I want to design the following inverter circuit, I know the general working of it, but I need some details. in the gates of the FET transistors used in H Bridge there is a design contain 4 diodes 2 resistors and tow capacitors connected to bjt then to the 4013 ic, I want to understand how this component works to determine the floating point in each transistor drain.
You can consider the gate of the IRF740 as being a capacitor. You put charge on it and the mosfets turns on, you discharge it the mosfet turns off.
To turn on the IRF740 you need to apply on the gate pin a voltage 10V-12V bigger than the source pin voltage.
For the low side mosfets (Q7 and Q8) it's easy, the source pin is connected to the earth/return, so you just need to apply a 12V DC on the gate and it turns on. The resistors R52 and R53 slow down the time to turn on the mosfet, limiting the charge current. The diodes D19 and D18 allow the gate to discharge (turn off) without passing to the resistor, making the turn off time faster than the turn on time.
The purpose of this time difference between turn-on and turn-off is the create a dead-time, where both mosfets, low side and high side, are shut down. We do this to minimize the risk of having both turned on at the same time creating a short circuit (called a shoot-through in the H-bridge).
To turn on the high side mosfets (Q5 and Q6) is more difficult, because the source pin voltage varies between 0V and +315V, and we need Vsource+12V to turn it on. So this circuit uses a technique called bootstrap capacitor. The capacitors C19 and C20 are charged do +12V via D14 and D15 and their charge is used to feed the gate and turn on the capacitor. You can have more details if you search for bootstrap capacitor for H-Bridges.
It may not be obvious, but the charge on capacitors C19 & C20 comes from the combination of the low-side FETs Q7 & Q8 and diodes D14 & D15.
Let's look at what happens when MOSFET Q8 is turned ON. That pulls the source of Q5 towards ground. C19 then charges through D15 to almost the full input supply voltage (1 diode drop plus however much voltage is across Q8).
When the flip-flop switches state, Q8 is turned OFF, sometime after that, Q3 turns OFF and allows the charge on C19 to turn Q5 ON.
At the same time, Q7 has turned ON and is charging C20 via D14 for the next half-cycle.