Question about LEDs and resistors 1/4, 1/5 ...

Hi, said.

*Electrons are repelled away from the NEG terminal and attracted to the POS terminal. *

I am [u]not[/u] quoting positive electron flow. I am quoting electron flow. I have defined my reference, if its the wrong way switch your unsigned variables to signed variables ..lol

Electrons are negative, a negative terminal has an excess of electrons, a positive terminal has a deficiency of electrons. Like polarities repel. Negative electrons repelled from negative terminal and attracted to positive terminal.

I agree that the positive electron theory worksbetter for some semiconductor explanations.

Tom..... :) I am also responsible it seems for the water spiraling down the plughole the right way down here in the southern hemisphere,and the wrong way in the northern. lol 8) (Happy Australia Day.. or should I sat Happy Ausstraya Day 26/1/2015, we also format the date the right way)

I agree that the positive electron theory worksbetter for some semiconductor explanations.

TOTAL RUBBISH

There is no electronic theory that uses positive electrons. A positive electron is called a positron, it is an anti partial and will annihilate an electron if it comes into contact with it.

What you are mixing things up with is a positive charge carrier which IS NOT IN ANY WAY a positive electron.

Hi,

What you are mixing things up with is a positive charge carrier which IS NOT IN ANY WAY a positive electron.

You are right and it tells you how often I have to use it...lol

Tom.... :)

Wow, this is getting deep. It has been so long since I have read a discussion on electron holes and current movement to this level.

I feel sorry for the OP who is getting a major lesson beyond what is needed.

so the LED will let current flows in one direction so if there is noise or repulsion it will block it.

Using the water analogy, it is a one way check valve, like a spring loaded valve. It will only allow flow in one direction. Prevent current from flowing in the wrong direction. Look up how a diode works. Light Emitting Diode = LED

This link may help.

How does a diode work

and resistor can be put before the LED and after, well this one is not clear why ?!!

In your circuit, it makes no difference as you have a loop.

Sticking with a water analogy, if you put a valve in a water line at the beginning or the end, you can still control the flow of water in the line. It isn't important where the valve goes.

In some circuits, the resistor placement is important due to other voltages and current issues. But that is another discussion.

ok so the current the same, or the LED let's say need 0.025 Amps, so the resistor's job is to make a voltage drop and not limit the current ? or the value of the resistor is defined by formula to let only 0.025 amps thru ?

ok about the loop and current same in the components of circuit but what about the source ?! it will always feed the circuit with 5V !

The resistor is one component in the circuit, it's job is to resist the flow of current so the the LED does not draw too much. It both drops voltage and limits the current there is no either or.

The power supply will give 5V as long as it is capable. That is as long as the chemical reactions are right or it is recieving some form of input power or the supply is not drawing greater than its maximum current. This last question is rather odd, what are you getting at?

let's say it's a current that comes from a DC adapter means won't stop until you turn it off, so the voltage of the source is 5V continues, so if the resistor isn't there, the LED will take 5V and boom burns ! so the resistor comes to make a voltage drop, and protect the LED from burning, so the drop will be on both sides of the resistor and the LED, mmm i guess i started to understand,

if i put the MM on the LEDs ends, i will find 2V, on the resistors 3V or whatever, mmm no other components but the source so if i put on the ends of the source i will find 5V, but if the resistor is not there LED will have 5V and burn....

if i put the multimeter between the - of the LED and the source will i get 5V on the wire ?

or V source = V resistor + V LED so no matter where i put the resistor the result is the same ! ;D

right ?

TomGeorge: should I say Happy Ausstraya Day 26/1/2015, we also format the date the right way)

Happy? It's raining here, I suspect the festivities will be badly affected.

I would say we format the date in a sensible way rather than the peculiar American convention. Whether it is the right way may be moot - because the European way - YYYY,mm,dd is more appropriate for sorting.

Hi, not raining Central Victoria,sunshine and 19.3DegC

firashelou, you are starting to get it, well done, something you can't see can be hard to explain sometimes.

Tom.... :)

firashelou: let's say it's a current that comes from a DC adapter means won't stop until you turn it off, so the voltage of the source is 5V continues, so if the resistor isn't there, the LED will take 5V and boom burns ! so the resistor comes to make a voltage drop, and protect the LED from burning, so the drop will be on both sides of the resistor and the LED, mmm i guess i started to understand,

if i put the MM on the LEDs ends, i will find 2V, on the resistors 3V or whatever, mmm no other components but the source so if i put on the ends of the source i will find 5V, but if the resistor is not there LED will have 5V and burn....

if i put the multimeter between the - of the LED and the source will i get 5V on the wire ?

or V source = V resistor + V LED so no matter where i put the resistor the result is the same ! ;D

right ?

By George you've got it. :)

thank you everyone :grin:

so I made the test, and I the the + of the multimeter on the GND of the LED, and the - of the multimeter on the - of the source, and I got 5V so I guess the only explanation for the resistor position is the Formula isn't it :) ?

so I guess the only explanation for the resistor position is the Formula isn't it

Don't understand the question.

so I made the test, and I the the + of the multimeter on the GND of the LED, and the - of the multimeter on the - of the source, and I got 5V

No you didn't. Try again with where you put the probes.

Grumpy_Mike: Don't understand the question. No you didn't. Try again with where you put the probes.

i wanted to see what voltage i will get after the LED, means between the GND on LED and the GND of the battery, here the GND of LED is considered the + where the current comes

there is no “GND of LED” a LED has an anode and a cathode
how did you wire the whole circuit ?
where exactly did you put the probes ?
what is “after the led” ?

Yep still not clear. You can measure three voltages here.

  1. From ground to supply - this should read 5V
  2. From LED anode to cathode - you should see the forward voltage drop of the LED
  3. From one end of the resistor to the other you will see a voltage equal to the current flowing times the resistance.

Adding up 2 & 3 will give you 1.

Vr + Vf = 5V no matter how arranged:

what I meant is this
the values are close up and down

PS: the lines are just pointing to the values like an arrow

schematic.JPG

The - 5V across a piece of wire is impossible. Or is that just 5V, either way that is not possible.

Either you have a faulty meter or you are not measuring what that diagram represents.

Grumpy_Mike: The - 5V across a piece of wire is impossible. Or is that just 5V, either way that is not possible.

Either you have a faulty meter or you are not measuring what that diagram represents.

no i'm sorry that's not a - it's just like an arrow pointing to it, maybe i shouldn't have put it there

Can you try again with that diagram. Are there four voltage measurements or three? What range is your meter on? 1.9V looks low for the voltage across the LED unless it is a digital meter on a 2V range and you are hitting the end stop.