We all know that YAG for diode pumping is usually doped at around 0.5% Nd.
What happens when one diode pumps a rod intended for CW lamp pumping? Obviously, there will be power loss, but how much?
Why do rods for diode pumping have lower doping?
We all know that YAG for diode pumping is usually doped at around 0.5% Nd.
What happens when one diode pumps a rod intended for CW lamp pumping? Obviously, there will be power loss, but how much?
Why do rods for diode pumping have lower doping?
This is the second time you have asked this question.
I mean this to challenge you, but for a engineering student, you sure seem very lasy. (pun intended) Old saying, "Get thee to a Library!"
For question one, I could be sarcastic and say "Why do you not get a copy of Silfvast and do the simple algebra for the gain equation?"
For question two, I could be sarcastic and say "Why do you not get a copy of Koechner and read it"
Go look at the ratio of adsorption for a purely 808 pump band vs a broad spectrum krypton lamp, and that will explain much. How much?, the math is left as a exercise for the student.
See page 263 of Silfvast, Chapter 9 Laser Pumping Requirements and Techniques.
ISBN 0-521-55617-1 Laser Fundamentals, William T Silfvast. I prefer the newer version with the whitelight HeCad spectrum on the cover. Silfvast invented the HeCad.
For question two, when you calculate the answer for question one, "Apparent will be Much" Yoda say..
Especially if you look at Silfvast's example 11.5 "Optimization of Output coupling for a Laser Cavity" on page 334, Chapter 11.
Kochener, Solid State Laser Engineering, in the more modern editions, directly answers this question
If your going to "Gain" posts spamming on this topic without researching, you will hit a point of "Saturation". There is more hint in this last sentence then you deserve.
Steve
With too high doping, the pump light is absorbed too soon, it's not pumping uniformly the rod, it will overheat the smaller pump abortion area and possibly crack the rod.