Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Please be kind, I am an electronics nub. This is in reference to getting an LED to emit photons.

From what I read (Getting Started in Electronics - Forrest Mims III and Make: Electronics) electrons flow from the more negative side to the more positive side.

In an example experiment (involving a primary dry cell, a SPDT switch, a resistor and an LED) it states that the resistor MUST be connected to the anode of the LED. In my mind, if the electrons flow from negative to positive, wouldn't the electron flow run through the LED before the resistor; thereby making the resistor pointless?

share|improve this question
While electrons flow from negative to positive, it's usually better to just play along with the convention that current flows from positive to negative so you don't confuse anyone when you talk about it. – Nick T May 2 '11 at 12:08
@NickT: Also physics.stackexchange.com/a/17131/176 – endolith Dec 16 '11 at 19:26
up vote 89 down vote accepted

The resistor can be on either side of the LED, but it must be present. When two or more components are in series, the current will be the same through all of them, and so it doesn't matter which order they are in. I think the way to read "the resistor must be connected to the anode" as "the resistor cannot be omitted from the circuit."

share|improve this answer

No it would not make the resistor pointless. Imagine if the resistor were so large it completely prevented electrons from flowing. Does it matter which side of the LED it's on? Either way, it will break the circuit and prevent current from flowing.

Don't think about individual particles traveling through the circuit. The charged particles are not "used up" by the LED. They go through it, and their motion is what carries energy from one place to another.

Think about all the particles moving at all points in the circuit at once, like a belt or chain. If you slow down the chain at one point, it slows down at every other point, too, due to the links pushing and pulling against each other.

I read Getting Started in Electronics as a kid, and I think it teaches ideas like this poorly. I had to unlearn everything in college and don't recommend it. Try this instead:

Try out this circuit. When you adjust the resistance, does it only slow down the charges before the resistor, or does it change the speed of all the charges in the entire circuit?

share|improve this answer
I love the "chain" or "link" analogy. I have been told other analogies, but none as good as those. – Kellenjb May 2 '11 at 13:04
A hand-crank on one end of a chain and a load on the other end. When you turn it, the entire chain moves, but it's not used up. Energy is transferred by the chain links pulling on each other, and this pulling movement travels very quickly from source to load, even if the chain links themselves move slowly. The only downside of the analogy is that the links only pull, they don't push on the other half of the circuit. Pipes filled with water works a little better. – endolith May 2 '11 at 15:39
I've used pipes filled with water some, but in that analogy I find that people tend to think of the water "going away" as soon as it comes out of the end. Guess it depends on who you are talking to and what you are trying to explain for what works the best. – Kellenjb May 2 '11 at 15:52
That Circuit Simulator applet is awesome ... played with it for 30 minutes now. :) – Spechal May 4 '11 at 5:58
@Hannibal: Why would there be an initial burst of current? Imagine lightly applying the brake to a bicycle wheel and then beginning to pedal. Is there a sudden burst of speed in the chain at any point? – endolith Jan 7 '15 at 19:58

Regardless of what side the resistor is placed on, it limits the amount of current that flows through the LED. It is usually a lot simpler to not think about what the electrons are doing and instead just think about it in sense of Resistance, Current, Voltage, and sometimes power.

In the case of an LED, if you connect a constant voltage source across the LED, the LED will act like almost 0 resistance, which will based off of V=IR (or V/R=I), will result in very large current, which causes the LED to "pop".

You have to connect a resistor in order to set the current that your LED is expecting.

share|improve this answer

The resistor doesn't need to be on the anode side, but it needs to be there (unless the voltage of the power supply is equal to or less than the voltage drop of the LED.)

After all, if you have a 9 volt power source, and an LED that drops 2 volts, then the other 7 volts have to get dropped someplace.

share|improve this answer

Look at the Forrest Mims III book again. It does not claim that resistors must be on the anode and has examples where they are on the cathode. In my 1988 edition of the book, series protection for LEDs is introduced on P. 69:

LED DRIVE CIRCUIT - Because LEDs are current dependent, it's usually necessary to protect them from excessive current with a series resistor. Some LEDs include a built-in series resistor. Most do not.

A formula is then given about how to calculate the resistance from the supply voltage and the LED's forward current. The accompanying diagram has the resistor on the anode, neglecting to explain that the choice is arbitrary.

However, on the same page, a "LED polarity indicator" device is introduced where two back-to-back LEDs share a resistor which is necessarily on the anode of one and the cathode of the other. In the "tri-state polarity indicator", the limit resistor is on the supply side, rather than ground side, too.

It's usually nicer in some sense (if there is a choice) to have the important device be connected to ground, and the surrounding paraphernalia, like a biasing resistors, to be on the supply side.

In high voltage circuits, the choice between supply-side or ground-side load matters from a safety perspective. For instance, should you place light switch on the hot side of the lamp, or on the neutral? If you wire the switch so that the light is turned off by interrupting the neutral return, that means that the light bulb socket is permanently connected to hot! This means that if someone turns off the switch before changing the bulb is not actually any safer; the main panel has to be used to actually break the hot connection to the socket. In a battery circuit, there is no safety ground: the minus terminal is arbitrarily designated as the common return, and the word "ground" is used for that common.

Whether a load device is ground side or supply side also makes a difference if the voltage from the device is being conveyed to some other circuit where it is used for some purpose. A 1.2V LED whose anode is connected to 5V will provide a 3.8V reading from the cathode, if current flows. If the cathode is grounded instead, then the anode will provide a 1.2V reading. So the placement of the resistor only doesn't matter if no such situation exists in the circuit: there is no third connection to the junction between the resistor and the LED which has an effect on some other circuit.

share|improve this answer

An LED does not require a resistor on the anode side, or the cathode side. It does need current limiting somehow, and a resistor is one way to do that.

Other ways of limiting the current:

  • use a current source instead of a voltage source
  • make the voltage source very close to the forward voltage of the LED, relying on the LED's intrinsic resistance to limit the current to a safe level
  • adjust the duty cycle of the voltage such that the LED never has to dissipate a destructive amount of heat

These are complex solutions to the current limiting problem. The series resistor is usually (but not always) the better solution.

share|improve this answer

If monitoring the current through the LED is important for you, put the resistor on the low side. So it will be easier to measure the current on each LED. By "easier" I mean, you fix one probe of the voltmeter to GND, and use the other one only to read the voltages on the resistors. So the current through the LED will be:

\$ I_{LED} = \frac{V_R}{R} \\ \begin{matrix} I_{LED} & : & \mbox{The current through the LED} \\ V_R & : & \mbox{The voltage on the resistor} \\ R & : & \mbox{Resistor series with the LED} \\ \end{matrix} \$

If you want to monitor the voltages on the LEDs, the you should connect the LEDs to he low side. So, you can read the voltages by fixing one of the probes to GND.

IF you don't care about the voltages or currents on/through the LEDs (e.g.; you are working with a digital circuit, or the LED is only an indicator), then it doesn't matter which side you connect the LEDs and the resistors.

share|improve this answer

it does not have to be on which side anode or cathode since it does not have a polarity. But, i do it in an anode side for a single LED and cathode side on a series LED. parallel connections on cathode side.

share|improve this answer
It's not because it doesn't have polarity. Another diode has polarity and can also be placed on either side. And parallel connections on cathode side isn't correct either: again both sides are ok. – Federico Russo Oct 7 '12 at 11:35

protected by stevenvh Oct 7 '12 at 11:36

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.