I'm working on a project that uses an LM4040 precision shunt in order to obtain a 2.5 V reference voltage (in particular, I'm using the LM4040-N-25). My circuit is similar to the one shown below (from TI's website):
In designing my project, I was basing my design on a prior project that used a 3.3 V input voltage Vs and an appropriately sized shunt resistor Rs. Unfortunately, I hooked Vs to a 5 V input and exceeded the maximum specified current for the device for some minutes (which is 15 mA, according to the datasheet). However, I only exceeded it by 1 mA, so I haven't let out any magic smoke.
What is the expected behavior of a precision shunt like this if it's failed? Currently, I am obtaining 4.3 V across my output and ground, indicating to me that the LM4040 is doing something, though I'm not getting the desired 2.5 V.