gate driver circuit for mosfet using optocoupler

In the LTspice simulation of two SiC MOSFETs in series, the proposed gate driver shows good voltage balancing performance as power loop current increasing. The analysis is given to demonstrate the working process of such a circuit structure, and the key parameters design and setting are focused. This gate driver only requires one standard driver circuit to drive two series-connected SiC MOSFETs by adding simple coupling circuits, and limiting snubber circuits are applied for voltage balancing, and thus, low cost, simple structure and high reliability are acquired. In order to solve the voltage unbalancing problem in such a series-connection application, a single voltage-balancing gate driver combined with limiting snubber circuits is proposed in this paper. The only problem with this is that if the microcontroller or optocoupler fails the mosfet stays on, but 5V failure means mosfet = off through 10k resistor.Silicon carbide (SiC) metal-oxide-semiconductor field-effect transistors (MOSFETs) are required to be connected in series to meet the high-voltage requirement, since the blocking voltage of a single device is limited.

if the transistor is on, the Vgs is = transistor Vforward? or about 0.7V? shutting off the mosftet since Vgsth is 2-2.6V.477mA, giving about 4.78 V to the gate once charged, activating the mosfet. If the transistor is off, the circuit is a voltage divider, 470 + 10000.I have thought about the common emitter config, and so far have come up with this schematic (attached). I put the R15 100 ohm resistor to limit the gate charge current, as to limit the ‘surge’ in the optocoupler transistor. The 470ohm resistor is to limit the current through the transistor to about 10mA or arounds there. When the optocoupler transistor is on, the mosfet gate charges through R15 and will quickly gain full charge and have no current, meaning we will have (5 - Vforward transistor), ( 5 - 0.7V) = 4.3V or so on the gate, Correct? When the optocoupler transistor is off, the gate of the mosfet goes to ground, through the 2 resistors, shutting it off, so power loss on the 5V or uC failure means mosfet = off. The reason I did the schematic that way (correct me if I am wrong I am still new at this) I think it would be equivalent to the one I put in my schematic, except for the base. The optocoupler I am going to use is a PS2501-4A, 16 pin, 4 channels.The mosfets I am going to use are similar but not the one on the schematic: Either FQP30N06L or AOI518 ( both logic level N channel, 2.6V max Vgs(th) ).

Gate driver circuit for mosfet using optocoupler