LM317 Driver DIY

[itsmei]

Hi, i am making a 1.6W 445nm laser, i have all the parts apart from the potentiometer. I want to make an adjustable laser by using a pot, i am using 1.5A through a 0.89 ohm resistor and a 10ohm pot. I know how to make the driver, i just need help to see if this is an ok pot:

Tyco Electronics | Passives | Resistors | Trimmers, Potentiometer and Rheostat | Wirewound |TW1100KA

It says that it is 1W, is this enough? I worked out that i need somthing like 23w, that can't be right, can it?

Thanks

 

[Jufran88]

If your running the LM317 at 1.5A your going to need at least a 2W minimum for both your resistor and your potentiometer. And you might want to get a 100 ohm pot for better adjusting the current.

Here is a website that calculates for the maximum current setting: *here*

 

[rhd]

Jufran88:

I'm not sure that's correct -

He'll need a 2W minimum for his resistor for sure. But for the pot (assuming the two are in series) I think he can get away with less.

At the pot's lowest, it may contribute 0.1 Ohms. With the 0.89 ohm resistor, that will essentially be 1 ohm of resistance, and 1.25A current.

RESISTOR POWER:
P = I^2 x R
P = 1.25^2 x 0.89
P = 1.39 Watts

POT POWER:
P = I^2 x R
P = 1.25^2 x 0.1
P = 0.16 Watts

At the pot's max of 10 ohms, you'll have really tiny current (could do the calculations if you want - but you still won't need a higher wattage pot)

If they're in parallel, the story is different - but I'm pretty sure that in series, most of the power will be dissipated by the resistor.

 

[Jufran88]

I had that mindset too, but HIMNL9 told me otherwise in the simpledrive2 thread and I'm still a bit confused about it.

But it seems at the 10 ohm pot (at max) the dissipated power is:

P = i^2*R
(current flowing through in series will be the same)
P = (1.5A)^2*(10 ohms) = 22.5W (which sounds preposterous)

So I used another power equation that would be a better fit

P=VI
(using the reference voltage and desired output current due to the fact that at the potentiometers max it would output 1.5A anyways)

P = 1.25V*1.5A = 1.875W that needs to be dissipated.

Can anyone else verify this?? :can:

 

[rhd]

If I stare at any more Ohms law, I'm going to go crazy.

I think the difference is that in the other thread, we had an RSET (or something like that) in parallel too.

 

[Jufran88]

I know how you feel. I was going crazy trying to figure out the calculations! :banghead:
In regards to RSET, it was just used as a current limiter at the lower end to avoid frying the potentiometer at higher current ranges so calculating it at the potentiometer branch would be the same with or without RSET.

 

[rhd]

Here's the way it works in my head -

Ohms are resistance. It's resistance of electricity that creates heat. Like friction is resistance, more resistance = more heat.

0.9 ohms is way more resistance than 0.1 ohms.

So in series, when they're both getting the same current, the one that does more of the resisting work (the 0.9 ohm) will get hotter than the one doing less resisting work (the 0.1 ohm).

But if they were parallel, the current gets its choice of where to go, and it will choose (as much as possible) the path of least resistance - the 0.1 ohm! So in that case, the 0.1 ohm would actually get more current, and have to end up dealing with more crap (sorry little resistor dude, it's mostly all on you now)

But here, they're series, so the current is the same for both resistors. The 0.9 ohm has more muscle to flex, and that means more heat, more power the dissipate, etc.

Now, what happens if the pot is up to it's highest resistance? Wouldn't that mean that it does MORE work than the resistor? Yep. Except that it also means your IC will cut the current way down to like 125mA - in which case, you don't have very much power to dissipate across EITHER the pot or the resistor - so we're still good to go on a lower rated pot.

So - at least in my head, where electricity is a chocolate river, and resistors are oompa loompas, it all adds up to needing a lower powered pot. If you ask me to properly defend the math, that's a different story

BUT - these are great!
Ohm's Law Calculators

 

[itsmei]

I am getting 2x 2w 0.4 ohm resistor in series, I will then add a 10ohm 1W pot, surely I have enough resistor wattage?

 

[rhd]

Yep -

Your pot won't ever need to dissipate more than 0.44 W in that scenario.

I think

 

[midias]

Here is a great excel sheet to calculate R I and W I made it for some driver I made. Adjust the value on R1 and R3 and the sheet will recalculate the PD though the pot and everything else you need.

Let me know if you have any questions I think this could be handy to calculate power. Feedback is always appreciated.

 

[Wolfman29]

So why is it that, using a 1206 SMD resistor in my 445, I am able to sustain the driver, etc.? The resistance on a 1206 SMD is nowhere near 2W, it's something like 1/8 W... yet it hasn't died on me yet. Maybe it's because I have the whole driver heatsank and only use short duty-cycles? Anyway, I am going to replace that 1 Ohm resistor with an array of four 1 Ohm resistors today (probably), but still... why does it work?

 

[midias]

So why is it that, using a 1206 SMD resistor in my 445, I am able to sustain the driver, etc.? The resistance on a 1206 SMD is nowhere near 2W, it's something like 1/8 W... yet it hasn't died on me yet. Maybe it's because I have the whole driver heatsank and only use short duty-cycles? Anyway, I am going to replace that 1 Ohm resistor with an array of four 1 Ohm resistors today (probably), but still... why does it work?

The 1206 is probably a 1/4W so I would not use it as it will heat up. Short duty cycles will keep it alive. Remember resistors gain resistance as they warm so eventually it will lower its own PD before it blows as long as you don't shock it too bad. Then go back to near normal resistance when it cools. If you provide a schematic I could tell you more.

 

[Wolfman29]

Just your normal LM1117 driver, using an SMD 10uF cap, discreet 1N4001 diode, and a single 1 Ohm resistor connecting the adjust and the output.

 

[midias]

Just your normal LM1117 driver, using an SMD 10uF cap, discreet 1N4001 diode, and a single 1 Ohm resistor connecting the adjust and the output.

I would measure the resistor after and before use I bet the chip is heating and changing value.

 

[Jufran88]

What current are you driving LM317?

 

[Wolfman29]

I'm not using an LM317, but an LM1117, if that was directed at me. I am driving it at around 1.25A though.

@midias: Alright, will do... but I may as well just replace it with an array of resistors just to be safe (if it fits on my chip!).