How is the reverse leakage of the circuit smaller than that of the Schottky diode? Let's reflect it through the conversion of the formula.

When there is a positive voltage between the input and output voltage, the op amp Schottky diode circuit turns on the MOSFET, as shown in the following formula:

VGATE=VOUT-(R2/R1)(VIN-VOUT)

Among them, vgate of Schottky diode circuit is the gate driving voltage of MOSFET, Vin is the input voltage and Vout is the output voltage. The input and output voltages can be converted into drain source voltage and gate source voltage of MOSFET, as shown in the following formula:

VDS = Vin-Vout, and VGS = vgate-vout

Where VDS is the drain source voltage and VGS is the gate source voltage. Combining these equations, it can be obtained that the MOSFET gate drive voltage is a function of drain source voltage:

VGS=-(R2/R1)VDS

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