How to power an LED, NEED CLARITY.

[Ivin]

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.

 

[Anon_LG]

Welcome to EP, I hope you continue with electronics as a hobby and continue experimenting

First of all voltage does not "travel" as such, it is potential difference measured between 2 points. Voltage will be dropped across a resistor however it is used to limit the amount of current flowing through a circuit.

I=V/R this can be re-ordered, see the resource here

so 16 Volts / about 53 ohms = 0.3 amp (300 miliamps)

You can work out what value of resistor you will need to make a certain amount of current flow through a component by

V/I = R (so thats the desired resistance) you plug in the numbers

V=16
I=0.3 (300 miliamps)

16/0.3 = (about) 53

You will not find a standard resistor with this value cut because an exact value is not needed a near by one can be chosen so use a 56 ohm, it is best to choose a higher value rather than lower. However if you use an individual resistor like this you will need a high power rating so I will provide the correct parallel equation and explanation.

If you were to use one resistor it would be taking about 3 and a half watts, this power rating can be quite expensive and so it is more convenient to put the in an arrangement called parallel to distribute the power between them. If you use the configuration below it will provide about 120 miliamps to the LED. You will need resistors with a 1 watt power rating. If you use a calculator and the parallel resistor equation you can calculate the configuration to suit your needs.

 

[Ivin]

Welcome to EP, I hope you continue with electronics as a hobby and continue experimenting

First of all voltage does not "travel" as such, it is potential difference measured between 2 points. Voltage will be dropped across a resistor however it is used to limit the amount of current flowing through a circuit.

I=V/R this can be re-ordered, see the resource here

so 16 Volts / 4.8 ohms = 0.3 amp (300 miliamps)

That was so incredibly fast...
Thanks so much. ill read that now.

 

[Anon_LG]

I am currently editing the above post, I just wanted to put it there so that you know a reply is coming

 

[Anon_LG]

The 4.8 ohms I put originally was wrong. Sorry.

 

[Ivin]

OH WOW! i really was not expecting such an awesome answer in such a short amount of time. this is so great, i really apreciate you taking the time! your post really helped me move on and im no longer at a roadblock, thanks!

im finding all of this alot of fun to learn.

 

[Anon_LG]

Thanks! Happy to help. If you want to do more resistor calculations for any projects Yenka is a good simulator, its free for use (outside of school hours) and extremely easy to use.

 

[Ivin]

Thanks i will definatly check it out, i was wondering if there was something like that available about 10 minutes ago

 

[Bluejets]

It's not a good practice to use a system as you have done here.
Either select a lower voltage battery preferrably closer to the lED value.
If that 16volts is required in some other section of a circuit then regulate down to what is required for your LED.
Linear reglators waste power and heat so no better off than with resistors.
UBEC type will do the job and at low cost.
Some 3Amp ones here that will drop your 16V down to 5V.
http://www.ebay.com.au/itm/Hobbywin...Radio_Controlled_Vehicles&hash=item20ea67bfca

 

[Ivin]

It's not a good practice to use a system as you have done here.
Either select a lower voltage battery preferrably closer to the lED value.
If that 16volts is required in some other section of a circuit then regulate down to what is required for your LED.
Linear reglators waste power and heat so no better off than with resistors.
UBEC type will do the job and at low cost.
Some 3Amp ones here that will drop your 16V down to 5V.
http://www.ebay.com.au/itm/Hobbywin...Radio_Controlled_Vehicles&hash=item20ea67bfca

But this LED does use 16V does it not? I could be reading The Spec Sheet wrong but isn't Forward Voltage the amount required for the LED to operate? In this case a range around 14.4v to a max of 16.2v. Thanks for looking at my question, the more eyes / brains the better!

 

[Bluejets]

Sorry...missed that.

 

[(*steve*)]

@Bluejets makes a good point in regard to the power lost as heat. However the higher this wastage the "better" resistors are. (sounds weird?)

The reason is that if you use a pure resistive method to adjust the current, where the overhead voltage is small (esp wrt the LED voltage) any change in the input voltage or the V_f of the LED will produce a larger than proportional change in current. For example a 5% change in voltage may produce a 15% greater current.

Since high power LEDs generally run "hotter" than small LEDs, protecting them from thermal runaway is important. Even if you don't get into a true runaway , you can significantly shorten their lives.

The solution is to use a constant current driver. I believe @Ivin was pointed at the LED resource. This lists a fairly simple constant current driver.

Now, *if* this LED had a forward voltage of 2V, and you wanted to run it from 16V, then yes, you would be more efficient to use a switchmode regulator to drop the voltage first, then use some form of constant current driver. Alternatively (and even more efficiently) you could simply use a switchmode constant current driver. (however I think the issue is moot in this case)

 

[hevans1944]

An LED operates as a forward-biased diode. The current through an LED is therefore a quadratically increasing function of the voltage. That means small changes in the forward voltage drop are produced by relatively large changes in the forward current. For this reason (among others) an LED works best with a constant-current power supply rather than a constant-voltage power supply, such as a battery. More on this later.

The spec sheet lists a range of 14.4 V to 16.2 V forward voltage on the LED at a current of 300 mA. That means the forward resistance of the LED, at 300 mA, can vary from 48 ohms (at 14.4 V) to 54 ohms (at 16.2 V). A particular sample is specified by the manufacturer to fall within this range. If you place a "dropping" resistor in series with the LED and the battery, that resistor must drop the difference between the LED forward voltage and battery terminal voltage. So, for a 16 V battery terminal voltage and 14.4 V LED forward-bias, the resistor must drop 1.6 V at 300 mA. The resistor required to do this would be 1.6/0.3 ohms or 5.3 ohms. The power dissipated in the dropping resistor would be about a half watt.

If you happen to have an LED that drops 16.2 V at 300 mA, it will not achieve that current with the 5.3 ohm resistor in series with it. That doesn't mean it would not emit infrared light, but the current would be reduced. To calculate how much it is reduced would require that you know the LED voltage drop as a function of current. Unfortunately the manufacturer does not provide that information. Still, it should be safe to use a 5.3 ohm dropping resistor because 14.4 V is a "worst case" minimum voltage drop across the LED. A larger LED voltage drop requires a lower-valued resistor (to achieve 300 mA ) until the value reaches zero ohms.

You could use the 19 V battery terminal output to drive an LED with 16.2 V forward voltage at 300 mA. Then the dropping resistor would have 2.8 V across it at 300 mA and a value of 9.3 ohms, dissipating a bit less than 1 W. Note this higher-valued resistor would also work for an LED with 14.4 V forward voltage drop but the LED current would be somewhat greater than 300 mA. Again, the manufacturer does not provide enough data to calculate how much greater.

A better solution than a dropping resistor in series with a voltage source is a constant-current regulator between the voltage source and the LED. If the compliance voltage of the constant-current regulator is at least as large as the largest expected voltage drop across the LED, the constant current regulator will provide the specified current of 300 mA.

A simple and affordable solution is to use the National Semiconductor LM317T voltage regulator configured with a 3.9 ohm 1/2 watt resistor connected to the OUT terminal. The other end of the resistor connects to the ADJ terminal and to the LED anode. The LED cathode connects to the negative terminal of the battery. Apply 19 V from the positive battery terminal to the IN terminal of the LM317T.

The LM317T always maintains a constant voltage, nominally 1.25 V, between the OUT terminal and the ADJ terminal by internally varying the conductance between the IN and OUT terminals. It does requires at least a 2 V differential between IN and OUT to maintain regulation, which means the OUT terminal must be 17 volts or less with a 19 V battery supply. Worst case is driving a 16.2 V LED at 300 mA: there will be 16.2 V + 1.25 V or 17.45 V at the OUT terminal, which is 0.45 V more than is required to maintain a 2 V differential between IN and OUT. The LM317T will regulate up to 1.5 A with a proper heat sink. Visit the National Semiconductor website to find and download a datasheet with application examples, including this one.

 

[Colin Mitchell]

The illuminator MAY have an internal regulator as the specifications say 14.4v to 16.2v @ 300mA and 16v is within this range.
You may not need a dropper resistor AT ALL.

 

[Ivin]

Wow! Thanks for the responses. I still have to read over the latest but I wanted to let everyone know that I really appreciate it.
...
Awesome, there is a lot more info/options now to research. It's late now, but I'll have a full day tomorrow to learn more, thanks!

 

[hevans1944]

From inspection of the pdf datasheet at high magnification for the OD-669-850 HIGH-POWER GaAlAs IRED ILLUMINATOR it is apparent that the device consists of nine series-connected LED "chips" in a 3x3 array. The series wire connections are clearly visible. There is no internal regulator, which can also be inferred by reading the reverse leakage and reverse voltage specifications.

Dividing 16.2 V by 9 yields 1.8 V forward-bias voltage per LED. Perhaps not by coincidence, multiplying 1.8 V by 8 yields 14.4 volts. I suspect the manufacturer allows for one (and only one) non-functional LED chip per packaged device. The now-functional LED chip is then shorted to allow the remaining eight chips to function normally.

I have a correction to my statement that the forward voltage is a quadratic function of current. At moderate currents, diodes obey the Shockley diode equation where the diode current is an exponential function of the junction voltage.

 

[hevans1944]

@Bluejets is a radio-controlled (RC) model enthusiast. It is understandable that he has extreme disregard for either dropping resistors or linear regulators. @Bluejets is absolutely correct that either one wastes power and produces heat. In a model airplane extra weight is a performance killer, and so is wasted battery power. However the recommendation to use a UBEC to drop the battery voltage down closer to the LED forward-bias voltage doesn't fly. The LED needs a constant current source, not a constant voltage source. A modern UBEC (or BEC: battery eliminator circuit) uses a switch-mode power supply to provide a lower operating voltage from a higher battery voltage. Presumably the higher battery voltage powers propellers for lift while lower voltage is required for on-board electronics such as receivers, electronic speed control, servos, cameras, etc.

If you could find a UBEC that provides a constant current output at 300 mA you would be all set. It would need to have a compliance voltage of at least 16.2 volts (more would be okay), and it would probably work just fine with the lowest terminal voltage from your battery. Some energy harvesting switch-mode supplies work down to a few tenths of a volt, but their output current is very low. They are typically used with a super-capacitor for intermittent applications, charging the super-capacitor from available environmental energy between uses.

Many high-power LEDs are pulse-driven, the pulse width being determined by the current supplied to the LED. The advantage is almost no power is dissipated in the switch device (typically a power MOSFET or IGBT) when it is conducting (providing current to the LED) because the voltage drop across the switch is very small when it is on. When the switch is off, no power is consumed because there is no current through the switch, although there is a voltage drop across the switch. Many such circuits store electrical power in the magnetic field of an inductor between on pulses, increasing efficiency at the expense of complexity.

The major losses in a switch-mode power supply occur during the finite times required to turn the switch on or off because the switch has both current through it and voltage across while switching. An advantage is higher peak power from the LED when it is on because the on current can be increased while the average current remains at a lower level, the ratio depending on the duty cycle of the pulse (on versus off time). This can be very useful for an infrared illuminator working with a CCD or CMOS imager where the on pulse is synchronized with the frame rate (or in the case of NTSC, the field rate). It might even be useful with night-vision or starlight viewers, because of phosphor decay time, if the pulse repetition rate is high enough. Unfortunately the cost and complexity is way beyond what is needed for a battery-powered infrared flashlight.

@Ivin: Congratulations on your budding interest in electricity and electronics. This is one of THE best forums on the Internet to get quality help. Oh, and make sure your LED has a proper heat-sink attached to dissipate the 5 watts of power it consumes.

 

[(*steve*)]

However the recommendation to use a UBEC to drop the battery voltage

Ah. Thanks. I had not previously come across this ETLA for SMPS.

If you could find a UBEC that provides a constant current output at 300 mA you would be all set.

True

Many high-power LEDs are pulse-driven, the pulse width being determined by the current supplied to the LED. The advantage is almost no power is dissipated in the switch device (typically a power MOSFET or IGBT) when it is conducting (providing current to the LED) because the voltage drop across the switch is very small when it is on. When the switch is off, no power is consumed because there is no current through the switch, although there is a voltage drop across the switch. Many such circuits store electrical power in the magnetic field of an inductor between on pulses, increasing efficiency at the expense of complexity.

Here you simply describe a SMPS (or UBEC in Bluejet's vernacular)

The major losses in a switch-mode power supply occur during the finite times required to turn the switch on or off because the switch has both current through it and voltage across while switching. An advantage is higher peak power from the LED when it is on because the on current can be increased while the average current remains at a lower level, the ratio depending on the duty cycle of the pulse (on versus off time). This can be very useful for an infrared illuminator working with a CCD or CMOS imager where the on pulse is synchronized with the frame rate (or in the case of NTSC, the field rate). It might even be useful with night-vision or starlight viewers, because of phosphor decay time, if the pulse repetition rate is high enough. Unfortunately the cost and complexity is way beyond what is needed for a battery-powered infrared flashlight.

I think you're conflating PWM with Switch mode regulation here. Whilst there's a relationship, they're not the same.

In addition I²R losses are higher if you increase peak current whilst maintaining average current.

The other issue is that LED efficiency typically drops with current, so operating the LED at ten times the current for 10% of the time will typically yield less total light.

Oh, and make sure your LED has a proper heat-sink attached to dissipate the 5 watts of power it consumes.

Definitely.

 

[hevans1944]

@(*steve*): Thanks for noticing. I guess I got a little long-winded and somewhat off-track there. All @Ivin needed to hear from me was in the last two paragraphs of post #13. I will wait for further questions from the OP before posting to this thread again.

You are absolutely right about LED efficiency going down with increased current. I was thinking about laser diodes, which are an entirely different component. So, never mind on pulse modulating LEDs for more peak power. Works okay for simple LED optical communication though, which is so far off-topic I should quit right here.

 

[(*steve*)]

I guess I got a little long-winded and somewhat off-track there.

And here was me thinking that was my job here