OK, as to the original question, let's be clear - you are applying a negative 0.7 V to the input pin when you press the button, which is somewhat out-of-spec and will to some extent be stressing the input protection diodes.You can correct for this by putting a 4k7 resistor in series with each input. Note that you show no pull-ups so you cannot reliably read the inputs unless you set the input pins to INPUT_PULLUP.
The Arduino does not require several seconds to start up - only milliseconds.You are getting confused with the bootloader which waits a few seconds to see if a download is required, but this happens on reset, not power-up.
Yes I use internal pullups of course, but what do you mean by "applying a negative 0.7V to the input pin"? Where this "0.7V" is coming from.
You are getting confused with the bootloader which waits a few seconds to see if a download is required, but this happens on reset, not power-up.
Simple, that's the voltage drop across the diode. So your GND is 0,7V higher then battery-. So by applying battery- to an input the input is -0,7V with respect to GND which is out of spec. That's why I said, use a Schottky (but you ignored). But a resistor in line with the input will work as well.
If you have a Uno (or Optiboot) Arduino, yeay. But he never told us what kind of Arduino he's using. A Pro Mini always waits with the default bootloader. And he told us just now he programmed the uC directly. He's skipping quite a lot of info...
Why would that be a problem? Just change the program.Also the equivalent to the RAW pin in VIN. The +5V pin is for 5 volts.
Don't remember what current it's acceptable to feed digital in with.
Well I will go against the gain here and say there is a way to safely wire a +12vdc digital signal to a arduino digital input pin. It requires that you wire it through a series resistor of a high enough resistance such that the positive clamping protection diode for the input pin will be current limited to at or below it's maximum rated continuous forward current rating, around 1 ma I think. The clamping voltage will hold the input voltage equal to chip Vcc + Vf of the clamping diode.
Vf for the diode is a miniumum of 0.8V and a maximum of 1.5V, both depending on the ambient temperature (see Datasheet, page 2)As it has a If of 10mA you need a resistor of (12V-1V)/10mA=1k1Ω in series with the diode.