What is the Resistor to use for 2 Green LED in series connected to 6V ?

[audioguru]

It takes NO time to learn about simple electricity (Ohm's Law) that is taught to 12 year old kids.
Ohm's Law is simple and a moron can learn it in 10 seconds, but most sober people learn it in 1 or 2 seconds.

 

[Reloadron]

Ohm's Law is simple and a moron can learn it in 10 seconds, but most sober people learn it in 1 or 2 seconds.

Sober? Now I know why it took me 10 plus seconds.

Ron

 

[carbonzit]

Once again I am thinking out loud here...
Assuming each device[Cellphone, Computer, TV] are all basically nothing more than a resistor to the power supply...more device connected to it....meaning more resistor in parallel, more resistor in parallel means same resistance but more current will be drawn...
When the current is no longer enough, everything grows dim...

Follow this line of thought a little bit further, and bam! you'll start understanding Thévenin's theorem (aka how to "Thévenize" a circuit).

Believe me, you're not that far from putting all this together. Like they say, it ain't rocket surgery ...

 

[Bracer]

If you don't learn simple Ohm's Law then you will be confused forever.

The LED always has about 3V across it when lighted so the current-limiting resistor has 6V across it when the battery is 9V. The value of the resistor and resistance of the battery determines the current with this LED. But other LEDs might be 1.2V to 3.6V so are different.
Ohm's law says that 6V across your 240 ohms produces a current in the resistor and LED of 6V/240 ohms= 25mA. The current in the LED and its brightness will drop as the battery runs down.

So the LED actually literally "drop" the voltage of the power source according to you.
The LED here is in a way a fancy resistor that gives off light [any load is a sort of resistor no ?]
But according to what I have learned thus far, resistors reduce the current [I = V/R] and only the current, and now you are saying voltage drop ?
So a resistor CAN be use to reduce the voltage of the power coming out of its other side ?

What is the formula for that ?

Output Voltage = Initial Voltage [Magic Magic Magic] Resistor ?

It takes NO time to learn about simple electricity (Ohm's Law) that is taught to 12 year old kids.
Ohm's Law is simple and a moron can learn it in 10 seconds, but most sober people learn it in 1 or 2 seconds.

Yes I see that you are smart and in an alternative universe I might even worship you for praising yourself that you are smart, but I am just going to ignore the superior complex here and try to learn.

 

[Reloadron]

So the LED actually literally "drop" the voltage of the power source according to you.
The LED here is in a way a fancy resistor that gives off light [any load is a sort of resistor no ?]
But according to what I have learned thus far, resistors reduce the current [I = V/R] and only the current, and now you are saying voltage drop ?
So a resistor CAN be use to reduce the voltage of the power coming out of its other side ?

What is the formula for that ?

Output Voltage = Initial Voltage [Magic Magic Magic] Resistor ?


Yes I see that you are smart and in an alternative universe I might even worship you for praising yourself that you are smart, but I am just going to ignore the superior complex here and try to learn.

Remember awhile back I mentioned terminology? Electronics is a language unto itself and when communicating in electronics it is important to get the terminology correct.

No a LED is not a resistor in a sense. A LED is a semiconductor device. Or we could say a LED is an active device and a resistor is a passive device. The Vf (forward working voltage is the voltage drop across the LED when the correct If (forward current) is flowing through the LED.

A resistor has what? Resistance correct? All of this basic material is built around Ohms law. Electronics being a language unto itself it has plenty of laws and rules like any language. There are countless math formulas involved with electronics, and Ohms Law is the first basic rule to learn. V = I * R or I = E / R or R = E / I are all variations of this formula. If I pass a current through a resistor and the resistance is a known I can calculate the voltage drop across that resistor. Get familiar with just basic ohm's law. Many electrical / electronic applications are built around it.

<EDIT> While yes, AG is smart, he is not in alternative universe, he is in Canada my northern neighbor. Same universe as the rest of us. </EDIT>

Ron

 

[Bracer]

Thank you so much for your time trying to teach this degenerate Reloadron.

The Vf (forward working voltage is the voltage drop across the LED when the correct If (forward current) is flowing through the LED.

So does this means the voltage that comes out from the other side of the LED is lower then the original source ?
If so, is this phenomenon unique to LEDs ? or even Resistor would cause a voltage drop by the time we measure the voltage on the other end.

If the answer is Yes then it's AWESOME ! But ohms law seems to be violated ?
Because V = IxR, not "Voltage of the other end = Initial Voltage [magic formula here that involve the load and current]"
Basically, what I am saying is there are no two V's in a formula, just one, so what's up with that ?
If we could use load to lower the voltage of a main, we can forget about 7805 or the likes ???

If the answer is no then I am ok with it too because V = IxR, Since Voltage has already been established by the main supply, it cannot change again, we are left with Current to change and these all obeys Ohm's law.

So is the answer yes or no ?
Can something that connects to the power supply [Resistor, LED, Peter Griffin] cause a "Change" in voltage on the other end ?

If I pass a current through a resistor and the resistance is a known I can calculate the voltage drop across that resistor. Get familiar with just basic ohm's law. Many electrical / electronic applications are built around it.

Doesn't that result in an infinite loop ?
Initial Voltage [Example Battery Voltage] = 9V
Resistor = example 30Ω

Resulting Current = 9/30 = 0.3AMP

So now we know this "circuit" will use 0.3 AMP.

Now base on this formula, we "extrapolate" the voltage drop.
V = I x R
V = 0.3 x 30 = 9V
The voltage drop of the 30Ω resistor is 9V ? There will be nothing left [Zero Voltage] to power anything on the other end.????

We could use any resistor I could name, but it seems based on this circular reasoning, the only voltage drop is the complete depletion of the voltage itself.

How is it that a Component like LED claim itself to have a voltage drop ?
V = I X R, The V is the main Battery Supply, the LED is the Load [R]...but the LED is not measured in ohms....mmm....but it does say what is its maximum AMP before it explodes....To reduce the Current, we have to put in a Resistor...Yes....that makes sense..then the causality is complete!, wait....but there is this thing call the "voltage drop"...what is that ? where did that come from ?

These ALL BOGS DOWN to a single problem.
I don't know the formula to use to calculate voltage drop nor could I presently "comprehend" such a phenomenon because according to Ohm's Law, there is only one V in the formula and the Battery's voltage has set the V in stone, it cannot be change.
If it can then where is the formula for it ?

[Again I am very very sorry that I have the brain of an ape]

 

[Reloadron]

Back around post #37 I made a little cartoon (Schematic) with meters showing voltage drops.

Next:

[Again I am very very sorry that I have the brain of an ape]

If you choose to believe that, I would suggest you look to more simple things than electronic theory. I have little patience for people who choose not to believe in themselves or their abilities. Got it? Personally I don't know any apes that are active and posting in forums, so it would be a first on that note.

Moving along, if I place a 1 ohm resistor across a 9 volt battery (it better be a big 9 volt battery with a high Amp Hour rating) the current will be 9 amps. The voltage across the 1 ohm load will obviously be 9 volts.

Do yourself a favor for awhile and forget LEDs!

Now I place three 1 ohm resistors in series across the same 9 volt battery. I have a R total of 3 ohms. Look at the attached circuit. What will the total circuit current be now? What will the voltage be if I measure A to B? A to C? A to D? B to C? B to D? C to D?

Ron

 

[carbonzit]

OK, let me take a crack at this. By the way, this is probably the stumbling block for most newbies learning about electricity and electronics, wrapping their head around all that confusing stuff about voltage, current, resistance, etc., etc.

**broken link removed**

For the purposes of this lesson, yes, you can think of a LED as a resistor. It does, in fact, have what's called "internal resistance". You can actually measure it with a meter if you like. (Of course, being a diode, its resistance in one direction is different from that in the other direction.)

Let's not worry about what exactly that resistance is for now. Your problem is to figure out what size resistor to use in this circuit to prevent the LED from being smoked. (The resistor's job is to limit the current through the LED to a safe level.)

So you apparently know something about Ohm's Law. Let's get the three forms of this law down. I'm going to use the traditional "E", which means voltage:

1. E = I * R
2. R = E / I
3. I = E / R

(you can see how these are all derived from each other; basic algebra, rearranging equations)

So since we can treat the LED as a resistor (because, like, I told you so up there), what we have is a circuit with a voltage source (the battery) and two resistors.

Now, we have to have some numbers to plug in to these equations to make them work. Gotta start from somewhere. So let's say we look at our LED datasheet and see the following:

Vf (forward voltage): 2.1V
If (forward current): 20mA

And let's say that's a 9-volt battery over there. So let's start by figuring out the voltages around the circuit.

By another law you're going to need to know (Kirchoff's voltage law), we know that the sum of all the voltage drops around a single closed path in a circuit is equal to the total source voltage in that loop. (Got that from a textbook; you may see it stated slightly differently in different sources, but that's the gist of it.) So on one side we have, what? 9 volts, right? So we know that the voltage drops (that is, the voltages across each part) on the right side have to add up to 9. We know that the voltage drop through (across) the LED is 2.1V, from our datasheet.

So simple arithmetic gives us 9 - 2.1 = 6.9. So we have the following data:

Vled = 2.1V
Vr = 6.9V

So far, so good. Now we need some "magic", as you put it. And where will that magic come from? From Ohm's Law, of course!

We know 2 things:

- the resistor wants to have 6.9 volts across it
- the resistor will have 0.02A running through it

(That second piece of information comes from the number from the datasheet, and from another law you need to know, Kirchoff's current law: The sum of the currents entering a node is equal to the sum of the currents leaving that node. Don't worry about "nodes" just now: just realize that this means, in our case, that in a simple circuit like this, the current (I) through the circuit is equal through all parts of the circuit.)

The thing we don't know is what the resistor R should be. Gosh, if only there was a formula to determine that ... hey, there is! Remember R = E / I up there? Let's plug in the numbers:

R = 6.9 / 0.02 = 345

Presto! we need a 345 ohm resistor.

See how that works?

 

[Bracer]

Thank you carbonzit !
These two new things that you have brought up, Kirchoff's voltage law, Kirchoff's current law explains so much, I was literally reading your post like a student listening to his teacher !
Thank you !

So the current through the entire circuit is the same…but the voltage on each side is different but all adds up to equates to the main supply's voltage !

THANK YOU!!!!!

THANK YOU!!!!!!!

THANKS!!!!

I can understand why you are angry at me Reloadron but based on what carbonzit has clarified in a way I do not understand before, let’s try your question.

**broken link removed**




Now this brings yet another question, since the voltage drop is dependent on the number of loads in the circuit to "spread the voltage around", how could the LED "claims" it have a certain voltage drop ?

Unless....unless the LED will die not just because of too much current...but too much voltage as well!

So!
A "typical" LED will die if more than 0.02Amp goes pass it OR since it have a voltage drop of 2 volts, it can ALSO DIE from too much voltage even when the current is crap low !

Yes ? No ?

If the answer is yes, my understanding is complete

 

[Reloadron]

I am not angry at you, hell I don't even know you. What I said was I get annoyed with people who underestimate themselves. You did fine picking the schematic apart. Good job!

Kirchoff's laws for voltage and current in series and parallel circuits follows ohms law.

So!
A "typical" LED will die if more than 0.02Amp goes pass it OR since it have a voltage drop of 2 volts, it can ALSO DIE from too much voltage even when the current is crap low !

Since a LED is a current device then yes, it will die if the current through it exceeds its rated current. If we increase the voltage accross a LED the current through the LED will increase. Eventually the LED is toast. The trick is just a matter of limiting the current through the LED and keep it within a safe operating range. See, you seem to now have a good basic understanding of what is going on.

Ron

 

[Bracer]

Yes, I can see that an increase in voltage automatically leads to an increase in current if the resistance stays the same.

It's complete.

It's Complete

 

[audioguru]

But an LED is not a resistor, it is a solid state diode.

If you feed 2.2V to a 110 ohm resistor the current will be 20mA.
Then if you feed 2.6V to the 110 ohm resistor the current will increase to only 23.6mA.

If the LED voltage is 2.2V at 20mA then if you feed it 2.6V its current might be 70mA and cause it to burn out.

 

[Bracer]

You are right, but had carbonzit and Reloadron not stood down to my level, I would not have understand the law.
I am grateful to them, really

I see your chart.
Fascinating...
Then is there a way to "calculate" the Current usage of a Typical LED Based on Voltage and erm...

[I can see the pointlessness of this question as nobody would care for the current an LED can draw once it's dead, ha ha ha but I am just curious as it does seems to go against ohm's law here, maybe because LED is not an Ohm which is what you are stressing here too ]

 

[audioguru]

You feed an LED with a current, not with a voltage. The current-limiting circuit or resistor in series with an LED sets its amount of current.

 

[Bracer]

That's not what I am asking Let me try rephrasing

Your Chart shows the Current Consumption goes up when the Voltage increase.
Is there a way to calculate the Level of current consumption by the LED based on the Forward Voltage other than looking at your chart ?

 

[audioguru]

Every LED part number has a different forward voltage and a different slope on a graph. Even when LEDs have the same part number and the same manufacturer their forward voltages are probably different. A forward voltage for an LED is a range of voltages.
Also, the forward voltage changes as the temperature changes.

Again and again, feed a current to an LED, not a voltage.

 

[kpatz]

Another way to look at how the "voltage drop" of a LED (or any junction semiconductor for that matter) works.

A resistor has a fixed resistance. A 300 ohm resistor is always 300 ohms (+/- some small tolerance). The "voltage drop" across the resistor depends on the supply voltage and how the resistance is divided up across the circuit, per Ohm's Law. The current flowing through the resistor is determined by I=E/R, where E is the voltage across the resistor and R of course the resistance in ohms.

To take a simple example of a resistor's voltage drop, take 2 300 ohm resistors in series and connect them to a 9V battery. Each resistor will drop half the voltage, or 4.5 volts each, and the circuit will draw 9/600 = .015 amps of current, or if you calculate the current through one resistor, you'd get 4.5/300 = .015. Now connect the resistors to a 12V battery. Now the voltage across each resistor will be 6V and the circuit will draw .020 amps of current (12/600 or 6/300 each resistor).

An LED, instead of having a fixed resistance, has a fixed voltage drop. In any circuit, an LED with a 2.1V rating will drop 2.1V in the circuit, regardless of the supply voltage. So, if you take a 2.1V LED and put it in series with a 300 ohm resistor and connect it to a 9V battery, the LED will drop 2.1V, leaving 6.9V across the resistor. The resulting current would be 6.9/300 = .023 Amp (23 mA). Connect it to a 12V battery, the LED will still drop 2.1V, leaving 9.9V across the resistor, causing the current to increase to 9.9/300=0.033 amps (33 mA, pushing the limits of most LEDs).

So, unlike a resistor which obeys Ohm's Law and only allows E/R amps of current through, an LED will "allow" nearly unlimited amounts of current through it if the supply voltage is near or above the LED's forward voltage. Of course, this invariably results in a smoke-emitting diode which is impractical in most circuits, so that's why you need a current-limiting resistor in series with the LED.

 

[carbonzit]

You are right, but had carbonzit and Reloadron not stood down to my level, I would not have understand the law.
I am grateful to them, really

Thanks, but I don't consider that stooping to your level; it's just the challenge of the teacher to make a difficult concept easier to grasp.

I see your chart.
Fascinating...
Then is there a way to "calculate" the Current usage of a Typical LED Based on Voltage and erm...

Well, as it turns out, Ohm's Law does not apply in this case! Shocking, no? But Ohm's Law only applies to devices that have linear resistance characteristics. If you look at audioguru's graph, you can see that a LED is a decidedly non-linear device.

You feed an LED with a current, not with a voltage. The current-limiting circuit or resistor in series with an LED sets its amount of current.

Not quite correct. One "feeds" a LED with both the correct voltage and current. You're correct in saying that one must limit the current to the LED's safe value, but one also doesn't want to apply too much (or too little) voltage across the diode.

To try to clear up your remaining confusion, Bracer, about how a LED can have a voltage drop and how to operate it at the correct voltage and current, look at audioguru's graph again. Notice how the LED curves go steeply upward as the voltage increases, meaning that for small increases in voltage, they draw a lot more current, until they reach the point where they go Poof!.

So what you, the designer, have to do is to pick a point on that curve--the LED's V/I curve--where the LED will be nice and comfy and safe. F'rinstance, for the red LED on that graph, let's pick a point somewhere near where it crosses the red line. Then you use the calculations I showed you earlier to set the size of the resistor(s) in the circuit to put the LED at that "operating" point (the voltage and current you picked from the curve on the graph). That way you can be sure that the LED is operating where it should be, with little danger of burning out, and with enough current and voltage to generate a nice amount of light.

Regarding why LEDs have a voltage drop, remember how I told you that you can treat them as a resistor back there? Yes, audioguru is correct that they are not actually resistors, but hey, you knew that anyhow. Nevertheless, they do have resistance. All electrical devices do; fundamental laws of physics and all. Even the wires we use to connect things in a circuit have resistance. Therefore, having resistance, it has a voltage drop, as all resistances do.

Does this help any?

 

[Bracer]

This helps tremendously carbonzit !

Audioguru is a bit harsh but I want to believe his heart is in the right place, I mean, he could have ignore this retarded thread, but he provided me a chart and he tried to explain to me from his knowledge-able point of view which I simple cannot relate due to my lack thereof.

An LED, instead of having a fixed resistance, has a fixed voltage drop. In any circuit, an LED with a 2.1V rating will drop 2.1V in the circuit, regardless of the supply voltage.

Thank you so much kpatz for explaining this voltage drop “phenomenon” to me, you’ve clarified the matter to me the fact that the voltage drop of the LED is of a constant type.

This is very important to me because as someone that JUST UNDERSTOOD how to calculate the voltage on the other side of a component thanks to Reloadron and carbonzit (You have no idea how much they have help me in my understanding of this area) and two new Kirchoff law introduced to me by carbonzit makes everything makes sense) makes me “thought” that voltage drop is DEPENDENT on the number of loads on a circuit and must be divided among them [not equally ofcourse as it would depend on each load’s internal resistance as Voltage Drop on the end of the component = Overall Circuit Current x component’s own resistance, seriously I am so grateful for them] and here you are with a BUMP! LED’s voltage drop is NOT dependent on the number of loads and it stands alone as the Alpha and the Omega of constant Voltage usage.
Thank you kpatz.
I now learned even more !
SO HAPPY!

Well, as it turns out, Ohm's Law does not apply in this case! Shocking, no? But Ohm's Law only applies to devices that have linear resistance characteristics. If you look at audioguru's graph, you can see that a LED is a decidedly non-linear device.

Yes I am shocked, you are right, you absolutely know where my plane of understanding is at the moment, now I learn something new from you again, that ohms law only works on resistance that possess a linear resistance characteristic.
That is very new to me, so….Ohm’s law is not God, doesn’t always work….

I LOVE you carbonzit

 

[carbonzit]

OK, Bracer, I think you're finally ready. I've devised a little test for you to show that you can now figure this out on your own. (You knew there'd be a test, didn't you?)

Don't worry, it'll be fun, since I'm making it part test, part Easter egg hunt.

So here's the test question. Lessee, turn to my teacher's handbook ... ah, here it is:

You're asked to put 3 LEDs in a device. They will be wired in series, and will be powered from a 12-volt supply.
The customer has chosen green LEDs, model LT E6SG-V2AB-36-1-Z. Find the proper size for the current-limiting resistor to use in this circuit.
As always, show your work.

So you need to first find these parts. Don't worry, I'm not playing with you; they're real devices. Find them, and find their specs.

Go to it. You can do this!