Battery Measurement Unstable results. Example Read vcc from megatiny

[Brendy1986]

Hello Attiny1626
I used the example of measuring batteries. I would like to measure the 18650 battery 3 - 4.2V.
I connected it from the computer for the test.
The voltage on the port monitor is higher than 5.08.
I have very variable results
How to calibrate it to get precise results? What do I need to change in the code?

#include <megaTinyCore.h>
/* This demonstrates both how the routines to measure temperature are implemented, with explanitary serial prints
 * and then calls the library functions provided that do the same thing (just faster and with less flash usage,
 * particularly on 2-series parts, where a shortcut was found in the mathematics
 */

#define RESULTCOUNT 4
int16_t results[RESULTCOUNT];
int32_t sum;
int16_t average;

void setup() {
  // put your setup code here, to run once:
  delay(1000);
  Serial.begin(57600);
  #if MEGATINYCORE_SERIES == 2
  Serial.println("On some parts, these results may be off by more than twice the spec'ed +/- 5 degrees.");
  #endif
}

uint16_t readSupplyVoltageExplained() { // returns value in millivolts to avoid floating point
  #if MEGATINYCORE_SERIES!=2
  analogReference(VDD);
  VREF.CTRLA = VREF_ADC0REFSEL_1V5_gc;
  // there is a settling time between when reference is turned on, and when it becomes valid.
  // since the reference is normally turned on only when it is requested, this virtually guarantees
  // that the first reading will be garbage; subsequent readings taken immediately after will be fine.
  // VREF.CTRLB|=VREF_ADC0REFEN_bm;
  // delay(10);
  uint16_t reading = analogRead(ADC_INTREF);
  Serial.print(reading);
  Serial.println(" (discarded)");
  reading = analogRead(ADC_INTREF);
  Serial.println(reading);
  uint32_t intermediate = 1023UL * 1500;
  reading = intermediate / reading;
  return reading;
  #else
  analogReference(INTERNAL1V024);
  //Serial.print(analogReadEnh(ADC_VDDDIV10, 12)); // throwaway reading just for good measure.
  //Serial.println(" (discarded)");
  int32_t vddmeasure = analogReadEnh(ADC_VDDDIV10, 12); // Take it at 12 bits
 // Serial.println(vddmeasure);
  vddmeasure *= 10; // since we measured 1/10th VDD
  int16_t returnval = vddmeasure >> 2; // divide by 4 to get into millivolts.
  /* The above method of doing the math is bad, because multiplication of a 32-bit value takes
   * 81 clocks. The smart way to multiply by 10 and then divide by four is to instead take the sum
   * of the original value leftshifted one place, and the original value rightshifted one place.
   * And here most of this can be donewith a 16-bit variable not a 32 bit one
   */
  if (vddmeasure & 0x02) {
    // if last two digits were 0b11 or 0b10 we should round up
    returnval++;
  }
  return returnval;
  #endif
}
void printRegisters() {
  #if MEGATINYCORE_SERIES!=2
/*  Serial.print("ADC0.MUXPOS: ");
  Serial.printHex(ADC0.MUXPOS);
  Serial.print("  ADC0.CTRLC: ");
  Serial.printHex(ADC0.CTRLC);
  Serial.print("  VREF.CTRLA: ");
  Serial.printHex(VREF.CTRLA);
  Serial.println(); */
  #endif
}



void loop() {
  int16_t reading = readSupplyVoltageExplained();
  //Serial.print("System voltage is: ");
  //Serial.print(reading);
  //Serial.println(" mV");

  delay(1000);
  //Serial.println("Repeating above without the explanations");
  reading = readSupplyVoltage();
  float volt = reading/1000.0;
  delay(1000);

  //Serial.print("System voltage is: ");
  //Serial.print(reading);
  //Serial.println(" mV");
  Serial.print(volt);
  Serial.println(" V");

}

5.38 V 5.39 V 5.37 V 5.38 V 5.36 V 5.37 V 5.36 V 5.41 V 5.38 V 5.38 V 5.40 V 5.38 V 5.42 V 5.39 V 5.41 V 5.36 V 5.37 V 5.38 V 5.41 V 5.38 V 5.37 V 5.36 V 5.37 V 5.36 V 5.38 V 5.42 V 5.46 V 5.47 V 5.44 V 5.47 V 5.42 V

[Modac]

Am I understanding correctly, that you are currently measuring the 5V USB voltage with the attiny? If so the variability you are seeing could just be the actual voltage. Have you checked the voltage with a multimeter?

[Brendy1986]

I made a big mistake. USB sends a lot of noise.
I connected it this way by adding a capacitor and the results are much better.
I measured the 18650 3.88V battery with a meter.
On the port monitor my results.

3.89 V
3.90 V
3.89 V
3.89 V
3.89 V
3.89 V
3.90 V
3.89 V
3.89 V
3.89 V
3.89 V
3.89 V
3.90 V
3.89 V
3.89 V
3.89 V
3.90 V
3.90 V
3.90 V
3.89 V
3.90 V
3.89 V
3.89 V
3.89 V
3.89 V
3.90 V
3.89 V
3.89 V
3.89 V
3.89 V
3.89 V
3.89 V
3.89 V

[Brendy1986]

I did some tests using a 3-4.2V linear stabilizer and it looks very good.
guys
Can I measure up to 12.90V without using dividers?

[hmeijdam]

No, you will damage a pin if you go over VCC + 0.5V. Above that the ESD protection diodes will begin conducting and will fry.

[SpenceKonde]

A lot of USB ports put out a few extra tenths of a volt, especially USB chargers, to improve the charge rate of power-hungry cellphones - with 5v, you can afford to lose 0.8v at most in both the charging circuit and the cable before you're at the limit of how fast you can charge. Even the devices that use a switching converter to charge the battery will still need like a third of a volt of headroom to get you the target output voltage. So what's the problem? You need 4.5ish and you have 5! But that doesn't consider the voltage drop in the USB cable. But many micro USB cables have non-negligible resistance because copper is expensive, such that the limiting speed of charging is poor. So people would complain and give bad reviews to chargers that put out the spec'ed 5v because they charged their phones slowly and with the same cables and a charger putting on 5.3V - that .3v would increase the maximum charge rate by something on the order of 50%. So by force of consumer demand, almost everything that says it puts out 5v onto USB port, it usually actually supplies 5.2-5.5. Otherwise people give it crap reviews and say it's slow to charge.

And yeah you always need to use a voltage divider. You cannot apply a voltage higher that the limit shown in the table posted above. The voltage dividers it has internally are only for VDD

[Brendy1986]

Thank you for answer