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Emulator by aoneill

A simple app to measure the battery level (%) and voltage (mV).

You need to add 2 Resistors to the Backpack.

GND -[330kOhm]- A1 -[100kOhm]- VBAT

I protected the PCB with some tape, because if you touch the Voltage Divider or A1 it influences the readout.


//importe the Gamebuino library and the gb object
#define ADC_PIN A1
#define ADC_PIN_OFFSET 165 //in mV (measured ADC from µC)-(Measured voltage on Pin with Multimeter)
#define ADC_PIN_MAXVOLT 3201
#define BATTERY_FULL 4190 //in mV measured when Battery Full on VBAT
#define BATTERY_EMPTY 3000 //in mV on LiPo dont go Lower!

/* Schematic: GND -[330kOhm]- A1 -[100kOhm]- VBAT When the Battery Voltage is 4.200V the Voltage on A1 is 3.223V */ void setup() { // put your setup code here, to run once: gb.begin(); pinMode(ADC_PIN, INPUT); }

int adcPinVolt = 0, batteryVolt = 0; byte batteryPercent = 0; void loop() { // put your main code here, to run repeatedly: while (!gb.update()); gb.display.clear();

adcPinVolt = ((3300 / 1023) * analogRead(ADC_PIN)) + ADC_PIN_OFFSET; batteryVolt = map(adcPinVolt, 0, ADC_PIN_MAXVOLT, 0, BATTERY_FULL); batteryPercent = map(batteryVolt, BATTERY_EMPTY, BATTERY_FULL, 0, 100); if (batteryPercent > 100) batteryPercent = 100;

gb.display.drawRect(10, 20, 60, 34); gb.display.drawLine(70, 31, 70, 42); gb.display.drawLine(71, 32, 71, 41); if (batteryPercent >= 60) gb.display.setColor(GREEN); else if (batteryPercent >= 30 && batteryPercent <= 59) gb.display.setColor(YELLOW); else gb.display.setColor(RED); gb.display.fillRect(11, 21, map(batteryPercent, 0 , 100, 0, 58), 32);

gb.display.setColor(WHITE); gb.display.println("Battery"); gb.display.setCursor(30, 30); gb.display.print(batteryVolt); gb.display.println("mV"); gb.display.setCursor(32, 36); gb.display.print(batteryPercent); gb.display.println("%"); delay(500); }

Last comments

1 year ago
Author :  Aurélien Rodot

Also the battery has a protection circuit which will turn off when the voltage is to low.

...theoretically. Experience proved otherwise ;)

1 year ago
Author :  ripper121

I dont think that it was a problem of the current draw of 8.6uA, because that will need years, the self discharge of the Battery is higher.

Also the battery has a protection circuit which will turn off when the voltage is to low.

1 year ago
Author :  DFX2KX

the self-dischage issue can be a serious issue if you don't know about it.

On that note, let me go plug my classic in before it burns out it's new battery... That explains a lot, actually.