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I've researched and it says that resistors limit the current flowing through the LED.

But this statement confuses me because we know that in a series circuit, the current is constant at every point, so how come a resistor can limit the current flowing?

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5 Answers 5

up vote 15 down vote accepted

LEDs have a fairly constant voltage across them, like 2.2V for a red LED, which only slightly rises with current. If you supply 3V to this LED without series resistor the LED will try to set for a voltage/current combination for this 3V. There's no current that goes with this kind of voltage, theoretically it would be 10s, maybe 100s of amperes, which would destroy the LED. And that's exactly what happens if your power supply can supply enough current.
So the solution is a series resistor. If your LED needs 20mA you can calculate for the red LED in the example

\$ R = \dfrac{\Delta V}{I} = \dfrac{3V - 2.2V}{20mA} = 40 \Omega\$

You may think that supplying 2.2V directly will also work, but that's not true. The slightest difference in LED or supply voltage may cause the LED to light very dim, very bright, or even destroy. A series resistor will ensure that slight differences in voltage have only a minor effect on the LED's current, provided that the voltage drop across the resistor is large enough.

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+1 because I once assumed that a LED would provide enough internal resistance and ended up with explosive shrapnel very nearly missing my eye. –  fluffy Mar 20 '12 at 22:53

The point is a LED is a diode anyway and diodes have very small internal resistance (in "forward" direction of course), so unless there's something else in series the overall resistance is very low and the current is barely limited and this barely limited current can damage the LED and overload the circuit that powers it.

So yes, you're totally right that the current is the same in each point of the circuit when elements are connected in series, but when you add a resistor you increase the overall resistance of the series and this decreases the current.

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Note that constant current around a loop is only for a relatively small subset of possible circuits. It is an OK assumption for this example but a dangerous one in general. –  Russell McMahon Mar 20 '12 at 11:21
    
@Russell McMahon: I don't get it at all. Which assumption do you mean? –  sharptooth Mar 20 '12 at 11:27
    
Re subset of circuits - anyhing with reactive components and AC or time varying anything will be able to have different currents at different places in a loop at any given time. An oscillator with eg series LC would probably be a useful example. You understand that such things can happen even if we don't usually put things in those terms but a raw beginner will have no concept of AC operation etc. –  Russell McMahon Mar 20 '12 at 11:39
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@RussellMcMahon If I intended it correctly, I have to disagree: no matter how fancy are the components, the current in a branch (set of components in series without other wires getting in or out), the current will be equal everywhere. –  clabacchio Mar 20 '12 at 12:23
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@PortreeKid see the comment in Russel's answer: you have to consider each component in the series as a whole, because what happens inside breaks the rule of a closed system –  clabacchio Mar 21 '12 at 13:32

Imagine that

  • You had a water powered motor whose speed was proportional to current flow.

  • The motor itself offered very little resistance to current flow - you had to control the current flow external to a pump.

  • You has a pump able to pump 10 litres per second through a 10 metre pipe to the motor then through the motor and then through another 10 metre pipe to the suction side of the pump. (Flow rate was related to the pressure that the pump made and pipeline resistance - ie NOT a positive displacement pump.

  • When the pump was operated you found that the motor ran MUCH too fast and that you needed to limit flow to about 1 litre/second.

To achieve the requirement you could place a reducing valve in the circuit to drop most of the pressure and to limit the flow. The valve worked to drop a certain amount of pressure across it at a given flow rate and as adjustable. (This is about how many rel water valves do work).

You could place the valve ANYWHERE in the circuit and it would achieve the desired result. It could be at the pump inlet or exit or at the motor exit or inlet or anywhere in either pipe.

This is a close analogy to you LED question. The current needs to be limited as it is too high without a limiter. The limiter may be placed anywhere in circuit.


With the Battery - resistor LED circuit

The LED has a certain defined voltage drop at a chosen current.
To be specific lets say that at 20 mA the LED drops exactly 3.00 Volt. This is typical of some modern LEDs.
If we wish to run the LED at 20 mA we MUST arrange for it to drop 3 V - not more and not less.
If we wish to use a 9V supply to operate the LED we N=MUST "get rid of" 9-3 = 6B somehow.
The resistor does this.
To drop 6V at 20 mA the resistor needed is R = V/I = 6 / 0.02 = 300 ohms.
In this example a 9V battery + a resistor + an LED will operate at 20 mA. The resistor can be placed before or after the LED. The current is dropped across it in either location.


It is not relevant to this question but extremely important to know that your statement that

  • "we know that series ciircuit, the current is constant at every point."

is incorrect.

There are many circuits where this is rue -but also many circuits where it is not true.
In DC circuits with only resistive components, such as this 1 LED, 1 resistor circuit, then it is true. BUT when there are reactive components present such as inductors and capacitors or certain other non linear elements then it is often NOT true.

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I disagree with the last paragraph: in a series circuit (one wire in - one wire out) the current will be the same at every point outside the components (treating them as black boxes). –  clabacchio Mar 20 '12 at 13:56
    
Yes, I'm a bit confused. @Russell, could you give an example for a series circuit where current is NOT equal through all elements? –  exscape Mar 20 '12 at 17:10

Always with the complicated answers ;-). Look at it this way. What happens when you put a wire across the terminals of a battery? In a perfect world you get infinite current which melts the wire. We call this a short curcuit. Because diodes are designed to have minimal forward resistance we get the same effect as a short. Put a resistor in there to provide something to resist against current to limit it down from infinity

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This may be understood and your question may have been rooted as a diode in a curcuit with other components that are limiting current by their resistance. While you may get a way with this - if anything changes in the curcuit, the LED is on it's own. Best to have its own R –  VariableLost Mar 21 '12 at 21:29
    
How can u say tht the wire is experiencing knfinite current? Why infinite in the first place? –  IvanMatala Mar 21 '12 at 22:31
    
A bit simplicistic for an engineering site, and a subset of sharptooth's answer...welcome anyway! –  clabacchio Mar 21 '12 at 23:42

Let's focus on what is important here: The LED (which is a diode) characteristic curve. Please look at this image from wikipedia. As you can see, for positive voltages across the diode its current increases exponentially. Imagine now you connect your LED to a power supply without resistor. You would have to set the exact voltage across the diode to get the exact current you need to light up the LED. If for any reason your power supply increases a little bit above the voltage you need then the current will be exponentially higher than before which may (it will!) damage your diode. So, how can a resistor help us with this problem? FEEDBACK! One of the most important concepts in electronics! Let's go back to our example, and add a resistor in series with the diode and the power supply. Now, everytime your power supply exceeds its nominal voltage the diode will increase its current exponentially again, but because the current got higher the voltage across the resistor will be higher too, which means, the voltage across the diode will decrease, compensating the power supply voltage increase.

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