How to protect/enhance an LDO so that it can sink current in the event of an external wiring fault?

Question

Background

I'm designing a signal acquisition board and one of its parts has several differential op-amps which are not very good from PSRR standpoint. So they need a very clean power supply, and since my main power rails (5 and 12V) come from a SMPS, I slapped a LM2931 LDO to get a clean 3.6V rail for the op amps.

Now I'm tempted to use this same rail for the supply of some 4 temperature probes. They are wired like this:

^{simulate this circuit – Schematic created using CircuitLab}

The problem

The thermistors are changeable/pluggable and are moved around often; therefore it's not impossible that external wiring fault gets me +5V or +12V in the "To ADC" node.

The ADC itself is protected, so it won't be toast. I've added a TVS parallel to the thermistor to guard against short transients (ESD as well).

However in a sustained +12V scenario, assuming the TVS fails open, current will overflow through R1 and potentially raise the +3.6V rail, since the LDO won't be able to sink that ~6mA of incoming current, ruining the other 3 temp sensors, and potentially damaging the LDO.

Desired effect

I want the acquisition board to be rugged against such faults. The wiring would get corrected eventually, but I don't want this to be an immediate show-stopper for the other measurements.

Proposed solution

One way would be to add a low-value resistor on the +3.6V rail to provide ample sinking capability, should it be needed.

However, I'm also thinking about adding an op-amp buffered second +3.6V rail, which would be capable of sinking current:

^{simulate this circuit}

The idea being that

1. The LM358 serves as a kind of isolation between the important +3.6V rail (that I really care about), and the thermistors power supply (not that high-priority).
2. The buffered 3.6V can sink current, if external faults arise.

Questions

Is the op-amp buffered idea worthy (would the second +3.6V line still be clean from noise), any caveats I might be missing? Some other approach entirely?

Answer 1

I would just use the load resistor in the single 3.6V LDO supply rather than adding additional active components.

The load resistor works by keeping your LDO in the active regulating mode. If the normal load ends up sourcing current back toward the LDO it simply stops outputting the same amount of current as coming from the load sourced current.

Adding the extra active component to provide the source/sink capability may be a decent cost trade off if you really want to minimize the power dissipation of the circuit. The load resistor I recommended above will have a constant power consumption and will add to the dissipation in the LDO.