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How do i know what resistor i need?

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Platupus999

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Dec 5, 2020
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So to build my driver i am going to use a LM317 and i also need a resistor to connect adjust and Vout however i am unsure how to calculate what resistor i need btw i am using a 50mw 405nm laser diode.
 





1.25 is your constant ( actually 1.2 to 1.25 ) for the zener so you divide your resistors into 1.25 to get your current when configured as a current regulator.

Put your resistors between your ADJ and Vout ( left and center pins ) and +batt on your right pin, you connect your laser diode + to the left ( ADJ ) pin. Batt ground to laser diode ground and of course switch one side of the battery.

NOTE* The lm317 mounting tab and center pin are common so if your case is tied to batt you need to isolate if you sink to the same case.

Make sure your resistors are rated to dissipate your drop, this depends on your supply voltage and how much current your laser will be limited to, for instance if your laser gets 100ma @ 5v and your supply is 8.4v your drop will be 3.4v @ 100ma = 0.34w so half watt resistors will be good.

I buy the good quality Texas instruments lm317's and I never use any caps, never have blown a laser diode and I have many set up this way with many hours on them.

LM317-Current-Regulator.png
 
RedCowboy said:
Make sure your resistors are rated to dissipate your drop, this depends on your supply voltage and how much current your laser will be limited to, for instance if your laser gets 100ma @ 5v and your supply is 8.4v your drop will be 3.4v @ 100ma = 0.34w so half watt resistors will be good.
Click to expand...

No, V-drop over the resistor will always be 1.25v or less, and that is what you base the resistor power calculation on. The rest is over the regulator. Also, remember that the regulator has a max allowed power dissipation too. Refer to the data sheet of the particular device you're using.

To get the resistance you want: V (1.25V) /A (0.1A)= 12.5 ohms. It is assuming you want 100mA through your diode.
 
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That's what I said, you divide your resistance into 1.25 so if he wants 50ma he needs a 25 ohm resistor.
 
Are you refering to my using the word " drop " when talking about the ammount of heat his resistor will need to be able to dissipate ?
 
It's all in my first post. You seem to think that the resistor takes all the excess power and has a varied voltage drop. Also you did not explain how to calculate the resistance at all...
 
It's in the image I = V/R

Yes the resistor would only dissipate 1.25 x 50ma
 
The image shows how to calculate the current, not the resistance.
 
That's just basic math but I suppose I could have explained it more thoroughly.
 
Your resistor = 1.25 divided by your current......... R = E/I .......... So divide 1.25 by your current 0.075


Did you get 16.66 ohms
If you want between 50 and 75ma then a 20 ohm resistor or two 10 ohm resistors in series will give you 20 ohms which means you get 62.5ma
 
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Platupus999 said:
yes i did and i have 17ohm resistors so i will use them
Click to expand...
Wow, Impressive how you have such an oddball value on hand! You must have lots of spare resistors. I recommend adding a 47uF 16v rated electrolytic on both input and output to reduce any possible voltage ripple.
 
Giannis_TDM said:
Wow, Impressive how you have such an oddball value on hand! You must have lots of spare resistors. I recommend adding a 47uF 16v rated electrolytic on both input and output to reduce any possible voltage ripple.
Click to expand...
my dad has a bunch of weird and wonderful resistors, i just searched around and found one eventually
 
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