I know a diode can short out and fail closed. Is it possible for a diode to fail in the open position, and what would cause this to happen?
Electronic – Can a diode fail to open position
diodes
Related Solutions
First, why will the zener fry? You may have a 12V supply powering your regulator, but your supply must have a fuse or circuit breaker somewhere to be safe. If this fuse or circuit breaker has a current rating which the zener can tolerate, then your PCB should be protected by the zener while the fuse shuts off, and no harm will come to the zener. If the current rating is higher than the zener can tolerate, well, you've made a mistake in choosing a current limiter.
What is the function of the 6V zener? If its function is to protect the 5V rail of your circuit, then I suggest that it's not going to do a very good job. Many 5V components have a maximum input voltage of around 5.5V or 6V, so a 6.2V zener won't help much. I'm not sure what your environment is like, but it's usually better to just shut everything down if your regulator fails than to try to let the zener run everything.
One common use of a diode for protection from error conditions is to use a reverse-biased rectifier diode across the input terminals. That way, if anyone connects the source backwards, the power will be dissipated in your designated diode. Make sure that this diode's forward voltage is less than the maximum reverse voltage on the protected components. In this configuration, when the source is plugged in backwards, nothing will work: The supply voltage will be at -0.7V. Presumably, you'd notice that the power LED was not on or that the circuit was not functioning, and correct your error.
Second, no, you can't say whether the result will be a short or open circuit. You've operated the device outside of the specifications, so anything could happen.
Inductors resist changes in current. To change the current in an inductor, you must apply a voltage. If you remove the voltage, the inductor will make its own voltage so that the current can keep flowing. Of course, making this voltage requires energy, and as that energy moves out of the inductor and into something else, the current will eventually reach zero, when there's no energy left in the inductor.
So, when the switch is closed for a while, there is a current in the inductor. When you open the switch, that current must keep going. Where can it go in this circuit, if its prevented from going through the battery by the open switch?
A diode allows current to flow only in one direction. It's no mistake that their schematic symbol looks like an arrow. If the switch is opened, there is only one place for the same current to keep flowing, and that's through LED1 and LED2. It will keep flowing here, even without help from the battery, until all the stored energy in the inductor has been transferred to the LEDs.
Best Answer
A diode fails closed due to overvoltage. This is called punch-through. This is is the principle used in ESD diodes. If they can't handle the voltage the PN juntion fails and short to ground, protecting any circuitry after them. On the image you see a small black dot where the voltage went through the junction. eg:
*Second picture added that shows this better on a BJT.
A diode typically fails to open happens due to over current. This is called metallization burnout and can occur from things like EOS (Electrical Over Stress). Image shown bellow. Over current causes excessive heating and literally burns the metal away. As mentioned above this is easy to demonstrate on LEDs as their current carrying capability is much lower than rectifier didoes.