## Ohm's Law / Limiting Variable Current to a Specific Value

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Didn't see a section to ask about basic electronicy stuff here. So not really sure where to ask this... As for me, I come from a software background, so some of the more subtle things with electronics elude me.

Pre-description summary:
1 of 2: I think I get Ohm's law, but it doesn't seem to mesh with my experience -- which means something in my experience must be wrong.

2 of 2: I want to build something, and I want to build it right, so I want to learn this well, how would you build it, where can I learn more?

Okay here goes:
1.) In my experience, matching power adapters with things (like a Nintendo for instance) you want to get the voltage as close to possible, the polarity correct (if it outputs DC), and a current rating that's either equal or higher. I usually do this in electronics as well, get the voltage correct, and either don't worry about current, or make sure your source can put out enough of it that it's not a problem (i.e. more batteries in parallel).

However, recently, I'm looking closer at data sheets, solenoids, motors, etc... and I'm seeing maximum current ratings -- I'd always figured devices would just take the amount of current they wanted and leave the rest, not like voltage where if you go too high you're shoving it down the devices throat force feeding it, current it just takes as much as it wants and leaves the rest. But I keep coming across things that say stuff like "Don't run more than 40ma through this motor or it will ruin it", etc... or microcontroller data sheets which say stuff like "don't draw more than 100 ma current from port X or you will fry the chip", how can you try to draw more current from the chip than what it wants to give? The chip pushes the current out the pin, the device doesn't pull the current from the chip, does it?

2.) I want to build a couple of things with motors, mini DC brushed ones, and steppers, and maybe solenoids, etc... and I don't want to just go and buy the \$50 "motor controller circuit board" or some random IC and be dependent on that -- I don't mind doing that for a project here or there, but I want to know how it works first -- I want to really "get it".

I think I understand Ohm's law... C(or "I") = V/R.

so if I run 5v over a 100 ohm resistor, I'm dissipating 50 ma (in heat?) across the resistor. So, if my source was putting out 50 ma too much, I could use this to shave off a bit, no? But isn't that bad for battery life? Also what if the next set of batteries puts out a different amount of current? Or what if I'm using a power adapter, and the user replaces it with one very similar, but a different current rating?

So I guess my question is, how do you do proper current limiting to a specific well known amount, with an unknown/variable current input?

And how do you step up current (say 1A, or even a lot more, like 10A) from an MCU pin (without a relay, obviously).

How about the hydrological analogy: volts is water pressure, amps is gallons per minute, resistance is the size of the hose. Power is volts times amps. If either one is zero, you dont have any watts. Same with a big pump. You need lots of psi and lots of gpm to get something wet.

Imagecraft compiler user

Quote:
how can you try to draw more current from the chip than what it wants to give?

Perhaps an example is the easiest way to explain this.

Put a digital I/O pin High.
Put a one ohm resistor on the pin, connected to ground.

According to Ohm's Law, I = V/R.
With a logical high of 5V, and a resistance of one ohm, the chip would try to source 5 Amps to the resistor.

It can't do that. So it pumps out as much current as it can. That exceeds the (typically) 40 mA output per pin, and can therefore damage the chip.

In practice the output voltage isn't held at 5 V, it drops down.

So, what's a designer to do?
If you want to keep the current being sourced by the pin under a specific limit, then you must insure that the load exceeds a given minimum.

For example, if you wanted to be sure that no more than 20 mA was being sourced to a load, with the chip at 5V, then one has to make sure the load is at least 250 ohms, (R=V/I 5/20mA = 250 ohms).

In practice the output voltage isn't really 5 V, but we'll save that fine point for another day.

One more common example, driving an LED:

Lets say you want to run the LED at 10 mA. You select this current level from reading the data sheet, past experience with this model LED, etc.

The voltage drop across the LED is given in the data sheet, also. For sake of example say it is 1.2 V when the LED is forward biased, (i.e. ON).

Circuit: I/O Pin (At 5 V) to Resistor to LED to ground.
If the LED has a fixed 1.2 V across it, and the circuit has a 5 V source, then there is 5-1.2 = 3.8 V across the resistor.

We decided we wanted 10 mA to flow through the circuit. R = V/I = 3.8 V / 10 mA = 380 ohms.

There is nothing critical about the 10 mA, (Not always true, but true enough for most circuits), so pick a "nice" value, say 330 or 470 ohms.

Then back calculate what the actual current will be.

The next step is to select a power rating for the resistor.

P(in Watts) = V(in Volts)* I(in Amps).

In the above LED example:
Using a 330 ohm resistor: I= V/R = 3.8 V / 330 ohms = 11.5 mA through the resistor and the LED.

The power dissipated by the resistor depends on the voltage across it and the current through it:
P = V * I 3.8 V * 11.5 mA = 43.7 mW

A 1/8 W resistor can handle 125 mW, (In theory).

It would be fine for this application.

A 1/4 W resistor can handle 250 mW, way more tha needed for this application.

Hopefully that will get you started thinking in the right direction.

JC

JC, what you say about the resistor and the LED, from my experience is correct, but I'm missing how the dots connect...

For example you say if I hook up a 1 ohm resistor to 5V it's going to draw 5 amps -- that implies that the load (the resistor here) decides how much current it wants to take, and it just tries to take it, if the microcontroller can't source that much, well too bad, it will just fry. (Which makes very little sense in the amps = gallons per minute analogy.) -- The implication here is that the load determines how much power it wants and takes it, regardless of what the source of power is capable of.

Secondly, you say the LED takes 10ma -- so why can't we just plug it right in, and have it only take 10ma? (I understand you can't, but I don't understand *why*.) The implication you make here with the resistor selection, etc is that the source determines how much power it wants to give, and the load must accept it...

These two notions are contradictory, are they not? Does the load determine how much power it wants to take, or does the source determine how much power it wants to give?

edit: Also if resistance is the "size" (diameter?) "of the hose"... how is hooking up a giantly wide section of hose (say a 1 ohm resistor) to a regular sized faucet (say an mcu pin) going to try to pull more water out of the faucet than it can source?

Quote:
:-) Moved.

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

The water pressure pushes the gallons per minute thru the hose. The resistor doesnt suck amps out of the power supply. The volts pushes the electrons through the resistor. If the power supply (the avr output pin at 5V) cant supply the needed current, we say the output impedance of the source is too high. You get a voltage divider between the 250 ohm internal impedance of the pin and the 1 ohm load resistor. Hopefully the engineers in Norway have designed the output stage so the magic smoke wont squirt out.

Imagecraft compiler user

Quote:
... The implication here is that the load determines how much power it wants and takes it, regardless of what the source of power is capable of.
Not exactly. The load determines how much power it wants and tries to take it. If the source of power is not capable of delivering that much power then it (the source of power) may (1) blow a fuse, (2) shut down, (3) deliver as much as it can for as long as it can before melting, and probably lots of other permutations.

Quote:
Secondly, you say the LED takes 10ma -- so why can't we just plug it right in, and have it only take 10ma?
An LED does not "take" 10mA. It may be designed to operate at 10 mA. It is up to the user to design a circuit that will limit the current through that LED to 10 mA. It's somewhat like a car, its accelerator, and the speed limit. Why can't you just get in the car, depress the accelerator all the way, and let the speed settle out at a value equal to or less than the speed limit? Depressing the accelerator all the way is roughly the same as running the LED without a series resistor.

Did you ever wonder why it takes so long to get an Engineering degree?

Don

V = I x R
This means that if you apply a current (lets say 10mA, or 0.010A) through a resistance (let's say 100ohm), the voltage across the resistance will be the product of said current by said resistance. So in this case, 100ohms x 0.010A = 1.0V will be measured across the resistance.

I = V / R
Current is a quotient of voltage over resistance. This means that if you push a voltage through a resistance, the current on the other side of the resistance will be limited to V / R. For example 5V through a 100ohm resistance will only let 0.050A through.

R = V / I
Resistance is a quotient of voltage over current. This means that if you have 5V, and want to have 0.010A, then you need a resistance of 5V / 0.010A = 500ohm.

```    / \
/ V \
/     \
/ I   R \
/_________\```

Pick two and you got the third.

Interesting fact. If there is zero resistance, then how do you make any of these equations work? :wink:

To follow the water pipe analogy, think of resistance as the pipe diameter. A smaller pipe exhibits more resistance to water than a larger pipe. If the water pressure (current) gets too high, the pipe breaks/explodes. Similarly if you try to pass too much current through a conductor, the conductor will melt. Where the analogy gets trickier is with voltage. Think of voltage as the density of the liquid flowing through the pipe. A denser liquid (higher voltage) would need less pressure (current) to make the same pipe (resistive conductor) explode.

To understand why you cannot just plug a LED to a source and be done with it you need to understand how a LED works. A LED is a diode. It will just pass current in one direction and block it in the other. It just happens that this particular diode emits light when current flows through. The LED is manufactured to emit a certain useful amount of light at a certain amount of current flowing through. It will try to pass as much current as you are supplying it, it's just that it physically can't. The diode has a resistance of its own, manifested through the metal leads you apply current to, and the semiconductor gap in between them in the diode's body -- they can only pass so much current before melting/exploding. Thus you add a resistor to limit the amount of current that will go through the diode.

Quote:
The next step is to select a power rating for the resistor.

P(in Watts) = V(in Volts)* I(in Amps).

In the above LED example:
Using a 330 ohm resistor: I= V/R = 3.8 V / 330 ohms = 11.5 mA through the resistor and the LED.

The power dissipated by the resistor depends on the voltage across it and the current through it:
P = V * I 3.8 V * 11.5 mA = 43.7 mW

One small remark on this.
Note that this is the power consumed by the resistor when the LED is functioning! In case of a failure the power dissipated will U^2/R = 25/380 = 66mW. You have to take that also into account. If for what ever reason the LED is shorted then you don't want to have a smoking resistor (eventhough this will make fault searching easier)
In this case it still can be done with the 1/8W resistor, but I have had occasions where in normal operation the resistor could be a 1/8W but in failure mode we needed a 1W resistor. when designing you always have to keep that in mind. And on a side note of that when you work with semiconductors in your circuit, it is not always such that the highest voltage will give most wattage, specially when using transistors.

Mikey83 wrote:

For example you say if I hook up a 1 ohm resistor to 5V it's going to draw 5 amps -- that implies that the load (the resistor here) decides how much current it wants to take, and it just tries to take it, if the microcontroller can't source that much, well too bad, it will just fry. (Which makes very little sense in the amps = gallons per minute analogy.) -- The implication here is that the load determines how much power it wants and takes it, regardless of what the source of power is capable of.

Secondly, you say the LED takes 10ma -- so why can't we just plug it right in, and have it only take 10ma? (I understand you can't, but I don't understand *why*.) The implication you make here with the resistor selection, etc is that the source determines how much power it wants to give, and the load must accept it...

Look at it another way, Mikey...

The current rating for a component that you're trying to drive will be *the current it needs to operate*. Allowing a component to take more current than that will cause it to die - usually due to overheating (there are some different causes which I won't worry about now).

Why overheating? Because the power dissipated in the component is the current *through* it multiplied by its internal resistance.

So looking at your LED, the spec sheet won't specify its resistance, but will specify an operating voltage and current. That voltage will be about two volts, and the current will be, say, 20mA. Now it's a non-linear device, because it has a semiconductor junction: if you feed it less than two volts it won't light up, and current won't flow through it. But when you get to two volts, things start to happen... the diode starts to conduct, current starts to flow, and pow, your diode just died!

What? How? Er?

Once the diode is conducting, it starts to dissipate heat across that junction. Power is current through times voltage across a component - in this case, the LED will try very hard to maintain two volts across itself. If the current is limited to 20mA, it can do this; the power is emitted and light and a certain amount of heat. If the current is less, the volts remain the same, and the light and heat are less.

But if your supply is not current limited, the LED will cheerfully glow brighter and hotter and eventually it's going to overheat the semiconductor and go pop.

So in your original example of a pin that can supply 40mA at 5v, driving a 20mA diode... that 40mA is enough to toast the LED hence the need for a resistor in line, whose job it is to dissipate the excess heat - per the calculations of one of the other posters upthread.

That resistor will never have more than 3v (=5-2) across it, and needs to hold that 3v at a current of no more than 20mA... V=IR, so R=V/I, R= 3/20e-3 = 150 ohms.

In practice you'd probably not want to run the LED that bright, but the principle's the same; at 10mA, the resistor would need to be 300 ohms.

Thanks guys for all the responses, and for bearing with me through all this! :-)

So, from reading through the answers, it seems it depends on the type of load then -- either the load will just take whatever the power source throws at it (and die a painful death via melting, or losing it's magic smoke -- examples are LEDs... what else?) -- where as other loads attempt to pull power from the power source (possibly causing the source to die if it pulls too much...? examples are resistors? what else?)

Still don't get how to determine which class a given random device falls into. Do these classes have names?

And still no one has offered any advice on regulating current from a power source of unknown current... I mean... my wall outlets can source up to 15A @ 120V A/C, and I know plugging in a light bulb doesn't just let 15A flow across it, so light bulbs must be the latter type that just pull as much as they need -- am I wrong in thinking you could plug a light bulb into an equivalent wiring that was capable of sourcing a different amount of current, and it would only draw the exact amount of power it needed? (I know this is the case for 10 amps, 15 amps, and 20 amps, what about 100 amps? seems like it should work just fine to me...)

Wikipedia seems to imply that there is some active transistor based way to regulate current (http://en.wikipedia.org/wiki/Current_limiting) -- but are all those active components entirely necessary...?

Put another way, if I known the volts the source is providing alone, how do I make sure only 20ma pass through a given junction if I don't know the current?

i.e. I have a motor that I want to run at 12v and 40ma... I have a 12v rail coming from a supply that can source up to X amps... (where X is unknown but far greater than 40ma...) how do I just limit it to 40ma?

Seems the AVR chips can do it -- somehow -- I can plug them into a 5V 1A DC power supply or a 5V 100ma DC power supply, and either way, the pin will supply 40 ma.

Also, thanks hugoboss for explaining the error I made about the water pipe analogy, I was thinking "if it can let 5A through it must be a wide pipe" but really I got it backwards, sounds like the 1 ohm "pipe" is very thin, and "pressure" builds up in the chip, that makes much more sense.

Quote:
Also, thanks hugoboss for explaining the error I made about the water pipe analogy, I was thinking "if it can let 5A through it must be a wide pipe" but really I got it backwards, sounds like the 1 ohm "pipe" is very thin, and "pressure" builds up in the chip, that makes much more sense.

That's not what I said though. You should really do yourself a favor and start at chapter 1 of some "Electronics for dummies" book. There are a lot of things you simply cannot understand by themselves, everything has a relation elsewhere. To understand Ohm's law implies you need to know the concept of thermodynamics and power dissipation and some basic particle physics.

Actually your wall outlets can source much more than say 15A! That is until the 15A fuse blows in your switchbox. Why have a fuse ( or other form of circuit protection)? If too much current flows, then the wire will go up in smoke.

As for your 40mA motor, the current it draws is based on its load ( assuming a permant magnet type of motor). If you stall the motor ( place too much mecanical load) it will draw 'infinite' current - in reality this is limited by the resistance of the windings and the motor will most likely overheat if the current isnt limited to a safe value. One could use a fuse, circuitbreaker, resistor or a constant current source. How do we know all of this? Get a degree! Or google lots. Each device has its traits - things like electric motors have various types of construction, so if you wikipedia 'permanent magnet motor' and compare with 'induction motor' you'll find out some fundamental differences. Manufacturers usually have datasheets for their products that outline the requirements. Break these rules and thi gs wont go to plan.
Get yourself a first year electronic engineering textbook and have a read. It will give you a foundation in basic electrical theory.

hugoboss wrote:
That's not what I said though.

hugoboss wrote:
To follow the water pipe analogy, think of resistance as the pipe diameter. A smaller pipe exhibits more resistance to water than a larger pipe. If the water pressure (current) gets too high, the pipe breaks/explodes.
Ahh crap, you're right, I had it right the first time, which again, doesn't make a lot of sense; 1 ohm resistor == super wide diameter pipe -- right? So why would hooking a super wide pipe up to a small faucet cause the faucet any damage? O.o

hugoboss wrote:
You should really do yourself a favor and start at chapter 1 of some "Electronics for dummies" book.
I have. For a couple of books even. Websites, and some blogs too. Like I said in my first post, if you can find a good resource that explains how to regulate varying or unknown current in an easy to understand way, I would love to read it. Please post links! (even to Amazon!) -- as so far, all I've found is that Wikipedia page.

Edit: Oh just noticed another response.

Kartman wrote:
Actually your wall outlets can source much more than say 15A! That is until the 15A fuse blows in your switchbox. Why have a fuse ( or other form of circuit protection)? If too much current flows, then the wire will go up in smoke.
Of course, I'm not saying I don't understand this. I mean this is why you can't just change a 10A breaker to a 20A breaker without checking to make sure the wiring is the correct gauge, if you try to pull too much current through it, your house will burn down. -- Just trying to figure out why a light bulb only pulls the exact amount of current it wants versus an LED will try to sink as much current as you pour into it...

A previous poster mentioned that an LED (being just a diode that emits light as well as heat, which we all know) WILL try to sink as much current as it can -- popping the fucker if you pour too much in...

....
....
....

AHHH THIS IS PART OF THE KEY:

DocJC wrote:
We decided we wanted 10 mA to flow through the circuit. R = V/I = 3.8 V / 10 mA = 380 ohms.

My error was that 3.8V + 1.2V == 5V, but 10 ma + 10 ma != 10 ma... 10 ma is constant across the whole series, correct? (got confused cause someone else was talking 20 ma, thinking 10+10 = 20...) -- so the volts are divided but the ma which is drawn from the source is constant... This correct, right?
.. It seems now... all devices just pull what they "want" to use... where want == resistance, is that correct?

So the problem with motors is that they may try to draw too much current if there is physical resistance (which can be a problem for both the motor AND the current source, yes?)

Is this correct?

EDIT: Crap ... though it seems like the problem is, back to my original understanding, not too much current, but too much voltage... (in the LED example) you're trying to divide the voltage so that only 1.2V is dropped across the LED, and the other 3.8V is dropped across the resistor... and because the LED only wants 10 ma... we use 10 ma to calculate the resistor for voltage dropping? Is that not correct?

The motor wanting more current than it can handle makes sense, but again, the LED, it seems is not an issue of current, but voltage. Guess I'll have to read that book.

edit: I mean... if I hook up a 2 volt 10 ma LED up to a 2 volt 1 A power supply.... it's not going to pop right? It's only going to draw 10 ma... it's that we're trying to drop 5V across something meant for 2V... right?

Last Edited: Thu. Feb 16, 2012 - 12:38 AM

Try something like this for a reference book:

BIRD, J. (2001). Electrical Circuit Theory and Technology (2nd ed.)

If you google that you will find that you can download the whole book as a PDF.

I edited/appended my previous post (forgot to refresh first), I think I get it a little better now.

Power is volts X amps. A 100 watt light bulb running from 120V is probably pulling .83 amps, and its resistance when heated up and running is V/I=120/.83=144 ohms. Now you make the calc for a 10 watt bulb. Hopefully it be more ohms, less amps.

Imagecraft compiler user

10 watts/120 volts = ~.083 amps (an order of magnitude less...)
120 volts/.083 amps = ~1.4K ohms...

... so a load rated for 1.4K ohms when hooked up to 120 volts is going to suck .08 amps... ?

...and when hooked up to 12 volts is going to suck .008 amps?

Quote:
. (in the LED example) you're trying to divide the voltage so that only 1.2V is dropped across the LED, and the other 3.8V is dropped across the resistor... and because the LED only wants 10 ma... we use 10 ma to calculate the resistor for voltage dropping? Is that not correct?

Correct!

JC

@JC -- so then... I see you can divide volts by pairing a resistor with a load... to get the voltage correct... -- but it seems current is not divided down to some lower value in this circuit. Is that wrong?

Well, this statement will get some rebuttals, but:

In this circuit, an I/O pin driving a resistor & LED, think of the voltage as a fixed entity.

Its 5 V, and you can't change it.

Now, to keep the LED in spec, we are going to run it at 10 mA.

To get 10 mA from a 5 V source you need to set the load, (the resistance), to R = V/I --> 5V/10mA = 500 ohms.

If you replaced the LED and its series resistor with a single 500 ohm resistor one would get 10 mA flowing out of the I/O pin, through the resistor, to ground.

The case of the LED is a bit of a special case, as when forward biased, it has a "fixed" 1.2 V (or whatever its Vf is), voltage across it.

So yes, divide the voltage, or drop the voltage, however you wish to view it, from 5 V at the pin to 1.2 at the Resistor LED junction, to 0 on the ground rail.

(uC Pin to Resisto to LED to ground)

As stated above, if you start with 5 V and have 1.2 V across the LED then you must have 3.8 V across the resistor.

If you have 3.8 V across the resistor, and 10 mA flowing through the path, (or loop, or circuit...), then use Ohm's Law to calc the value of the resistor needed.

If you make the resistor smaller, then I=V/R, a smaller R gives a greater current, and vice-versa.

Think of current as the flow of electrons. All of the current, (electrons, water, etc), leaving the I/O pin flow through the resistor and through the LED to the Ground rail.

JC

Well Mikey, you seem to be a Quick Study. Next week we'll do dBs. Always fun.

Imagecraft compiler user

@Bob Gardner thanks, I think... Still a bit confused though...

@JC so our 120(?) ohm load (the LED)... at 5v will sink ~42 ma? -- I guess that explains why hooking up "too much voltage" (in my experience) has been bad for a device, but not too much current... -- the difference is being able to source 1A of current, versus the device trying to sink 1A of current?

So... it seems the constant here (per device, per situation) is the resistance of the load. If I know the load's resistance, I can find how many amps it will try to suck at any given voltage... right? :idea:

I read online, with one of the motors that I have, it was "safe" to run it at double or even triple the specified voltage (12 volts), so long as you never put more than the specified amount of current into it... (~40 ma if memory serves) It seems that if I drop the current with a resistor, it will drop the voltage as well here, is that correct? So how would one go about supplying 24v or 36v but limiting the current to 40ma. :?:

There are essentially three kinds of loads, classified by their predominant characteristics between inductance, capacitance, and resistance. Each have different behaviors, and each are partly the other two.

A motor is primarily an inductive load, a microcontroller is generally regarded as a capacitive load, and a resistor is, well, a resistive load. They all behave differently and have different source requirements.

It's never "safe" to exceed the manufacturer's specifications for any given component, including motors. There is usually a leeway you can play with if you know what you are doing, but you generally want to stay within specification.

Read the book you just illegally downloaded ( :wink: ), you will learn about all those things, and much much more along the way.

Quote:
It's never "safe" to exceed the manufacturer's specifications for any given component, including motors. There is usually a leeway you can play with if you know what you are doing, but you generally want to stay within specification.
Hence the quotes. :lol: ...and this is why I'm suddenly motivated to learn what the heck I'm doing before I go messing about! ;)

Good information on the three types of loads, that is awesome, I think it points me in the right direction for now.

As for illegally downloading the book... it came from a .edu -- which seems legit to me. ;) I'll be checking it and my other references, and maybe scouring wikipedia a bit for that information. :D

Quote:
I read online, with one of the motors that I have, it was "safe" to run it at double or even triple the specified voltage (12 volts), so long as you never put more than the specified amount of current into it... (~40 ma if memory serves) It seems that if I drop the current with a resistor, it will drop the voltage as well here, is that correct? So how would one go about supplying 24v or 36v but limiting the current to 40ma.
As you realise, doubling the voltage will double the current, so unless the resistance of the motor decides to increase, then you will violate the requirement of 40mA. since the motor is most likely thermally limited, ie the more current that flows, the hotter it will get. If you do this for only short periods, the motor may survive. Here we introduce the concept of 'average'

Two things will mess up a motor... revving so fast the windings fly off, or so much current and heat that the insulation melts and the windings short against each other. If it will run all day at 12V, it will run like a bat out of hades at 24V and get hot as a rocket. If you shut it off before it melts, you're golden.

Imagecraft compiler user

hugoboss wrote:
There are essentially three kinds of loads, classified by their predominant characteristics between inductance, capacitance, and resistance. Each have different behaviors, and each are partly the other two.

But... an LED is a non-linear device. Until a supplied voltage reaches its junction voltage, it presents a very high resistance and effectively no current flows. Once the voltage supplied exceeds that junction voltage, it will cheerfully pass all the current it can. Then it will go pop because it's overheated.

The trick is to ensure that it gets the right voltage across its junction while limiting the current to a level within its capabilities...

In a resistor-LED series circuit, the same current *must* flow through all the components since it has no-where else to go. If there is sufficient voltage across the pair to exceed the junction voltage, then current will flow. Voltage in excess of that junction voltage must be across the resistor, so if the supply voltage is only slightly greater than the junction, there won't be very much across the resistor, so the current won't be very high. Increase the voltage and the current will increase - but there'll still be only 2v across the LED.

The value of the resistor you calculate by knowing what the voltage across it will be (in the examples upthread, 5v - Vled) and the current required (10mA, say) and doing the R=V/I calculation.

OK Mikey83, to get at the problem of your understanding can you do three things for us.

a) tell us what current means.
b) tell us what voltage means.
c) tell us what power means.

Don't get it out of a book, because that just tells us that the author knows what it means. We need to know what is in your head!

Can you please delete the expletive in your previous post. Whilst I might use it in private, I avoid it in public, lest I offend someone! It is not considered acceptable on AVRFreaks!

Charles Darwin, Lord Kelvin & Murphy are always lurking about!
Lee -.-
Riddle me this...How did the serpent move around before the fall?