You need to know the forward voltage (Vf) and current requirement (A) of the LED. Then, calculate if the 1.5kΩ resistor results in enough (and not too much) current for the LED. AND as dl324 said, the LED may load the output line to the point the MCU cannot detect the output of the PIR… That means the resistor in the series combo needs to drop (9 - (2*3)) = 3 volts, and the resistor on the lone LED needs to drop (9 - 3) = 6 volts. Current is continuous through a wire, so 15mA is the current. Using those values, you use Ohm's law which is very basic. This will calculate the values for the resistor: V = IR -> R = V/I. LED Series Resistor Calculator. Note: When you select a resistor for this purpose, choose a device with a power rating between 2 and 10 times the value calculated below in order to avoid excessive resistor temperatures. Supply Voltage. V. Forward Voltage. V. Forward Current. mA. Resistor Value. This total current is equal to the sum of the three LED currents. For example, with a 6 V battery, 3 V LED (on average), and a 1000 ohm resistor, the LED current would be 6V − 3V 1000Ω = 1mA 6 V − 3 V 1000 Ω = 1 m A. The circuit is really only a bad idea if you are operating the LEDs near their max current. In your case, answer is NO, you do not need resistor. No need for a resistor. The general rules for these kinds of things are: Don't exceed the rated voltage (in your case 12 V DC - or 15 V DC, it is not entirely clear to me). Use a power supply that can provide at least the rated current (in your case 85 mA). Figure out what resistance it will require to hit that current. Use Ohm’s Law and math it - voltage source minus forward voltage at the desired current, then V/I. We will hopefully have a tool for this ASAP. Make sure the resistor has a high enough power rating. Review everything to make sure it makes sense. So connect that size resistor at each bulb housing, or divide the resistor Ohms value in half, double the Wattage, and connect one to each side. In the quoted numbers, V=14, and the resistor replaces one bulb at 21W, so you get 196 / 21 = 9.8 Ohms, >21W. The 10W value is a bit off, but that's possibly the nearest available value in a 10W When it comes to turn signal bulbs, LED models need a resistor (or flash controller) in order to function properly and at their maximum efficiency. One LED resistor is needed for each turn signal bulb. Many LED turn signal bulb models feature built-in resistors to make the upgrade a seamless one. Stock incandescent bulbs drain more energy from So, the WS2812 does not need a resistor on its input. Full stop. The datasheet even has an exemplary schematic without one. It's still a good idea to have one between your 5V data source and your first WS2812, simply because: These things are often used in a long chain, which can very quickly change very sharply in current draw. N5i4IVC.

do i need a resistor for led