How to power an LED, NEED CLARITY.

[Ivin]

We'll it's very apparent that I have a lot to learn yet. But I'm having a good time doing it. Especially when people are willing to help out. This project is fun because it give me a goal to reach. And I'm looking forward to the day I read this thread and know everything that's being mentioned.

@hevans1944 thanks for all the info, and the solution. Now I just need to figure out how the LM317T does what it does. I will be sure to follow your advice. And I will check back here when I reach my next roadblock.

 

[(*steve*)]

Just remember that for the LM317 solution you require a voltage source that is between 2.5V and 4V higher than the required LED voltage (it depends on the required current). If you find that the current is significantly less than calculated, you may have sufficient voltage for the LM317 to maintain regulation.

 

[elebish]

SPECS

HIGH-POWER GaAlAs IRLED ILLUMINATOR
Spec Sheet
http://optodiode.com/pdf/OD669-850.pdf
V72 Laptop Battery
Output: 5V/2A USB, and 12V/4A, 16V/3.5A, 19V/3A
http://www.voltaicsystems.com/v72
If I wanted to power this LED From this Battery what kind of resistance would I need?


That’s the question, here’s the filler.
I know this should be a super basic question, I’ve really only started seriously trying to understand electricity two days ago, but yesterday I really struggled with this, and this morning I still haven’t wrapped my brain around it. I’ve chosen “Make: Electronics” as my starting point in this world of electricity.

If my battery is capable of outputting 16V/3.5A then I need to resist the current to a level that doesn’t exceed my LED to avoid frying it. My voltage should not need to be affected because 16v is within the limits of the LED.

But here’s where I’m confused:
from what I understand when Voltage travels in a circuit it is affected by resistors, by increasing or decreasing the resistance you adjust voltage pressure or the “potential difference”
But if you add the voltage drops across the devices in the circuit, the total is the same as the voltage supplied by the batteries.
In this case I don’t think I need to limit my voltage if I use the batteries 16V/3.5A output setting.

Now comes the Amperage, the continuous forward current of this LED is 370mA, that’s a lot lower then what the battery is capable of putting out. So I need to add some resistance to reduce the current.
in this circuit I want to produce the most IR light from my LED as possible but I don’t want to fry it, so would 300mA be a good amount to supply the LED with?

But how do I reduce the Amperage so much without affecting the voltage?
How do I apply Ohms Law here?

Lets say I set the battery to output 19V/3A. (Because I’m not sure if I can use a resistor if my voltage is already perfect but my amperage isn’t.)

Then the difference of voltage between my led and my battery is 3V, this is the number I will use in my ohms law

I only need 300mA to flow thru the LED. So .3A is the number ill use in ohms law.

R = V/I
R = 3/.3
R = 10Ω

That just looks weird to me its such a small amount of resistance.
am I actually supposed to sumbtract the A the the LED with use with the A the batterie supplies? Like we do with the voltage?
so would I really be:

R = V/I
R = 3/2.7
R = 1.11Ω

I thing thats weird to!

But let’s say I wanted to run the 16V/3.5A setting on the battery.
Then my formula would look like:

R = V/I
R= 0/.3
R = WHHHAAAAT… you cant do that!

Do you see how I’m confused? How can I put an LED in a circuit when there is no resistance? It will fail…

What am I doing wrong, and why doesn’t my brain understand this, can you help?

I’m sorry for hopping on your fourm and asking for help so soon without really getting to know anyone, But I’m lost hopefully this confusion was decently explained and you can help re-wire my neurons.

This isn’t a School project it’s just a bit of self-directed confusion.

I also realize the theory must be the same for a regular led and a say a 6v bat. but this is more fun.

Some notes to remember:
Current can not flow without a source of voltage (potential). All purely resistive loads will draw current according to the voltage applied. Too much voltage = too much current, so potential is paramount to proper operation. Current is critical in a LED.
As for the LED, one must know what the current is required for proper brightness. Most LEDs are about 2 volts dc. Voltage above that will have to be dropped by a resistor before it reaches the LED. The resistor can be either on the anode or the cathode side of the LED. Too much will blow the LED quickly. Use ohms law to determine the voltage to be dropped by the resistor and to calculate the wattage of the resistor. Since the LED and the resistor will be in series, the current will be the same through both. Observe polarity! Example below of a typical LED.
1.8 to 2.0 volts dc @ 20 ma. Using 5 volts as the supply, a drop of 3.1 volts is required, assuming we'll use 1.9 volts for the LED. Ohms law---E/I is 3.1 divided by .02 = 155 ohms. EXI = P, so resistor wattage is .062.

E=energy expressed in voltage
I=current expressed in amps
R=resistance expressed in ohms
P=power expressed in watts

Ed.

 

[hevans1944]

@hevans1944 thanks for all the info, and the solution. Now I just need to figure out how the LM317T does what it does. I will be sure to follow your advice. And I will check back here when I reach my next roadblock.

Don't follow my advice or the advice of anyone else blindly. Understand the circuit and check the math. My advice in post #13 was incorrect. The LM317T will not be operating within its specifications if you try to drive an LED with 16.2 V forward voltage at 0.3 A. It will work within specs if the forward drop is only 14.4 V. See below.

You really don't need to know how the LM317T works. You do need to know its operating parameters and limitations. Treat it as a three-terminal "black box" device that always tries, by some internal (and unknown to you) means, to maintain 1.25 V internal reference voltage between the OUT pin and the ADJ pin, provided there is at least 2 V (but no more than 40 V) between the IN pin and the OUT pin.

The power supply positive terminal is connected to the IN pin. The power supply negative terminal is connected to one side of the load which is the cathode of your LED. The other side of the load, the anode of your LED is connected to the ADJ pin.

Connecting a fixed-value resistor between ADJ and OUT means there will be constant current equal to 1.25/R flowing through that resistor and to whatever load is connected between ADJ and the negative power supply terminal, even if that load is a short-circuit. Especially if that load is a short circuit. There is a maximum resistance that can exist between ADJ and the negative power supply terminal that will allow constant current operation. This is the resistance that causes the compliance voltage to reach it maximum value. This value will be the power supply voltage minus the minimum 2 V IN-to-OUT differential voltage minus the 1.25 V dropped across the current-setting resistor.

For a 19 V supply and 2 V minimum between IN and OUT, the maximum output of the LM317T will be 17 V. Subtract 1.25 V for the drop across the resistor and you obtain a maximum compliance voltage of 17 - 1.25 = 15.75 V. This is less than the 16.2V maximum forward voltage across your LED at 0.3 A, so the advice I gave in post #13 is incorrect. Bad arithmetic on my part. The circuit may still drive the LED at reduced current if, at that reduced current, the forward voltage of the LED decreases to 15.75 V.

So, what to do about the failure to operate the higher voltage LED within LM317T specs? You could increase the resistor value from 4.166 ohms (a non-standard value) up to 4.7 ohms (a standard value for 5% tolerance resistors) which would reduce the LED current to about 0.266 A. At this reduced current the LED forward voltage drop might decrease enough to bring its terminal voltage down to 15.75 V. If not, choose a larger resistor until the LED forward-biased voltage is equal to or less than 15.75 V. Or you could raise the power supply voltage above 19 V to restore the necessary 2V differential between IN and OUT. That probably isn't a practical solution for you since it would require another battery pack wired in series with your existing pack. It would also bring the IN-to-OUT voltage dangerously close of the 40 V limit if both packs are set to 19V.

So maybe a fixed resistor between your 19 V battery and the LED is the most simple solution after all. You need to determine whether your LED has 16.2 V or 14.4 V forward voltage at 0.3 A. For the higher voltage, try an ohms value of (19 - 16.2)/ 0.3, which is 9.33 ohms if I calculated correctly. Nearest standard value with 5% tolerance is 9.1 ohms. Power dissipation in 9.1 ohm resistor at 0.3 A is 0.819 watts, so use a 1 watt or greater resistor. It is always okay to substitute a higher wattage resistor for a lower wattage resistor of the same ohmic value. The current will be somewhat greater than 0.3 A because of the lower valued resistor.

If the forward voltage is 14.4 V, the current with a 9.1 ohm resistor will be (19 - 14.4)/9.1 or 0.5 A, which may be too high. In which case, increase the value of the resistor to 16 ohms. The power dissipation in the 16 ohm resistor at 0.3 A is 1.44 watts, so choose a 2 watt or 5 watt resistor. If I were playing with this I would start with the 16 ohm resistor and measure the forward voltage drop to see if I had the lower voltage or the higher voltage LED. If you have the higher voltage LED, the current will be below 0.3 A and you can safely decrease the resistor to 9.1 ohms. I hate this approach because of the large amount of power wasted in the resistor, but there is also power wasted with the LM317T approach.

Please let us know if your LED survived the advice you received here.