For a 9VAC-supply, I want an indicator that it's turned on so I hooked up a 1k-resistor and a yellow LED (3mm) in series, parallell over the 9V-source.

The LED lights up in yellow, but it's slowly changing to bright orange over time. While Fluke-ing it under operation it says 3.5V. I think that's a bit high since a loose LED shows 2.3V in diode-measurement on my Fluke.

My guess here is that the LED got too much current and got damaged, hence the bright - and wrong - color. 9VAC is roughly 13V û, meaning 1k gives 13mA through it. Datasheet says "ok up to 20mA, recommended 10-15mA".

Was 1k too low, or what caused the LED to burn?

  • \$\begingroup\$ Measure the actual current through the resistor and led. \$\endgroup\$
    – Passerby
    Jun 1 '17 at 19:19
  • 3
    \$\begingroup\$ What's the reverse breakdown voltage? \$\endgroup\$
    – Oskar Skog
    Jun 1 '17 at 19:19
  • \$\begingroup\$ 13 V is more than the 5 V it can take in reverse. Did you burn it out? Add a diode in series. \$\endgroup\$
    – winny
    Jun 1 '17 at 19:20
  • \$\begingroup\$ Right. I forgot about the reverse voltage. Thanks. \$\endgroup\$
    – bos
    Jun 1 '17 at 19:38
  • \$\begingroup\$ you can put 3 of your LEDs in series to avoid the burnout... \$\endgroup\$
    – dandavis
    Jun 1 '17 at 21:16

The maximum reverse voltage on most LEDs is specified as 5V. Your 9V RMS supply has a peak voltage that is almost 13V, maybe more when the LED is not loading the supply. Some LEDs will break down with that voltage (although many will not) and that can cause excessive dissipation.

The color change you observed is indicative of extreme dissipation in the LED chip, causing the die temperature to rise to destructive levels.

Put a diode such as a 1N4148 in series with the LED. And check the resistor value with your multimeter to make sure it is what you think it is.


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