# How do resistors react in an LED-Array? Answered

Hey guys.
I've been reading myself through a lot of articles on websites about how resistors and LED's work,
but what I still don't understand is why, when calculating the required resistors for an LED-Array, this website ( http://led.linear1.org/led.wiz ) takes maximum 4 LED's for each resistor at 9V and a diode forward voltage of 2.0 - is it to prevent it become unstable or something like that? And frankly, I think that I miss understand resistors, although I've read a lot of articles about it.

So, what I think resistors do is this:
Since resistors are used in a simple 1-LED-circuits that has more voltage than the diode forward voltage, it simply lowers the voltage to  the desired forward voltage ( 2.0 for a Red-LED as far as I remember ) - so my point is, when it always is lowered, why would you need more resistors when the voltage is lowered from the first LED on? Perhaps I am very bad at explaining my logical problem here, but I hope that someone actually understands something out of this mess. xS

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## Comments

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The resistor limits the amount of current flowing to the LED. The Forward voltage is how many volts it takes a resistor to light. The problem with LEDs are they want to pull as much current as possible. The more current the LED gets the Brighter it gets. So it will pull current until the power source can't give anymore or the LED pops. So you have to limit the current to the LED. The Resistors only act as current limiters. They don't change the voltage.

You have to have a resistor on with every LED. If you put one at the first LED in the series then the current is limited by that resistor and all the LEDs are forced to use what little current is available. They won't be able to divide it evenly among themselves since each one is trying to take as much as it can get. So as you get further down the string the LEDs be get dimmer and dimmer. The first couple may work alright but the rest may not light up at all.

Do you have a good understanding of Ohms law and the calculations regarding it all? If not look into it if you want a better understanding of why things are the way they are. Or just except that you need a resistor on each and the calc is giving you the best possible answer and move on.

When working with a series/parallel array the LEDs in series need 2 V to light. If you have a 9V source then you can only string 4 LEDs in series. As you should know LEDs in series are splitting the voltage evenly between them and the current stays the same.

When you string LEDs in Parallel its the current that gets divided among the LEDs and the voltage stays the same. So an series/parallel array allows you to string many more LEDs on a single power source. You take the number of LEDs you want to use and divide them up into multiple strings of LEDs in series.

Even with a basic understanding of Ohms law you can see how the math gets pretty complex when trying to figure out the resistor values for a series/parallel array.

Hmm, the second section of your comment enlightened me a bit more, thanks.
So, let me understand how I would solder it all together, because the schematic from the LED-Array-Calculator is giving me some chills down my spine already ...
So once again we goto http://led.linear1.org/led.wiz and use these parameters:
Source voltage 9V, diode forward voltage 2.0, diode forward current (mA) 20 and number of LEDs in your array 20.

Now, if we look at Solution 0, the only solution, we see a dot with a "+9V" above it - now, thats already something that is weird to me, why isn't it connected to the other dot, so it's a circuit? Well, whatever, let's proceed.
Then we see it split up - How the heck am I supposed to solder so many points to one contact? That's a really freaky thought, and I guess it looks messy, too.
Then they have resistors at the end, which is okay, but then it all goes back to one dot - you gotta know, I am new to electronics, but even I know that this isn't a complete circuit. It looks very weird to me, and I have no idea how this would work out on a physical circuit-board. ( I am aware that a schematic isn't a physical reflection of the circuit, but the circuit in a schematic has to be connected to loop, doesn't it? )

The left side of the schematic with the +9V and the right side is the ground. Your use to seeing a Schematic use a battery symbol for the power input. But when the power source is unknown the schematic will look something like that.

The schematic is a complete circuit. 9V goes into the LEDs, through the resistors and out to the ground just like it should. Here is probably how your more comfortable seeing it.

And you can connect it to a Circuit board exactly like that. Often times with simple circuits like this the PCB layout can exactly match the Schematic. Or solder in wires where the LEDs would go so you can mount the LEDs however you want. I also attached the Fritzing file i use to create the Schematic. Fritzing is free and allows you to play around with circuit layouts on a breadboard. It also allows you to make a schematic and then layout everything on a PCB so you can print it off and etch your own board.

Perhaps you do not understand what an LED light emitting diode is....

Any Diode is a semiconductor and will accept or Hog all the voltage and
Current a power supply can deliver.

This means
1. at 2 volts the LED draws 0.02 Amp
2. at 3 volts the LED draws 0.20 Amp
3. at 4 volts the LED draws 2.00 Amp
4. at 5 volts the LED draws 20.0 Amp
Yes, the response is exponential and non-linear.

Obviously you can not expect a single LED to absorb 100watts
without destructing itself.

Now, Enter the linear Resistor in series to limit the current and voltage
and we can use this non-linear LEDiode where it will do useful light
output.

A