Whats with mA vs mW stats?

[dnar]

`I often see LD graphs charting current vs voltage which is ok. Then I see the mystical mA per mW spec stated...

This figure is totally meaningless without considering Vfwd!

What is more meaningful in terms of efficiency and thermal output is mW (optical output) per mW (power input).

mW input is easily achieved as mA * V input.

Lets compare apples with apples.

For example, the new red 635 diodes I have seen stated 1W output per 1A input "like 445s". Well actually it's nothing like the 445's when you consider Vfwd and therefore power input...

/Off my soapbox.

Maybe I am just grumpy because it's the end of the week and I have not had a day off in 18 days...

[LaNeK779]

Maybe it is as simple as "if you feed it 300 ma, you can expect 250mw out the other end".
No doubt technically you are right, but imagine a guy without an LPM, and only a DMM at hand. I find it usefull.

[norty303]

I thought mA vs mW was a test of linearity assuming that your driver delivers linear current increase in relation to the modulation voltage.

[steve-o]

I think Dnar was referring to the laser diode's electrical characteristics/ properties.
Let's say that you have laser diode "A" which will output 250 mW of optical power with 2.8 volts input (to the diode) and the resulting current draw (by the diode) is 500 mA. Then there is diode "B" which outputs 250 mW with 3.1 V at 450 mA. The power input to the diodes are almost exactly the same for both diodes (E*I) however it would appear that diode B is more efficient by looking at current only instead of voltage times current (power.)

Sorry to hear about your excessive working Dnar.. hope you get a day off soon :]

[dnar]

I think Dnar was referring to the laser diode's electrical characteristics/ properties.
Let's say that you have laser diode "A" which will output 250 mW of optical power with 2.8 volts input (to the diode) and the resulting current draw (by the diode) is 500 mA. Then there is diode "B" which outputs 250 mW with 3.1 V at 450 mA. The power input to the diodes are almost exactly the same for both diodes (E*I) however it would appear that diode B is more efficient by looking at current only instead of voltage times current (power.)

Sorry to hear about your excessive working Dnar.. hope you get a day off soon :]

Thats it.

Yeah, 3 days off now! Packing the laser in the car and heading far far away to a mates place. Lots of laser action and drinking Jack till out heads explode!