# LED series Resistor- Noob question

Can someone please explain to me, like I'm 5, why this circuit does not work?

Every resistance calculator I use says that there is not enough voltage to calculate a resistance. If I understand correctly, wouldn't the LEDs light but just at a lower brightness?

Using R = V/I, The source voltage is 5v, the LED voltage is 6 divided by the current of 20mA equals -50ohms. All of the charge is used up by the time it reaches the arduino right?

The simple reason is that LEDs are not light bulbs. An LED is not a passive device and does not follow Ohms law.

To a first approximation if the voltage across an LED is below the forward voltage then no current will flow. In that situation the voltage across the LED pins is whatever the supply voltage is.
If the voltage across an LED is over the forward voltage then an unlimited current can flow and the voltage between the pins is fixed at the forward voltage.

This approximation isn't perfect but 90% of the time it's good enough.

So with 1 LED and no resistor you get 3V over the LED and then 2V left over with zero resistance, I=V/R = 2/0 = infinite current flow. The LED is dissipating 3*infinity watts of energy and instantly vaporizes. Clearly this isn't exactly what happens in the real world but the result is enough to tell you that bad things will happen.

With 1 LED and a 200 ohm resistor you get 3V over the LED and then 2V extra with 200 ohms resistance, I=V/R = 2/200 = 10mA. The LED is within its operating current and will light up.

With 2 LEDs and no resistor you get 3V on one LED and 2 V on the second (or 2.5 on both if you prefer). Either one or both LEDs are below their forward voltage and so no current flows. The LEDs remain off.

The two LEDs in series have a forward voltage of 6v which is more than the Arduino's supply voltage.

Forward voltage is the bare minimum amount of voltage required for the LEDs to start conducting electricity (ie: to allow the current to pass through)

Read this: https://www.baldengineer.com/led-basics.html :

Forward Voltage

The Forward Voltage defines the amount of voltage required in order to conduct electricity. Any voltages below this amount will cause the LED to remain “Open” and non-conductive. This also means any components in-series with the LED will not have current flowing through them either! Once the voltage dropped across the LED reaches the Forward Voltage, it will begin to conduct electricity. Not only that, but the LED will only drop its Forward Voltage at any given time.

For example, consider a LED with a Forward Voltage rated at 3.0V. Now what happens if you attach the Anode to the Positive (+) Terminal of a AA (LR-6) Battery and the Cathode to the Negative (-) Terminal? Will the LED do anything? No! The AA (LR-6) Battery only has a nominal voltage of 1.5V. Until you add a second battery, the LED will not light up.

LEDs won't necessarily light at lower brightness (or conform to Ohm's law at all), it's explained here. If an appliance apparently does provide a "dim" LED mode, it's usually using a clocked signal (and blinking rapidly).

• I understand, thank you for the link. If i run on this configuration, would it damage my LEDs or Arduino? May 4 '17 at 15:44
• Probably not, as the other Andrew noted above, no current will flow.
– jvb
May 4 '17 at 19:06

Can someone please explain to

With the right leds, the circuit can work.

Where it may fail:

1. Two less with combined forward voltage greater than the one applied to them. The meds wouldnt light up or wouldn't light up brightly, depending on the leds used.

2. The two leds with combined forward voltage much less than the supply which has low output resistance. The two leds may light up very brightly, or so briefly ending up in led heaven permanently. If you can capture them in high frame rate code on the glow before their death is quite pretty.