1
\$\begingroup\$

Complete newb here; sorry this may be a basic question but trying to learn. I made a simple circuit with nothing except an Arduino power source a 100 Ohm resistor in series with a 10mm red LED. Weirdly it gets dimmer if I remove the resistor and my question is why is that occurring? I thought reducing resistors would brighten the LED. The datasheet is attached. The max current is 80mA and the LED worked fine after and this phenomena is repeatable so I don't think I burned the LED out. This phenomena doesn't happen if the power source is 3.3V in which case I don't notice much brightness difference between the two. This phenomena also doesn't occur with a same sized blue LED by the same manufacturer. FYI, there's no code, I am running the power direct from the Arduino Uno's 5V and 3.3V pin and gnd pin. I read the Sparkfun tutorial on resistors and am confused.

https://www.sparkfun.com/products/8862

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
4
  • 2
    \$\begingroup\$ You've probably already damaged the LED. Not broken it completely. Just damaged it. If you have a second one laying around, try hooking both up separately with the same circuit and turn out the lights and compare them. (Use perhaps 68 Ohms each instead of 100?) If I'm right about the more permanent nature of the damage, the fresh LED will shine a little bit brighter. Worth an experiment. \$\endgroup\$ – jonk Feb 12 at 3:35
  • 1
    \$\begingroup\$ (1) I suspect your LED is sort of super bright with a built in resistor, so that you can directly apply 5V and get say, 70mA without any problem. If you supply 3V3 the current might be smaller, but might not be much smaller. (2) the built in resistor might be much bigger than 100R, so with or without the series resistor does not make much difference to human eyes. \$\endgroup\$ – tlfong01 Feb 12 at 5:05
  • 1
    \$\begingroup\$ References: (1) Red LED 5mm with Resistor (2V ~ 5V, 2000 ~ 3000 MCD, 620 ~ 625NM) Product Sheet - SparkFun $9/25pcs sparkfun.com/products/14560, (2) YSL-R531R3c-5V Red LED 5mm with Resistor Datasheet - YunSun cdn.sparkfun.com/assets/c/9/b/9/c/YSL-R531R3C-5V.pdf, (3) YSL-R531R3c-5V Red LED 5mm with Resistor Spec - YunSun imgur.com/gallery/GkKqwEo. \$\endgroup\$ – tlfong01 Feb 12 at 5:39
  • \$\begingroup\$ The LED seems to behave normally after I put the resistor back, FYI. Also, I put a link to the LED I used which is different than these links in the comments, FYI. \$\endgroup\$ – rfii Feb 16 at 3:51
2
\$\begingroup\$

No resistor = way too much current

Way too much current = LED is running way too hot

When an LED gets too hot, it's ability to emit light goes down.

Touch it, bet it's really warm if not hot.

Another thing, when the temp of an LED changes, it's output wavelength changes. It gets longer. So your red LED may have shifted to infrared

OR

The LEDs are overloading the power supply and its responding by folding back it's output voltage

Either way,, blue behaves differently because it has a higher voltage requirement than red. This is probably limiting the over current condition.

They behave similar at 3V instead of 5 because there is less difference compared to Vf, so they aren't getting as much over current.

.

Got an ammeter??? Measure the current you're putting through them.

\$\endgroup\$
2
  • 4
    \$\begingroup\$ I don't think the wavelength shift is that pronounced that it'd get to infrared before the LED melts. \$\endgroup\$ – Hearth Feb 12 at 3:39
  • \$\begingroup\$ The blue one was <30mA at 5V and <20 mA at 3.3V with the resistor and only slightly higher without the resistor. The red one was closer to 120mA 5V and 80mA at 3.3V with the resistor, but I didn't test the current without the resistor bc I was worried the abnormal behavior meant I was damaging it. This was being powered by the 5V pin (not a GPIO pin) on an Arduino Uno powered by USB. Do you still think I was drawing too much current from the supply? \$\endgroup\$ – rfii Feb 16 at 3:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.