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FrozenGate by Avery

LM317 driver whit to high voltage

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bestrider14

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hi,

i new here and i would make my laser but a make the lm317 driver but i have a little bug. when i test i have 3.3volt on the bank test and i need 3v how a can reduce hit and in theory 1.25v/5ohm = 250mA but i had 295mA :S



sorry for my bad englis :P
 





That is theoretical volt drop of 4 1N4001 diodes.
Don't worry, just check current and hook up laser diode (always discharge capacitors before soldering laser diode)

Volt drop of red diodes is between 3 and 3.3V, and voltage is not that important.
However, LM317 will adjust voltage by diode characteristics.
 
Yes, there is tolerances of resistor and even chip itself that makes current different.
We can ignore lm317 tolerance, error of Lm317 is very, very small.

Two resistors in parallel in reality always give smaller value, and also there is resistor tolerance that is probably big... that is probably reason why current is higher.

Maybe resistor with smaller tolerance will solve problem.

Also there is possibility that Lm317 is defective.
I had similar problem recently.
 
Last edited:
OP, I have pretty much the exact setup, except the only capacitor I had on hand was a 10uF electrolytic, not the best filter. (LPC-815 LOC looks just like those, probably the same thing)

In reality, it works. Haven't tried simulating it.

One thing I will say, is when you attach your driver to your laser diode, touch both legs of the capacitor at the same time with something metal to deplete any residual charge stored in it.
 
Check your resistors... They say they are five ohms or what ever but they probably have a 10% tolerance so your not getting five ohms the 1.25/5=250mA is a perfect scenario if it was .5ohm under it would give you 277mA's..., and thats still in the resistor tolerance value... if it was a 20% resistor it could be as high as 312mA's and still be within the tolerance of the resistor... so you need to actually measure your resistance and you will probably get your answer
 
What CURRENT you need for your laser diode, and what is the VF (forward voltage) of it ?

In that schematic, 5 ohm gives you 250mA, but keep in mind that the LM317 have 2,5V of internal dropout, that added to the 1,25V of dropout on the setting resistor, give you 3,75V typical of total dropout (can also be a little bit more) ..... if you have, as axample, a diode that require 3V FV, then you need a MINIMUM theorical of 6,75V of power supply ..... calculate at least one more volt for dispersions, tolerances and so on, you need at least 7,5 / 7,75V at the input ..... if your batteries are not completely charged, that can be one of the problems.

For reduce the dropout, you can try with an LM1117 (NOT 117, 1117, 3 "1" in the part number ..... LM117 is a different thing) ..... LM1117 have an internal dropout of 1,3/1,4V, so also a pair of li-ion batteries not completely charged may be enough .....
 
bestrider14 said:
but if I use a bigger ressistor i dont have 250mA anymore :S so a will check it because its true I have 295mA. I will give some new.

EDIT: i can't grow my ressistance whitout change my mA same if I have 150mA i have over 3v :S. So if you say that is not a probleme if a have 295mA this is a probleme ?

this is my LED Brand New 660nm 200mW Mitsubishi CW Red Laser Diode/DIY - eBay (item 250601073082 end time May-20-10 04:14:41 PDT)

tank you
Click to expand...
These diodes look like LOC that can work on about 380mA without problem if you have good heatsink.

If you have problems with resistors tolerance then buy a pot and adjust current to your desired current. To calculate power dissipation for pot use this formula .. P=1.25 x I
Don't buy small pots, you could burn them.

There is one more thing....how do you measure current?

@HIMNL9
Lm1117 usually comes in SOT-223 package and if you power up it with 2xLiIon batteries and don't provide good cooling then current will drop due to overheating.
 
No, they are available also in standard TO220 package ..... you have to considerate, anyway, that the maximum power dissipation of this package without an external heatsink, is 1W (more with heatsink) ..... and the dissipation from the regulator is given from (Vdropout+Vdiff)*Iload, where Vdropout is the INTERNAL dropout of the IC (without the dropout of the Rset), Vdiff is the difference between the input and output voltages minus the TOTAL dropout (including the Rset dropout), and Iload is, ofcourse, the current needed from your load .....

For keep as example the 250mA of the above posts, if you have a 3V FV laser diode and you use 9V of power supply (cause the 7,2V from Li-Ion is a bit low), with a LM317, the internal dropout is 2,5V, so, (2,5+2,25) *0,25 , that make 1,31W of dissipation (need an external heatsink, otherwise it go in overtemp protection) ..... with the LM1117, you can power the same 3V laser diode with a pair of Li-Ion batteries (typical 7,2V output), so the dissipation of the IC is (1,4+1,5)*0,25 = 375mW .....
 
Yeah, there are also LM1117 in To220 package, but hard to find.

My last red build uses Lm317 and 2x18350 batteries and I must say that current stays the same even on lower voltages.
I tested Lm317 and current starts to drop only when input is less than 6V.
That means that Lm317 need about 3V to work. Lm1117 need less to work.

LM317 or LM1117, doesn't matter too much...

I've never seen LiIon battery under 3V....I recharge them when voltage drops to about 3.3V.
 
ok now I have 245mA I a have change my ressistor by 2 10Ohm 5% in // + 1Ohm 5%. Now how i can drop my voltage to 3v ?

tank you
 
This is voltage regulator....
It adjusts voltage by itself.

When you connect red laser diode, voltage will adjust by itself between 3 and 3.3V. (Depends on diode)
 
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