Protecting ATMega88 Input From Battery Charger When ATMega88 Is Unpowered

My circuit has a battery charger and an ATMega88. The ATMega needs to sense the state of the battery charger - but if the battery is too low, the ATMega may not have power.

I know it's bad to drive a CMOS input more than 0.3V above VCC - you can power chips through the protection diodes, cause latch-up, etc.

I see three possible solutions:

1) Use a dual-supply bus transciever. http://www.ti.com/lit/ds/symlink/sn74lvc1t45.pdf This adds more cost than I'd like.

2) Put a 100K resistor inline with the signal, so that "not much" current flows into the powered-off circuit. I think this shouldn't cause latch-up or burn anything out, but could still power-up some circuitry and cause undesired operation.

3) Use divider resistors to step the signal down to

The ATMega has ESD/protection diodes on its inputs that will shunt any input above VCC to VCC (minus diode drop). These will effectively protect an unpowered chip against voltage at an input as long as the current is limited to a value that will not destroy the ESD/protection diodes. I believe the current limit is in the datasheet.

I assume from your outline that the signal you are attempting to send between the two devices is a digital signal.

Option 2 will work, but if you have 5ma of drive current on the charger side of the signal you might consider using an inexpensive optoisolator (transistor output).

Option 3 will work electrically, but it doesn't really offer any advantage over option 2 and if the signal is more than a simple on/off dealing with the ADC and decoding the data stream will probably be more trouble that it is worth.

Also, the battery charger must be powered by something (often an LDO powered through diodes on the Batt+ and VIN+). Consider powering the ATMEGA from the same LDO or if the charger doesn't have an LDO, see if there is a power bus you can use to source power to the ATMEGA through an LDO. If you take this strategy, you should still use series resistors on the digital line as with a second LDO there will be a delay in the power for the ATMEGA and you might still have a momentary period where the charger is driving the output and the ATMEGA is not powered up yet. Since this interval is pretty short, 1K is a typical value for the series resistor in this case (at least that's what I use).