Electrical – How does forward voltage work


I had a small circuit with a few LEDs and a 10k ohm resistor, powered by a 5V pin from an Arduino.

I noticed that every time I added another if I had just one LED, then the resistor would drop most of the voltage, while the LED would only drop 1.6V. As I added another LED, now having two in the circuit along with a resistor, the resistor started to drop less voltage.

  • So my question is, if I kept on adding LEDs, why does the voltage dropped by the resistor decrease, is it because the forward voltage does not change?
  • What would happen if I added 10 LEDs, that have a forward voltage of 1.6V, only using an input of 5V, from my arduino, and a resistor?
  • Would the resistor create no voltage drop?
  • Would the forward voltage be reduced due to only providing 5V?
  • Or would only a few LEDS light up?
  • Why does voltage drop reduce for resistors and not LEDs?

Best Answer

The current through a diode vs the voltage is a very steep curve. A small change in voltage results in a huge change in current. When you add another diode, it takes approximately the forward voltage to turn on and then after that it takes very little change in voltage for a huge change in current. A resistor on the other hand is a gentle linear curve. A change in voltage results in a very similar change in current on a resistor.

Because of these different dynamics, you get the effect you see. Diodes consume their forward voltage and not much more, and resistors absorb all the changes in voltage. If you have so many LEDs that result in the requirement of 10V of forward voltage but you only give the series 5V, then they will either not light at all or will very very dimly glow.

The forward voltage is nothing more than a simplification of the diode curve which is actually exponential in nature. The forward voltage just gives us a good approximation.