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

That chart is intense Reloadron, thanks.

Thanks for taking the time to explain the manual way of retrieving the specs of the LED, even though you say it isn't an exact science, it does serve to answer my quest here.
Thanks

You know I just realize that the formula:
(Battery_Voltage-LED_Forward_Voltage)/LED_Forward_Current
Didn't take into consideration the Batteries or the Power source's current.

hmm...so does this formula automatically assume a benign current is used here ?
I am pretty sure that this formula will work fine with a 6 Volt battery but if say a hypothetical 6 volt with 1000000000~Whatever Amp, the LED will die.
So I guess this formula is lacking here ?

What is the formula that takes into consideration the source current then ?

You don't need a formula.
Instead you need the datasheet for the LED, common sense and Ohm's Law.
If the battery voltage does not drop when it is loaded then the value of the current-limiting resistor is simply calculated with simple arithmatic if you look at the forward voltage and recommended current of the LED on its datasheet.

Example:
6.0V battery. 2.0V LED. 20mA is recommended.
Then the resistor has 4V across it and 20mA in it so its value is 4/0.02= 200 ohms. Simple.

What is the formula that takes into consideration the source current then ?

Click to expand...

You only care about the LED current. It matters not what the source can provide. With the proper series resistor the LED will be limited to drawing its rated current. If you look at the formula Vsupply - Vled / Iled you have a version of ohms law to solve for R.

Ron

Ok so, say we have:
One 9 small volt battery
Red LED :: Forward Voltage: 2 Volts :: Forward Current: 0.020Amp.

The resulting calculation would be 350Ω.

Now all is fine and dandy here say I use one of those ¼ watt resistors.

But imagine, what if I were to connect 10 Billion 9 volt batteries in parallel.

The question here would be, would the resistor still be a ¼ watt 350Ω ?

Common sense tells me that I would have to use a higher watt rating resistor.

Now here is where the Achilles heels of ohms law is.
Because the logic of Ohms law stats that V = I X R.

Since we already know V and R, current cannot be change just because I added 10 Billion 9 Volt batteries in parallel.

It seems according to ohms law, the current will always be I = V/R, hereby 0.025~Amp regardless of how many Zillion 9V batteries are connected in parallel.

The problem here is, we all know this to be completely untrue, when you connect batteries in parallel, their voltage is the same but the current goes up.

But it seems in this paradox, ohms law here states that it have no other value to be then 0.0257.

Am I missing something here ?


Another example, I recently use a smaller battery that have the voltage I need but when connected to my circuit, the LEDs are dim like a joke, unlike the bigger battery that I use, now everyone including me would be able to sound like a genius here by saying things like "Ooo this is because the current is not high enough" or "there is not enough current in the small battery".
But do you guys see the irony of this reply if we were to respect ohms law ?
The batteries voltage is fixed.
The circuit [hence its resistance] is fixed.
Who gives the size of the batteries any right to be of a lesser current ?
The batteries shouldn't have any say left because its I is now a derived value from I = V/R.

But we all know this is not true because the battery's current does matter, so what's going on here ?

Try to think of it more in watts .. the LED will consume 20mA at 2v = 4 milliwatts the resistor will consume 20mA at 7v = 140 milliwatts at total of 160 milliwatts consumed, now if the battery is say 9v and can deliver a consumer 100 miliwatts, it stands to reason that 2 batteries would be required to do the job (wouldn't be for long) . The long and hard of it is, if the consuming device requires more current that the battery can provide the voltage will drop.

Cheers Ian

The battery in my car can supply 700A to the starter motor when the engine is very cold.
But the clock in my car draws only about 5mA from the same battery all night long.

If you connect a zillion 9V batteries in parallel then the available current is increased if you have a load that will draw that much current.

In my attempt to understand this simple law, I had limit my vision to just Volt, Resistance and Ohm.

But thanks to your suggestion Ian Rogers, I have attempted to bring "watt" into the equation and now everything is "starting" to make sense to me.

It feels good that everything is united, that they are not their own separate theories.

I am thinking aloud here, Say we have 6 Volts, Red LED Forward Voltage: 2 Forward Current 0.02Amp.
(6-2)/0.02 = 200Ω

Current = Voltage/Resistance :: Hence Current = 0.03 Amp
(What exactly is this 0.03 now ? is this the amount of current used by the LED in Total ? Or the Amp taken by the entire circuit ? I don't think this is the case as we are just dividing it by one of the component's resistance in the circuit, LED itself have some resistance in which this calculation has not taken it into account.)

W = V x A :: Hence Watt = 0.18 Watt.
W = V² / Ω :: Hence Watt = 0.18 Watt.
W = A² * Ω :: Hence Watt = 0.18 Watt.

So beautiful...the equations are connected, Watt, Resistance, Current, Voltage...all related...so beautiful...no strays...

======================================================
I think I am really starting to understand here...

audioguru said:

If you connect a zillion 9V batteries in parallel then the available current is increased if you have a load that will draw that much current.

Click to expand...

All this thinking is starting to make this sentence makes sense....you are saying that the circuit will only draw the amount of current that it "needs" to and no more.
So if I were to connect a Device that needed 6V whos own internal overall resistance is say 200Ω which would mean the device itself needs only 0.02Amp.

EVEN if I were to connect this "device" to a power supply of 6 Volt with a ZILLION Amps, everything is still going to be ok ?
Simply because of the charity behavior that the device will only take what it wants and the power supply will not kill it with an over current ?

By this measure one could not, theoretically, "destroy" a device by giving it an overpowering amount of current, because the device will only "take when it needs" WHEN it comes to current right ? (Voltage is another matter, over-voltage WILL kill a device because that is how CMOS chips get destroyed.)

Now if this logic is true...how is it that Human beings could be zapped to death by putting his finger in the main ? People say it's because of the Current but if the "device", in this case the Human body, will "Only take what it needs", there could theoretically be no case where it will take "more than what it needs" hereby killing the Human body no ?
But we do KNOW it will kill us...so what's going on here ?
This Result (death) seems to imply that a device just "take it all in", instead of the "The device just takes what it needs, the main's huge Amps is there and is good for supplying it if the device require it" charity theory.

Bracer said:

I am thinking aloud here, Say we have 6 Volts, Red LED Forward Voltage: 2 Forward Current 0.02Amp.
(6-2)/0.02 = 200Ω

Current = Voltage/Resistance :: Hence Current = 0.03 Amp

Click to expand...

Current = Voltage/Resistance is Ohm's law and it only applies to resistors. (Ok, there has been a big discussion on this forum about where it does and doesn't apply. I'm not getting into that)

Here the voltage across the resistor is only 4 V so the current is 4 / 200 = 0.02 A

Bracer said:

EVEN if I were to connect this "device" to a power supply of 6 Volt with a ZILLION Amps, everything is still going to be ok ?
Simply because of the charity behavior that the device will only take what it wants and the power supply will not kill it with an over current ?

By this measure one could not, theoretically, "destroy" a device by giving it an overpowering amount of current, because the device will only "take when it needs" WHEN it comes to current right ? (Voltage is another matter, over-voltage WILL kill a device because that is how CMOS chips get destroyed.)

Now if this logic is true...how is it that Human beings could be zapped to death by putting his finger in the main ? People say it's because of the Current but if the "device", in this case the Human body, will "Only take what it needs", there could theoretically be no case where it will take "more than what it needs" hereby killing the Human body no ?
But we do KNOW it will kill us...so what's going on here ?
This Result (death) seems to imply that a device just "take it all in", instead of the "The device just takes what it needs, the main's huge Amps is there and is good for supplying it if the device require it" charity theory.

Click to expand...

Well a person isn't designed for electricity at all, so there is no correct voltage for a person. It happens that currents over about 50mA can be fatal. The voltage needs to be about 100 V to produce that sort of current though a person, but it does depend wildly on the conditions. Skin resistance can be a lot or a little.

Devices designed for a voltage take what they need. Cars have starter motors that take 200 to 500 A, and clocks that take 0.00001 A (give or take a lot). They are both run from the same car battery.

The current rating of supply is usually designed around the maximum load. Car batteries are that big because of the starter motors. The clock in a car will be quite happy running from a little 9 V battery for weeks, if its lights aren't on.

The current of a supply can only be too big if something goes wrong. If a car clock shorts out, it will take more current. Then it will take more than it needs, and the current that it takes depends on what current the supply can produce.

If a shorted car clock is run from a little 9 V battery, you get a warm, dead 9 V battery in an hour or so. The little 9 V battery can only supply 0.1 A or so.

If a shorted car clock is run from a car battery, you get a melted car clock in seconds, because the car battery supplies 1000 A.

That is why big power supplies have fuses. In a car, the clock is supplied from typically a 5 A fuse. That and the thin wires limit the current to less than 50 A if the clock shorts out, and the 5 A fuse cuts the current after 0.1 seconds, and there is little extra damage.

The only difference in behaviour of the clock etc is when something goes wrong. Most of the time the clock runs fine connected to a huge battery, and the 5 A fuse that feeds it does not blow.

EVEN if I were to connect this "device" to a power supply of 6 Volt with a ZILLION Amps, everything is still going to be ok ?
Simply because of the charity behavior that the device will only take what it wants and the power supply will not kill it with an over current ?

By this measure one could not, theoretically, "destroy" a device by giving it an overpowering amount of current, because the device will only "take when it needs" WHEN it comes to current right ? (Voltage is another matter, over-voltage WILL kill a device because that is how CMOS chips get destroyed.)

Click to expand...

That's about it. If I connect a 6 volt 200 mA motor to a 6 volt battery rated at 1 AH or if I connected to a Zillion AH 6 volt battery it matters not. However, if I connect that same motor to a 12 volt battery what will happen? Assuming the battery can supply the current the motor will draw twice its rated current and in short order be toast.

Ron

Thanks Diver300 and Reloadron, after this long thread of question and answers.
I can honestly say this is the FIRST time in my life in 15 years since I first saw ohm's law that everything makes sense to me.

I love you guys

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...

But Ian Rogers, raise an interesting concept I do not yet comprehend:

Ian Rogers said:

The long and hard of it is, if the consuming device requires more current that the battery can provide the voltage will drop.

Click to expand...

I x R = V
IF more current will drop the voltage...hmm...
In what universe does an increase in "I" lead to a drop in V ?
Unless the Resistor drops to 0.~, but we all know the resistor's value stays the same.
I x R = V.
Say:
0.48Amp X 500Ω = 240Volt
Now we increase the Amp:
0.5Amp X 500Ω = 250Volt.

Now we can go on forever, but as you can see, mathematically, in what universe does an increase in current drawn led to a voltage drop.

Don't get me wrong, common sense wise, it makes PERFECT sense, I mean, the energy HAVE to come from somewhere, and the battery is just a battery, not a magically device, if more current is required from it then it can give, it must give something else, like Voltage or its sister bunny I don't know, but Ohm's law wise...the math doesn't support this no ?

Next question, how do we "know" the overall resistance of a Device/Circuit ?
Because knowing its overall resistance will give an answer as to how much current it requires

Bracer said:

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...

Click to expand...

More resistances in parallel means less resistance, if you take all the loads that are connected as one resistor.

Bracer said:

I x R = V
IF more current will drop the voltage...hmm...
In what universe does an increase in "I" lead to a drop in V ?

Click to expand...

It is because you are looking at different voltages.

If you consider a battery as producing a voltage, lets say 12 for a car battery, and the battery has an internal resistance. For a simple model you can consider it as just a small resistance in series with a perfect 12 V supply. Let's guess that it has a resistance of 0.01 ohms.

So you take 10 amps from the battery, and the voltage across the resistance obeys ohms law, and so it is 10 x 0.01 = 0.1 V.

That means that the voltage that you can actually measure, (because the 0.01 ohms isn't a real resistor) is 12 - 0.1 = 11.9 V.

Now operater the starter, and take 300 A. Now the voltage drop across the resistance is 300 x 0.01 = 3 V and the voltage on the terminals of the battery is 12 - 3 = 9V.

The internal resistance of the battery obeys Ohms law, but the battery does not, because Ohms law only applies to resistors.

This is so interesting to me!
A simple resistor as a voltage regulator…

Because I had this idea that the main supply set the voltage and end of question.
I thought the resistor limits the current because the V in ohms law has been set by the main supply and the R has been set by the resistor itself, we only left current to calculate…

Now you are saying we “cascade” this theory to find out the drop in voltage on the other hand! Wow!

Let me think out loud.

Say 9 Volt Battery.
300 Ohms Resistor.

9/300 = 0.03Amp

Ok, so now we now the resistor takes 0.03Amp to “drive” itself [in a matter of speaking].

Now according to you, we “cascade” this theory to the other end.

Resistor x Current.
300ohm x 0.03Amp = 9…

hmm…
Ooo stuck…ooo geez…

Main supply 9v minus away the resistor voltage usage of 9V and we get 0…

There is theoretically nothing left on the other end!

Ok I am stuck.

I am stuck because I am given the impression that resistor can drop the current but not the voltage but here I am hearing “voltage” drop after a load.

Please help.

A 9V battery is small so it has a fairly high internal resistance. Its internal resistance is spec'd to be 1.8 ohms when a 9V alkaline battery is brand new. Then if its load is 500mA the battery voltage is only 9V - (1.8 x .5)= 8.1V.
When the battery runs down a little then its internal resistance is 9 ohms. Then with a 500mA load its output is only 9V - (9 x 0.5)= 4.5V and drops quickly.

Let's go easy.

I am stuck because I am given the impression that resistor can drop the current but not the voltage but here I am hearing “voltage” drop after a load.

Click to expand...

We need to work on the terms a little. The resistor does not drop the current, it actually serves to limit the current. The result of a current flowing through a resistor is a voltage drop.

Take a look at the attached circuit and tell us what is going on? The battery in this circuit is huge, really huge, even bigger than a bread box. For all purposes my battery can supply infinite current. The volt meters have absolutely no loading effect, their input impedance is infinite.

Ron

So....the batteries internal resistance is its own ultimate undoing....

Please allow me to ponder on your reply a bit, I really really appreciate your teaching me Diver300.
You are really really nice.

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.

Just focus on learning and understanding Ohms Law right now as it pertains to DC circuits. Pay attention to what AG has posted and think about it. Note in the play circuit I posted how things work and why the meters read what they read. Like anything else, it takes time to learn and understand.

Ron