Re: gEDA-user: Laser diode operation?

Would a string of these work for each laser? The Battery would be 5 vdc.

              Battery(+) o----------------------+--------+
                                                    |          |
                                                   _|_     __|__
                                              PD /_\     _\_/_ LD
                                                     |         |
                                               Ipd| |      Ild| |
                                                   v |        v |
                                                     |           |
                                              R1   |        |/ E
                                          +--/\/\--+---------| Q1 PNP
                                           |   1K            |\ C
                                          /                     |
                                   VR1 \<-+                 |
                            S1    10K / |                  |
                            Power      \ |                  |
                                 _|_      | |                  |
             Battery(-) o----o o-------+--+--------------+

In this case, the power output is determined by the equation:

             Vbatt - (Ipd  * (R1 + VR1)) = Vld + Vbe1

Or:

                           Vbatt - Vld - Vbe1
                    Ipd = --------------------
                                R1 + VR1

Where:

Vbatt = battery voltage under load.
Ipd = total photodiode current.
Vld = voltage across the laser diode.
Vbe1 = Base-emitter drop (.7 V) of Q1.

Since Ipd is proportional to optical power output, like LP-LD1 and LP-LD2 (above), brightness is dependent on battery voltage. In this case, it is a much more non-linear relationship as Vld and Vbe1 set a threshold of about 2 to 2.5 V below which there will be nothing and then output will increase based on Vbatt/(R1 + VR1). The circuit operates on 3 V but 4.5 V seems like the minimum to get any decent output.

On Sat, Aug 30, 2008 at 11:23 AM, Kai-Martin Knaak <kmk@xxxxxxxxxxxxxxx> wrote:

On Fri, 29 Aug 2008 23:15:03 -0400, Robert Butts wrote:

> 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.

It is. You need to protect the diodes from over current rather than over
voltage. Laser diodes will die in micro seconds, when exposed to over
current. A proven design is to use an adjustable voltage regulators as a
current limiter. See the bottom of page 18 of the data sheet by national
for this rather simple circuit:
http://cache.national.com/ds/LM/LM117.pdf
Each laser diode needs its own protection.

---<(kaimartin)>---
--
Kai-Martin Knaak
http://lilalaser.de/blog

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