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FrozenGate by Avery

Where to buy high power 12mm module?

Diachi, you actually pull the driver from a 532 module and solder it back?:o
 
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Diachi, you actually pull the driver from a 532 module and soldered it back?:o
Just what I was thinking...what the hell?

Let's pull the engine out of the car to test the fuel pump? Okay!? Logic?! :(:crackup:
 
Well some members here are awesome at soldering and probably wouldn't think twice.
now as for me:cryyy:
 
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Just what I was thinking...what the hell?

Let's pull the engine out of the car to test the fuel pump? Okay!? Logic?! :(:crackup:

Diachi, you actually pull the driver from a 532 module and solder it back?:o


I don't do that, but that's the proper way to test how much current the diode is getting generally speaking.

Measuring before the driver could be wildly inaccurate. Best to measure power consumption and work it out from there, but you still need to make some guesses to get the LD current - so still not a particularly accurate number.
 
Is there any way that's better than before the driver but that doesn't involve removing the diode?


Only if the driver has current test points, which a cheap one like that almost certainly won't. Measure before the driver, we can guesstimate the LD current, it just won't be all that accurate. Just need input voltage and current.
 
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What is the procedure to "guesstimate" the LD current draw using the input voltage and current?
 
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What is the procedure to "guesstimate" the LD current draw using the input voltage and current?

Calculate input power using P=IV, guesstimate driver efficiency (this is partly where the error in LD current appears) and use that to find output power, guesstimate LD voltage (IR LDs as found in 532nm modules are usually 1.8-2.2V when running at spec), use that value again with P=IV (Knowing power and voltage) to calculate output current.

As you can see, it's not all that accurate. You've got a couple variables that you can only make an educated guess at, although, you could measure the voltage across the LD to find that number accurately. Even in that case, you still need to assume driver efficiency. I usually just assume ~80% and go from there, tack on a margin of error for the final answer and it gets you close enough much of the time.

That's why the better solution is to use a test load or built in current sense resistor, that's much more accurate than what is essentially an educated guess.
 
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To be honest Craz, on some exoticts or foreign;)cars yes you do have to pull a engine to do that, crazy as it sounds..
 
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