I just started to learn electronics and LED in particular. What I've read in literature is that you MUST have a resistor, otherwise your light LED will die, since the LED current is ~20-30mA, and the current from the 4AA batteries is definetely >3A (it burned the 3A fuse instantly). My circuit shown in the picture my circuit

Why the LED works and not burned down instantly? What could it be? And does it work for most of LEDs, or I just happen to have the one with higher power dissipation, or something like that?

  • \$\begingroup\$ Share the LED part number or detailed specifications. \$\endgroup\$
    – AKR
    Commented Sep 16, 2015 at 4:45
  • \$\begingroup\$ Have you checked how much current is being drawn with a meter? \$\endgroup\$
    – HandyHowie
    Commented Sep 16, 2015 at 7:15
  • \$\begingroup\$ @AKR, unfortunately I don't know thr specifications. These LEDs were in bulk with arduino controllers, so it was just included there without any specifications... \$\endgroup\$
    – chetmik
    Commented Sep 16, 2015 at 12:52
  • \$\begingroup\$ @HandyHowie, I was afraid, that if current is high (which is supposed to be without a resistor, theoretically), it could destroy my multimeter... But seems like the current is pretty small there (thanks to the SpehroPefhany answer), and I'm also going to check it, as he did. \$\endgroup\$
    – chetmik
    Commented Sep 16, 2015 at 12:55

2 Answers 2


LEDs, particularly cheap lousy LEDs, don't behave like ideal diodes. Rather, there is a resistive component in series with something that's like an ideal diode (well, ideality factor not 1.0 but something that follows the Shockley equation).

We would expect that an ideal LED voltage would drop as the temperature increases, so abuse would tend to make it die even faster, but that's not what really happens. As the die overheats, the resistive component increases in resistance faster then the LED voltage drops. The relatively high resistance factor and the positive tempco of that resistance is what lets the cheap flashlights get away with paralleling many LEDs. This is abuse of the part and will greatly shorten the life of the LED, as well as giving you much less light for your expensive battery energy. That's for moderate overvoltage- but 6V is almost double the normal Vf for a blue LED. There may be other permanent damage effects to the semiconductor that decrease the forward current.

Here is a cheap generic blue 5.0mm LED with 6.0V applied. The current rapidly drops to only 2-3 times the recommended maximum. After about 15 minutes, it's down to less than 50mA. Light level also drops greatly as it heats. It appears to have been permanently damaged by the heating and/or overcurrent and no longer outputs full brightness- at 3.3V it only produces about 1/3 of the current it used to, and about 1/10 the brightness.

enter image description here

Remember that there is usually a significant difference between the parameters that guarantee proper operation and long healthy life and those that will instantly kill the part. Just like that person who smokes like a chimney and never exercises and texts whenever they are driving can go on for many years, so can an abused LED (but probably minutes or hours rather than years).

  • \$\begingroup\$ Now that makes sense to me. Thank you, @SpehroPefhany, for the detailed answer and explanation! \$\endgroup\$
    – chetmik
    Commented Sep 16, 2015 at 12:57
  1. The internal resistance of the LED, as detailed by @SpehroPefhany
  2. The internal resistance of the battery.

All real power sources will have an internal resistance that will limit how much power you can get out of them. You can model the battery as a an ideal voltage source in series with a resistor.

(The answer to the important question of when you get the most power out of a battery is: when the external resistance in the circuit is equal to the internal resistance. This known as Impedance Matching.)

  • \$\begingroup\$ The internal resistance of the battery is important in some cases, but the OP already proved that short-circuit current is greater than 3A, so the internal resistance is not high enough to be relevant. \$\endgroup\$
    – MikeB
    Commented Jan 29, 2021 at 11:06

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.