# A Transformer-Less Buck-Boost Grid-Tied Inverter with Low Leakage-Current and High Voltage-Gain
## 4 Loss Analysis  
Loss analysis is performed to compare the efficiency of the investigated transformer-less buck-boost inverter with the traditional grid-fied system. Since the working states are symmetrical, and the mains is a periodic sine wave, only the quarter of period needs to analyze and calculate. Assuming that the inverter always operates in CCM can simplify the analysis, and the mains voltage \(v_{s}(t)\) is able to be presented as in Equation (12):  
\[v_{s}(t)=V_{P}\sin\omega t. \tag{12}\]  
If \(R_{L}\) is the equivalent resistance of the output load, the output load current \(i_{s}(t)\) will be presented as in Equation (13):  
\[i_{s}(t)=\frac{v_{s}(t)}{R_{L}}=\frac{V_{P}\sin\omega t}{R_{L}}. \tag{13}\]  
Regarding the timing diagram shown in Figure 4, the inverter works in the step-down mode in the interval of time \(t\) = 0-\(t_{m}\). Within one high-frequency switching period, the average current of each inductor is able to be presented as in Equations (14)-(16):  
\[i_{L1,Bu}(t)=i_{L2,Bu}(t)=i_{s}(t)=\frac{V_{P}\sin\omega t}{R_{L}}\, \tag{14}\]  
\[i_{L3,Bu}(t)=i_{L4,Bu}(t)=i_{s}(t)=\frac{V_{P}\sin\omega t}{R_{L}}\,\ \text{and} \tag{15}\]  
\[i_{L8,Bu}(t)=i_{s}(t)=\frac{V_{P}\sin\omega t}{R_{L}}. \tag{16}\]  
The inverter works in the step-up mode in the interval of time \(t\) = \(t_{m}\)-\(T_{ac}/4\). Within one high-frequency switching period, the average current of each inductor is able to be presented as in Equations (17)-(19):  
\[i_{L1,B_{0}}(t)=i_{L2,B_{0}}(t)=\frac{1+\text{N}\cdot d_{B_{0}}(t)}{1-d_{B_{0 }}(t)}\times i_{s}(t)=\frac{(V_{P}\sin\omega t)^{2}}{V_{in}R_{L}}\, \tag{17}\]  
\[i_{L3,B_{0}}(t)=i_{L4,B_{0}}(t)=i_{s}(t)=\frac{V_{P}\sin\omega t}{R_{L}}\,\ \text {and} \tag{18}\]  
\[i_{L8,B_{0}}(t)=i_{s}(t)=\frac{V_{P}\sin\omega t}{R_{L}}. \tag{19}\]  
When the peak voltage of \(v_{s}(t)\) is higher than the input voltage, the conduction loss of each inductor is able to be obtained as in Equations (20)-(22):  
\[P_{\text{con},L1}=P_{\text{con},L2}=\frac{2}{T_{ac}}\sum_{n=1}^{N_{\text{Im} }}\left[i_{L1,Bu}(\frac{n}{N_{s}}\cdot T_{ac})\right]^{2}\cdot R_{ESR,L1} \cdot T_{5}+\sum_{n=N_{\text{Im}}}^{N_{\text{L}}/4}\left[i_{L1,B_{0}}(\frac{ n}{N_{s}}\cdot T_{ac})\right]^{2}\cdot R_{ESR,L1}\cdot T_{5}\Bigg{]}, \tag{20}\] \begin{table}
\begin{tabular}{c c c c c} \hline \hline \multirow{2}{*}{**Component**} & \multicolumn{2}{c}{**Positive Half-Cycle**} & \multicolumn{2}{c}{**Negative Half-Cycle**} \\ \cline{2-5}  & **Step-Down Mode** & **Step-Up Mode** & **Step-Down Mode** & **Step-Up Mode** \\ \hline \(S_{Bal}\) & switching with \(d_{B_{0}}\) & remaining on & remaining off & remaining off & remaining off \\ \(D_{1}\) & switching & remaining off & always off & remaining off \\ \hline \(S_{Ba2}\) & remaining off & remaining off & switching with \(d_{B_{0}}\) & remaining on \\ \(D_{2}\) & remaining off & remaining off & switching & remaining off \\ \hline \(S_{Ba}\) & remaining off & switching with \(d_{B_{0}}\) & remaining off & remaining off \\ \(D_{3}\) & remaining on & switching & remaining off & remaining off \\ \hline \(S_{Ba2}\) & remaining off & remaining off & remaining off & switching with \(d_{B_{0}}\) \\ \(D_{4}\) & remaining off & remaining off & remaining on & switching \\ \hline \(S_{1}\) & remaining on & remaining on & remaining off & remaining off \\ \(S_{2}\) & remaining off & remaining off & remaining on & remaining on \\ Gain & \(d_{B_{0}}\) & \((1+Md_{B_{0}})/(1-d_{B_{0}})\) & \(d_{B_{0}}\) & \((1+Md_{B_{0}})/(1-d_{B_{0}})\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: The voltage gain and the condition of each switching component within one AC cycle.