PWM Drive Efficiency

Here we consider a half-bridge circuit, which could be used in a motor drive or synchronous switching regulator.

The first thing to consider is the IR loss from the FET switches, which is simply the RMS current on the $R_{ds}$.  The more ripple current the higher is loss even with the same average current.  This loss is independent of the switching frequency.

The switching loss is consisted of the gate drive and the turning on/off the FETs.  To turn on an FET, the gate capacitance has to be charged.  It is easier to use the total gate charge given in the MOSFET data sheet than dealing with gate capacitance, $C_{gs}$ and $C_{gd}$.  The total energy required to deliver the gate charge $Q_g$ is $\int{VI}dt = V Q_g$.  Note that we do not need to know the exact function of $I(t)$, which is depended on the gate driver and the gate capacitance.  The gate capacitor is charged to $CV^2/2$, but the energy delivered by the supply is $CV^2$, so half of the power is dissipated by charging and the other half is lost during discharging.  So the total power for two FETs switching at Fsw is $2VQ_g F_{sw}$.

The loss in turning on/off the FETs is depended on the rise and fall time.  The current is flowing as the voltage across the switches rises or falls.  Assuming the average current $I$, the power loss is $V I (T_r + T_f) F_{sw}/2$.  The rise and fall time are on the order of 100ns.  This is usually the dominant loss.

To prevent the shoot-through current, a dead time is added between the turning on one FET and the turning off the other.  During the dead time, the current continues to flow because of the inductance.  The conduction goes through the diodes, either the FET's body diodes or external diodes.  So the current flows through one diode drop, this is usually greater than the Rds drop.