Is it because it can exceed the reverse breakdown voltage of semiconductors?

For instance in low voltage devices say 1-5v is the reverse voltage rated a lot lower than that for damage to occur?

Also I watched a video where a car battery was accidentally connected backwards and the main 80 amp fuse blew yet no other damage occurred. Makes me wonder what failed that drew so much current?

  • \$\begingroup\$ I believe there are circuits where a diode is connected from negative to positive. So if you connect the battery backwards, it's essentially a short circuit. \$\endgroup\$ – Bradman175 Jun 4 '16 at 13:38
  • \$\begingroup\$ Why would it be connected that way? I thought diodes used as protection would be connected to stop the flow of electricity in a reversed polarity incident \$\endgroup\$ – ohmmy Jun 4 '16 at 13:41
  • 1
    \$\begingroup\$ I think that you are correct. Reverse polarity produces great currents because of breakdown of semiconductors. Also some semiconductors (MOSFETs) have a parasitic diode in antiparallel that will conduce if you apply reverse polarity to them. \$\endgroup\$ – Claudio Avi Chami Jun 4 '16 at 13:52
  • \$\begingroup\$ What are the general breakdown voltages? Because some devices already use low power voltages and I imagined semiconductors could handle that rating in reverse no? \$\endgroup\$ – ohmmy Jun 4 '16 at 14:13

So speaking to the semi-conductor / Silicon side of things. The expectation is that people who use the device will operate it correctly, so reverse wiring really isn't designed into the product (with some exceptions). Most of the the reverse bias (forward conduction in the protection diodes) arises from parasitic structure in the die, and as such they will easily conduct at very very low resistances and if you try to drive those structure hard the metal layers or bond wires are going to lose big time. Overheat, melt and be irrecoverably damaged i.e "magic smoke".

The examples that need to handle this reverse bias and survive will be designed to do so. But typically you don't design to protect people from their own stupidity unless it's endemic.

One concern that will always come up is the event of latch up with parasitic SCRs in the substrate. These can be activated by reverse bias or very high \$ \frac{\partial V}{\partial t} \$ spikes. However, as much as this is a reality in the bare process, a lot of time is spent in ensuring that this cannot happen if the product is designed properly. In short during qualification of a new process this is analyzed and studied extensively and then parameters are put in place that are used to ensure that Latch up cannot happen in the final product. These are manifest in the DRC (Design Rules Check) and have to do mainly with spacing of structures and implants.

Some designs may require that these DRC be waived, if this is the case then this would be noted in the data-sheet. But for most products this is not going to happen.

  • \$\begingroup\$ That was a very long answer for - the body diode in the semiconductors will pass an almost unlimited current at a voltage (~0.7 V) which is lower than the normal supply voltage (say 12 V) and with no other mean of limiting the current, the semiconductor(s) and/or fuse will give. :-) \$\endgroup\$ – winny Jun 4 '16 at 17:22
  • \$\begingroup\$ @winny yep, but let's keep that as a secret between us ... ;) Most people here don't know/understand the structure so it's better to give a more general response. On the flip side a lot of people here always mention latch up, which is 99% of the time not possible due to design. \$\endgroup\$ – placeholder Jun 4 '16 at 17:25
  • \$\begingroup\$ Ah yes, secondary breakdown. The NiCd-memory effect of transistors. :-) \$\endgroup\$ – winny Jun 7 '16 at 7:06

Is it because it can exceed the reverse breakdown voltage of semiconductors?

That's one of the reasons, but you'll have to look at the device as something much more complex. For example, think of a transistor that would normally be "off", but negative biasing puts it into conducting mode.

In that mode, it might carry a much higher current than it was designed for, leading to the destruction of the device.

Then you've got protection diodes, whose job is to break down in case of ESD, or in case you suddenly stop supplying current to a motor, and that causes, by the inductive properties of that motor, a negative voltage spike of very large amplitude (but only for a very short duration), which you need to "short away" as quickly as possible. Those diodes would normally be in "non-conducting" mode, but if you wire them up reversely, they will constantly conduct electricity, overheat and die.

More examples:

In classical switch mode power designs, there's often a diode antiparallel to the excited coil; in normal bias without the switch mode controller active, it doesn't let any current through. If you plug that circuit in in reverse, the diode will constantly carry high current and evaporate.

Electrolytic Capacitors (typically, the round cans) are voltage sensitive and will quickly be destroyed by reverse voltage, reducing their oxide dielectric layer, leading to them gassing, high currents flowing, liquid blowing up, fires, the four apocalyptic horseman and more Justin Bieber albums.

  • \$\begingroup\$ How would negative biasing put a transistor in conducting mode? I'm also wondering what's a general reverse breakdown of some transistors and ic chips \$\endgroup\$ – ohmmy Jun 4 '16 at 16:03
  • \$\begingroup\$ @ohmmy the base-emitter and base-collector voltages decide in what mode a transistor is in. bipolar Transistors aren't fabricated symmetrically, but in principle, there's no big difference between the B-E and the B-C junction, and at a certain potential difference they become conductive, hence "turning the transistor on". That's just how transistors work. \$\endgroup\$ – Marcus Müller Jun 4 '16 at 16:05
  • \$\begingroup\$ "general reverse breakdown" doesn't exist. These depend on a) what you consider forward, and b) on the technology used. In principle, they can range from nearly zero to something like 6V for a single IV/IV or III/V semiconductor junction. \$\endgroup\$ – Marcus Müller Jun 4 '16 at 16:06
  • \$\begingroup\$ Really what you're asking is quite a bit like "why can't I put water in my fuel tank?", and we're telling you, yes, that's because it's water and then the engine draws water and instantly takes damage, and you're still a bit like "but it's still just a liquid just like fuel!". \$\endgroup\$ – Marcus Müller Jun 4 '16 at 16:08
  • 1
    \$\begingroup\$ Also, really, many diodes in ICs are used in reverse mode to establish a given effect – reversing them will make them go up in smoke. ICs are far more complex than just one type of transistor and one type of Diode combined in a manner where the diode is always forward-biased. \$\endgroup\$ – Marcus Müller Jun 4 '16 at 16:09

I watched a video where a car battery was accidentally connected backwards and the main 80 amp fuse blew yet no other damage occurred. Makes me wonder what failed that drew so much current?

Nothing 'failed'. With the battery connected in reverse the rectifier in the alternator would have been forward biased, causing a very high current to flow because it is connected directly to the battery. Luckily rectifier diodes can handle very high peak currents, so provided the fuse opened quickly they should have been protected.

Here is a typical alternator circuit. The arrows on the rectifier diodes show the direction of current flow. Normally the diodes steer current from the stator coils into the battery, but block current from going from the battery back into the coils. However if the battery is connected in reverse then its positive terminal is connected to chassis ground, so it can push current straight up through the diodes.

enter image description here

Most solid state devices have components that are polarity sensitive, whether by design or as an inherent part of their makeup. If the device is not specified to handle reverse polarity then you must assume that it can't.

Don't plug a power supply into a device it was not designed for without first verifying that the voltage and polarity are correct. Never connect an electronic device directly to a car battery without a fuse!

  • \$\begingroup\$ Ah that diagram really helped me get that aspect. Thank you \$\endgroup\$ – ohmmy Jun 4 '16 at 20:56
  • \$\begingroup\$ Never connect to a car battery, that is in a car without a fuse, and a real good filter rectifier. The noise that comes out of a car battery, when in a car, is horrendous, and the voltage can fluctuate 8v to 15v. (may be more) \$\endgroup\$ – ctrl-alt-delor Jun 4 '16 at 21:20

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.