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I am using a blue LED (InGaN) in a simple circuit to show me when my garage door is opened by looking through a window.

The door is usually closed and may get opened once per day for a couple of minutes but sometimes longer.

The LED will work for 1-2 months then stop. I replace it and then the same thing happens.

The circuit includes a blue LED in series with a 4.7K Ohm resistor, a 120VAC/24VAC transformer connected to 120 VAC source, a normally-open magnetic switch that closes when the garage door opens, and a fuse. The picture included does not show the transformer or the fuse.

As far as I can tell, I am not exceeding any design limits of the LED including temperature.

Any ideas?

enter image description here

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    \$\begingroup\$ Note Vrev =-5V max . Use two LEDs back to back to block neg V \$\endgroup\$ Jan 28 at 20:11
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    \$\begingroup\$ Sure you are. As Tony said, you are applying more reverse voltage than it can handle. Why are you using 24 Vac in the first place for the LED? \$\endgroup\$
    – winny
    Jan 28 at 20:13
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    \$\begingroup\$ @winny - 24Vac is a common voltage supply for garage door opener remote control receivers. Presumably it is already available. \$\endgroup\$ Jan 28 at 20:18
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    \$\begingroup\$ using 24VAC because I had a transformer that was readily available. Sounds like I i'm over driving it. Thx \$\endgroup\$ Jan 28 at 20:27
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    \$\begingroup\$ Yeah, the problem is the resistor is dropping 0 volts in reverse current. Because it's a diode, there is no reverse current, so resistor voltage drop is 4700 ohms x 0 amps = 0 volts drop. 24V RMS - 0V drop = 24V RMS against the diode. \$\endgroup\$ Jan 30 at 1:58
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Any ideas?

Yes. You are feeding the LED with an AC supply. That means that every second half-cycle it will be reverse biased. Most LEDs can withstand about 5 V in reverse. You're applying 24 V AC which will have a peak voltage of \$ 24\sqrt 2 \ \text V \$ so I'm surprised you got any length of time out of it.

You have at least four options:

schematic

simulate this circuit – Schematic created using CircuitLab

  1. Add a diode in series.1
  2. Add a bridge rectifier.
  3. Add a diode in reverse parallel with the LED.1
  4. Add another LED in reverse parallel with the LED.2
  • 1 LED will be half brightness for a given resistor value as only positive half-waves will light the LED.

  • 2 Each LED will be half brightness for a given resistor value as only positive half-waves will light one LED and negatives the other.


Technical note

Strictly speaking, the first diagram relies on most of the reverse voltage being dropped across D2 while D1 remains at < 5 V. What happens in practice will depend on the reverse leakage current of both devices and the one with the lower leakage current will experience the highest reverse voltage. This would require a dig through the datasheets for both devices.

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    \$\begingroup\$ On SE sites you hit the "Accept" button on the left if there's an answer that solves the problem. Otherwise the system keeps bringing the question up for an answer. You can also upvote other answers. \$\endgroup\$
    – Transistor
    Jan 28 at 22:28
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    \$\begingroup\$ Presumably in option 1 you'd need a diode with Vrev>=30V. In particular, a LED is by definition a diode, and you can't reliably solve the problem by putting two of these blue LED's in series. (Although the failure rate would likely drop sharply) \$\endgroup\$
    – MSalters
    Jan 29 at 13:23
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    \$\begingroup\$ Some 1N400x would work fine. \$\endgroup\$
    – Lundin
    Jan 29 at 13:51
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    \$\begingroup\$ I just remembered that when I graduated in '75, I put a GaAs LED with my 24V doorbell switch and it got dimmer after a few yrs when we moved out. It blinked when released from the flyback solenoid then stayed on. Tough little diode. \$\endgroup\$ Jan 29 at 14:14
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    \$\begingroup\$ @MSalters that's going to be any normal diode though, no need to look for a special one, feel free to double-check it if you're worried \$\endgroup\$
    – user253751
    Jan 29 at 19:09
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schematic

simulate this circuit – Schematic created using CircuitLab

The solenoid current will produce a pulsed current when released so a greater R value is preferred. Most of the voltage drop is across R with V=LdI/dt. (which can reach HV )

  • next time you have a spare LED to use this to also protect against ESD damage in circuit or from small flyback currents.
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    \$\begingroup\$ From very small flyback currents. I would never recommend an LED in place of a proper flyback diode. The latter are designed for purpose, LEDs are not. Yes, they are both diodes, but flyback diodes can take significantly more punishment - and in a flyback application they will receive that punishment. \$\endgroup\$
    – J...
    Jan 29 at 15:40
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    \$\begingroup\$ I used to correct my Prof's all the time. I guess this is payback senior's moment ..doh .. at least someone is paying attention, kudos. \$\endgroup\$ Jan 29 at 20:58
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Got anything else to monitor? Detect 2 things on 1 cable

Which works because this is AC.

schematic

simulate this circuit – Schematic created using CircuitLab

Note that we're depending on each diode to be working to protect the other from reverse current. If D3 failed with SW1 closed, no current would flow, so R1 would drop no voltage, and D4 would get smacked with 24V RMS. The only way to fix that is another guard diode right next to each LED.

Also, you'd need to make the diodes different colors, and they might have different forward voltages, so a resistor above one of them might be necessary.

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Gotta get some voltage drop in the series resistor when the diode is back biased. Another way not mentioned above is to hang a parallel resistor across the diode. Calculation of the values is left as an exercise for the student. :-)

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