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Minimalistic 1000mA driver?

gksudo

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Making this as simple as I can.

I'm not a super pro on this circuit. Please provide feedback and I'll add to/change.

My goal? Be as minimal and cheap as possible. Hoping to make it so small that it needs no board, just parts soldered to parts with lots of heatsinking.
 
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Cyparagon

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It'll work, but the cap will do more good on the output. It doesn't need to be nearly as large, either.
 

ped

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And shouldnt the second band on the resistor be orange? :thinking:


0.1uF cap on the output.
 

Johnyz

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Just keep in mind, that the heat pad of the LM317 to Vout pin and even if the 445 diode is case isolated, you would be still risking accidental short circuit and killing the diode. Even touching the brass bar would somehow alter the current to the diode... So, make sure to be using a mica pad (all electronics parts store have them) in between the chip and the brass bar. Also, thermal-gluing the resistor to the brass bar will help too.
 

Johnyz

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Even if the diode is case isolated, case to either pin is still only a 0.4mm distance. Just to be safe you know... wire can bend a pin from pressure etc. The main issue would be that you would alter the diode current just by touching the laser's body. Anyway, I assumed he is looking for a driver for his brass bar labby...
 

Cyparagon

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Even if the diode is case isolated, case to either pin is still only a 0.4mm distance. Just to be safe you know...
Many, many builders use the case pin for battery negative which is at neither diode positive or diode negative potential. I have never heard of this causing problems.

And what about red laser diodes? They're case negative, so there's no way around it. Somehow you still never hear about shorting them.

The main issue would be that you would alter the diode current just by touching the laser's body.
How?
 
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Johnyz

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Your own body has a resistance of about 10K/cm, so by touching the body (connected with Vout of LM317) you would be essentially adding a resistor to ground.
 

Cyparagon

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There's no potential TO ground. The output of a power supply is isolated.

Even if that were the case, 1.25V/10kohm is 0.1mA which is a 0.01% change in current.
 
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