I have just ordered LEDs with resistors.

I am going to fit the leds to my car interior panels to come on when the doors open and the led's are to act as lighting under the doors.

I will be running 5 led's per door.

I will wire the leds in series to each other but do I need to put a resister in front of each led? or do I even need resistors in the circuit?

  • 1
    \$\begingroup\$ Could you share a link to the specific product you purchased, please? \$\endgroup\$ Nov 12, 2012 at 9:50
  • \$\begingroup\$ What is the expected temperature range for system? Because if you want it to work in a car, the difference of temperature between cold winter and summer bellow the sun is significant and could have to be taken into account in the resistor value computation. \$\endgroup\$
    – Blup1980
    Nov 12, 2012 at 9:50
  • \$\begingroup\$ Note typical Vf (forward voltage) per LED. Red x 5 may be OK. White or blue x 5 will not be. \$\endgroup\$
    – Russell McMahon
    Nov 12, 2012 at 12:44

2 Answers 2


The voltage across a LED is more or less constant, hence the current varies widely when you vary the voltage. So yes, you almost always need a resistor (or another current limiting device) in series with a LED.

In your case, assuming standard 1.8V/20mA LEDs, and the LEDs are put in series, the 5 LEDs will drop 1.8V * 5 = 9V. At the maximum battery voltage of 13.8V this leaves 13.8V - 9V = 4.8V for the resistor. To get 20mA this resistor must be 4.8V / 0.02A = 240 Ohm. That is not a standard value, 270 will do.

At the nominal battery voltage of 12V this will result in a LED current of ( 12 - 9 ) / 270 = 11 mA. This is unavoidable with this type of setup.

Alternatively, you could put the LEDs in parallel with a series resistor for each LED. Now the resistors must be ( 13.8 - 1.8 ) / 0.02 = 600, 'round' up to 680. The nominal current will be ( 12 - 1.8 ) / 680 = 15 mA.

To get less current variation in the current you can use an LM317 in constant-current mode, check the datasheet.


You just need one resistence for the entire series. You need to compute the value considering the voltage drop cause by all diodes ( just by adding each voltage ) and considering the current you need for one diode.


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