This picture shows some part of a high frequency (about 250kHz) MOSFET gate driver circuit.I searched for various types of gate drivers in google and didn't encounter anything like this. Can someone explain this configuration.
What is the purpose of the diodes?.
why mosfet Source isn't connected directly to the gate driver supply ground?.
what is the purpose of two 2k resisters and C12 capacitor connected between +12V and the ground?.
In here +12V and the GND are the gate driver power supply terminals.
Best Answer
In such cases, it is useful to redraw the circuit: as @Oldfart says in their comment, we have the following situation
simulate this circuit – Schematic created using CircuitLab
The source of the power MOSFET is connected to the ground via three of the four diodes \$D_1, D_2, D_3, D_4\$ packaged as a couple of double diodes: the resistors \$R_{21}\$ and \$R_{22}\$ bias this series connections of diodes with a approximately ten milliamperes, in order to let the source see few ohms of differential resistance (at least in the small signal regime) through ground and (perhaps this is the real reason of their presence) set a threshold of \$V_\mathrm{th} = 3V_\gamma+V_{GS_\mathrm{ON}}\$ for the gate driver signal voltage, where \$V_\gamma\$ is the threshold voltage of the diodes and \$V_{GS_\mathrm{ON}}\$ is the gate source turn-on voltage.
Edit
What happens if the source of the power MOSFET is connected directly to the ground, bypassing all the diodes and the capacitor \$C_{12}\$ with a short circuit? From my point of view, it is unlikely that this stage is used in some analog circuit, thus the only consequence I see is that, in this case, you simply have \$V_\mathrm{th} \simeq V_{GS_\mathrm{ON}}\$: this means that you obviously cannot set this threshold by changing the bias of \$D_1, D_2, D_3\$. Without knowing more details of the circuit, this is the only hypothesis I can do about how the circuit would work.