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?
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9\$\begingroup\$ If it burns long enough, sure. \$\endgroup\$– Ignacio Vazquez-AbramsAug 2, 2013 at 2:16
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2\$\begingroup\$ Thread on a similar subject. And here, somebody argues that the dominant failure mode is short. \$\endgroup\$– Nick AlexeevAug 2, 2013 at 3:22
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7 Answers
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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 MOSFET.
A diode typically fails open 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.
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\$\begingroup\$ +1 I can't make much of the first picture, but the second one looks "open" indeed ;) \$\endgroup\$– RevAug 2, 2013 at 6:31
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\$\begingroup\$ Added a better image of the same problem but on a BJT, not a diode. \$\endgroup\$– EasyOhmAug 2, 2013 at 6:37
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\$\begingroup\$ So the junction is already permanently damaged where the black dots are? It looks intact. \$\endgroup\$– RevAug 2, 2013 at 7:01
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\$\begingroup\$ Where the black dot's are the gate and drain are shorted together - while the rest looks intact current will flow through the short instead. \$\endgroup\$– EasyOhmAug 2, 2013 at 7:18
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2\$\begingroup\$ @arudino.tyro That would probably cause an open due to burnout. \$\endgroup\$– EasyOhmJul 8, 2015 at 22:57
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Diodes fail open all the time. Sure, they are of the light emitting variety, but they are still diodes.
Why do they fail open? Thermal Runaway, cause by increased voltage increasing the current through the led junction, causes Heat to literally fry the diode junction. It then goes poof, no more junction. It physically can't conduct any more electrons.
This is separate from the led bond wires melting or breaking.
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A diode will typically fail open if one of the bond wires from the leads to the actual semiconductor opens like a fuse.
To verify the contention made in a recent answer to another question, I applied a high voltage pulse without current limiting, using a capacitor bank, first to a couple of no-name LEDs, once forward and once in reverse bias, and later did the same to a couple of 1n4148 diodes.
All the diodes failed open. The LEDs were more spectacular, with the cases cracking open, but the 1n4148 diodes failed with the same end result.
So yes, a diode can fail open.
answered Aug 2, 2013 at 5:55
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\$\begingroup\$ I would argue that a bond wire failure is different from a diode failure, in the same way that a lead breaking off does not mean the ic has stopped functioning. It's a trivial difference in practical terms, yes, but a difference a manufacturer testing would make. \$\endgroup\$– PasserbyAug 2, 2013 at 6:44
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\$\begingroup\$ @Passerby Agreed. The end-user couldn't care less though - it sure beats the more typical "it's broken" :-D \$\endgroup\$ Aug 2, 2013 at 6:49
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\$\begingroup\$ If you can still remember this experiment, was there any visible damage to the 1n4148's? I can surely burn some myself, but maybe you can save me the trouble. \$\endgroup\$– FizzAug 31, 2015 at 20:12
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\$\begingroup\$ @respawnedFluff no visible damage to the 4148 but cracks visible on the LEDs. \$\endgroup\$ Aug 31, 2015 at 20:14
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As far as I know, the only way for a diode to fail open is when it fails destructively, i.e.:
- Leads or bond wires get destroyed (this very often happens with LEDs, which have comparatively long and thin bond wires from the lead to the silicon/sapphire)
- For Schottky diodes: the metallization-silicon interface gets heated beyond the range of temperatures where they thermally match, and heat stress peels them loose
For low and medium energy situations, the dominant failure mode of diodes is to fail closed. Open failure is usually a subsequent event, not a seperate failure mode.
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\$\begingroup\$ I can confirm this claim: I fried 2x'1N4148' diodes and 1x'1N914' diode by letting
6 Ampsflow across them. I set the measurement up so that as soon as there was no voltage drop across the device the supply was shut off. The all failed short immediately, without any significant amount of heating of their package. I.e. the bond wires can't have melted away. I tried frying a few '1N4001's as well, but they are tough beasts. They heated to over 200°C but did not fail, despite the fact that they released quite a bit of smoke and their package developed cracks. \$\endgroup\$– pfabriAug 11 at 15:03
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By "fail open" I assume you mean fail in such a way that it doesn't conduct in either direction. It is possible for a diode to fail open. I've applied too high a voltage to a diode and had it fail open.
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\$\begingroup\$ Thank you as I have seen this condition also , I have someone who claims that a diode can not fail in the open position, I was trying to get the information to explain what would cause this to go to an open fail, Thank you for the info \$\endgroup\$ Aug 2, 2013 at 2:31
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\$\begingroup\$ I have only ever seen diodes fail open in 15 years of general fault finding. It seems to be where a large amount of energy is applied causing the diode to pop. Where as diode shorting seems to be from small amounts of energy but at high voltage. \$\endgroup\$– SpoonAug 2, 2013 at 7:30
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Sure. Just a little while ago I pulled an IN749A Zener out of a Tektronix T922 oscilloscope. It had been said to be a failure-prone part in this scope and it affected triggering, which is what's jacked up so that's why I went after it first thing. I pulled the diode and went at it with a DMM, and sure enough it was reading many megohms in both directions. In the circuit it sits in series with a 620-ohm resistor between the -8VDC supply and ground, and the node in between the two is apparently used as a 4.3V reference so the diode is reverse-biased into its Zener voltage drop all the time. I guess this poor IN749A could only live for so long like that (the scope was built in 1977)
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I repair and design vintage electronics and every failure mode ends up being weird. Just pulled out 2 1n4002 diodes out from a bipolar power supply of a 1970's Maestro ps1a phaser that looked to be slowly cooking for a while. The positive 12v supply would only show .7v under load of the op amp and transistor circuit (which probably has a failure yet to be found) but when isolated worked properly +12v and tested properly except a varying forward drop off voltage. .5 to .6v on my DMM. After pulling them out and testing new ones sussessfully, I tested the forward resistance at 100k. Hope this helps.
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\$\begingroup\$ Welcome to EE.SE. What do you mean by "forward resistance" of a diode? \$\endgroup\$ Aug 24, 2018 at 19:02
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\$\begingroup\$ Using a DMM on the resistance testing mode, the diode's forward conductance threshold can be measured in a resistance format. Unfortunately I think it is an unreliable method due to low currents and voltages of the DMM but still interesting and useful in diagnosing bad components that are not working like others. I was hoping I could directly measure the added resistance of a failed diode but Interestingly the failed diodes forward voltage threshold measurements were less than working diodes. This makes me think the failed diode is now operating like a tunneling diode. \$\endgroup\$ Aug 29, 2018 at 6:58
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\$\begingroup\$ Working diodes will register 800kohm forward conductance and infinite ohms backwards. A germanium diode will register 1k ohm forward and 300k backwards restistance. A 13v Zener diode shows infinite ohms on the 13v side and 8M ohm on the .6v side. \$\endgroup\$ Aug 29, 2018 at 6:59