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### Topic: LM317T Driving me nuts (Read 9261 times)previous topic - next topic

#### SgtOneill

#30
##### Jul 04, 2011, 04:56 pm
Hmm yeah thats a potentiometer symbol (or whatever kind of variable resistor), i remember it from a book I was reading before.
but if thats a potentiometer whats this symbol (R2)?

when i made that circuit to test it, i used a potentiometer.
Either its a different symbol for the same thing, or I used the wrong part

#### floresta

#31
##### Jul 04, 2011, 05:01 pmLast Edit: Jul 04, 2011, 05:04 pm by floresta Reason: 1
That's a 'rheostat', it only has two terminals.

Quote
when i made that circuit to test it, i used a potentiometer.
Either its a different symbol for the same thing, or I used the wrong part

You can connect a potentiometer so that it works like a rheostat except that you will burn burn out the potentiometer version sooner as you approach the zero ohm end of the travel.

Don

#### Grumpy_Mike

#32
##### Jul 04, 2011, 05:28 pm
Quote
You can connect a potentiometer so that it works like a rheostat

To do this you connect the wiper (the center) to one of the ends, hence the first symbol is a pot wired up as a variable resistor or rheostat.

#### SgtOneill

#33
##### Jul 04, 2011, 05:41 pmLast Edit: Jul 04, 2011, 06:10 pm by SgtOneill Reason: 1
rheot-what? oh boy

Circuit Version 2.0:

Hopefully the calculations are ok
Im just not too sure about the resistors capability to handle the Watts that go throu them (because im not sure how to calculate them).
Using W = V^2 / R
W = 7.2V^2 / 0.88ohm
W = 51.84V / 0.88ohm
W = 58.9W

if this is correct, regular resistors are going to catch fire really fast =D

Not even the 5W resistors i bought would handle that.
Hopefully its just me , not doing the calculations the right way.

PS:
According to this calculator: http://diyaudioprojects.com/Technical/Voltage-Regulator/
my Voltage Regulator is only handling 1.774Watts wich seems ok. (image edited)

however, according to this video: http://www.youtube.com/watch?v=IjJWWGPjc-w&feature=player_detailpage#t=188s
Power = (vIn - vOut) * iOut;
Power = (12v - (3.6v+3.6v)) * 1.4a
Power = 6.72W
wich is ok i guess, with a decent heatsink.

Maybe the 1.774Watts is the power that the R1 / R2 are going to have to handle. and that seems alot better than 60Watts

Also added a 0.1uF capacitor as suggested by the datasheet

#### Grumpy_Mike

#34
##### Jul 04, 2011, 06:07 pm
Quote
Hopefully its just me , not doing the calculations the right way.

Yes it is. The voltage to use is the one across the resistor. You have 7.2V out of the regulator but if your figures are correct 7.2V across the LEDs. This gives you zero volts across the resistors so you can use anything. This of course is rubbish.
You don't need R2, this is a constant current supply. You need to know the voltage you have to drop across R1 to make the regulator output 7.2V

Your regulator is dropping 12V to 7.2V that means it has a voltage of 4.8V across it. With a current of 1.4A this gives a power dissipation of 6.72W.

#### SgtOneill

#35
##### Jul 04, 2011, 06:19 pm
Ok, new calculations done and image updated:

#### Grumpy_Mike

#36
##### Jul 04, 2011, 06:29 pm
So you still insist in having R2 in circuit.
The power dissipated in the resistors can be calculated by W = I2R
So you have R1 with 1.72W and R2 with just 10W

Therefore R2 alone will drop 7.14V and there won't be enough voltage to light the LEDs. Of course it won't happen like this because you won't get 1.4A flowing round this circuit.

#### SgtOneill

#37
##### Jul 04, 2011, 06:33 pm
ah damn it, forgot to remove R2, sorry.
Removed and Updated the imagem =)

#### Grumpy_Mike

#38
##### Jul 04, 2011, 06:39 pm
Looks better. I would still put a 0.1uF capacitor from the output of the LM317 to the ground. Also the cathode of LED2 need to be connected to the ground by that I mean the other end of C1.

Good luck in finding a 2W 0.88R resistor.

#### SgtOneill

#39
##### Jul 04, 2011, 07:32 pm
well, i guess that's going to be a problem. (find the 2W 0.88ohm resistor).

#### SgtOneill

#40
##### Jul 04, 2011, 08:26 pm
I think i can get my hands on a 0.82ohm 7Watt resistor.
That should do the trick. ill get 1.5amps, which isn't too bad.

#### Grumpy_Mike

#41
##### Jul 04, 2011, 08:34 pm
Remember you can put resistors in series or parallel to get closer to the value you want.

#### robitabu

#42
##### Jul 04, 2011, 11:56 pm

Remember you can put resistors in series or parallel to get closer to the value you want.

And if you use two big resistors in parallel you will cut down the current flowing through each one; hence you may get away with much lower wattage on each resistor.

#### floresta

#43
##### Jul 05, 2011, 03:41 am
Don't get too concerned about the exact value of the series resistor.  It's value was determined from the nominal value of the reference voltage which may or may not be 1.24 volts and a desired LED current that we pulled out of a hat.

Don

#### robitabu

#44
##### Jul 05, 2011, 08:49 am

Don't get too concerned about the exact value of the series resistor.  It's value was determined from the nominal value of the reference voltage which may or may not be 1.24 volts and a desired LED current that we pulled out of a hat.

Don

By that I think Don means you can use different R1 values if it comes handy, so long you stay in the reference values for LM338 and the LED.

E.g. using R1=0.5Ohm you get I = 1.24/R1 = 1.24/0.5 = 2.48Ampere which is totally acceptable to your LED (that uses to work nicely even at 2.8Ampere) and to the LM338 that can sustain 5Amps continuosly. Since I don't know if a 5W 0.5Ohm resistor even exists, I'd suggest you use two 5W 1Ohm resistors in parallel in order to get an equivalent R1 of 0.5Ohm.

Anyway, a quick note on you calculations. Pay attention to the LED's datasheet. It states each LED drops 3.6 volts with a forward current of 2800mA (not with 1400mA!). In case you limit the current to 1400mA, each LED will drop 3.3V only.

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