I'm curious about the nature of reverse voltage failure modes. Is it literally the amount of reverse voltage that does the damage, regardless of current flow? Or is it that reverse voltage beyond a certain limit causes reverse current flow - perhaps at unregulated levels - which in turn rapidly builds heat and causes damage that way.

I'd really love to know if there's any answer to these questions in a general sense, or perhaps a small number of different answers depending on the scenario (like cmos fails one way, but ttl fails another.)

In case there aren't any such simple answers, here are two specific parts I'm interested in understanding:
MCP6542:
http://www.microchip.com/wwwproducts/en/MCP6542

SS495a:
https://www.digikey.com/catalog/en/partgroup/ss495-series/12446

I use both of these chips on a board we produce quite a lot of, and I failed to include reverse voltage protection in my original design. The 5 boards that I know have been exposed to reverse voltage (-5VDC) all seemed to handle it just fine, and appear to work properly when fed proper voltage, but I'm not sure if they should be trusted now. The 5V supply that feeds them is limited to 80mA, and I've wondered if that current limit plays a role in the boards' survival.

 

It can be either/or voltage or current depending upon the IC internal design.
Commonly it's likely current since a reverse voltage can forward bias the isolation substrate diode with the current being limited by the external circuit.
It's possible that the 80mA limit prevented the ICs from being zapped but I wouldn't use them if you are concerned about failure in the future as it may have degraded the IC.

A P-MOSFET can be used as a near ideal diode to block the application of reverse voltage to a circuit, as shown here.

 

Failure modes are different for CMOS and TTL ICs.

CMOS outputs are typically push-pull. Both output devices have a parasitic (body) diode that will conduct when the supply terminals are reversed. The voltage must be high enough to forward bias both diodes and the current needs to be greater than what one of the diodes can handle.

In this diagram, the diodes, excluding D3, on the MOSFET source/drain terminals are part of the manufacturing process.

 

Thank you both for your insights and explanations!

I might try that MOSFET circuit. Quite simple, yet very clever! I'll have to see how bulky it gets on the board compared to a simple diode. Space is somewhat tight on this design. I'll also have to double check a few things, but I think the circuit in question can tolerate a Schottky diode drop with no problems, so the MOSFET might be overkill in this case.