gEDA-user: Laser diode operation?

Rob Butts r.butts2 at
Thu Sep 4 00:07:02 EDT 2008

Referrer to the following datasheet for the laser diode I'm using:

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?
On Sat, Aug 30, 2008 at 9:03 PM, John Griessen <john at> 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 at
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <>

More information about the geda-user mailing list