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?

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    \$\begingroup\$ 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. \$\endgroup\$
    – Nick T
    Commented May 2, 2011 at 12:08
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    \$\begingroup\$ @Nick: electronics.stackexchange.com/questions/11734 \$\endgroup\$
    – endolith
    Commented Jun 6, 2011 at 19:20
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    \$\begingroup\$ @NickT: Also physics.stackexchange.com/a/17131/176 \$\endgroup\$
    – endolith
    Commented Dec 16, 2011 at 19:26

9 Answers 9


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."

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    \$\begingroup\$ Or more simply, it's read as "the resistor must be connected to the anode", not "the resistor must be connected to the anode". \$\endgroup\$ Commented Nov 1, 2016 at 6:02
  • \$\begingroup\$ Is it a correct assesment that the current starts at 0A and increases until it basically hits the limit of available passthrough? Because surely otherwise the order in which you place the components would matter, wouldn't it? just FYI, I'm an absolute beginner in electronics. \$\endgroup\$
    – xorinzor
    Commented Feb 21, 2018 at 16:20
  • \$\begingroup\$ We can describe circuits with equations well enough (with "lumped element" approximations, which do break down after a point) using rules like Ohm's law and Kirchoff's laws. Then we can solve the equations and find the voltages across components and the currents in the loops. Sounds like you're entering philosophical territory, asking "How does the resistor know? What's the mechanism?" Physical systems can arrive at the solution to the equations, and without a pocket calculator. How? I'm a hobbyist, so I can't say what's going through an electron's mind. Maybe a physicist can. \$\endgroup\$ Commented Feb 22, 2018 at 17:19
  • \$\begingroup\$ @xorinzor Not exactly: electrons move at roughly the speed of light, so there's not exactly a "speed-up" or "warm-up" time. It's more or less 0 to "full" (whatever full is) instantaneously. That being said, no matter what, if current is flowing through two or more elements connected in series, then that current is the same value for all elements in that circuit. \$\endgroup\$
    – John Doe
    Commented May 25, 2018 at 21:36
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    \$\begingroup\$ @JohnDoe Electrons do not move anywhere close to the speed of light but rather mosey along at a rather leisurely pace called the drift velocity. However, electromagnetic waves (which carry voltage and current information throughout a circuit) do indeed move at the speed of light. \$\endgroup\$
    – pr871
    Commented May 25, 2018 at 21:53

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?

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    \$\begingroup\$ I love the "chain" or "link" analogy. I have been told other analogies, but none as good as those. \$\endgroup\$
    – Kellenjb
    Commented May 2, 2011 at 13:04
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    \$\begingroup\$ 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. \$\endgroup\$
    – endolith
    Commented May 2, 2011 at 15:39
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    \$\begingroup\$ That Circuit Simulator applet is awesome ... played with it for 30 minutes now. :) \$\endgroup\$
    – Spechal
    Commented May 4, 2011 at 5:58
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    \$\begingroup\$ @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? \$\endgroup\$
    – endolith
    Commented Jan 7, 2015 at 19:58
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    \$\begingroup\$ @endolith Ahhh that actually makes perfect sense. I thought there would be a huge initial burst of energy but when I think about it some more you make perfect sense. Thanks for the answer and clearing this up for me!!!!!! (highfive) \$\endgroup\$
    – Hannibal
    Commented Jan 7, 2015 at 20:30

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.


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.


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.


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.


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.


Functionally it doesn’t matter. The elements (LED and load resistor) are in series so the current flowing through them will be the same regardless of the order in which they're hooked up.

That said, if the LED is driven on the low side I have a preference to put the load resistor from VDD to LED anode. Reason? Connected this way, if there’s a short to GND at the anode (from, say, an errant scope probe) then it won’t kill the LED. Conversely, if the LED anode is tied to VDD and the load resistor to the cathode, shorting the LED cathode puts the full supply across the LED, which makes a nice popping noise...


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.

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    \$\begingroup\$ 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. \$\endgroup\$ Commented Oct 7, 2012 at 11:35

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