Electronic – arduino – 3.7V Rechargable battery as arduino power supply

This is several questions and should probably be broken apart, but nevertheless

1. According to this TP4056 datasheet, a 1.2 kΩ R_PROG will give a charging current between 950 and 1050 mA. This is a bit much for a lithium-ion battery, which typically prefers a charging current no more than 1 C (e.g. 900 mA for a 900 mAh battery). If you have a datasheet for your battery, it should say the recommended maximum charging current. You might want to go to 0.5 C (450 mA) to reduce battery aging. A 1.43 kΩ resistor would give a current below 900 mA, and a 3 kΩ resistor would give one below 450 mA.

2. The output of a switching regulator like that is typically fairly well-regulated (hence the name “regulator”). There are some graphs of the line and load regulation on this MT3608 datasheet.

3. You almost certainly do not need or want a linear regulator on the output of the boost regulator, as it would reduce the efficiency and therefore battery life a lot. The only common reason to add a linear regulator to the output of a boost regulator is to reduce the output noise, but the only really noise-sensitive component you mentioned is the XBee, and it will need a little 3.3 V regulator anyway (and feel free to use a linear regulator for that).

4. It looks like that USB port is directly connected to the Vin hole, so you can't connect USB power while a battery is connected, or you stand a good chance of exploding the battery. USB is already 5 V nominal, though, so you don't need to connect it to the input of the boost regulator—you can just disconnect the boost regulator and hook USB power up directly to your circuit.

No, this will not effectively charge the added battery. The voltage regulator for the car will only provide a high enough voltage for charging main battery needs to be charged. When the main battery is charged the voltage will be reduced to prevent overcharging the main battery.

Even worse, the diode you added in series will reduce the voltage to the secondary battery by about 0.7 V. So even if the regulator for the main battery is providing an adequate charging voltage the voltage to the secondary battery will be too low to charge it properly.

To properly charge the secondary battery you need a dc/dc boost converter, perhaps to 15 V, and then a proper lead-acid charging circuit for the secondary battery.