## 4 The Efficiency of Hybrid Power Stage under Light-Load Conditions
### 125 kW Si IGBT Converter vs. 10 kW SiC Power Stage under Light-Load Conditions
#### 4.1.2 Switching Losses of IGBT- and SiC-Based Light-Load Setup  
In order to derive the switching losses under the light-load condition, the commutation of both devices (IGBT and MOSFET) has to be studied in detail. For the sake of simplicity switching losses under no-load operation are determined. An occurring switching sequence that is representing the switching losses for one the IGBT device is illustrated in Figure 11. Assuming the initial condition of the negative input current (\(x=0\)) and the low-side IGBT is being activated, results in the low-side diode as the current conducting device as the IGBT cannot process negative currents while gated. During the transition from negative to positive values, the input current is commutated from the low-side diode (\(D_{\text{n}}\)) to low-side IGBT (\(S_{\text{n}}\)) at nearly zero voltage of \(V_{\text{CE}}\). Therefore, negligibly small switching losses emerge during this crossover (Figure 11--yellow circle). When \(S_{\text{n}}\) is turned OFF, and \(S_{\text{p}}\) not turned ON due to the existing dead-time, the current from \(S_{\text{n}}\) is commutating to \(D_{\text{p}}\). Therefore, turn-off losses of the low-side IGBT \(S_{\text{n}}\) must be considered (Figure 11--purple circle). When the high-side IGBT \(S_{\text{p}}\) is activated, the input current will be still free-wheeling through the high-side diode \(D_{\text{p}}\), due to its positive nature. Only when the input current reverses its sign again, the current will commutate from high-side diode \(D_{\text{p}}\) to high-side IGBT \(S_{\text{p}}\) at approximately zero voltage (Figure 11--yellow square). If the \(S_{\text{p}}\) is turned OFF, the input current will transition back to the negative diode \(D_{\text{n}}\). This commutation process results in turn-off losses of the high-side switch \(S_{\text{p}}\) occurring at full DC-link voltage (Figure 11--purple square). For MOSFETs, the sequence is very similar. However, as MOSFETs can conduct current in both directions through the channel during on-state, some additional considerations must be foreseen. For the initial condition, it is considered that the input current is of negative nature, and the low-side MOSFET \(S_{\text{n}}\) is already turned ON and conducting current. When \(S_{\text{n}}\) is turned-OFF, the input current is  
Figure 11: Derivation of switching losses of semiconductors under no-load condition. already positive in sign. As the high-side MOSFET \(S_{\text{P}}\) is still in OFF state due to the applied dead-time, the current is commutating from the low-side MOSFET to the high-side diode \(D_{\text{P}}\) (turn-off losses of low-side MOSFET). After the dead-time has elapsed, the input current transitions from the parasitic diode \(D_{\text{P}}\) to the MOSFET channel of \(S_{\text{P}}\). When the high-side MOSFET is turned-OFF again, the input current already reversed in sign again. Thus, the current is handed over to the low-side diode \(D_{\text{n}}\) during the envisaged dead-time (turn-off losses high-side MOSFET) and transferred to the low-side MOSFET when turned ON. Therefore, the total switching losses of each (high-side and low-side) IGBT and MOSFET can be calculated to:  
\[P_{\text{sw}}\Big{(}V_{\text{DC}},R_{\text{g}},T_{\text{j}},f_{\text{sw}} \Big{)}=f_{\text{sw}}\Bigg{(}\Bigg{(}1-\frac{M^{2}}{2}\Bigg{)}\cdot\frac{ \Delta I_{\text{L,max}}}{2}\cdot k_{\text{off}}\Big{(}V_{\text{DC}},R_{\text{g }},T_{\text{j}}\Big{)}+d_{\text{off}}\Big{(}V_{\text{DC}},R_{\text{g}},T_{ \text{j}}\Big{)}\Bigg{)}, \tag{23}\]  
under the consideration that the turn-off energy approximately follows a linear characteristic in the respective current region of interest.