I made a bulk purchase of red laser diodes off Ebay. There are three bags together unmarked. 25 of them are 3.3V., another 25 are 4.5V and the other 25 are 5V. But like I say they're not labelled so I don't know what spec correpsonds with which batch. They're all in brass hosts with adjustable lenses. Is there a simple way to tell which is which - establish safe working current and voltage for each batch? I don't mind blowing a few in the process if necessary as I can afford to lose some. thanks.
You might try starting with a 12 volt supply and a resistor to limit the current. Use a resistor of say 2000 ohms and lower it as needed. Then measure the voltage across the diodes. That should tell you the voltage ratings of the diodes.
If you know the current rating of the diodes you can calculate the resistance by subtracting the voltage rating of the diode from the power supply and applying Ohm's law.
Noramlly the power supply voltage does not make much difference if over the voltage of the dioide as the diodes are a current device.
OK, so if I gradually increase the current through the laser diode to the point where the voltage across it increases no further, I've determined the voltage rating of that diode?
Next question is, how far beyond that point is it safe to go to get the greatest combination of light output and longevity?
That is about it. The LEDs act similar to a zener diode.
If you put say a 1000 ohm resistor in series with one and use a varitable DC supply of 12 volts or more you can see this. YOu may have to use a lower value of resistor if the LED requires a lot of current.
Measure the current in the circuit and the voltage across the diode. Start with zero volts. As you raise the power supply voltage at first almost no current will be drawn, less than a miliamp in most cases. Then it will almost jump to a milliamp or more and the led will start to produce light. You may not see the light at this time. As you continue to raise the power supply voltage the current will start to go up very fast and the voltage across the diode may go up about half a volt or so. The more current the more light up to the point it burns out. Just keep an eye on the current so as to stay with in the current rating of the diode.
This will tell you about what the voltage of the diode is. Then you can sort the ones you have by the voltage range you gave.
Without any kind of paper work it would be difficult to say how much current they can take. The supply voltage does not make any difference. You just add more resistance as to keep the current to a safe level.
That is the next obstacle, though: I don't have any current ratings for these diodes.
Got that bit sorted out, anyway!
The art seems to be in getting the most light output sustainably. It would be great if there was some sort of formula such as "'burn-out current' minus 10%" or such like, but life's not that simple I guess. :-/
On Thu, 14 Apr 2016 13:47:34 -0400, Ralph Mowery wrote: [...]
So, just to get this clear, there is basically no difference in the dynamic behaviour of a laser diode (wrt current and voltage) and an ordinary run-of-the-mill rectifying diode such as a 1N4001 for example? The laser diode is very static sensitive; the ordinary diode is not. But aside from that.... nothing?
I would say more like a zener diode than the rectifying diode. Not much effect before the device goes into zener type action and it starts to draw lots more current.
I'm certainly no expert in this area, Ralf, but I don't quite see how the zener analogy works here. These laser diodes require forward voltage to operate and (this bit is the main difference from 1N4*** series diodes) that forward voltage is typically much higher than the 0.6 - 0.7V - less for Schottkys - anything up to about 5V in fact. But the 'knee' shape you would see with a curve tracer will be the same; just further to the right on the x axis than with a conventional diode.
Those voltage numbers don't make much sense, unless there is some circuitry driving the diodes already. Are these bare diodes?
The best analogy for a laser diode(LD) is an LED. The forward voltage is mostly set by the band gap.. color of light. There should still be an increase in the voltage as you increase the current (~60 mV for a decade increase in current... (I think))
I'm guess you've got diode with a built in controller (in the brass case.) Thye probably need ~10's of mA to lase. I'd hook one up to a power supply with a bit of resistance to limit current. (say a volt at 10 mA.. 100 ohms or so.) Turn up the voltage and see what you get.
Yes, I believe so. The only other component is what looks like an SMD capacitor across the diode leads just before it goes into the brass housing. I guess this is to iron out any voltage spikes?
That's what I'm seeing under test, actually. And despite the description, all the diodes I@ve checked so far go into forward conduction at 3.3V.
I'm pretty sure these ones have no controller. Another batch from another supplier just arrived. I'll check them tomorrow...
No, it's a cap (unless it's a diode with a junction capacitance of 10nF!)
There is, but the current starts to increase hugely at 3.3V but the voltage then seems to stick around that level.
You owe me a beer. There's nothing but a lens in there. :) It's hardly surprising, though. These things are like DIRT cheap so I wouldn't expect any driver.
OK, it is Friday. Can you stop by for Lunch and we'll grab a beer and a burger?
The numbers don't make sense... but I don't know cheap diode lasers either. If it's a bare diode laser then you if you drive it with a current source. (a resistor and voltage source would work.) then you should see the laser output increase linearly with current.... I was just measuring the power out of my diode laser. Here's a crappy graph.
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The good news is I can get about 30 mW which is enough to make a MOT with. (MOT- Magneto-optical Trap.)
Maybe we're at cross-purposes here. I was discussing voltage across the laser diode vs. the current through it. Not sure at all what you've graphed a photodiode's response into the equation here?
Yeah I'm saying the voltage is not relevant to the light output. (if it's bare LD) (The bottom (X) axis is the LD current, the horizontal (Y) axis is the current from a photodiode looking at the laser. (about 0.6A/W so 20 mA is ~30 mW.) (It just happened to be something I was doing today.)
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