LM317 Driver DIY

[midias]

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!).

Yea just put 4 1206 1/2W 4 ohm resistors in parallel

 

[Wolfman29]

Was actually going to do a 2x2 array of 1 Ohm resistors (my stock consists of 1, 3.3, and 10 Ohm resistors), so I only have those and the 3.3s to work with for this application. But... if I just paralleled three 3.3 Ohm resistors, I get a little less power, but it would take up a little less space.... Hmm. How does one go about calculating a resistor array's power distribution?

 

[midias]

Was actually going to do a 2x2 array of 1 Ohm resistors (my stock consists of 1, 3.3, and 10 Ohm resistors), so I only have those and the 3.3s to work with for this application. But... if I just paralleled three 3.3 Ohm resistors, I get a little less power, but it would take up a little less space.... Hmm. How does one go about calculating a resistor array's power distribution?

Amusing all R in parallel

1.25 / R individual = Ir

Ir^2 * Rindividual = PD

So for 3 3.3 ohm 1.25/3.3 = .378 A

PD = .378^2 *3.3 = .4734W

To check 3 resistors do PD total = 3 * .4734W = 1.42W

1.25 / 1.1 = 1.136

PD = 1.136^2 *1.1 = 1.42W check

1.25/3.3 = 1.136

 

[Wolfman29]

I am confused. What does PD stand for? And so what you are saying is that each 3.3 Ohm resistor needs to be able to handle roughly .5W of power?

 

[midias]

I am confused. What does PD stand for? And so what you are saying is that each 3.3 Ohm resistor needs to be able to handle roughly .5W of power?

Power dissipation. And yes for 1.1A of output power over 1.25 V you will need to handle ~1.5 W total. Personally when I design I always try to double or triple the PD needed because I don't want hot things but do whatever you want. So if I have a design that needs a resistor to handle an 1/8w I put in a 1/4w

 

[Wolfman29]

Hmm.... My supply is only 1206s, so I may have to use many multiples... I may just try using a 3x3 array of 1 Ohm resistors... that would give me nine resistors in total (lots, but lots of PD, too), and then I should be able to handle it all. And it would be compact if I literally stacked the resistors on top of one another in order to parallel them =p

 

[midias]

Hmm.... My supply is only 1206s, so I may have to use many multiples... I may just try using a 3x3 array of 1 Ohm resistors... that would give me nine resistors in total (lots, but lots of PD, too), and then I should be able to handle it all. And it would be compact if I literally stacked the resistors on top of one another in order to parallel them =p

A 1206 can handle form 1/8 to 1/2 W depending on who makes it. I would make an excel sheet and look at different combos. If the quick numbers I ran are right that should be about .17W per resistor

 

[itsmei]

For the LM317 driver I am making how many Volts do you recommend, I currently have 2x 18650 batteries which are in series at 3.7v each to give 7.4V. When fully charged they are 4.2V each so that makes 8.4V. Is this voltage ok?

Thanks

 

[Cyparagon]

I didn't read the whole thread. Apologies if this has been said already.

you might want to get a 100 ohm pot for better adjusting the current.

Wrong. This is a 445 circuit. At the pot's maximum value of 11Ω, the current will be 110mA, which is below threshold. If he had a 100Ω pot, the thing wouldn't even lase until you turned the pot to 96% max.

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

You're forgetting that current is a function of the resistance the regulator sees. If you have 11Ω, the current will NOT still be 1.5A, it will be 110mA. Power dissipated is ≈130mW.

 

[Jufran88]

I didn't read the whole thread. Apologies if this has been said already.

Wrong. This is a 445 circuit. At the pot's maximum value of 11Ω, the current will be 110mA, which is below threshold. If he had a 100Ω pot, the thing wouldn't even lase until you turned the pot to 96% max.



You're forgetting that current is a function of the resistance the regulator sees. If you have 11Ω, the current will NOT still be 1.5A, it will be 110mA. Power dissipated is ≈130mW.

The LM317 isn't just for 445nm LD, you can use it for any LD you just need to apply the needed voltage for it to work properly, and at the pots maximum the current is limited to the current setting resistor, which the pot isn't calculated for max output but calculated for minimum output. If you look at ROG8811's site:

Laser driver - it can be done

He explains each component used and that the potentiometer needs to be a low value, I explained to him a 100 ohm pot would be better maybe because of personal preference and DDL first implemented it, but you could get away with a 10 ohm potentiometer. Although, setting the current could be difficult to get it to a specific current. Just as using a 1K ohm potentiometer would be difficult to set the current.

And what I'm trying to say is that current is indeed a function of resistance, however I meant each resistor in series will receive the same current but output a different current because series circuits will receive the same current. Sorry if I didn't word it correctly.

For example:
445nm LD Vf = 4.2V
Voltage needed for LM317 to work = 3V
Total voltage needed to properly set the current = 7.2V
(you could put additional head room for longer runs = 1.2V, total voltage used = 8.4V, but gotta remember any voltage not used by the LM317 will be dissipated into heat.)

R1 = current limiting resistor = 11 ohms
R2 = potentiometer = 100 ohms

So, the potentiometer will max out at the current limiting resistor which will be:

1.25/11 = .113A = 113mA

So at the output current the it would need to dissipate for R1:

P=I^2*R
P = (.113A)^2*(11) = .140W = 140mW

So, the problem was how much would the potentiometer need to dissipate?

I had a discussion about this in the simpledrive2 thread with HIMNL9:

Quote:
Originally Posted by Jufran88
@HIMNL9:
Sorry, I'm really a novice at this, but if you had a 2W rated resistor in series with the trimmer and RSET was 2W rated would it allow it to dissipate more heat together? Or it just depends on individual resistor watt rating?

It's matter of current, and the current flowing in elements connected in serie is always the same, where for elements connected in parallel it change with their values ..... and also, the voltage of the elements in parallel is the same, so the powers are different in the two branches, depending from the current that flow in them (P=V*I, common Ohm laws) ..... so, the purpose is to keep the maximum current that flow in the trimmer branch, at a level that, for 1.25V, it does not cause a power dissipation greater than the one that the trimmer can hold, and let the rest of the current flowing through the fixed RSET ..... so, for a 500mW trimmer, this is done using a serie resistor of 3 ohm, that keep the current in this branch at a maximum of 1.25/3=417mA, causing a maximum power dissipation of 520mW (the trimmer can hold easily 20mW extra power), and this in the trimmer branch is the variable part of the current ..... where the fixed value RSET determine the rest of the current, that is a fixed value, and can be greater than the one in the trimmer branch, ofcourse, and when this happens, also the power dissipation is greater on the RSET, that must be dimensioned opportunely. (sorry, can be explained better, i think, but English is not my main language)


EDIT:

@anselm, remember about power dissipations, too .....

For a linear driver, all the unused power will become dissipated in heat from the regulator .... if you use a test load that take part of this power (causing a dropout and simulating the LD FV), the regulator under test will need to dissipate less power ..... just as example:

Assuming you are testing the current on an 1A driver for a 445nm diode, that have a FV of 4.5V, powered from a 9V input voltage ..... if you use a test load that emulate your LD FV, it take and dissipate 4.5W, and the driver need to dissipate the remaining 4.5W ..... if you connect your DMM directly on the output of the driver, and your DMM internal resistance cause, say, 0.1V dropout, your driver need to dissipate the remaining 8.9W, and maybe it fry or give you false readings going in thermal protection shutdown .....
__________________

I used the power equation:

P=I^2*R

which I calculated for potentiometer resistance but that failed and I got 22.5W which is obviously wrong. But I thought well resistance is a variable in a potentiometer so I used another power equation:

P= EI or P=VI (joules law)

where V is the potential difference or Vref and I for the current being outputted 1.5A

So that's how I got:

P=VI
P= (1.25V)(1.5A) = 1.875W

But I wasn't too sure of this value, so just to be on the safe side and not burn out any components I suggested 2W resistors and potentiometer especially driving currents passed 1A.

Anyways, I am sick :barf:. All started today and got worse at work.

 

[Cyparagon]

The LM317 isn't just for 445nm LD

I never said it was. "This is a 445 circuit" as in "there is a 445nm diode being used in this circuit"

I explained to him a 100 ohm pot would be better maybe because of personal preference and DDL first implemented it

Maybe for red, but I'll say again, this is a 445nm laser diode. The current requirements are much larger. And there's no point setting the current to anything less than 200mA (or whatever the threshold is) since it won't lase.

P= (1.25V)(1.5A) = 1.875W

The only way you can get 1.5A is by setting the pot to 0. So all of that 1.8W will be on the resistor

 

[rhd]

It's way simpler than all of that. If you've got two resistors in series (or a resistor and pot, whatever), your current across BOTH will be the same at any given point in time.

So lets test some values. R = Resistor, P = Pot, TR = Total Resistance, ok?

A) R1 11, P1 0.1 = TR 11.1
B) R1 11, P1 11 = TR 22
C) R1 11, P1 100 = TR 111

The power needed to be dissipated by the pot is given by:
I^2 x P1

Where I is the current, as determined by:
1.25 / TR

So to check the pot's required wattage for each scenario, it's just:
(1.25 / TR)^2 x P1

A) (1.25 / 11.1)^2 x 0.1 = 0.0012 W
B) (1.25 / 22)^2 x 11 = 0.035 W
C) (1.25 / 111)^2 x 100 = 0.012

You might want to create and excel table to figure out where exactly the max dissipation for that pot is. I think this is correct:

	R1	P	RW	PW
1.25	11	0.1	0.139497606	0.00126816
1.25	11	0.5	0.129962193	0.005907372
1.25	11	1	0.119357639	0.010850694
1.25	11	2	0.101701183	0.018491124
1.25	11	3	0.087691327	0.023915816
1.25	11	4	0.076388889	0.027777778
1.25	11	5	0.067138672	0.030517578
1.25	11	6	0.059472318	0.032439446
1.25	11	7	0.05304784	0.033757716
1.25	11	8	0.047610803	0.034626039
1.25	11	9	0.04296875	0.03515625
1.25	11	10	0.038973923	0.035430839
1.25	11	15	0.025425296	0.034670858
1.25	11	20	0.017885016	0.03251821
1.25	11	25	0.01326196	0.030140818
1.25	11	30	0.010224569	0.027885187
1.25	11	40	0.006608035	0.02402922
1.25	11	50	0.004619054	0.0209957
1.25	11	60	0.003409542	0.0185975
1.25	11	70	0.002619646	0.016670477
1.25	11	80	0.002075534	0.015094795
1.25	11	90	0.001684884	0.013785413
1.25	11	100	0.001394976	0.012681601
				
		MAX:	0.139497606	0.035430839

 

[rhd]

The only way you can get 1.5A is by setting the pot to 0. So all of that 1.8W will be on the resistor

I don't think he can EVER get 1.5A. The circuit he referenced here has the pot in SERIES with the resistor. He's never going to escape the 11 Ohm R1.

He's never going to get above 114mA out of his circuit. If this is for 445, it's silly.

If the resistor and pot were in parallel, that would be different.

 

[itsmei]

I am getting confused, are you relating this to me? Are you saying that me using 2x 0.4ohm 2W resisters in series with a 10ohm 1W pot will not let 1.5A through?

 

[rhd]

Sorry - no I was replying to Jufran88

 

[itsmei]

Just to finalize my question will this pot and 2x 0.4ohm resistors in series to get 0.8ohm work? Will i get 1.5A and is the pot the right wattage?