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The death of a M140

Kalsongen

Kalsongen

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ARG said:
1ohm resistor = 1.25A
A single 9V battery cannot supply that amount of current IIRC, the internal resistance would be too high. Your problem is likely your voltage source.



Remember to short the leads with the PSU turned off before connecting the diode.



Dafuq? Am I missing something? OP is using an LM317 as a current regulator, so the reference voltage would be 1.25V then. If you're trying to calculate the lowest resistor value you can use you're doing it wrong.

httpdiyaudioprojects.comTechnicalVoltageRegulatorLM317CurrentRegulator.png


V = 1.25V
I = ?
R = 1ohm

I = V/R = 1.25V/1ohm = 1.25A

I don't see anything wrong with his setup.
Click to expand...

Can i try with a 400w 12V PSU ? connected to cc driver ofc?

-victor
 





ARG

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Kalsongen said:
Can i try with a 400w 12V PSU ? connected to cc driver ofc?

-victor
Click to expand...

Yes. You can use the 12V rail as the input to the driver.

Don't use it for long though, the driver will dissipate a lot of heat, more than your heat sink can probably handle continuously.
 
Kalsongen

Kalsongen

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ARG said:
Yes. You can use the 12V rail as the input to the driver.

Don't use it for long though, the driver will dissipate a lot of heat, more than your heat sink can probably handle continuously.
Click to expand...

Ill see if the laser outputs any light, if so ill fix a better psu

thanks

Victor
 
chipdouglas

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ARG said:
1ohm resistor = 1.25A
A single 9V battery cannot supply that amount of current IIRC, the internal resistance would be too high. Your problem is likely your voltage source.
Click to expand...

this is funny. Something most seasoned members have drilled into new members heads, gets buried In the see of posts. there should be a sticky about not using 9v batts for anything over 100ma.

edit... just send it to a trusted member with regulated ps.

Michael.
 
Last edited:
Kalsongen

Kalsongen

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ARG said:
Yes. You can use the 12V rail as the input to the driver.

Don't use it for long though, the driver will dissipate a lot of heat, more than your heat sink can probably handle continuously.
Click to expand...

Something must be broken inside the diode, the voltage dont dissapear when the psu i turned off, is it ok to Ohm meter the Diode?

Still only 2,25V supplied to the diode :O I think that the VR is broken

-Victor
 
Last edited:
ARG

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Can you check the current delivered to the diode? That would be more informative to help troubleshoot the issue.
 
ARG

ARG

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Yeah, sounds like your LM317 is dead. What's the voltage across the set resistor?
 
Kalsongen

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ARG said:
Yeah, sounds like your LM317 is dead. What's the voltage across the set resistor?
Click to expand...

Should i get the lm317hvst or what its called ive typed the name in a previus post. Ill. check that hold on
 
ARG

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Yeahhh. Somethings not right if it's not 1.25V

A 1.5A rated LM317 should be fine.
 
Zeebit

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Are you referring to the LM317HV? No. Its simply rated for a higher input voltage but its just the same.
 
DTR

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Kalsongen said:
they have got CC power supplyes on my school. Do i crank the volts to the max and slowly increse the Amps?
Click to expand...

I would not set the voltage to max. Keep it under the max forward voltage of the diode. Around 4.8V will be more than sufficient to test it and set the current to zero. This will limit the surge potential. If set to max voltage the surge potential is very high and can LED the diode. After doing that power the unit off then short the leads. Connect the diode power on and slowly turn the current up.:beer:

 
Last edited:
DashApple

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Kalsongen said:
So 12vdc Will be okay for a longer time then?

Thx, Victor
Click to expand...

If the supply is 12 Volts and the regulator is set up at 1 Amp CC ( 1.25 Ohm resistor ) and the diode drops around 4.7V - 4.8V at 1 amp

The resistor will drop 1.25V with a power dissipated of 1.25 Watts .

The regulator will drop the remanding 5.95 Volts and dissipate 5.95 Watts .

As long as the regulator is properly heat sinked and cooled it can be run no stop .
 
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