Electrical – DC-DC Buck Converter Issue

You have Vcc connected, to Vs, putting 15V on the output. Vs needs to float with the capacitor and FET. D1 Anode must connect to VCC only.

When the FET is off its Source voltage goes down to Ground and C1 charges up to 15V through D1. When the FET turns on its Source voltage rises to 24V, but the capacitor stays charged (for a little while) so the voltage at the other end of C1 goes up to 24V+15V. Thus the driver always has 15V to work with, even when the FET is turned on and its Gate needs to go above the supply voltage. The entire driver output circuit moves up and down with the FET, as the FET produces the square wave which goes into L1.

RG damps the tuned circuit formed by the FET's Gate capacitance and wiring inductance in the gate drive circuit. It needs to be large enough to prevent excessive ringing, but small enough to avoid slowing the FET's switching time too much. D3 bypasses the resistor while the FET is being turned off, to make it switch off faster. Since inductance depends on the circuit layout (wiring lengths etc.) the optimum value of RG may have to be determined by experiment.

Without the FET in circuit VS may gradually rise as the output capacitor charges, unless you put a small load on the output to keep it down. However without a Gate load the driver output draws very little current, so it could take a long time for the voltage to rise.

You don't need to use the low-side driver, so just leave it disconnected.

The characteristics of a passive network like this are very much dependent on the load impedance it sees. Obviously, the step-up scenario has half the load impedance of the step-down scenario, and this provides greater damping (lower "Q").

If your converter had feedback and dynamic control of the switching duty cycle, you could pretty much eliminate these effects.