Go Down

Topic: LM317T Driving me nuts (Read 5806 times) previous topic - next topic

floresta

#15
Jul 01, 2011, 12:20 am Last Edit: Jul 01, 2011, 12:24 am by floresta Reason: 1
Quote
(i have now 2 LED's in parallel).

You can't do that either (and expect them to survive).  They don't work like resistors.

Quote
How could i limit the current to 2amps? how many ohms on that current limiting resistor? also, since i have 2 LED in parallel, 1 resistors for both is enough? or should i have 1 resistor for each LED?

Have you looked through the forum lately?  This thread is only a few days old: http://arduino.cc/forum/index.php/topic,65285.0.html


Don

SgtOneill

#16
Jul 01, 2011, 01:31 am Last Edit: Jul 01, 2011, 01:47 am by SgtOneill Reason: 1
Oh boy... i must be torturing these LED  :smiley-red:

I mounted them on my motorcycle (it was the whole purpose since the beginning).
Everything seems to work fine however, as you said floresta, without current limitation these guys go nuts.
I added a heatsink to the voltage regulator but it gets really hot really fast, even faster supplied by the motorcycle battery. I try to keep them on only for a minute or two.

I need to fix this current problem. So far its a weak prototype.

How come i cant have the LED in parallel? wouldnt that share the amps between them and keep the voltage equal for both of them?

Thank you once again =)

floresta

Quote
How come i cant have the LED in parallel?

You must limit the current separately for each LED that means each one must have it's own resistor and then they, by definition, will not be in parallel.

Quote
wouldnt that share the amps between them

The current is the dependent variable.  Each LED will have it's own current determined by the supply voltage, the series resistance, and the particular characteristics of the particular LED.  The current for both LEDs will be provided by your supply.

Quote
and keep the voltage equal for both of them?

They are like snowflakes.  No two are exactly the same and no two will have exactly the same forward voltage.

Take a look at this thread, it is the one I meant to link to before.  http://arduino.cc/forum/index.php/topic,64978.0.html

Don

robitabu

Quote from: SgtOneill

Oh boy... i must be torturing these LED  :smiley-red:


Just use the LM338 and build a Current Limiting Circuit with that. Driving LEDs with voltage can be frustrating (you are experiencing that right now). The key for success is driving them with current. Take a look at LM338's datasheet, you'll find the answer.

Grumpy_Mike

Quote
the caps are optional

No they are not, the data sheet is sales talk, you need caps. Many beginners fall into this trap. Without caps on any regulator you will not get a stable output and you will continue to be driven nuts.

SgtOneill

#20
Jul 01, 2011, 09:21 pm Last Edit: Jul 01, 2011, 09:53 pm by SgtOneill Reason: 1
I made a circuit drawing so we can easily visualise whats happening.

I bought 1.2ohm 5Watt resistors and i'll be applying them as soon as I finishing the circuit =)

@floresta: the circuit has some problems you already pointed out, but I want to fix them all, one by one, in the circuit drawing first and than on the workbench =).
About the LED's being parallel... You said they shouldn't be like that, but I cant see how to put them in series would work properly (im not doubting you, just trying to understand it better)...
Would it be something like:  Voltage Reg. vOut ---> RESISTOR1 ---> LED1 ----> RESISTOR2 ----> LED2 ----> GND

Heres the schematics so far: http://dl.dropbox.com/u/1739057/Project.png

robitabu

It should be like this:  Voltage Reg. vOut ---> RESISTOR ---> LED1 ----> LED2 ----> GND

Just consider Led1 + Led2 as a single Led that needs tyq

floresta

#22
Jul 01, 2011, 10:38 pm Last Edit: Jul 01, 2011, 10:58 pm by floresta Reason: 1
The LEDs in your drawing are NOT in parallel.  The LED/resistor combinations are in parallel and this is OK, but your values do not make sense.

Let us assume that the voltages are correct.  The voltage across either of the resistors will be 4.9v (8.5 - 3.6) and therefore the current through each resistor (and corresponding LED) will be 4.08 A (4.9 / 1.2).  This is not too healthy for the LED with an absolute maximum current rating of 2.8 A.  It's not too healthy for the resistor either since it's power dissipation will be 20 W (42 x 1.2).  It's not healthy for the regulator either since it is trying to supply 8 A.

Let's try a different approach.  Assume that the LED current is indeed 2.8 A.  The voltage drop across the resistor will be 3.36 V (2.8 x 1.2).  The drop across the LED would have to be 5.14 v (8.5 - 3.36).  This is so much higher than the forward voltage you have assumed that the values just don't go together.

Since you have already purchased the 1.2 ohm resistors let's try to work with them.  Lets assume an LED current of 1.4 A and a forward voltage of 3.3 V (A guess).  The voltage drop across your 1.2 ohm resistor will be 1.68 V (1.4 x 1.2) and therefore you must adjust your regulator to deliver about 5 V (1.68 + 3.3).  Each resistor will dissipate 2.35 W (OK).  Your regulator, however, may be in trouble due to it's power dissipation of about 20W (7 V x 2.8 A).  At the bare minimum you will need a big heat sink.  

To do the design you have to start with the desired LED current, I would start with the an LED current of about half it's absolute maximum rating.  Next, determine the typical forward voltage at this current from the data sheet (or take a guess) and then calculate the required resistance using the regulator output voltage minus this typical forward voltage and the previously selected current.  Don't forget to go back and check out the power dissipation of each component.  You will probably find a problem somewhere at these current levels.  You would be much better off with a current limiting circuit as proposed by 'robitabu'.


Don

SgtOneill

ok then obviously I need to remake the whole circuit (not a big deal :D)

not sure what you mean with "current limiting circuit" but tomorrow ill google it and re-draw the circuit , posting the image here so we can re-evaluate the situation.
I want to make it as perfect as possible because theres alot more going to be involved in this project.

I want to connect an arduino with a clock, gps, temperature, 3-axis accelerometer, LCD screen and a OpenLOG to register all the values and store them. This will need a PERFECT voltage or i'll just blow everything up ;)
Lets leave that for later.


Thanks once again floresta, I shall remake the circuit ;) Thats some fast and precise calculations u did there, impressive O_O


PS: I dont need to use the 5Watt resistors just because i bought them, rather have a decent circuit than be limited by a bad decision :)


floresta

Quote
not sure what you mean with "current limiting circuit" but tomorrow ill google it

You don't have to Google it, you were already told exactly where to find the information.  Go back and reread the posts in this forum thread.

Don


SgtOneill

I was checking the lm338 and there were some examples.
These were the two i found to be useful for this project.





Precision Current Limiter seems to be a good choice for the LM338 implementation.
Probably will limit the current to 1.4Amps and work from there.

got to remake my circuit but this was a busy weekend. maybe tomorrow!

robitabu

#26
Jul 04, 2011, 07:13 am Last Edit: Jul 04, 2011, 07:15 am by robitabu Reason: 1

I was checking the lm338 and there were some examples.

Precision Current Limiter seems to be a good choice for the LM338 implementation.
Probably will limit the current to 1.4Amps and work from there.


This is the one you need! SImple and effective to your case.
If you limit the current to 1.4Amp you're good to go with your led (ref. datasheet). You can even connect both leds in series (always at 1.4Amp) if you provide enough voltage to the LM338. And then increase the current even more since the LM338 is capable of providing a maximum 5Amps total (with proper heatsink).

SgtOneill

just a quick question, i dont recognize that component on R1, the arrow pointing at a resistor? How's that suppose to be implemented?

robitabu


just a quick question, i dont recognize that component on R1, the arrow pointing at a resistor? How's that suppose to be implemented?


Well, that's a variable resistor, the arrow pointing on R1 represents the third connection of a variable resistor (they have three pins; ref. http://www.pictutorials.com/resistors.htm), the one that moves along the resistant material and makes it change its value.

This circuit is an example of a "variable" current regulator; R1 is the variable resistor (could be a slider, a trimmer or whatever); by changing its value you change the result of the formula Iout=Vref/R1 and you get various currents.

In your case, you need a specific current value, so you don't need a variable resistor; you have to calculate R1 in order to get the needed Iout and use that R1 in your circuit.

Grumpy_Mike

If you are tempted to use a variable resistor remember that if you are putting 1.4 A down it then you will need it to be a very high wattage. What ever size you choose there is a danger of burning it out when it is at the end of it't travel. Still without capacitors you don't stand mich chance of success anyway.

Go Up