5
\$\begingroup\$

I am trying to connect on a breadboard 4 red LED's in parallel (Forward voltage: 2V, max current 20mA). To do so, i followed the schematics below and used a 50 ohm resistor (put two 100 ohm resistors in parallel) and a 6V voltage source.

enter image description here

I connected them on the breadboard like this. The problem is that the LED's are burning and i can't understand why. Any help is welcome :)

enter image description here

enter image description here

\$\endgroup\$
2
  • 4
    \$\begingroup\$ +1 Kudos for providing the photos, as you can imagine there would be no ideal answers with just the schematic. \$\endgroup\$ Commented Jul 3, 2023 at 15:04
  • 1
    \$\begingroup\$ if I had a dollar for every time I've seen the "why can't I use a single resistor for multiple LEDs to limit the current" and all of it's mutations... :Z \$\endgroup\$
    – user213769
    Commented Jul 4, 2023 at 16:29

2 Answers 2

14
\$\begingroup\$

The problem is that the LED's are burning and i can't understand why. Any help is welcome

One LED will have a slightly lower forward voltage than the others so...

Initially, the LED with the slightly lower forward voltage will hog nearly all the current and burn. Then, the next LED (with the slightly lower forward voltage) will hog nearly all the current and burn.

Can you see where this is going? Use individual resistors for each LED please.

The other problem is this: -

enter image description here

\$\endgroup\$
29
\$\begingroup\$

The most serious problem is that you have the LEDs connected directly across the input power, bypassing the resistor.

enter image description here

There is also a more subtle issue with matching of Vf as Andy has described, but that's not what's killing your LEDs right now. Ideally you should use individual 200 ohm resistors (one per LED) rather than one 50Ω resistor, particularly if you are running near the maximum LED current. In practice you'll probably get away with that circuit under relatively benign conditions (not too hot Ta, red LEDs so power dissipation is less than for blue or white).

\$\endgroup\$
8
  • 13
    \$\begingroup\$ @Aleat Your calculations are right. But you've wired it wrong. It doesn't match the circuit you drew. \$\endgroup\$
    – Simon B
    Commented Jul 3, 2023 at 21:47
  • 2
    \$\begingroup\$ If you wired it right, it would more-or-less work, however the comments made by Andy are valid and the illumination might be visually uneven and (worse) one LED might draw excessive current (and even more as it heats) which is not good for reliability or even-ness of lighting. That's because the 2V you've used for the calculation is approximate and if one LED is lower than the others it will tend to take the lion's share of the current. Fortunately cheap LEDs have a lot of non-ideal behavior (resistive) because of the tiny die etc. so they'll probably survive okay. \$\endgroup\$ Commented Jul 3, 2023 at 21:59
  • 5
    \$\begingroup\$ I should also note that this is a case where simulations can be deceiving. If you run a simulation with the same model for each LED, they will be mathematically identical and everything will look fine, but in the real world, one LED might be 1.8V and the next 2.1V, at the same current, so the 1.8V one will hog the current. \$\endgroup\$ Commented Jul 4, 2023 at 1:15
  • 2
    \$\begingroup\$ @Sphero yeah exactly, in the simulation try specifying four different manufacturers of LED. \$\endgroup\$ Commented Jul 4, 2023 at 1:23
  • 3
    \$\begingroup\$ @Aleat Drawn (very) crudely, this is what you actually wired instead of what you intended to. Do you understand the issue now? \$\endgroup\$
    – Mast
    Commented Jul 5, 2023 at 6:50

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.