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FrozenGate by Avery

445nm Labby, Over-Current?

That's interesting. Why is it that I am able to get .94A out of my wall-wart powered diode, though? It's supposed to be limited to 500mA, but it seems to be going over....

Am I measuring correctly? I just plugged it in, turned off the switch, and connected my multimeter to the two leads of the switch to complete the circuit and measure the current across it. Is that a good way?

And how can I test how much current a battery is able to supply?

By the way, I would only need around 5 or so D cells =p Would still be a ridiculous amount, and that's why, once I get money, I am going to get a couple of 18650s.
 
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Um, if your measuring .94A at the power switch going into the driver, that's how much current the driver is using, not how much the diode is getting. To measure how much the diode is getting you really need to make a test load to attach to the driver in place of your diode and measure the voltage drop across the 1 ohm shut in the test load.

And the rating on the wall wart is the maximum the manufacturer recommends. Not the absolute maximum it'll put out, though usually most are fused internally at some value not an extreme amount higher than the rated output.. but not always. Just depends on how cheap it was made.
 
Oh... alright. Hmm. What if I put the multimeter between the driver and the diode? I have a test-load, but I don't have a powerful enough resistor, so they keep on blowing >.> All my 1 Ohms have blown... hehehe. That's a shame though - it could be running much weaker than I thought it was *still*?!

Okay, so I just wired a resistor I just salvaged from an old computer PSU that measured around .9 Ohms, and I got a reading of 716mV from my laser. That seems about right - around 800mA to the laser, around 140mA to the driver, eh?
 
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How much of that .9 Ohms was your meter leads? What resistance does your DMM read with the leads shorted together? You have to subtract that from what ever you get measuring the resistor to find the true value of the resistor when dealing with low resistances.
 
Yeah your resistor is probably .8 or so would be my guess. Which would put the output around 860mA. Try this, measure what voltage your wall wart is actually putting out when the laser is on. they you can find the efficiency of your driver.
 
So something weird just happened... my diode doesn't want to output a good beam/dot anymore. I hope I didn't kill it... there is still a dot on the wall, so I assume I didn't LED-ify it... but I really hope I didn't just blow a 50$ diode... -.- I didn't put any current in it that was a lot, nor any voltage that was a lot. No idea what I could have done to kill it.

FUCK! I think I killed it. Just disconnected it from everything and ran it solely from a 9V (which would cause it to lase) and it didn't produce a big dot. FUCK FUCK FUCK. Pardon the language... -.-
 
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Depends. In general, the more you mess with them, the more likely you are to kill them. 445's are fairly sturdy but bad connections or drivers will still kill them, as will ESD. Generally when prototyping a driver, an actual diode usually isn't connected until it's sure things are working properly. That's the purpose of test loads. to simulate a diode electrically without risking an actual one.

and don't feel too bad. I've killed two $60 12x 405's within a week before. It's all part of the learning process.
 
God damnit. Well, that's $50 down the drain... guess I can't play around with lasers any more until I have more money....

Wow... I know what I did... I connected my power supply on the driver to the diode and the LD leads on the driver to the power supply... frack!
 
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Yeah... just no more laser fun for another week or so. But now I am curious - I reversed the driver in my box, and my test load is still reading 6.5V or so. Is it possible to kill an LM317 regulator? Because none of the other parts have burnt out.
 
If it's not heatsinked properly, definitely. However, why are you worried about voltage on the test load? the Vf on it should be roughly 1.1 x number of diodes if they're 1N400X diodes. You should have 4 to simulate a 445nm. So your Vf should be around 4.4V.

When dealing with LD's or LED's for that matter.. Current is what matters most. Proper drivers will limit current and let the voltage float to wherever it needs to be to maintain that current. This is why you set pretty much all drivers people make by a certain current limit.. Not a voltage limit. Hooking a diode to a purely voltage limited source generally equals a dead diode.

It would be easier to help you if you would post pics of everything.
 
Well, the reason I was worried about the voltage was because the voltage translated to amperage =p Should have mentioned that, sorry. Apparently, I had 6.5A running through it. That would definitely kill a diode. I just have no idea how I managed to get amperage in the first place... =p
 
While you're saving up for a new diode, you can perfect your driver.

Make yourself a test load, but use a better resistor. Also, easy rule of thumb: two resistors paralleled is product over sums. Paralleled means that both leads of the resistors are connected to each-other, basically splitting the current between them. Example:
two 2Ohm resistors paralleled=(2*2)/(2+2) = 4/4=1Ohm. This allows you to use cheaper lower wattage resistors that are more easily available.

If you are fairly confident in your math, you can parallel those 1Ohm resistors you have, 2 of them paralleled would be 0.5Ohms, but it would increase the power they can handle from 1/4W to 1/2W. From there, you can use Ohm's law to figure out the current.

Ohms Law: V=IR, where V is volts, I is current, and R is resistance.
Note: I is current, in Amps. 1000mA=1A.

This can be rearranged mathematically so that V/R=I. Since we usually use a 1Ohm resistor for simplicity, V/1=I, or V=I is what we usually use. We set the meter to mV because 1000mV=1V.

If you were using a 0.5ohm resistor in your test load, you would have to actually divide the number you get on the meter by 0.5 (multiply by 2 is easier) to get your current.

Anyways, I think that's what you should do right now is try to perfect your driver. Then, when you get your diode, all you have to do is discharge the caps on your driver and hook up your new diode.
 
That's exactly what I have been doing! In fact, I built a DDL driver that can fit in an Aixiz module once I get my SMD LM1117 or LM317 (whichever is easier to get for me) and a smaller 10uF cap, preferably SMD as well =p

I set my rkcstr driver to run at 120mA this morning for a blu-ray diode and I think I am going to run a red LPC on my DDL driver... can't decide what current to set it to, though.

Anyway, I am proud of my 9mm x 12mm DDL... it's pretty damn impressive, considering I just used a through-hole board.

EDIT: Oh, yeah, I haven't actually got a pot on my DDL driver... and I have two SMD resistors on my board... a 1.5Ohm and a 10Ohm. Works out to be about 1.3 Ohms when paralleled (which is what they are), so I figure that, for a red laser, I will get something like... well, 1.25V reference voltage, and I am getting about 950mA, which is in line with the calculations (1.25/1.3)*1000. I think I will drive it at 380mA, as is recommended at HTDz, so...

I need two resistors adding up to about 3.3 Ohms or so... I think? Sound right? 1000*(1.25/3.3). Yeah, that's right. Anyway, what two resistors should I use? I want to balance the load on them evenly, because I will be using SMD resistors and I don't want them blowing. So, err....

6.6 Ohm resistors, I guess, would be best? Yeah, the math checks out. So I guess that's my goal now: get two 6.6 Ohm resistors!
 
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