Timeline for What's the difference between a laser diode and ordinary LED?
Current License: CC BY-SA 3.0
7 events
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Oct 4 at 22:09 | comment | added | D.A.S. | I tried to find corroboration to my experience with -ve incremental resistance and knowing laser controllers tend to ramp up slow while specs are based on a 25'C heatsink for lumen/mA, I conclude the sharp current slope with a NTC operating at much higher Tj and much lower MTBF than 50kh that the fig 7 here shows an irregular slope near/before max rated power. The lag/lead loop compensation may have amplified this behavior I saw. newport.com.cn/medias/sys_master/images/images/hc9/hd1/… | |
Oct 4 at 19:38 | comment | added | MicroservicesOnDDD | Thanks. Very interesting! How about regulating the input power? The Joule-Thief does this. (Wikipedia page here). The 1K resistor replaced with a potentiometer, or a digipot, and given a constant input voltage, will produce a constant output power. (I'm not sure that I understand the magnitude of the NTC behavior.) | |
Oct 3 at 21:44 | comment | added | D.A.S. | I agree the details of his answer are interesting. The important difference that protects the LASER is that the lasing has a low negative incremental resistance nonlinear behaviour due the NTC unlike the +ve Rs of the diode which has low NTC like regular diodes, so it is critical to include the PD optical current feedback to prevent burnout in milliseconds if otherwise unregulated. This regulates the optical power better than a current limiter. | |
Oct 3 at 18:20 | comment | added | MicroservicesOnDDD | Thank you, @VoltageSpike, for your Excellent answer! | |
Jun 13, 2020 at 13:21 | comment | added | D.A.S. | Ask Voltage Spike, he can explain it better. I know that bulk Rs causes heat losses Pd=I^2*ESR which increase with current and above some threshold LED's lumens/watt is maximum and then declines with rising current. @MicroservicesOnDDD | |
Jun 12, 2020 at 22:47 | comment | added | MicroservicesOnDDD | Why the significantly lower efficiency? I would imagine part of it is due to a certain amount of loss with each reflection, and that with so many reflections, it keeps multiplying. But I'd rather know than guess. | |
Jan 16, 2017 at 1:01 | history | answered | D.A.S. | CC BY-SA 3.0 |