I am making a simple circuit with 2 LEDs but it doesn't work - the LEDs do not turn on when I supply 3.3V power to the circuit. (I've tried connecting one LED and it worked.) The resistor is 330 Ohm.

Any thoughts or ideas, please?

enter image description here

  • 4
    \$\begingroup\$ Welcome to EE.SE. You have missed two critical pieces of information - the power source and the resistor value. Please add these in your question rather than in the comments. \$\endgroup\$
    – Transistor
    Aug 9, 2016 at 20:39
  • 3
    \$\begingroup\$ Schematic would help too and the indication of the LED polarity on the picture as well, as it is not clear from it. \$\endgroup\$
    – Eugene Sh.
    Aug 9, 2016 at 20:41
  • \$\begingroup\$ Are you using 3.3V by chance? (That will then also immediately be your answer) \$\endgroup\$
    – Asmyldof
    Aug 9, 2016 at 20:45
  • 5
    \$\begingroup\$ Every diode has a characteristic called its forward voltage. If a diode's forward voltage is 1 volt, for instance, the "voltage drop" over that diode will be 1 volt. Now, the forward voltage of LEDs is usually about 1.8 volts. You have 2 LEDs in series... so 2 1.8 volt drops. 1.8 + 1.8 = 3.6... is that more or less than 3.3 volts? \$\endgroup\$
    – John M
    Aug 9, 2016 at 20:54
  • 1
    \$\begingroup\$ Whenever you are building lab stuff with LEDs, always measure the forward voltage with a multimeter (or read the datsheet). LED characteristics are typically specified at an ideal current (usually 20mA), so if you want maximum brightness, you should strive to drive that current through it. The value of the series resistor is obtained from supply minus forward voltage, then Ohm's law: divide with the desired current. Also note that because of the forward voltage, putting two LEDs in series isn't a great idea if you want them to have the same brightness. \$\endgroup\$
    – Lundin
    Sep 9, 2016 at 7:51

1 Answer 1


Chances are fair that at least one of your diodes is in backwards.

LEDs are directional, but figuring out the direction might not always be simple. There is an ANODE, which needs a more positive voltage than the other side, the CATHODE, to turn the LED on. enter image description here

If you look at the LED "bulb" part, there is often a flat side on the base somewhere. That is the cathode. The Anode needs to go to the more positive voltage, and often has a longer lead on it than the Cathode.

Another possibility is that your resistor is too big. Assuming about a 2V drop for each diode (this will vary largely by color), there will be 1V across the resistor, assuming you're driving things with 5V. If it is 330\$\Omega\$, that leave you about 3mA, which is not enough to light an LED. Try about 100 \$\Omega\$.

We also don't know how you're powering things.

You should get in the habit of drawing real circuit diagrams using the symbol as shown, so people know what you're talking about.

  • \$\begingroup\$ I agree - this could be an issue but in my case wasn't. The direction of both LEDs were right. My problem was with the power supply as it was too low. Thank you for your answer. :) \$\endgroup\$
    – mk202
    Aug 9, 2016 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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