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Old 03-12-2010, 12:01 AM #17
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Default Re: I know it's DANGEROUS, but...

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Originally Posted by Bluefan View Post
bearded galaxy: the internal differential resistance of the laser diode is incredibly low. It's a diode, these have an exponential relation between voltage and current. On your average diode, the current doubles with each 0.025V rise in voltage. The operating point of your LD may be around 2V. The battery is 3.7V. The limit will be the internal resistance of the battery and the resistance of the wires. Wires probably only have a few milliohm resistance. The only thing limiting the current is the internal resistance of the batteries, which is probably strongly related to how full it's charged and the temperature. And it's a far too high current
jayrob used the same battery and had an LPC directly connected and said it drew like 700mA or so. So, he just used a 2ohm resistor and said he had about 370mA. I'm using about 3 ohms, and I think I'm getting more than he was, but I'm unsure. I'd like to know what the internal resistance is, on an average LPC (which also factors in, not just the battery and the wire's internal resistance. Otherwise, there would have been incredible amounts of current across the LD, and it would've died instantly. The current one of my batteries gives, when shorted, is enormous. Also, since the diode has a Vf, the current is like this: (Vbattery - Vf) / resistance. Remember, jayrob got 370mA with a 2ohm resistor. Ohm's Law would then conclude that the voltage over the resistor is about 0.74V, give or take. Besides, what you've said is mostly information I already knew and considered (at least, the accurate information). BTW, I measured the voltage across my resistors, and it's about 1.6-ish, when the battery is charged to 3.9V-ish.

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Running in duty cycles this long only prevent long term thermal effects, nothing more of the many possibilities. Another thing waiting to happen: some batteries may build up charge like a capacitor, making them capable of delivering a pulse far above the short circuit current, frying a diode even faster. Drivers have a reason, stability and reliability.
Again, nothing I don't already know. Besides, now, my current is pretty well limited. I sure wish I had an power meter for my laser, though...


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Old 03-12-2010, 04:42 PM #18
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Default Re: I know it's DANGEROUS, but...

I know this is a bit off topic for this thread but it still pertains to it. How do you measure the mA that the battery will put out if directly hooked up to the diode?

Thanks,
Justin
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Old 03-12-2010, 05:52 PM #19
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Default Re: I know it's DANGEROUS, but...

With an AmpereMeter.....

Jerry
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Old 03-12-2010, 06:05 PM #20
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Default Re: I know it's DANGEROUS, but...

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With an AmpereMeter.....

Jerry
Well I have a DMM and I would like to know a procedure if at all possible would you hook the battery up to a teast load and measure the mV across the resistor like normal?

Thanks,
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Old 03-12-2010, 06:20 PM #21
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Default Re: I know it's DANGEROUS, but...

If you want to know how much current a battery will supply connected
directly to a Laser Diode (which I highly do not recommend) then use
a Dummy Test load with the correct number of diodes and a 1 Ohm
resistor to mimic the type of diode you want to use connected to your
battery of choice...

Set your DMM to read Millivolts and check the voltage drop across the
1 Ohm resistor... 1mV=1mA...

The higher the voltage of the battery... the higher the current..


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Old 03-13-2010, 01:49 AM #22
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Default Re: I know it's DANGEROUS, but...

Thank you very much Jerry but I have done that and all I read across the resistor is 2mV that doent make sense the battery I am testing is a new Energizer CR2 and I am using a rkcstr test load set to a blue ray. Any thoughts and thanks!

Justin
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Old 03-13-2010, 03:03 AM #23
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Default Re: I know it's DANGEROUS, but...

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Thank you very much Jerry but I have done that and all I read across the resistor is 2mV that doent make sense the battery I am testing is a new Energizer CR2 and I am using a rkcstr test load set to a blue ray. Any thoughts and thanks!

Justin
Just one CR2? That won't be enough voltage to lase a blue diode. You'd have to use 2 CR2s.
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Old 03-13-2010, 03:31 AM #24
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Default Re: I know it's DANGEROUS, but...

But using two would that only put out 4mA lol I guess thats why the flex drive is a boost driver lol. I wanted to run Jays CR2 build off just the battery and not have to buy another flex drive the one I have now make a high pitched sound and only puts out 50 to 70mA its unstable and just jumps no matter what setting and how much I turn the pot I think its broke lol btw its a V5 flex drive and thanks

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Old 03-13-2010, 09:39 AM #25
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Default Re: I know it's DANGEROUS, but...

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But using two would that only put out 4mA lol I guess thats why the flex drive is a boost driver lol. I wanted to run Jays CR2 build off just the battery and not have to buy another flex drive the one I have now make a high pitched sound and only puts out 50 to 70mA its unstable and just jumps no matter what setting and how much I turn the pot I think its broke lol btw its a V5 flex drive and thanks

Justin
No, let's use math: rckstr test load set for blue = 6 diodes. 1 diode = -0.6V(give or take). 1 CR2 = 3V. 6 * -.6V = -3.6V. So, you can't get an accurate reading because you don't have enough voltage to overcome the voltage requirements of the diodes. Get a second CR2 and try it then. You'll measure at least a couple volts over the resistor, I'm guessing.
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