I am building a non-isolated buck converter with a N channel MOSFET(2SK4017), I am using the FAN7382 High and low side gate driver. I just want to verify if my wiring is correct before I construct the circuit.
Since I only need the high side output would it be sufficient to ground pins 3, 4 and LO? Do I still need to connect Vcc to the supply?
I will be using a fast recovery diode, Rboot=5 Ohm and Cboot= 100nF.
In my application it will be 12V instead of 15V and the PWM signal driving HIN will have a frequency of 10kHz.
Will this design work?
Cheers!
Best Answer
It is OK to use only the high side output and use a freewheeling diode at the low side. But keep in mind that, a significant amount of losses in Buck Converter occurs on the freewheeling diode. If you use a MOSFET with sufficiently low Rds,on resistance on the low side too, there will be less losses.
You must always connect the COM pin to the ground, because it is the GND pin of the IC. If you desire to use only high side, you should ground the pins LIN pin. But never ground the LO pin! It is an active output pin. It might be shorted and damage the IC. Even if it doesn't damage it, there may be a power loss on it.
Yes, of course. Vcc is always needed for powering the internal circuitry of the IC, and for boot-strapping the high side.
My suggestions: