Sarge wrote:it's really the re-reading that gets you, not the actual effort the laser gives per read
Rereading a disc because of read errors causes more wear and tear to the mechanical parts that move the laser, rather than to the laser diode itself. The increased effort (extra current) to emit light is what wears out the diode. The laser diode does so because the data it is reading requires more light due to lesser reflectivity, a common problem with cheap dyes burned too quickly. I say this based on papers I've read.
I was trying to find a free PDF version of
"Optimum Reflectivity Design of Laser Diode Facets and Recording Medium for an Integrated Flying Optical Head" a paper by Hiroo Ukita and Yoshitada Katagiri, written in 1993 for the Japan Society of Applied Physics. I can't find it online free anymore. I remember that paper did a lot to shape my views on diode damage via lesser reflectivity on low quality substrate.
But here's some info:
https://assets.newport.com/webDocuments ... ion_IX.pdf "time to failure is usually defined as the time at which the forward current has increased by 20% to 50% of its initial value""For this analysis we have defined end-of-life as a 20% rise in laser drive current (Iop) over the initial value. Using this definition, the aging trend for
each laser diode was extrapolated to end-of-life using a linear regression of laser drive current vs. aging time.""Longer lifetimes can be achieved by designing drive circuitry that can accommodate a larger increase in current than the 20% increase that was used to define end-of-life in this analysis"Sarge wrote:But you're right, the potentiometer will adjust the power of the laser, which can temporarily resuscitate a system, but will also make it wear out that much faster.
That resuscitation is a byproduct of the diode now having enough current to output enough light to achieve the reflectivity it needs due to low quality reflective surfaces (i.e. cheap dye on a cheap disc burned too quickly). However because the current has been amped higher, it does indeed wear out the diode that much more quickly.
So that's just what I believe based on research I've done over the years. But don't take me as saying you're all wrong, just explaining my stance!