Some things I understand just fine others I'm pretty new to. Can someone explain how they would convert the circuit found here: http://www.zen22142.zen.co.uk/Circuits/Misc/staticdet.htm ... To replace the meter with one of the Arduino's analog inputs? I'd really appreciate it.
This circuit produces both positive and negative values. The arduino is designed to measure the relative value (positive) of one voltage (the input to the pin) with respect (Vref). Negative voltage input to the Arduino is not a good thing.
Well not totally... The potentiometer (if I understand things correctly) constitutes a voltage divider. So you are measuring a voltage difference between that median voltage and the voltage flowing through the FET. In the case of this circuit the supply voltage is 9V and if one assumes that the potentiometer is centered resistance-wise... one would assume that a voltage flowing through the FET below 4.5 volts would be considered a "negative" result... Or am I mistaken?
No, I think you're right. I'd need to look at the specs for the FET, but it does look like the pot in the middle is acting as a voltage divider, and the meter is really showing mid-scale when the analog pin would be putting out a value of 512.
If the FET operates at 5 volts, or you include a suitable voltage divider circuit, you could use an Arduino in place of the meter.
Plug the wire on the left into the analog pin, and the wire on the right into ground. Call analogRead to get the value.
So, I'm taking it that the new schematic would look something like this:
I assume the 2.2K resistor would no longer be valid when taking the changed voltage into account (i.e. from 9v to 5v)? So what might the value be (and why)?
The value of the 2K2 resistor is largely irrelevant with regard to voltage of the supply, what happens is that the resistor and the FET form a voltage divider, so if the FET is at 2K2 due to the charge on the gate then you get half the voltage (2.5V) into the arduino.