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ArcticMyst Security by Avery

Lm317 Help

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Mar 2, 2011
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I am working on a simple driver for a a140 445nm blue diode.

I am using a simple constant current setup.
I have a set of resistors hooked up to a LM317,
set up so I can Adjust the current from 300ma to 1A.
Problem is the smallest pot radio shack had was 25ohm. So When I adjust current it is like an E^2 graph.
(Current starts at 300ma, I turn the pot half way and current is 350 3/4 of the way and its 450, then all of a suddent it jumps to 1a)

Does anyone know where to get a pot about 5 ohms or even 3 ohms?

I am currently using 2W resistors is that nessasary or would 1/4w ones work fine?

acording to my calculations there would be 1w max going through my resistors right?
 





anselm

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Hello and welcome to the forum!:wave:

You can "tune" your potentiometer by soldering a resistor onto it's pins (in parallel).

Suppose your potentiometer gives you a range from 0.5Ohm to 25Ohm.
Now suppose you add a 4Ohm resistor in parallel.

Now your potentiometer has a range of 0.44Ohm to ~3.45Ohm ;)
1/Rt = 1/R1+1/R2+....+1/Rn

1/Rt = 1/0.5 + 1/4 = 9/4 <=> Rt = 4/9 = 0.44_
1/Rt = 1/25 + 1/4 = 29/100 <=> Rt = 100/29 = 3.448...

Now add about 0.6Ohm worth of resistor in series and your pretty close to your goal.

I have done this very same thing in my 445nm project, 2nd link in my signature.;)

Stick around, read the stickies, and most of all be safe, it's a class IV laser we're talking 'bout.
 

rhd

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Even if your pot is linear, once you parallel it with a static resistor, I think you'll end up with an exponential pot.

At 25 ohms (100% of your pot's max), with a 4 ohm resistor in parallel, you'll get 3.45 ohms (your max resistance now)
At 20 ohms (80% of your pot's max), you'll get 3.33 ohms (96.6% of your max resistance)
At 15 ohms (60% of your pot's max), you'll get 3.16 ohms (91.6% of your max resistance)
At 10 ohms (40% of your pot's max), you'll get 2.86 ohms (82.9% of your max resistance)
At 5 ohms (20% of your pot's max), you'll get 2.22 ohms (64.4% of your max resistance)
At 0.2 ohms (1% of your pot's max), you'll get 0.19 ohms (5% of your max resistance)
 
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Yes My 25 ohm Pot in parallel with 3.3 ohms works, but I get a very uneven spread of current.

Anyway I have a working prototype:
img0057ct.jpg

img0058lt.jpg

img0062sz.jpg


Lol at my battery setup -- Lantern battery in series with a cr123 battery.

Can you recomend a lens for this.
The one I have now is acrilic (I think)
The guy who sold it to me said It would work upto about 1.4watts
 
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WOW!! Pocket portable, and classy looks! Love your heatsink and host!

Just kidding. Kudos for a valiant effort! You do not need to worry much about the wattage in the current limit loop (it is a sensing, not shunting, path) 1/4 wat will more than suffice. If you want low R linearity, you will probablly have to go with a trim pot or 10 turn (or 20 turn). Check somebody like JameCo or DigiKey. Good Luck and keep at it.
 

anselm

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Even if your pot is linear, once you parallel it with a static resistor, I think you'll end up with an exponential pot.
but I get a very uneven spread of current.
That's cool, because brightness doesn't increase linearly with current!
So things kinda even out one each other!
:):)

Mixing batteries like that, just can't be good, especially when you're drawing high currents.
+1 for the successful effort though!
 
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WARNING WARNING WARNING!!!

Justinjja,
I'm supprised some of the 'gurus' didn't catch this. I LIED! I was thinking voltage regulator mode, not current regulator mode. In Vreg mode the current in the Adjust leg is in the 100uA (micro AMP) range and a 1/4 watt resistor is way more than sufficient.

When you are using a 317 as a current regulator, the R is in the current path and you must use higher wattage resistor(s). Apply Ohm's DC power calc.s to determin wattage of the output path (and thus the resistor(s) in the output path):
Power in Watts= Current in amps squared times Rt in Ohms, or Voltage in volts squared divided by R in Ohms, or Voltage in volts times current in Amps. Pick the one which serves you best, they are all the same.
300mA @ 4 Vdc = .3*4= 1.2 Watts ,, 1/4 W will work for a short time, but will get real hot.
1 A @ 4 Vdc = 4 Watts ,, 1/4 Watt will go brown to black, scorch the PCB and burn open fairly quick..

Sorry man, did not mean to miss lead you. I'm an old fart and mind does not always correctly engage with fingers.... Mind does not seem to engage at all sometimes....'
But, I eventually get there, I think, where am I anyway?
 

rhd

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Slight correction here -

You're both right and wrong.

YES, in current regulating mode, the resistor IS in the current path, so you need a heftier resistor.

NO, it's not 300mA @ 4 Vdc (DiodeCurrent x DiodeVoltage) that you use to determine the Wattage of resistor required.

It's actually DiodeCurrent x 1.25 Vdc (regardless of DiodeVoltage) that you use to determine the required resistor wattage.

*Disclaimer* I don't entirely understand the "voodoo magic" of these wonderful ICs. But in essence, the current regulation comes from the LM317 trying to maintain a voltage of 1.25 accross ADJ and OUT. The resistor is what manipulates the amount of current across those two pins that is required in order to achieve this.

However, it's STILL just 1.25V running across those two pins, and across the resistor. While I don't quite grasp how the Voodoo Magic gets the diode it's required voltage, I DO know that it isn't shooting that magic voltage across the resistor.

So for 300mA, you're looking at 0.3 x 1.25 = 0.375 Watts (in other words, a 1/2 W resistor is fine)
 
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RHD thanks for the correction, But I don't understand (look around, you'll find more than enough evidence of my addle headedness... so, please bear with). The IC is trying to minimize the current in the adjust path at the nominal 1.25 reference voltage (National Semiconductor LM117 family data sheet page 8, Application Hints), BUT,, The Rt value is in the output current path, not the adjust path. Doesn't it have to handle the output current of the regulator (which would be with respect to the LASER diode and not to the reference/adjust loop? If you adjust the pot for a 300mA output, that means that the regulator is delivering 300mA across the LASER diode (of dummy load), not the adjust loop. I'm NOT saying that you are wrong, I just don't understand your logic (like your numbers much better than mine, but don't totally understand them).

Justinjja,
Coherent Light's links to the calculators look like a good idea/shortcut (haven't tried the yet though), but it is always nice to be able to figure out how things work and what is happening. Sorry if it seems like we are highjacking your thread, but we may BOTH be learning something here.

Update: I just tried out coherent's suggeted links and they bear out RHD's numbers. I still don't understand the logic, but (assuming the calculators are correct, as the do appear to be) must agree with and humblly accept RHD's correction.
 
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rhd

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Ok, I can probably explain better if I use Ohm's law. If someone else jumps in here and corrects me - trust them. Really, I feel like I get by on the foggiest of recollection from high-school physics classes, and not much else.

In terms of the IC setting the current
- To get the 1.25V drop accross ADJ and OUT, the regulator has to increase the current to 312.5 mA because:
V / R = C
1.25 / 4 = C
C = 0.3125 A

In terms of the resistor wattage you need
- To get the wattage rating of the resistor:
C^2 x R = W
0.3125^2 x 4 = W
0.09765625 x 4 = W
W = 0.390625

It's kind of cool that you can get the same WATTAGE figure much easier by just doing a simple CURRENT x 1.25 operation. But that's more abstracted from what's going on, so hopefully this helps clear it up?
 
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Lol at my battery setup -- Lantern battery in series with a cr123 battery.

The CR123A will drain first, then be reverse charged in series by the 6V lantern battery...which may push it over the edge, start a fire, and destroy a perfectly good investment.
Why not parallel a couple 9Vs instead? simple, square, easy to mount, practical to carry.
 
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RHD, (justin, if you don't like us hi-jacking your thread, just tell us to shut-up and go start our own. RHD, you'll have to do that, I'm new and don't exactly know how to yet), Now back to physics class. Got ch'a and agree with the Ohm's laws calculations. The part that is giving me problems (old, brain gets stuck in one gear and cann't shake it...) is that you are (and, b.t.w. everybody else,, so I must be in error) using the voltage of the reference loop (never goes to ground, goes to a comparator in the regulator IC) to calculate the wattage in the output leg. But, the work in the output leg is done across the resistance to ground (the LASER Diode).
I know it's a nit and everything shows me to be wrong, but I just cann't get the concept of why down. Thanks for letting an old fart ramble and it is educational to the noobs.
 

rhd

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The way I'd conceptualize it:

You care about wattage because you want a resistor that can handle dissipating the amount of power you're expecting it to.

That's a function of the current across the resistor multiplied by (and ONLY multiplied by) the voltage that is ACTUALLY DROPPED by the resistor. That's what the resistor must turn into heat. Thats always 1.25v.
 




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