Understanding voltage drop

[pme720]

I am working with LEDs, and trying to get a better understanding of how current drive, LED voltage curves and voltage drop all come together. Assume a high powered XLAMP 4550 125mA green LED, data sheet here: http://www.cree.com/products/xlamp_docs.asp.

So I can make this circuit:


+--i|i|i|i|----|LM317|----|R1|-+--|LED|--+
| +------------+ |
| |
+----------------------------------------+

I am no expert a courier circuit schematics, so this is supposed to show a DC (with whatever voltage makes this conversation interesting, see below) power supply, LM317 with a resistor to the ADJ pin for constant current output, and my green LED.

According to the datasheet, the LED's voltage drop is 3.6V at 125mA draw and 3.1V at 35mA draw.

scenario 1
If R1 = 10 Ohms and E = 5.5V I get 125mA constant regulated out of the LM317 (drops 1.9V at 200mA per LM317 National Semi datasheet) which leaves 3.6V for my LED. So my LED gets his 125mA, 3.6V and lights at full brightness. 1.9V x .125A = .2375W dissipated between LM317 and R1, so they are also happy. And I don't need any other resistor in series with the LED, because his current is well under control via the LM317. All correct?

scenario 2
R1 = 10 Ohms and E = 5V. I still get 125mA out of the LM317, but only 3.1V are left for the LED. What does he do? This is my confusion about "current driven" devices like LEDs. The more current, the brighter the LED burns. But there is still a voltage component, the voltage drop curve. So in this example, does the LED light up as though he has 35 mA because at 3.1V that is his current draw? If so where does the extra 90mA in the circuit go? Burned off by the LM317 and R1? Or is the LED at full 125mA brightness, despite the lower voltage?

scenario 3
R1 = 10 Ohms and E = 6V. 125mA to the LED, regulated smooth. But now there is an extra .5V on the circuit, and the LED isn't going to eat it because his voltage drop at 125mA is 3.6V. So this means the LM317 and R1 must dissipate the difference, correct (in this case 2.4V x .125A = .3W which is still OK)?

Final scenario
Take out the LM317 and R1. Make E = 2.9V, just enough to light up the LED. What happens? LEDs are "capacitative loads" not "resistive loads", correct? There is nothing on this circuit limiting current draw(except the wires, whose Ohm value will rise quickly as they get hot). But the LED is not at full brightness (I know, I've done it). What is happening in this short circuit? Is the battery discharging as quickly as it can? Why isn't all that current making the LED burn like the sun?

Thank you for making it through this long post. I wanted to be complete in my descriptions.

 

[Ralpht]

A couple of things first.
The LM317 will give you a max of 1 Amp with correct heat sinking.
The LED has a Max forward voltage drop of 4 V. It needs that voltage across it to fully conduct.
The LED has a MAX forward current of 125mA. any higher and it will overheat and fail, any lower and it will not glow as bright.
LEDS are current operation devices.

We will use the 3.6 volts in your questions as the Max voltage for the LED and ignore the 4V max allowable.

The current thru the LED mus never exceed 125mA otherwise it may/will cook itself.

Using your calculations with a 10ohm resistor - I =125ma (Max) and V=3.6V

I=V/R = I = 3.6/10 = 0.36A = 360mA. Using a 10ohm resistor will destroy the LED.

Wth a fixed voltage drop across the LED of 3.6V, to get the max allowed current of 125mA for the LED -

R=V/I = R=3.6/0.125 = R=28.8ohms. Assuming you are using 5% tolerance resistors, the nearest is 27ohms so to be safe the next highest value you can use is 30ohms. If you are using 1% tolerance resistors then you can get 28ohms. The resistor wattage will be 0.43W so a half watt resistor is needed.

In your scenarios excess voltage or current is used up as heat.

The LED is a diode and it will act accordingly. It will have some capacitive loading effects but will also have resistance and a bit if inductance as well.

In your final scenario, without the resistor, the current limiting effect is the diode itself because you are not giving it the voltage it needs to fully conduct in the forward direction (2.9V instead of 3.6V). It will then limit the current flowing because its not turned on fully and its own internal resistance has a greater effect- it will glow but not at full brightness.

The LM317 will supply for eg. 5V. The LED needs 3.6 volts and will take it, so the remainder will be dropped across the resistor and wasted as heat. The resistor is used to control the current flowing into the LED and is calculated so the LED can not get more than it's max of 125mA.
If you set up the LM317 to give more voltage, ie 6V, the LED will still take 3.6 and the rest, 2.4V will ve dropped across the resistor and wasted as heat.

The LED , being a current device, must have a fixed voltage across it to operate correctly and the current is limited to its required value by the resistor in series with it.

I hope that has at least partially helped with your questions.

 

[westaust55]

LED Driver circuit

I would drop the voltage regulator and just use an NPN transistor to switch the LED as required. See the attached file for a schematic.

With the transistor driven into saturation rather than as a regulating pass device it will generate negligible heat.

My scheamtic is based on an LED capable of up to 500mA but being used at 125mA

Edit:
Found what I think is the right data sheet. 125mA is the maximum forward current so should back off a little to say 100mA which suggests a 22 Ohm resistor.
Then the data also indicated MAX forward volt drop of 4 Volts ! Then its back down to a 12 Ohm resistor.

Pity this site does not permit uploading a spreadsheet - could give you a working example as opposed to a pdf.

 

[pme720]

Ralph: The LM317 is not used as a voltage regulator, but as a current regulator. The LM317 regulates voltage on its ADJ pin to 1.25V, the reference voltage. Per the device data sheet, if you put a resistor (my R1) on the OUT pin and tie it back to the ADJ pin it outputs constant current = Vref/R1. Since Vref is always 1.25V (+/- .05V thats what the LM317 does) that gives Iout = 1.25/10 or .125A to the LED. See here for a better explanation: http://www.reuk.co.uk/Using-The-LM317T-With-LED-Lighting.htm.
Also, I know excess power is dissipated as heat, be it excess voltage or amperage. I want to know WHERE that is happening (I assume in the LM317, since everything downstream is getting controlled current). That is the part I am unsure of right now. You point about the LED being "fully conductive" is very helpful, thank you.

West: I appreciate the schematic and will keep it for further reference. You seem to miss the point, though. I am not interested in how to improve to circuit. This is a very simple circuit that I understand how to construct and control. I want to further my understanding of the relevant devices and Ohm's law, which is why I asked the question.

Thank you both for your responses.

 

[Ralpht]

Hi again pme720,

westaust55 is spot on. He was smart and suggested backing off to 100mA max current. You won't notice much difference in light output. his circuit also looks to be the way to go as well.

But at 100mA and 4V drop the resistor is actually a bit higher because its reducing the current flow a bit more so must be of a higher value, not around 12ohm as suggested by westaus55. (sorry mate) Nearer to 40ohm.
At 3.6V and 100mA, the resistor will be about 36ohm.

But never down to 10ohm, you will have a very bright light for about a second, then a nice burnt smell.....

 

[sghioto]

The 10 ohm resistor is the correct value as stated in pme720's post. He's using the LM317 as constant current regulator. However it will require an input voltage of at least 7.85 volts to pushs .125 amps through the LED. The LM317 requires a minimum voltage drop of 3 volts, there's 1.25 volts across the 10 ohm resistor and 3.6 volts across the LED.

Steve Ghioto

 

[pme720]

Steve: I appreciate your response, and am confused. The national semiconductor LM317 I used has a datasheet available here: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=LM317AMDT-ND
(scroll down and click where it says datasheet)

On page 6 of the sheet, middle graph on the right entitled "dropout voltage". It shows a line at 2V or below for an output current around 200mA. I read that to indicate about a 2V drop on the regulator. Am I reading this graph incorrectly? Is it the correct graph to reference? I found no other source in the datasheet to indicate voltage dropped (I could've missed it!)

If there is another easier/better/suitable source for this info, please send me a link.

Thanks again for the help.

Never mind, I forgot to include the drop over the resistor to the ADJ pin. Thanks!

 

[Ralpht]

Just got your post after I responded to wesaust55.

Actually I noticed you were using the 317 as a current regulator and as sghioto says 10ohms is (maybe) OK in that way but I took it as a hard and unnescessarily complex way to go about it for such a simple circuit. (my opinion only).

As I alluded to in my longer post, any excess current or voltage is gone as heat. The where is - regulator in effect will dissapate most of the heat especially as in your case its acting as a constant current source, but also remember that the resistor , because it has voltage drop across it will also dissapate some as well. Not as much as in the 317 in your'e case but worth remembering.

Don't forget that the LED will dissapate heat as well.