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Driver test load confusion`

gksudo

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Hello all. I've searched many results and threads on laser driver test loads, but can't find what I need to know.

I will be using one of DTR's 1W 445nm diodes in a laser project. I've built an LM317 driver.

The circuits I've seen vary from the senkat red diodes, to blu-rays. Since there's apparently a variation in the circuits and I couldn't find a 1W 445nm driver test load, I wanted to be absolutely 100% sure I will be measuring a driver that is correct.

If you could post at least the parts and possibly a thread with the EXACT type of circuit to use, that'd be helpful.
 





jimdt7

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You must use 6 diodes like blue-ray ! But how much current do you want ?
 

gksudo

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You must use 6 diodes like blue-ray ! But how much current do you want ?

Ok thanks!

Need 1250mA. But I understand the diodes work up to 1amp. I'm sure they can take it for a little while till I measure.

Do you have a quick link or pic of the schematic you refer to?
 
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test%20load01.jpg

IM A NOOB !! SO CORRECT ME IF IM WRONG.

I think the above is the scheme for a correct test load ..
 

jimdt7

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Yes the schematic is correct ! You must connect your diy driver leads where the blue-ray should ! Also at 1,25A the lm317 will get very hot ! Do you have a heatsink for it ? I recommend you to use an lm350 which has maximum of 3A so you wont need heatsink at all !
 

gksudo

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Yes the schematic is correct ! You must connect your diy driver leads where the blue-ray should ! Also at 1,25A the lm317 will get very hot ! Do you have a heatsink for it ? I recommend you to use an lm350 which has maximum of 3A so you wont need heatsink at all !

The lm317 will have 2 heatsinks plus dedicated 12 volt 1.5 inch fan
 
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Test load is not necessary for a linear driver such as this, especially if you're cooling it as much as you are. Just hook up your ammeter to the output.
 

gksudo

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Test load is not necessary for a linear driver such as this, especially if you're cooling it as much as you are. Just hook up your ammeter to the output.

Don't I still have to measure the output to the diode?
 

gksudo

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Important question! I'm at the store now and need some help.

If I"m correct, the test load diodes need a certain voltage drop, but I find the same 1n4001 to have a different voltage drop depending on where I shop. What's the proper voltage drop for DTR's 1W 445nm diodes?
 
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I think you missed Cyparagon's reply twice. LM based linear drivers will output the same amount of current no matter what you hook up to them. (if they are used as current regulator. Iirc they can be used as voltage regulators as well)
 

gksudo

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I think you missed Cyparagon's reply twice. LM based linear drivers will output the same amount of current no matter what you hook up to them. (if they are used as current regulator. Iirc they can be used as voltage regulators as well)

I saw his reply, but I want to be sure everything is fine before I assume it's going to work. I only have a pocket sized digital volt meter at the moment.

Aside from this driver, whats wrong with asking for an answer to this general question?
 
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There's nothing wrong with asking. It's ignoring the answers that I have a big problem with.

By "voltmeter" do you mean "multimeter"? Can it measure current? If so, put it on the output. If not, you can simplify the test load by just using the 1 ohm resistor.
 

gksudo

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Its a very small meter that only measures, resistance, continuity, voltage, frequency, and capacitance. Does not measure current.

What exactly do the diodes do in the test load? Can you describe what "voltage drop" is and how it relates to a test load for a driver?
 
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Laser diodes require more voltage to operate than regular diodes do. To create a more accurate test load, one must put more diodes in series so that as a group, they behave electrically similar to a laser diode.

I'll say it again. You do not need the diodes when working with a linear driver. We already know the lm317 works. If you want to measure the current with your voltmeter, place a 1 ohm resistor on the output and measure the voltage across the resistor. The voltage reading will be equal to the current.
 




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