Lm317t

I have a 6v battery ranging from 5.5 to 6.5v. and it's 4.5AH
I have 16 LEDs I would like to run off it using a LM317T.
Each LED is 1.5-1.6v

What I would like to is run a series of 2 LEDs x 8 paralle.
So each pair would need around 3.2volts.
The symatics on the back of the package for the LM317T are confusing to me. If some one knows how to limit this chip could you put it in laymans terms?

I wouldn't mind if someone wanted to post a diagram of the project. But I still want to learn.

These are the specs on the LEDs
Voltage: 1.5V~1.6V
Current: 140mA, Peak(pulse): 700mA
Power: 200mW
Size: 10mm
Wavelenghth: 850nm

Thank You.

Use a series resistor to power the LEDs, not a voltage regulator.

The formula to calculate the resistor value can be found using Google.
https://www.google.com/search?clien...ing+resistor&sourceid=opera&ie=utf-8&oe=utf-8

Hero999 said:

Use a series resistor to power the LEDs, not a voltage regulator.

The formula to calculate the resistor value can be found using Google.
https://www.google.com/search?clien...ing+resistor&sourceid=opera&ie=utf-8&oe=utf-8

Click to expand...

The voltage drops pretty quick, so I need to keep the voltage stable from the battery. I'm still using a series resistor. Right now I'm testing with a 7805 5v regular. Everything I've read said you need 2-2.5v head way. Right now I'm running at 6.46V from the battery, and it's keeping at 5 volts still on the regular. Have to see what happens when the battery drops more. Right now I'm running 3 in series with a 2.2 ohm resistor.

If that makes any sense. In my head it does.

I'm still getting confused on Voltage, Current, and how resistors work with both of them. Smacking my head on the keyboard. Do resistors limit voltage? or Current? or both?

If your LEDs are 1.6V then 2 in series need 3.2V.
The LM317 needs to have an input voltage that is about 2V higher than its output and the current-limitinging resistors need about 1V.
So the minimum battery voltage is 6.2v but your battery cannot supply it.

Use a single LED in series with a current-limiting resistor. Then the LM317 needs a minimum input voltage from the battery of 4.6V.
The LM317 can drive a few single LEDs with their own resistor if it has a heatsink.

audioguru said:

If your LEDs are 1.6V then 2 in series need 3.2V.
The LM317 needs to have an input voltage that is about 2V higher than its output and the current-limitinging resistors need about 1V.
So the minimum battery voltage is 6.2v but your battery cannot supply it.

Use a single LED in series with a current-limiting resistor. Then the LM317 needs a minimum input voltage from the battery of 4.6V.
The LM317 can drive a few single LEDs with their own resistor if it has a heatsink.

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Right now I'm testing with the 5v regulator which is pretty basic, and I can understand. What do I need to get 1.6 or 3.2 volts out of the LM317? Do I need the capacitors like on the diagram on the back? I don't care if the voltage fluctuates as the battery drops. Is there a basic diagram explaining the minimum componets and how they connect?

Use a 5V low dropout regulator with a dropout of less than 500mV. Connect the LEDs in series pairs of two each with the appropriate series resistor.

LEDs need a current-limiting resistor in series with them to prevent them from blowing up.
If the LED is 1.6V then use a 2.0V supply. Use a 3 ohm resistor for 133mA.

If two LEDs in series are 3.2V then use a 4.0V supply. Use a 6.2 ohm resistor for 129mA.

Hero999 said:

Use a 5V low dropout regulator with a dropout of less than 500mV. Connect the LEDs in series pairs of two each with the appropriate series resistor.

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Eh? I thought the low drop out only helped you on the load side with quickly recalulating the voltage drop from a non fixed voltage source. Doesn't my battery still need to be at 6.5-7v to keep the 5v on the load side? The Battery drops to 5.5v pretty quick.

This is why I'm using the LM317T So I can set my voltage to around 3.5v and then use a current limiting resister in series with 2 LEDs. or set it to 2v and use a resister and one LED.

I just don't think the Low Dropout will work with only a extra .5v overhead.

So back to the original question, how do I use the LM317 chip? Just looking for some one to put it in laymans terms.

audioguru said:

LEDs need a current-limiting resistor in series with them to prevent them from blowing up.
If the LED is 1.6V then use a 2.0V supply. Use a 3 ohm resistor for 133mA.

If two LEDs in series are 3.2V then use a 4.0V supply. Use a 6.2 ohm resistor for 129mA.

Click to expand...

K, this is what the plan is, Just have to get my voltage source to drop that low and stay steady. So still trying to figure out how to use the LM317 to do that. Just don't know if I need capacitors and I think there is 2 resisitors on there, but it doesn't really say which one controller how much voltage you want. there is a * on the package that say some things are optional. But it's just all confusing to me.

I'm also using a Acid Lead 6V battery. Because I need it portable and rechargable. So that is my power source.

The datasheet for the LM317 says it needs a 0.1uF input capacitor 9use a ceramic disc type) and a 1uF to 10uF output capacitor (use an electrolytic type.

The datasheet shows the simple calculation for its two external resistors to set its output voltage. The resistor from its output to its ADJ pin is 120 ohms and has 1.25V across it so its current is 10.42mA. Then the other resistor is calculated with Ohm's Law. For a 3.2V output then the resistor from the aDJ pin to ground is 187 ohms. If you use 180 ohms then the output voltage is 3.13V which is close to 3.2V. If the resistors have a 5% tolerance then the voltage could be plus or minus 10%.

Thank you, I picked up the capacitors today. The diagram on the back of the LM317, says the 1uf is optional and improves transient response.

It also says the .1uf is needed if the device is far from the filter capacitors.

What the heck is "far" in reference to your components? 1 inch? 10 inches? Now if the other 1uf capacitor is optional, and the .1 is only needed if device is far from the filter capacitors..... what the heck are the filter capacitors? Since it sounds like both of those capacitors may not be needed?

I picked them up so when I do start testing it, I have them if they really are needed, but that is just confusing.

Did you have a link to the reference material you are looking at for the LM317. I just want to make sure I'm looking at the same thing you are.

I still don't understand what those capacitors are really doing in this diagram. They don't look like they are doing anything to me. They go from the Vin to the Vout, they bypass the entire chip?

Thank you for you patience and putting up with me in advance.

Now I also have another question about resistors. If I'm using a 1 ohm resistor for each 16 LEDS say 1/4 watt. Can I replace that will a 10w or so 1 ohm resistor and run all the leds off of that paralle? Is there really much of a different there?

16 of your LEDs draw more than 2A which is more current than an LM317 can supply.

If you connect 16 LEDs in parallel and use a single 1 ohm current-limiting resistor then the LEDs will burn out one after the other and the resistor will also burn out. The LM317 will get too hot.

If you connect two LEDs in series and in series with a 6.8 ohm resistor then the current is 126mA and the LM317 can power 8 or 10 of these strings.
The LM317 will need a small heatsink.

audioguru said:

16 of your LEDs draw more than 2A which is more current than an LM317 can supply.

If you connect 16 LEDs in parallel and use a single 1 ohm current-limiting resistor then the LEDs will burn out one after the other and the resistor will also burn out. The LM317 will get too hot.

If you connect two LEDs in series and in series with a 6.8 ohm resistor then the current is 126mA and the LM317 can power 8 or 10 of these strings.
The LM317 will need a small heatsink.

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There is something Odd about these LEDS. I can run 4 of them right off the battery with no resistor Well 4x4=16 of them. They work find and don't get hot at all??? They run around 12+ hours, off a 4.5Ah battery. Which the math doesn't add up for that either. I've ran it over 100 hours with no problems. Just trying to upgrade it to keep the light the same for a longer period of time.

Right now I I'm running 12 of them off a 5v regulator 3 series x 4. it's a 1amp regulator. I'm using a 2.2 ohm resistor for each series. Needless to say it ran for over 24 hours still giving light to the LEDs. 16 hour very bright. then slowly dimmed. The regulator only gets luke warm to the touch with no heatsink. The leads on the LEDs don't get any heat.

10pm I was running at 6.38v on the battery, and 4.68v on the chip.
10pm the next night I was running at 5.12v on the battery and 3.76v on the chip. The chip got cooler after 24 hours, but that is expected, and the voltage dropped just do to not having enough overhead voltage, and that was expected also (this is why I'm trying to use the LM317).

But the math just doesn't add up. 4.5Ah on the battery, it should only last around 2-4 hours??? Depending if I'm using the 12 or 16 config, but I'm getting 24+ hours. I think it was close to 35 hour before the lights stopped working.

I started at 1.44 volts across the LEDS and 24 hours later I was at 1.25volt. Now this is a fair amount less current going accross them, but still doesn't seem enough to run them over 24 hours.

So since the LM317 is .5 amp higher then the 7805 Chip I'm using now. I shouldn't have a problem. The math just doesn't work, so that's why I've been having to experiment to figure it out.

But the math just doesn't add up. 4.5Ah on the battery, it should only last around 2-4 hours??? Depending if I'm using the 12 or 16 config, but I'm getting 24+ hours. I think it was close to 35 hour before the lights stopped working.

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This is normal. If you have 3 LEDs running off the 7805/2.2hm: they would need between 4.5 - 4.8V across them before they drew their rated current. Since the 7805 can't supply much current with only 1V across it so it's output drops to 4.68V, the LEDs get much less current than the 91-227ma calculated (With only 4.68 from 7805 and a Vf of 1.5-1.6V they would draw between 10-82ma) and the battery lasts longer. Look at the spec sheet for your LEDs, and you'll see a curve that shows the typical relationship between LED voltage and the current through them.

a 4.5Ah battery will supply 0.45A for 10 hours. It might supply 4.5A for 1/2 an hour.

The battery is not supposed to be deeply discharged since then its life is shortened.

You had 4 strings of LED and each string might have been uing 120mA so the total current was 480mA. Then the LEDs would be bright for about 8 hours then get dimmer and dimmer gradually.

I think you will blow up your LEDs. Their current was way more than the 140ma max allowed continuous current.
Measure the voltage across the current-limiting resistor and use Ohm's Law to calculate the current.

audioguru said:

a 4.5Ah battery will supply 0.45A for 10 hours. It might supply 4.5A for 1/2 an hour.

The battery is not supposed to be deeply discharged since then its life is shortened.

You had 4 strings of LED and each string might have been uing 120mA so the total current was 480mA. Then the LEDs would be bright for about 8 hours then get dimmer and dimmer gradually.

I think you will blow up your LEDs. Their current was way more than the 140ma max allowed continuous current.
Measure the voltage across the current-limiting resistor and use Ohm's Law to calculate the current.

Click to expand...

All that was a little confusing. I have 4 strings of 4. So 16 LEDs. 16 x 120ma so that is 1.92A. Almost 2A. straight off the battery. Running for 10 hours of Brighter light dimming over time. After 10hours there is still hours of light from the LEDs, just not for what I need it for.

I get these LEDS from Ebay, and I can not get a spec sheet for them. There is also nothing to ID them, and I've search tons of online distrubtors and no one carries them. This is all the info I have for them (down below).

These are really high powered IR LEDs. 16 of them easily light up a 25x20 foot long room using nightshot, Like you have all the lights on.


Now another test I did with them I used a Lithum Ion 12v Battery and 8 in series, No resistor. They get very hot. Have ran for 10+ hours, (just using 8 of them), None ever blew though. But this was just a test. Using the 6v battery and no resistor, I've been using it and recharging it for 100 hours or so using 16 LEDS, and they do not get hot. It's just odd, running far longer then what they should.


When I got to test some more I will check the voltage on the resistor. Then try to figure out how to do that math with Ohms law.


Voltage: 1.5V~1.6V Color: Water clear Angle: 15~30 degree
Current: 140mA, Peak(pulse): 700mA
Power: 200mW Size: 10mm Wavelenghth: 850nm

ReallyConfuzed said:

All that was a little confusing. I have 4 strings of 4. So 16 LEDs. 16 x 120ma so that is 1.92A. Almost 2A.

Click to expand...

No.
We don't know what is the current of each string. If it is 120ma then 4 strings have a total current of 480mA.

After 10hours there is still hours of light from the LEDs, just not for what I need it for.

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As the battery runs down its voltage drops which reduces the current and brightnes so the battery lasts a little longer than calculated.

I get these LEDS from Ebay, and I can not get a spec sheet for them. There is also nothing to ID them, and I've search tons of online distrubtors and no one carries them. This is all the info I have for them (down below).

Click to expand...

I never buy from E-Bay because then you have no spec's.

Now another test I did with them I used a Lithum Ion 12v Battery and 8 in series, No resistor. They get very hot. Have ran for 10+ hours, (just using 8 of them), None ever blew though. But this was just a test. Using the 6v battery and no resistor, I've been using it and recharging it for 100 hours or so using 16 LEDS, and they do not get hot. It's just odd, running far longer then what they should.

Click to expand...

A lithium Ion battery has a much higher voltage when it is charged then the voltage drops a lot as it runs down. If you discharge it too low or at too high a current then it might catch on fire.

You are destroying the LEDs by using a current that is much too high and maybe no heatsink.

When I got to test some more I will check the voltage on the resistor. Then try to figure out how to do that math with Ohms law.

Click to expand...

The math for Ohm's Law is simple. Current= The voltage across the resistor divided by the value of the resistor. The current in the resistor and in everything in series with it (one LED or many LEDs) is the same amount of current.

Ok, here is the current setup. 2.2 ohm with 3 LEDs x 4. (total of 12) Running off the 5v regulator that is rated for 1A.

The volts across the 2.2 ohm is .32 so .145A x 12 = 1.72A

Now this 7805 chip has Internal Thermal overload protection and Internal short circuit current limit. Even though it is working.

But on the final product I can run 2 regulators to split the load just for saftey.
So it would be 8x.140=1.12A on the LM317 which is rated for 1.5A.

I'm really starting to understand this more, so it is helping. Next will be to test the LM317 and get the right current to the LEDs.

Thank You.