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FrozenGate by Avery

Quick Diode Test without driver?

Xer0

Xer0

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Dec 2, 2008
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As these can take up to 2A without instantly dieing; can you recommend me a kind of Battery which i can use to quickly (for a very short pulse to see if its working) so they dont overheat. I just want to test the harvested diodes fast, dont have a good driver here atm. some Button/Lithium cells maybe?
 





Shouldn't be a problem to test them with a lithium battery (4.2V).

Greets
 
An button cell? Cuz an 18650 could pump in more than 3A...
 
pHeneX said:
Shouldn't be a problem to test them with a lithium battery (4.2V).

Greets
Click to expand...
If you test them with only a Lithium Battery at 4.2 Volts without any
current limiting you risk blowing your LD's..

Do you really want to risk blowing a $30-$50 LD by not using a $5.00
Driver...:thinking:


Jerry
 
Some guys wanted to resell diodes with me, but they lack the skill to solder an DDL driver, and i dont want to send them some to little-timbuktu ;) So i hoped for a kind of battery, which is so weak, that the maximum currenty would be just enough for see it glimming.
 
You can damage the diode by just shorting it to batteries, and you don't want to send damaged diodes to customers. Just solder a 2 ohm resistor across a LM317 and there you go, a driver to test diodes. Yes I know you said your friend lacks the skills, but it is really easy.
 
Just to add to that, if you use a 2 ohm resister make sure it's at least 1 watt rated as 2 ohm resister and a LM317 will output 625ma. Be sure the battery input voltage is at least 4.5vdc but no more then 5.5vdc or so.
 
Lm317 need FV+~3V to work...
You cannot give it only 5.5V....
LM317 requires at least 7.5V to work with 445nm diode.

2x18650 should be fine.

btw... you can calculate power of resistor when you multiply current with 1.25V.

For 2ohm resistor, you need 0.78W resistor..so 1W should be fine.
 
The LM317 drops like 3V or so so you have to have at leas 7.5V. Any more voltage than that just gets turned into heat in the LM317.

Edit: ReNNo beat me to it...lol.
 
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You're really looking for a non-permanent (i.e. soldering) method for testing these, right? Just buy some machine-pin headers and solder the driver leads to that.

Then before you test, short the two leads, plug the diode in, and turn on the driver. When you're done, pull the diode out of the sockets and you're done. The headers are reusable too.

Batteries are designed to output as much current as the device needs, so you're not going to be able to guarantee any maximum current rating from your batteries (maybe a minimum rating), nearly dead or not. Again, why would you risk a $50 part with some low-quality and dangerous test methodology that can be prevented with $5 worth of parts? It just boggles my mind.
 
Just wondering here, is it unsafe to use machine pin headers in place of solder permanently on the diode? As in making a quick disconnect module swappable laser?
 
LM317, LM317, LM317...........did I say LM317 yet? Take a look at styropyro's advice. It's ridiculously easy.
 
Xero;

I would just use a 1/2 ohm resistor in series with a 4.2 Volt 18650 battery.

It will drop the voltage an additional 1/4 volt @ 500ma (leaving ~3.85 volts).

Larry
 
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Larry has it nailed. There are a few of us who understand Ohm's law and direct drive !!!

HMike

Make sure the resistor can handle .5 to 1 Amp of current. Likely a wire wound.
 
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Yeah but that's only assuming the diode will draw 500mA at 3.7 volts.

It's never a good idea to not use a driver but it should work.

-Tony
 
LarryDFW said:
Xero;

I would just use a 1 ohm resistor in series with a 4.2 Volt 18650 battery.

It will drop the voltage about 1/2 volt @ 500ma (leaving 3.7volts).

Larry
Click to expand...

I'm confused. Is Xplorer877 correct that you are assuming the diode will only draw 500mA at 3.7 volts?
 
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