Could someone please confirm that I have this correct? If I want to read 4V full scale on a 3.3V Arduino Analog input, I can set the reference to external, and attach the input via a 6.5K resistor. With the internal resistor in the Arduino, it should be a voltage divider where the ADC will see 0-3.325? Calculated as 4 * 32/(6.5=32)= 3.325 . Have I got that right? (I hate blowing things up and I just KEEP on doing it...) Thanks everyone.
Input voltage should not exceed Vcc by more than 0.5V.
With 3.3V, that means 3.8V. Above that, the input clamp diodes conduct up to 1mA, above that and you eventually blow the diode.
I think you missed my question. With this voltage divider, voltage should never go above 3.325V when reading 4VDC. Source = 0-4V--->Divided Output=0-3.325V . Just asking if I've done it correctly.
With the internal resistor in the Arduino
What internal resistor?
You need an external voltage divider on the ADC input, with two resistors. Unless you add a small capacitor from the junction to ground, the parallel resistance of the two should not exceed about 10K ohms.
If an Arduino had an internal PULL_DOWN resistor, you would only need one external resistor to your 4volt source. But the Arduino only has internal PULL_UP resistors. Can't use that in a voltage divider, unless you want to read negative voltages. So you need TWO external resistors with a 7:33 ratio, e.g. 10k/47k Leo..
OK everyone. Thanks. That's the problem with the Internet; you read everyone else's idea of "fact" and can then be completely wrong. That's why I wanted to run it by you first. You saved many headaches trying to use the pullup resistor for this. What I read was that one must be careful of putting voltages on a pin through a resistor BECAUSE of this pullup resistor creating a divider. Got it - not true. And I move on....