Electronic – TP4056 module parallel output

The setup in your answer is a bad one.

First you parallel two batteries, which you shouldn't, because their voltages are never exactly the same. Their low internal resistance will cause a current from one battery to the other. So feed three LEDs from 1 battery, and the other three from the other.

Then you place all LEDs parallel, which means that you lose (12 V - 3.2 V) x 20 mA = 176 mW per LED in its series resistor, while the LED itself uses only 64 mW. Total power loss is more than 1 W. That's because the large voltage difference between battery and LED. The best way to get a longer endurance is to keep losses in the series resistors as low as possible. So better place two times three LEDs in series, so that their total current is 40 mA instead of 120 mA. The power loss in the series resistors is then (12 V - 3 x 3.2 V) x 20 mA = 45 mW per 3 LEDs, or 96 mW in total. That's less than 10 % of the power loss for all LEDs in parallel.

Then your batteries will last 100 mAh / 40 mA = 2.5 hours or 150 minutes. This is pretty optimal. The batteries' capacity is 1200 mWh, and the LEDs consume 384 mW, so with an ideal conversion you can get a little over 3 hours out of them. But the most efficient conversion using a switching current regulator will get you maybe 85 % efficiency, and then you only gain 9 extra minutes.

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An alkaline battery's voltage quickly drops by 10-15 %, and then remains more constant for a great part of the discharge cycle. So either you calculate the resistors for a larger current at the start, and 20 mA for the rest, or for 20 mA at the start, and a lower current later on. The latter solution will give you a longer battery life, but a bit less brightness.

jippie suggests to use a switcher anyway to get more out of the batteries, and it's a thought. You'll have to place the batteries in series to get 24 V to allow a voltage drop as high as possible. The larger Vin/Vout ratio of the switcher will make it less efficient, but overall you should get some extra time from the batteries.

Check the datasheet for the input current on a I/O pin It depens on the exact type of controller on your Arduino. It is probably called Input Leakage Current I/O pin and will be near 1μA.

Here is a typical ATmega datasheet used on Uno / Duemilanove / ... , you can find the parameter on page 304. Check the datasheet for your specific controller for accurate details.

If you use a 3.7V battery and a step up converter to supply 5V to the controller, you can connect the battery directly to your controller analog input. However when the output voltage of the step up converter drops below the voltage of the battery (for whatever reason) then your entire Arduino will be fed through the analog input pin and that is what you don't want. All input pins have protection against overvoltage which enables this behaviour, but the diodes are not rated for continuous currents. In conclusion: it is best to include a series resistor between battery and input pin.

But now your measurement relies on the accuracy of the 5V power supply. Depending on which exact controller you have, there are various internal reference voltages available which are far more accurate than the power supply voltage. If you add an extra resistor to ground (R2) you can use such a reference to accurately measure the input voltage. This is called a (resistive) voltage divider. With the given ratio, the voltaga on the input pin will be 1V when the battery voltage is 3.7V:

\$V_{measurement} = \dfrac{R2×V_{BATT}}{R1+R2}\$

^{simulate this circuit – Schematic created using CircuitLab}