What's the term for this?

Say I have a laser diode rated for continuous duty at a Vf=4.5V @ 500mA but I want to push it to a higher power but at a lower duty cycle. What's the fancy term for that?

Reply to
Chris
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Duty cycle?

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Rick
Reply to
rickman

Ash.

Reply to
Clifford Heath

Pulsing.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

No, that's not the thing here. It's something similar to "de-rating" if that makes any sense?

Reply to
Chris

Pulse mode.

Your putting in the same amount of Joules over the same period as continuous duty mode.

It doesnot sound like you can use the other terms such as Pulse Pumping or mode-locking in your application.

Cheers

Reply to
Martin Riddle

Why do you want to do that?

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

To help the economy by blowing up lasers, of course. I expect he'll discover why that is a much less good idea than it seems.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

That's what I was thinking.

Reply to
jurb6006

A different take on the "broken windows" theory, I suppose.

Reply to
krw

At the risk of net-cop Phil Hobbs overhearing and reporting me to the Feds, I want to know how much I can exceed the rated constant duty V & A if I operate the diode at a much lower duty cycle (say 15%). Is there a formula for this sort of thing?

Reply to
Chris

Don't mind Phil. He's just one of these naturally nosy types who likes to interfere in everyone's business. Just be thankful you don't live next door to him. ;-)

Reply to
Julian Barnes
[about pulsing a laser diode]

Lots of 'em, but they use info you won't find in the spec sheet. I've seen wires blow up at modest overcurrent (not melt, the OTHER failure mode) in a short pulse. Before you get a workable formula, you'll see it, too.

Reply to
whit3rd

It's much akin to secondary breakdown - localized heating.

Reply to
krw

I was inspecting one under the stereoscope earlier today - admittedly one of the dirt-cheap Chinese 5mW ones, but those bond wires.... good for electrons in single file only, I'd say.

Reply to
Cursitor Doom

'taint bond wires that set the limit.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

It's the facet on the chip, right?

Reply to
tom

mA

In principle, what you want to know is the thermal mass of the active junct ion, and how hot you can get it before it stops being an active junction.

The problem with laser diodes is that the active junction is very small, an d you can blow it up in microseconds.

Phil could probably quantify that statement for particular diodes - he's be en using laser diodes for quite a while.

As was mentioned in a recent thread, it's well known that it's remarkably e asy to blow up a laser diode with a brief episode of above specification cu rrent. In principle the absolute maximum current rating is what you are loo king for, and a well specified laser diode would tell you how long that abs olute maximum current could be sustained.

A quick google search gave me this

formatting link

which just specifies the absolute maximum current as 45mA and tells you not hing more.

You could use the lasing wavelength as a way of measuring the junction temp erature, and get the thermal mass of the junction by watching how it's temp erature changed with time, but the time constants involved are going to be small - think microseconds - so the measuring set-up might be interesting.

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Bill Sloman, Sydney
Reply to
bill.sloman

Do mind Phil Hobbs. He know more about laser diodes than anybody else who posts here. If you did live next door to him, you might be able to learn a lot.

Julian Barnes would be a less attractive neighbour - he does have a tendency to sound off about stuff he doesn't know much about.

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Bill Sloman, Sydney
Reply to
bill.sloman

Laser diodes have two failure modes. One relates to thermal heating, which would allow for short duty cycle high currents. The second concerns maximum lasing light output. That one can be a killer, and it's why many laser diodes have three pins to provide for a built-in photo-detector.

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 Thanks, 
    - Win
Reply to
Winfield Hill

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