## 3 Proposed Power Losses Model
### Proposed Switching Losses Model
\begin{array}{l}E_{\text{ON}}=\frac{1}{2}V_{L}I_{L}\tau_{on}(T)\\ E_{\text{OFF}}=\frac{1}{2}V_{L}I_{L}\tau_{off}(T)\end{array}  \tag{41}\]  
The switching losses based on the current-phase period can be expressed as:  
\[P_{igbt\_cur\_sw}(t)=\frac{1}{NT_{sw}}\times\sum_{n=1}^{N}\left(E_{\text{ON}} (nT_{sw})+E_{\text{OFF}}(nT_{sw})\right) \tag{42}\]  
where \(E_{on}(nT_{sw})\) and \(E_{off}(nT_{sw})\) represent the turn-on and turn-off energy of IGBTs in \(n_{th}\) switching period, respectively. Substitute Equation (41) into Equation (42):  
\[P_{igbt\_cur\_sw}(t)=\frac{1}{Ts_{sw}}\times\frac{\big{(}\tau_{on}(T)+\tau_{ off}(T)\big{)}V_{L}}{2}\times\frac{1}{NT_{sw}}\sum_{n=1}^{N}I_{c}(nT_{sw})T_{sw} \tag{43}\]Taking \(T_{sw}\) as the time derivative, Equation (43) can be also expressed as:  
\[P_{igbt\_cur\_so}(t)=\frac{1}{T_{sw}}\times\frac{\Big{(}\tau_{on}(T)+\tau_{off}(T )\Big{)}V_{L}}{2}\times\frac{1}{T_{c}}\int_{t=0}^{T_{c}}I_{c}dt \tag{44}\]  
Considering that \(I_{c}=I_{p}\cos\theta\) and only the half period of the current flows through the IGBT, Equation (44) can be simplified as:  
\[ \begin{array}{l}P_{igbt\_cur\_on}(I_{p},V_{L},T)=\frac{1}{T_{sw}}\times \frac{\tau_{on}(T)V_{L}}{2}\times\frac{1}{2\pi}\times\int_{-\frac{\pi}{2}}^{ \frac{\pi}{2}}I_{p}\cos\theta d\theta=\frac{\tau_{on}(T)V_{L}I_{p}}{2\pi T_{sw}} \\ P_{igbt\_cur\_off}(I_{p},V_{L},T)=\frac{1}{T_{sw}}\times\frac{\tau_{off}(T)V_ {L}}{2}\times\frac{1}{2\pi}\times\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}I_{p} \cos\theta d\theta=\frac{\tau_{off}(T)V_{L}I_{p}}{2\pi T_{sw}}\end{array}  \tag{45}\]  
where \(P_{igbt\_cur\_on}(I_{p},V_{L},T)\) and \(P_{igbt\_cur\_off}(I_{p},V_{L},T)\) denote the turn-on and turn-off power losses of IGBTs based on the current period. It can be seen that the IGBT switching time constant is an unknown parameter. However, it is not necessary to solve this value and only the data at the related reference operating point are required. As shown in Equations (46) and (47), the reference switching energy are measured based on the reference dc-link voltage \(V_{L\_ref}\) and reference current \(I_{L\_ref}\) at two reference temperatures. \[ \begin{array}{l}E_{on}\Big{(}I_{ref},V_{ref},150\Big{)}=\frac{1}{2}V_{ref}I_ {ref}\tau_{on}(150)\\ E_{off}\Big{(}I_{ref},V_{ref},150\Big{)}=\frac{1}{2}V_{ref}I_{ref}\tau_{off}(150 )\\ E_{on}\Big{(}I_{ref},V_{ref},25\Big{)}=\frac{1}{2}V_{ref}I_{ref}\tau_{on}(25 )\\ E_{off}\Big{(}I_{ref},V_{ref},25\Big{)}=\frac{1}{2}V_{ref}I_{ref}\tau_{off}(25 )\end{array}  \tag{46}\]  
Combining Equations (45) and (46), the switching losses based on the reference data can be expressed:  
\[ \begin{array}{l}P_{igbt\_cur\_on}\Big{(}I_{p},V_{L},T\Big{)}=\frac{E_{on} \big{(}I_{ref},V_{ref},150\big{)}}{\pi T_{sw}}\times\Big{(}\frac{\tau_{on}(T )}{\tau_{on}(150)}\Big{)}\times\Big{(}\frac{V_{L}}{V_{ref}}\Big{)}\times\Big{(} \frac{I_{p}}{I_{ref}}\Big{)}\\ P_{igbt\_cur\_off}\Big{(}I_{p},V_{L},T\Big{)}=\frac{E_{off}\big{(}I_{ref},V_{ ref},150\big{)}}{\pi T_{sw}}\times\Big{(}\frac{\tau_{off}(T)}{\tau_{off}(150)} \Big{)}\times\Big{(}\frac{V_{L}}{V_{ref}}\Big{)}\times\Big{(}\frac{I_{p}}{I_{ ref}}\Big{)}\end{array}  \tag{47}\]  
Defining \(m=\frac{\tau_{on}(T)}{\tau_{on}(150)}\) and \(n=\frac{\tau_{off}(T)}{\tau_{off}(150)}\), these two ratios can be measured based on the double-pulse experiment and the Inf Infineon datasheet to obtain the fitting relationship with the temperature. The fitting formulas are obtained:  
\[ \begin{array}{l}m=\frac{\tau_{on}(T)}{\tau_{on}(150)}=\Big{(}\frac{T+273}{1 50+273}\Big{)}^{\frac{ln[\frac{E_{on}(I_{ref},V_{ref},25)}{E_{off}(I_{ref},V_{ ref},150)}]}{ln[\frac{250}{25}]}}\\ n=\frac{\tau_{off}(T)}{\tau_{off}(150)}=\Big{(}\frac{T+273}{150+273}\Big{)}^{ \frac{ln[\frac{E_{off}(I_{ref},V_{ref},25)}{E_{off}(I_{ref},V_{ref},150)}]}{ln[ \frac{250}{25}]}}\end{array}  \tag{48}\]  
Thus, the proposed switching losses model of IGBTs can be expressed as:  
\[