Electronic – Using battery and USB power at the same time

Both the boost regulators you have listed have an EN input. This is used to enable or disable the regulator, and can be very simply controlled by the USB's input power:

^{simulate this circuit – Schematic created using CircuitLab}

Under normal battery operation R3 pulls the EN pin high, so the boost regulator is enabled. When you connect the USB current flows through R1 and turns Q1 on. this pulls EN low. R2 is used to ensure Q1 switches off when there is no USB connection allowing R3 to pull EN up again.

You may want to still keep the two diodes to prevent any back-flow of current through the regulator that isn't currently in use.

Note: I chose the resistors with ball-park figures. You may need to tweak to both ensure the EN pin gets the right voltages / currents to activate, and keep your quiescent current through R3 as low as you can while the USB is connected.

Driving FETs or IGBTs is effectively charging and discharging a capacitor at whatever your switching frequency is. The faster you charge and discharge the gate capacitance, the more efficient the switch is since there is a shorter period of partial conductance.

Any FET or IGBT will have a charge in coulombs listed for the gate at different drive voltages. Typically there is a plot like below. Given your driving ~15V into the gate of your AUIRG4PH50S IGBT, every charge and discharge you need to move that amount of charge. $$ q=130nC $$

Since the Amp is defined as follows the math gets very easy. $$ 1 Amp = \dfrac{1C}{1 Sec} $$

The Average Current your driver will use is your switching frequency times the amount of charge moved every period. For your application with a 1Hz switching frequency you will use ~ 1hz * 130nC = 130nA per IGBT. $$ I_{avg} = f_{sw} * q_g $$

The Peak Current your driver will use is just ohms law. You have resistance in your drive, as well as a series resistor with the gate, and some dynamic gate resistance in the IGBT that is often given in the data sheet. There are some higher level complexities also depending on voltages across the transistor and miller capacitance etc. For a good first order approximation, you can just use whatever series resistor you have between the driver and the gate. Right now that looks like nothing, but you may want to consider adding some to help protect the driver, reduce ringing and allow slew rate control on the transistor (how fast it turns on and off to control EMC, ringing etc.) $$ I_{pk} \approx v_{driver} / r_g $$

Other Notes:

1. IGBTs have a constant voltage drop ~1-2 Volts, so for low voltage applications like this they are not good. Use a FET instead.

2. While your average drive current is very low, you need a lot of capacitance on the driver board to buffer the large (10+ Amps) it sounds like you will be pulling for the few nano or micro seconds it takes to charge and discharge your gates during switching.