Hello
I just try to test INA219 not used from 2 years ans with the adafruit schematics + ino code I get a strange answer :
16:19:19.478 -> Bus Voltage: 0.01 V
16:19:19.478 -> Shunt Voltage: -1.35 mV
16:19:19.511 -> Load Voltage: 0.01 V
16:19:19.511 -> Current: -13.50 mA
16:19:19.511 -> Power: 0.00 mW
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I think no
Please show your schematic (exactly yours rather than "similar to page anywhere in the net"). And insert the code to the message using the code tags
void setup(void)
{
Serial.begin(115200);
while (!Serial) {
// will pause Zero, Leonardo, etc until serial console opens
delay(1);
}
uint32_t currentFrequency;
Serial.println("Hello!");
// Initialize the INA219.
// By default the initialization will use the largest range (32V, 2A). However
// you can call a setCalibration function to change this range (see comments).
if (! ina219.begin()) {
Serial.println("Failed to find INA219 chip");
while (1) { delay(10); }
}
// To use a slightly lower 32V, 1A range (higher precision on amps):
//ina219.setCalibration_32V_1A();
// Or to use a lower 16V, 400mA range (higher precision on volts and amps):
//ina219.setCalibration_16V_400mA();
Serial.println("Measuring voltage and current with INA219 ...");
}
You cannot connect the LED without a resistor, else you will burn either the LED or the Arduino pin. Probably, something in your scheme has already broken, and therefore does not work.
My mistake indeed, I did not put the resistor as I had a green LED that should accept 3.3V (I thought the resistor was only there to drop the voltage for the red LED...)
But with a resistor in serie or parallele, the ina219 still read the same very low voltage...
To be sure the arduino was not damage, I test on another arduino and get the same result...