The optimum amount of light to be put on the detector (for the
feedback) depends on the following factors:
- You want to use as little as possible if you are short of output
power.
- For the detector, you ideally want something in the order of 1 mW so
as to get the best signal-to-noise ratio of your error signal.
This may require some compromise, unless your laser anyway produces way
more than 1 mW.
The feedback loop must be carefully designed if you want to achieve
optimum noise suppression. The feedback amplifier could of PID type,
ideally with adjustable P, I and I parameters. You may find a
commercial device to do this job.
If your feedback is supposed to use the drive current of the pump diode
as a control, you will also have to take into account the low-pass
filtering which is introduced by the finite upper-state lifetime of
your gain medium. For erbium, e.g., this may limit your control
bandwidth to something in the order of a few hundred Hertz.
---------------------------
Dr. Rüdiger Paschotta
RP Photonics Consulting GmbH
www.rp-photonics.com
Sam Goldwasser - 29 Dec 2004 17:21 GMT
> The optimum amount of light to be put on the detector (for the
> feedback) depends on the following factors:
[quoted text clipped - 13 lines]
> your gain medium. For erbium, e.g., this may limit your control
> bandwidth to something in the order of a few hundred Hertz.
We do this with Nd:YVO4 microchip lasers. A photodiode gets about 10
percent of the beam power and the feedback loop varies the pump diode
current. Although a PID scheme may be optimal, a very simple circuit
will get something like 20 dB of reduction in noise.
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> ---------------------------
> Dr. Rüdiger Paschotta
> RP Photonics Consulting GmbH
> www.rp-photonics.com