Protecting microcontroller from inputs rising faster than Vcc

Question

I have a microcontroller (ATMEGA328) that I'm using to do analog data acquisition. The problem is that I want my ADC values to be referenced to a regulated external source that is ~3V, but that source rises faster than my power supply. Since the microcontroller has ESD protection diodes like shown, I think it is likely to blow up when the circuit starts. Is that true? If so, how can I create a low impedance path (<32 Ohm) between ARef and the external source, but still protect the circuit?

The external source is part of an irreplaceable system, so I don't want to just put a Schottky to VCC since that might overdraw the external regulator.

I thought perhaps an n-channel MOSFET with the gate attached to VCC between the pin and the reference?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

It really depends on your VREF voltage source.

If you have an enable, then I would tie the enable to the 5V source, so that it is only enabled when the 5V source is higher than 3V. You can do this using a voltage divider, kind of like an undervoltage lockout configuration.

If you don't have an enable, then you can still do undervoltage lockout, or you can just have the 5V supply enhance a MOSFET.

Another possibility, if your 5V supply has a "power good" output, is to tie the power good to the enable of the 3V supply.

Answer 2

Assuming that Vref is not changing much you can put a grounded capacitor (e.g. 100nF) at the Vref input and connect it via a current limiting resistor (e.g. 10k\$\Omega\$) to the external Vref. The capacitor gives you low impedance. The resistor limits the current in case that Vref > Vcc.

If for some reason that is not an option you can insert a unity gain OpAmp that has a high impedance input and a low impedance output.