If Voltage less than and relay switch

Projects Programming
1

This is my code for my project. The code works as when the voltage is greater than 10.68volts the first relay is active, but when its low than 10.68v it switches to relay 2. the relays are connected to two different batteries but the same load. The code does not work as needed, both relay are on continuously when input in arduino. but it should be like one relay must be on at a time

// Voltage sensor data starts
#define ANALOG_IN_PIN A0
const int analogInPin1 = A0;  

// Floats for ADC voltage & Input voltage
float adc_voltage = 0.0;
float in_voltage = 0.0;

// Float for Reference Voltage
float ref_voltage = 3.27;
 
// Integer for ADC value
int adc_value = 0;
//voltage sensor data ends

// Floats for resistor values in divider (in ohms)
float R1 = 30000.0;
float R2 = 7500.0; 

//set point 
int setPoint = in_voltage > 10.68;    // Initial value of setPoint (analog value)
int setPoint1 = in_voltage < 10.68;

// motor pins starts
int motor1pin1 = 2;
int motor1pin2 = 3;

int motor2pin1 = 4;
int motor2pin2 = 5;

//relay data starts
int relay1 = 6;
int relay2 = 7;  
//relay data ends   

void setup() {
  // Motor data starts
  pinMode(motor1pin1, OUTPUT);
  pinMode(motor1pin2, OUTPUT);
  pinMode(motor2pin1, OUTPUT);
  pinMode(motor2pin2, OUTPUT);

  //(Optional)
  pinMode(9, OUTPUT); 
  pinMode(10, OUTPUT);

  //relay data starts
  pinMode(relay1, OUTPUT); 
  pinMode(relay2, OUTPUT);
  //ends

  //start serial Monitor
  // Setup Serial Monitor
  Serial.begin(9600);
  Serial.println("DC Voltage Test");
}

void loop() {

  // Read the Analog Input
  adc_value = analogRead(ANALOG_IN_PIN);
   
  // Determine voltage at ADC input
  adc_voltage  = (adc_value * ref_voltage) / 1024.0; 
   
  // Calculate voltage at divider input
  in_voltage = adc_voltage / (R2/(R1+R2)) ; 
   
  // Print results to Serial Monitor to 2 decimal places
  Serial.print("Input Voltage = ");
  Serial.println(in_voltage, 2);

  
  // put your main code here, to run repeatedly:

  //Controlling speed (0 = off and 255 = max speed):     
  //(Optional)
  analogWrite(9, 100); //ENA pin
  analogWrite(10, 200); //ENB pin
  //(motor drive start)
  
  digitalWrite(motor1pin1, HIGH);
  digitalWrite(motor1pin2, LOW);

  digitalWrite(motor2pin1, HIGH);
  digitalWrite(motor2pin2, LOW);
  
  if (analogRead(analogInPin1) > setPoint) {
  relay1 = 1;
  digitalWrite(relay1,LOW);
  delay(4000);
}
  else {
  relay1 = 0; 
  digitalWrite(relay1,HIGH); 
  delay(1000);  
}  
  if (analogRead(analogInPin1) > setPoint1) {
  relay2 = 1;
  digitalWrite(relay2,LOW); 
  delay(4000);
  
}
  else {
  relay2 = 0; 
  digitalWrite(relay2,HIGH); 
  delay(1000);  
} 

  // wait 10 millisecond before the next loop  
  delay(1000);                       
}
2

Please post your wiring diagram.

3

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4

You can see the relay is active (both led are on) which Is not right. When voltage is less than 10.68V then it should switch from relay 1 to relay 2.

5

What's going on here:

  if (analogRead(analogInPin1) > setPoint)
  {
    relay1 = 1;

Why are you changing the relay pin?

6

These lines DON'T work like you think!

7

I was thinking of two different relay switching mechanism. So instead of just using one relay for both I used two.

8

:sweat_smile: Really I suppose here is where I went wrong. No. Clue how to correct that. Do you have anything that can help me out

9

Which pins are those two relays attached to?

10

int relay1 = 6;
int relay2 = 7;

Relay 1 is attached to pin number 6 of digital pwm and relay 2 is attached to 7

11

But you have changed relay1 to 1 or 0 before trying to operate it with the digitalWrite.

12

?? I can't understand can you share some. Example or correct the code by your input

13

What do you think is happening when you say relay1= 1 or 0 when relay1 was originally 6?

Steve

14 
if (in_voltage > setPoint) {
  relay1 = 6;
  digitalWrite(relay1,LOW);
  delay(4000);
}
  else {
  relay2 = 0; 
  digitalWrite(relay1,LOW); 
  delay(1000);  
}  
  if (in_voltage < setPoint) {
  relay2 = 7;
  digitalWrite(relay2,LOW); 
  delay(4000);
  
}
  else {
  relay2 = 0; 
  digitalWrite(relay2,HIGH); 
  delay(1000);  
}

Still no change, should i change 0 with something else; Sorry for such a lame discussion guys i am a beginner at this.

15

Don't change relay1 or relay2 at all.

16

it worked for this.

if (in_voltage > setPoint) {
  relay1 = 6;
  digitalWrite(relay1,HIGH);
  delay(4000);
}
  else {
  relay2 = 0; 
  digitalWrite(relay1,LOW); 
  delay(1000);  
}  
  if (in_voltage < setPoint) {
  relay2 = 7;
  digitalWrite(relay2,HIGH); 
  delay(4000);
  
}
  else {
  relay2 = 0; 
  digitalWrite(relay2,LOW); 
  delay(1000);  
}

Now its switching according to setpoint.

17

The setpoint must be kept such as when voltage is < 10.68 volts it should change its relay. though the setpoint should not be like this.

18

No, it really didn't. It might help if you change the names. Make it relay1pin. The pin is the one that the relay is attached to, if you change it, you will be trying to digitalWrite a pin with no relay on.

Actually, relay1 should be const so that the compiler will complain when you try to change it.

You need logic that turns the relays on or off as necessary. Nowhere is there any reason to change the pin number.

19

So I did the change like this for constant relay action;

if (in_voltage < setPoint) {
  relay1 = 6;
  digitalWrite(relay1,HIGH);
}
  else {
  digitalWrite(relay1,LOW); 
}  
  if (in_voltage > setPoint) {
  relay2 = 7;
  digitalWrite(relay2,HIGH);   
}
  else {
  digitalWrite(relay2,LOW); 
}
20

Better, but now you're setting relay1 to the value it already had, which is rather pointless.

Also, no need for two ifs. If the voltage isn't less than the setpoint it must be greater or equal. Why check again?

As a bonus though, I expect it works now.