Measuring DC voltage with Arduino Mega

[WrichikBasu]

I created this simple project to measure a DC voltage using an Arduino Mega. The circuit diagram is as follows:

The power jack supplies 9 V ideally. With my multimeter, I measured it to be 10.16 V.

The code for measuring the voltage is:

#include <Wire.h>
#include <hd44780.h>                       // main hd44780 header
#include <hd44780ioClass/hd44780_I2Cexp.h> // i2c expander i/o class header

hd44780_I2Cexp lcd;

const int LCD_COLS = 16;
const int LCD_ROWS = 2;

const float R1 = 4.55 + 0.099;  // ideally 4.8 kohm
const float R2 = 1.166;  // ideally 1.2 kohm

void setup() {
    lcd.begin(LCD_COLS, LCD_ROWS);
    lcd.setCursor(0, 0);
    lcd.print("Voltage: ");
    lcd.setCursor(0, 1);
}

void loop(){
    float reading = (analogRead(A0) * 4.9) / 1000;

    float voltage = (reading * (R1 + R2)) / R2;
    
    lcd.setCursor(0, 1);
    lcd.print(voltage);
    lcd.print("V");
    delay(5000);
}

To get an accurate reading, I replaced the resistance values in the code with what I measured using the multimeter.

But, as you can see below, there is still a difference of around 2 V.

Any idea how I can rectify this?

[LarryD]

What do you measure with the DVM across the 1.2k resistor ?

[WrichikBasu]

I get 1.257 V.

[LarryD]

@WrichikBasu

float reading = (analogRead(A0) * 4.9) / 1000;

And assume your 5v power supply is 4.9v ?

And

V(4.8k) = 8.883v

[WrichikBasu]

The DC jack is providing 10V, which is what I am trying to measure. I am powering the Arduino using a USB cable from my laptop. So, yes, the power supply to the Arduino is 5V.

[LarryD]

V(4.8k) = 8.883v ?

[WrichikBasu]

Yes, I am getting 8.87 V.

[LarryD]

If you indeed have 4.8k and 1.2k.

(10.14 * 1.2k) / 6k = 2.028v across 1.2k

(10.14 * 4.8k) / 6k = 8.112v across 4.8k

---

float reading = (analogRead(A0) * 4.9) / 1000;
should be:
float reading = (analogRead(A0) * 5v) / 1023;

[LarryD]

Also, is there AC ripple on the 9v DC ?

[WrichikBasu]

@LarryD

If you indeed have 4.8k and 1.2k.

As I said in the OP, the resistors are not exactly 4.8k and 1.2k; I am using the correct value in the code (correct as in measured by my DVM).

@LarryD

float reading = (analogRead(A0) * 4.9) / 1000;
should be:
float reading = (analogRead(A0) * 5v) / 1023;

With this, the reading on the LCD becomes 9.97 V, far closer to the real value. Thanks!

@LarryD

Also, is there AC ripple on the 9v DC ?

There could be some minor ripples, but I can't measure it with my oscilloscope, as I use that same adapter to power the DSO1838.