Help explain a conceptual roadblock?

[rhd]

Would someone be willing to help me clear up a bit of mental fuziness over some conceptual stuff? I'm having trouble understanding exactly the purpose of a driver when it somes to powering small laser diodes (like ones in a TO18 package)

From what I know of electronics, it is generally the case that a load (like a light bulb) needs to be supplied the correct voltage, but it will only draw the ammount of current it can utilize.

This seems to be the wrong logic to apply to laser diodes however. It seems that laser drivers mainly focus on limiting the current, and don't explicitly specify a voltage range. This leaves a couple (probably related) gaps in my understanding:

1) Why does current need to be regulated at all? Wouldn't a diode simply utilize whatever it could handle - like a light bulb?

2) If you had a 3.7v battery (like an 18650), and a diode that could run on 3.7v, why would ANYTHING need to be boosted/regulated about that setup. Why could one not be connected directly to the other.

If I've missed an obvious sticky, apologies. I know this is a very basic question.

 

[Mohrenberg]

From what i understand, LED's and laser diodes's output are determined by the current and not the voltage. supply a diode with 5v and 1000mA or 7v 1000mA would bring about the same results.
Also IIRC, as a diode heats up (which they all do) it draws more current. So if it's unregulated you get a kind of runaway effect. It keeps heating up and drawing more current until it kills itself (i think this is true....because it's in my head )
I'll wait for one of these electrical geniuses to chime in to give a more thorough answer.

 

[Benm]

In terms of electronics, (laser)diodes are not linear resistors. Their current/voltage curve jumps up very steeply, as is the case with LED's to a somewhat lesser degree.

You could, for example, see values like these:

3.60V - 10 mA
3.70V - 100 mA
3.80V - 1000 mA

Lets say your device can handle 200 mA of current: You would have to feed it something like 3.728 volts exactly to run at that current. 10 mV below and you dont get full potential, 10 mV above and its fried.

So you may think: "well, as long as i get a precise voltage source, i'm fine!". Unfortunately the current/voltage curve also depends on temperature, and in such a way that it will draw -more- current at a given voltage with an increase in temperature. This will lead to a runaway increase in current and temperature until the laser diode is destroyed.

 

[rhd]

Thank you both very much.

I appreciate the explanation - it makes sense to me now!

 

[bobhaha]

Also they serve as voltage "dampeners" in a way. They smooth out voltage ripples and cut out the "peak" voltage supplied by the battery.

If you connect a battery to a load it will "peak" giving a massive serge of voltage that will soon settle down. This can kill a diode quite easily. So capacitors are used to stop that.

Also working off what Benm said... in the reverse, when you use a battery the potential voltage lowers with use. Therefore to stay at the same current level drivers are needed to boost the current in exchange for higher voltage input. Which is used to keep the diode at the nominated current range.

 

[JLSE]

This will lead to a runaway increase in current and temperature until the laser diode is destroyed.

Have any of us actually witnessed this? or is it just theory concerning LD's?

The perfect example that comes to mind is 1n4001 diodes..
Yes they will draw more as they heat, but this is ignoring the fact
that our LD's are mounted in metal, and usually in a good heatsink.

The only difference I see is if an LD is regulated and you fall
asleep with your laser on, it wll get hot, but not draw more current,
and may die from the heat.

Repeat the same scenario with an unregulated current, or voltage
regulator like the 7805, and increasing current really only becomes
an issue if you run the laser hot.

If we were using LD's with no mount and no H/S I think it would
be more of an issue.

Even if you try to run a LOC in a pen with 2x AAA's @ 3v, your
lucky if you see 200mW. The current drops as the Batt's struggle to
keep up the voltage.

Even with little heatsinking in a pen, the laser just drops out along with the
voltage and current and of course the mW goes down.

I have a pen that I made in 2007 that is a LOC and it performs the same
as the day I made it. All that governs the power from the batt's is a cap..
No resistor, as the current the LD sees is already too low..

Ive used good batteries in this laser from Duracell to the cheap AAA's that
come with DX products.

 

[LSRFAQ]

I have a pen that I made in 2007 that is a LOC and it performs the same
as the day I made it. All that governs the power from the batt's is a cap..
No resistor, as the current the LD sees is already too low..

Ive used good batteries in this laser from Duracell to the cheap AAA's that
come with DX products.[/QUOTE]

In your case the limiting factor is the internal resistance of the battery, and that is the only thing limiting it. Placing the same diode on a bench supply at the same voltage will explode the diode in about a millisecond, because the lab supply has nearly zero internal impedance.

There are a few diodes that run in ranges that are just about matched to a two or three cell battery, otherwise all the Kip-k&y resistor limited videos would result in scrap metal.

See this thread:

http://laserpointerforums.com/f67/h...es-how-measure-internal-resistance-57576.html

Steve

 

[Benm]

I havent seen any laser diode destroyed by the thermal runaway effect, but that is mainly because i am aware of it.

If you want proof, just send me any laser diode and i will feed it constant voltage regardless of current use. I will make sure that the supplied voltage is constant and 'hard' - so it remains exactly the same regardless of what the laser diode does.

This could work out for sub-optimal currents, but if you run the diode at or near its limits, COD is a certain outcome.