See how long it lasts (or how quickly it fails) if you feed it its max of 100mA.

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Uncle $crooge

 

I'm not being a smart ass. I want to know what i am doing wrong. What should i be looking for? If you would like, i will take a picture of the back, and then you can tell me what i should be looking at. Why does it say "Forward voltage: 100mA" under "Electrical characteristics"?

 

No. It says, "Forward Current= 100mA but only if the ambient temperature is 25 degrees C or less.

100mA at 1.2V is 120mW which is a lot of power for a little plastic-cased LED. 120mW will make it get hot inside. If the LED is mounted without free air flow or is used in summer then it will be in an ambient temp that is more than 25 degrees C and won't be able to dissipate 120mW so its current must be reduced.

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Uncle $crooge

 

Yes, but isn't 25°C room tempurature? So then it should run at 100mA? What would you suggest running it at? And why would you run it at that? Please bear with me, i am still trying real hard to understand.

 

The listed forward current is the maximum (EDIT: *CONTINOUS*)allowable forward current, and you want to stay under the max, so try not to run it at 100mA. Run it at less than that depending on how bright you want it, and the hotter the ambient temperature is, the lower this maximum is. So that's another reason to run it at under 100mA.

You know how a diode only allows current to flow in one direction right (this goes for all diodes, LED, zener, regular diode, schotky, blah blah blah)? Well the diode can only block so much voltage in the reverse direction, once you exceed the Reverse Voltage the diode will start to conduct and be in breakdown mode. Once this happens the voltage across it stays constant. Same thing with the diode when the voltage is in the forward direction, once you exceed the forward voltage the diode will start to conduct current and the voltage across the diode remains constant at the Forward Voltage. If you connect a diode right across a battery in the direction to allow current to flow (regularily, not reversed in breakdown) then it appears pretty much like a piece of wire and is a short circuit and overheats and fries. The reverse current is the leakage current that will still flow through the diode when the voltage is reversed and is not exceeding the reverse voltage- the diode doesn't perfectly block current going in the other direction, a small amount of leakage will occur.

You place a resistor in series with the diode to limit the current across it so it is not a short circuit. Remember how the diode voltage is constant once it starts to conduct? Well you just do some math with a resistor in series with it. Pick how much current you want to flow through the diode. Now you should know the supply voltage you are using (whatever it may be). You know that the voltage across the resistor changes with current, but you also know that the diode is taking a fixed amount of voltage away, so that must means the resistor has the remainder "excess" voltage from the power supply across it. Therefore you know the voltage across the resistor (Vsupply-Vforwarddiode), now just use V=IR to pick the resistance you need to get the current you want, given that you know the voltage across the diode (which you do).

I like to think of voltage as how much energy each individual electron has (higher voltage, each electron is vibrating faster). And I think of current as how many electrons there actually are flowing. The energy in total is, of course, dependent on both. Think of water, more water is the same as more current, and hotter water (or higher pressure) is the same as higher voltage. Current has to do with amount of electrons, while voltage has to do with the energy carried by a single electron.

You know V=IR right? What a resistor does it as current flows through it, a voltage drop develops across the resistor. The resistor "resists" the flow of electrons. When electrons have more energy, they can push harder and more of them can flow since more of them can break "barriers". When electrons pass through a resistor, they get a bit tired and weaker (they "lose some voltage", causing a voltage drop to form across the resistor), but remember that even though they get weaker, there is still the same number of electrons flowing (so current stays the same before as after, this makes sense since if you have 1000 electrons going in, you must have 1000 electrons coming out). NOw that the electrons have worked and pushed through the resistor, they are more tired and since they are more tired, not as many can push hard enough to break barriers, therefore less electrons flow. (I know this makes it sound like 1000 electrons go into a resistor, and some are tired out enough that they can't get out, so only 700 electrons come out, but as I said, the same number in means the same number out, the current on either side of the resistor is the same). Think of it as the electrons from the starting point use some binoculars to look down the circuit to see how many obstacles (resistors) there are, and only the ones that have enough energy travel the obstacle course, and the ones that are too weak don't bother. The other alternative is to think of the entire circuit has one great big wall, rather than many consecuive walls. Electrons try to push against the wall to get out of the battery, the ones that can't get stay inside the battery and don't contribute to current flow. The ones that are strong enough to push through the wall leave the battery travel through the wall and return to the battery causing current flow.

 

It is summer and my air conditioner isn't perfect. It is frequently 30 degrees C. Also, the LED is going to be mounted in something that will block cooling air flow to it.

Not if its ambient air is more than 25 degrees C.

I would probably use 80mA max because the performance isn't much less but the reliability will be greatly improved.

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Uncle $crooge

 

But under "Absolute Maximum Ratings", it says "forward current: 1.2A"! I am so confused! GOD!! I need to get this straight.

I will say this one last time;
On the data sheet, there are two catagorys; Electrical characteristics, and Absolute maximum ratings. I am guessing the colum under Electrical Characteristics are the ratings that are ideal/recomended. I know that the Absolute maximum ratings mean the MAXIMUM. The electrical characteristics are as follows;

Radiant power output (100mA): 16mW min.
Forward Voltage: 1.2v
Forward current: 100mA It says this under Electrical Characteristics, NOT ABSOLUTE MAXIMUM RATINGS.
Viewing angle to 1/2 intensity: 45°

Now under the ABSOLUTE MAXIMUM RATINGS, it says;
Forward voltage (20mA): 1.6V
Reverse voltage: 5v
Forward current: 1.2A This is under: ABSOULTE MAXIMUM RATINGS, NOT Electrical characteristics.
Reverse current: 10uA
Wavelength: 940nm

Now, unless i am missing something, it says, under the RECOMENDED ratings, that the ideal conditions for the Infrared LED, is 1.2v@100mA. THE ABSOLUTE MAXIMUM SAYS: 1.2A(AMPS).

There is no where else that says anything other than: 100mA, and 1.2A.

I don't understand why you think 100mA is the MAX? It says the maximum is 1.2A? Now you have me so confused, ahh! I guess there is only one way to find out who is right, eh? I will test it right now, then post the outcome.

 

Oh, ok. I posted my reply before i saw AUDIOGURU's. 80mA makes scence. Ok, i will try it at that. Thanks. Don't worry, i didn't try it yet. Thank you.

UPDATE: Ok, i looked for a 47ohm resistor, and couldn't find any. I will have to get a bunch of resistors, then i will try it. Thanks though!

 

My mistake, I didnt read the info carefully enough. 1.2A is the absolute maximum current rating that the diode can handle (momentary spikes, not continous), exactl as titled, while 100mA is the continous current rating (at room temperature, decreases as temperature increases obviously). YOu will find most devices have a continous current rating which will cause them to generate heat enough just as fast as it can be dissipated so the device temperature remains constant. Beyond that, more heat is generated than is dissipated so the device will continue to heat up and fry itself. So yeah, most devices have a continous rating, and a momentary/spike/instaneous/peak rating (for voltage and current).

They don't always list it as "continous ratings". Usually they separate peak ratings from regular (continous ratings). They may not label continous ratings as such, but they will always label peak ratings as absolute maximums. So you can usually pick out which is which.

You got into the habit of poring over datasheets...that's good. After a while you will sort of understand what is what exactly (uC sheets- or databooks in some cases are by far the worst).

I was typing more analogies at the end of my last post when you replied so I hope those help.

 

RadioShack doesn't make anything, they just buy it and sell it. They don't have any engineers. Their non-technical people print stuff that is missing all the details and doesn't make any sense.

Look at the datasheet of an IR LED from a manufacturer. It has very detailed maximum ratings that even include the max allowable micro-seconds duration of the 1.2A pulses. The datasheet has derating curves showing the max allowable current when the ambient temp is higher than 25 degrees C.

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Uncle $crooge

 

Yes, thank you dknguyen. Your analogies really helped. I am still alittle confused though, but not as bad. I was able to figure out(on my own), what type of resistor i needed to run an LED off of 9v, and it worked awesome! I think i know enough now to start using resistors. Once i learn alittle more about electronics (i.e. diodes, capacitors, and transistors), i will ask more about resistors(like power disipation, and the like). Thanks guys!

 

Yes, your calculation is correct. The 47 ohm resistor will come very close to dropping about 4.8 volts with 100 mA of current flow. Exactly right.

Yes, the power dissipatoin is the number of watts the circuit will use up.
The total power dissipation is the sum of the power disipated in the resistor plus the power disippated in the LED. In the resistor, it is all turned into heat, but in the LED, some is heat and some is InfraRed light energy. I believe that the conversion efficiency of an IR LED is fairly poor, so most of the energy will become heat. Anyway, the total power drawn from the battery is P=VxI = 6 x 0.1 = 0.6 watts. The resistor accounts for P=V^2/R = 4.8^2/47 = 0.49 watts, so the LED is taking 0.6-0.49=0.11 watts. So the resistor will get pretty warm while the LED will get warm too. This points out that you absolutely must use a resistor rated at at least 0.5 watts and for best reliability, perhaps use a 1 watt rated resistor.

The circuit would only absorb as much as it needed. If the power supply is correctly designed, it would then only convert and deliver as much as the circuit asked for. The value of 2 watts tells you the maximum capability of the supply. Power supplies are usually constant voltage sources, so the way they work is that they provide the fixed voltage that it is designed to (or the voltage you ask for if it is a variable supply) and then hold that voltage steady while you take as much current as your circuit needs (as determined by ohm's law). So the voltage is fixed and the current is pushed into your circuit by the voltage of the power supply. When your circuit is first hooked up and turned on , the current will increase instantaneously until the voltage drop across your circuit equals the voltage available from the power supply. At that point, the power supply voltage is balanced by the voltage drop of your circuit and everything remains nice and steady at that point.

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RadioRon

 

I was scanning through the thread and found this statement and felt I must comment. LEDs and ordinary diodes and transistors do indeed obey ohm's law and to say that they don't indicates that you don't understand ohm's law and are misleading and perhaps confusing our students on the board. Ohm's Law only states that V=I*R. It does NOT state that if you calculate R at one current then that R will remain constant at other currents. This is simply the behavior of a resistor. Ohm's law holds true at any current through a diode, for example, where you can calculate what the resistance is, AT THAT PARTICULAR CURRENT. A diode is simply one of many electronic devices where the relationship of current to voltage is not a straight line. In other words, its resistance varies as current through it varies. Same with transistors.

Off the top of my head, I can't recall any device that doesn't obey ohm's law. Can anyone suggest a device that doesn't and why?

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RadioRon

 

Thanks RadioRon. That cleared up some things. Thanks again.

 

I cannot agree. Ohm's law only states that the "relationship" between current and voltage across certain material remains constant, so V=I*R is not ohm's law but a direct result of applying ohm's law as ohm's law implies that for centain material, the proportional constant is called R so one can work out either the current or the voltage if the other is known, or work out R if both voltage and current is known.

These statements can not be true as active components does not has a constant linear relationship between current and voltage so ohm's law does not apply from the start.

A common filament lamp bulb does not obey ohm's law between the range from zero to normal working current because its resistance has changed many times. But it will still obey ohm's law if it's element is immersed in liquid nitrogen and being kept at a constant temperature.

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L.Chung