user: Laser diode operation?

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 <john@xxxxxxxxxxxxxx> 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

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