Impressive. I'd like to take a look at your product when it becomes available or you're looking for someone to test it. I'm not worried about my probe being plugged in because I am the only one working with it.
The main reason I'm using DC is because once everything is tested and working properly, I am going to deploy a large number of these and can't afford the significant price of an AC probe for each one.
I will know the target resistance value of the water solution. You're saying it is most accurate when the resistor is close to what the EC of the water should be correct?
Target EC should be between 700 (max) and 800 ohms and will alert around 1500 (min) ohms - if all of my calcs are correct. Would you mind looking at this code and telling me where I'm going wrong? It seems like I'm missing a resistor value (146~?). cF is actually 4.956093/1024 because that's the value I'm getting when I ground the voltage divider. - correct me if I'm wrong. And I have a 2200 ohm resistor at the end of the probe.
/*
Print output for voltage divider
*/
int v1 = A1;
int v2 = A0;
float cF = 0.0048399345703125; // 5/1024; // ADC factor is 5V
int vdPin = 7; // Pin to voltage divider
int sD = 5000; // delay in milliseconds to loop sensor
void setup() { // Function to setup serial
Serial.begin(9600);
pinMode(vdPin, OUTPUT); // output digital pin
}
// the loop routine runs over and over again forever:
void loop() {
digitalWrite(vdPin, HIGH); // turn sensor on
delay(1000); // Let sensor voltage settle
// read the input on analog pin 0:
float av1 = analogRead(v1);
float av2 = analogRead(v2);
// Convert the analog reading (which goes from 0 - 1024) to a voltage (0 - 5V):
float avc1 = av1 * cF;
float avc2 = av2 * cF;
Serial.print(cF,6);
Serial.print(" - v1: raw(");
Serial.print(av1);
Serial.print(") ");
Serial.print(avc1,6);
Serial.print("|v2: raw(");
Serial.print(av2);
Serial.print(") ");
Serial.print(avc2,6);
Serial.println();
digitalWrite(vdPin, LOW); // turn sensor off
delay(sD);
}
Attached is a diagram of my circuit.
