Measuring current draw with LTC4150 + ESP-12E


My LTC4150 Coulomb counter has finally arrived!

For testing, I hooked up the unit to the spare ESP-12E I have lying around:


All the jumpers on the LTC4150 are soldered (SJ1 = interrupt-driven counting; SJ2, SJ3 = 3.3V circuit).

The code for driving the Coulomb counter is as follows:

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#include <Time.h>
#include <TimeLib.h>

const int BATTERY_CAPS = 2300;
const byte INT_PIN = D1;
const float INT_TO_COULUMB = 0.614439;

bool trigger = false, init_done = false;
unsigned long total_time = 0, total_interrupts = 0;
volatile unsigned long num_interrupts = 0;
volatile unsigned long time1 = 0, time2 = 0;

void debug(const char *format, ...) {
  char buf[256];
  va_list ap;
  va_start(ap, format);
  vsnprintf(buf, sizeof(buf), format, ap);
  va_end(ap);
  Serial.println(buf);
}

void handleInterrupt() {
  if (time1 == 0) {
    time1 = millis();
    init_done = true;
  } else {
    num_interrupts++;
    time2 = millis();
    trigger = true;
  }
}

void setup() {
  Serial.begin(115200);
  pinMode(INT_PIN, INPUT_PULLUP);
  attachInterrupt(digitalPinToInterrupt(INT_PIN), handleInterrupt, FALLING);
} 
 
void loop() {
  if (init_done) {
    Serial.println();
    init_done = false;
  }

  if (trigger) {
  cli();
    trigger = false;
    unsigned long interval = time2 - time1;
    unsigned long ni = num_interrupts;
    time1 = time2;
  num_interrupts = 0;
  sei();
  total_time += interval;
  total_interrupts += ni;
    float ma = (ni * INT_TO_COULUMB) / (interval / 1000.0) * 1000.0;
    float ma_avg = (total_interrupts * INT_TO_COULUMB) / (total_time / 1000.0) * 1000.0;
    float lifetime = BATTERY_CAPS / ma_avg / 24.0;
    debug("\ninterval = %ldms; num_interrupts = %ld; ma = %fmA; ma_avg = %fmA; lifetime = %f days",
      interval, num_interrupts, ma, ma_avg, lifetime);
    time1 = time2; 
  }
  else {
    Serial.print(time1 == 0 ? '#' : '.');
  }
  delay(10*1000);
}

When the load is a 1K resistor, expected current draw is ~4V/1K = ~4mA. The following output was observed:

interval = 150862ms; num_interrupts = 1; ma = 4.072855mA; ma_avg = 4.072855mA; lifetime = 26.598871 days
interval = 150775ms; num_interrupts = 2; ma = 4.075205mA; ma_avg = 4.074029mA; lifetime = 26.591200 days
interval = 150829ms; num_interrupts = 3; ma = 4.073746mA; ma_avg = 4.073935mA; lifetime = 26.591818 days
interval = 150710ms; num_interrupts = 4; ma = 4.076962mA; ma_avg = 4.074691mA; lifetime = 26.586882 days
interval = 150850ms; num_interrupts = 5; ma = 4.073179mA; ma_avg = 4.074389mA; lifetime = 26.588854 days

When the load is a 47K resistor, expected current draw is ~4V/47K = ~0.08mA:

interval = 8380827ms; num_interrupts = 1; ma = 0.073315mA; ma_avg = 0.073315mA; lifetime = 1477.645142 days
interval = 8124328ms; num_interrupts = 2; ma = 0.075630mA; ma_avg = 0.074454mA; lifetime = 1455.033325 days
interval = 7995918ms; num_interrupts = 3; ma = 0.076844mA; ma_avg = 0.075234mA; lifetime = 1439.949219 days

I aborted the test after 3 readings because the current draw is so low it was taking too long to get 5 readings. But I think the readings are consistent enough to conclude that the power meter circuit gives reasonably accurate readings.

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