2
\$\begingroup\$

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?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
0
2
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ I don't have enable lines, but I imagine similar applications would. Glad to hear I'm not crazy on the MOSFET idea. \$\endgroup\$ May 12 '16 at 14:30
2
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ The first solution works well for input pins, but the AREF pin has the shown 32k resistor, which would form a divider with the 10k, so not so good. \$\endgroup\$ May 12 '16 at 14:29
  • \$\begingroup\$ @Paul Foster: You didn't specify the range of the analog voltage at the input of the ADC. If its upper limit is below Vref = 3V * 32k / (10K + 32k) = 2.28V then there is no problem having a voltage divider. \$\endgroup\$
    – Curd
    May 12 '16 at 20:37

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.