# Design of Direct Current Microgrid Converter with Cost-Effective Low-Voltage Battery Storage System
## 8 Simulation Results  
After defining the operating modes and cycles of the inverter, we can then start calculating the circuit components for the inverter simulation scheme. For the calculations, we will consider the most unfavorable condition, which will be the inverter function in buck-boost mode (variant (b)); this has a constant 80% duty cycle ratio for the buck inverter. From the dependence of the output voltage on the relative rise times, we can then express the relative rise time required for the boost converter to achieve the desired output voltage. We consider the ripple current for an output current value of 4 A.  
\[\text{D}_{\text{2 min}}=\frac{(D_{1}\cdot U_{in})-U_{out}}{U_{out}}=\frac{(0.8 \cdot 12.6)-12}{12}=0.16=16\% \tag{1}\]  
\[\text{D}_{\text{2 max}}=\frac{(D_{1}\cdot U_{in})-U_{out}}{U_{out}}=\frac{(0.8 \cdot 9)-12}{12}=0.4=40\% \tag{2}\]  
\[\text{L}_{\text{min}}\;>\frac{(U_{immax}-U_{out})}{f_{sw}}\cdot D_{2 min}\cdot \frac{1}{I_{ripple}}=\frac{(12.6-12)}{250\times 10^{3}}\cdot 0.16\cdot\frac{1}{4 \cdot 0.3}=0.32\ \upmu\text{H} \tag{3}\]  
\[\text{L}_{\text{min}}\;>\frac{U_{limmin}^{2}}{f_{sw}}\cdot\frac{(U_{out}-U_{ limmin})}{U_{out}^{2}}\cdot\frac{1}{I_{ripple}}=\frac{9^{2}}{250\times 10^{3}} \cdot\frac{(12-9)}{12^{2}}\cdot\frac{1}{4\cdot 0.3}=5.625\ \upmu\text{H} \tag{4}\]  
\[\text{I}_{\text{ripple}}=\frac{U_{out}}{U_{in}}\cdot\frac{(U_{out}-U_{in})}{f_{ sw}\cdot L}=\frac{12}{9}\cdot\frac{(12-9)}{250\times 10^{3}\cdot 6.8\times 10^{ -6}}=2.35\ \text{A} \tag{5}\]  
From the calculation, we obtain the minimum inductor values needed for both modes of operation; we choose the nearest larger value, namely L= 6.8 \(\upmu\)H. The current carrying capacity is deliberately chosen to be as large as possible to reduce the DC resistance of the inductance winding. Thus, we selected an inductance type XAL1010-682 with saturation current \(\text{I}_{\text{sat}}\)= 21.5 A and DC resistance Rdc= 8.1 m We determine the size of the output capacitor from the relationship for the response when the full load is disconnected, corresponding to an output power of 100 W, I\({}_{\text{step}}\) = 8.33 A. Then, according to the required output voltage ripple and the calculated current ripple, we can determine the maximum equivalent series resistance of the output capacitors. \[C_{out}=\frac{L\cdot\ (I_{step}^{\ \ 2})}{2\cdot\ U_{over}\cdot\ U_{out}}=\frac{6.8 \times 10^{-6}\ \cdot\ 8.33^{2}}{2\cdot\ 0.2\cdot\ 12}=97.59\ \upmu\text{F} \tag{6}\]  
\[ESR_{max}=\frac{U_{ripple}}{I_{ripple}}=\frac{0.1}{2.35}=0.0425\ \Omega \tag{7}\]  
For the simulation model, we choose a capacitor of 220 \(\upmu\)F with ESR = 22 m\(\Omega\). From all the calculated parameters, we can then construct the desired inverter model in LTspice, Figure 10, also with the considered shunts for the input and output current measurement. From the simulated efficiency plot of Figure 11, the most efficient form of operation is variant (c); thus, the variant where the transistor of the unnecessary converter is permanently open, i.e., the transfer function of the given converter, is almost equal to one. The least efficient is variant (a); thus, the relative switch-on time for both types of inverters are the same. The consequence of the increase in losses can be seen in the graph with current waveforms Figure 12 at the same output power. The mean value of the current is the smallest in variant (c), and the highest value is in variant (a). This causes greater switching and especially conduction losses on all the active resistors in the current path, i.e., the  
Figure 11: Simulated efficiency of the converter in different modes of operation. Figure 10: LT-spice model with calculated values of component. shunt resistors, the resistance in the switched state of MOSFET-type transistors, and the DC resistance of the inductor winding.