## 3 Enhanced SoC-Balancing Strategy with Saturation
### Selection of the Current Limits in CDRs  
If the enhanced strategy is used for parallel applications, \(i_{\text{satu}}\) cannot be too low. It should satisfy the following:  
\[\frac{i_{\text{satu}}}{i_{\text{sum}}}>\frac{SoC_{10}}{SoC_{10}+SoC_{20}+SoC_{30}} \tag{16}\]  
Otherwise, the equilibrium point is not reachable before the SoC reaches zero. When \(i_{\text{sum}}<2i_{\text{satu}}\), only one ESU may be saturated, namely, ESU#1. The other two ESUs are equalized at the very start. Then, the two currents correspond to  
\[i_{2}=(i_{\text{sum}}-i_{\text{satu}})\times\frac{SoC_{3}^{n}}{SoC_{3}^{n}+SoC_ {2}^{n}},i_{3}=(i_{\text{sum}}-i_{\text{satu}})\times\frac{SoC_{3}^{n}}{SoC_{3} ^{n}+SoC_{2}^{n}} \tag{17}\]  
After ESU#2 and ESU#3 are balanced, \(CDR_{11}\) should continuously decrease until reaching \(1/3\) and the equalization of the three ESUs is acquired. The slope of \(CDR_{11}\) is calculated as  
\[\left\{ \begin{aligned} CDR_{11}^{\prime}=\frac{nSoC_{1}^{n-1}[ ASoC_{2}^{n-1}+BoC_{2}^{n-1}]}{(SoC_{1}^{n+1}+SoC_{2}^{n}+SoC_{3}^{n})^{2}}\\ A=SoC_{2}SoC_{1}^{\prime}-SoC_{1}SoC_{2}^{\prime},B=SoC_{3}SoC_{1}^{ \prime}-SoC_{1}SoC_{3}^{\prime}\end{aligned} \right. \tag{18}\]  
where \({}^{\prime}\prime\) is a time-derivative operator. The following is introduced:  
\[\left\{ \begin{aligned} SoC_{2}=SoC_{3}\\ SoC_{2}^{\prime}=SoC_{3}^{\prime}=-(i_{\text{sum}}-i_{\text{satu}})/(2Q) \\ SoC_{1}^{\prime}=-i_{\text{satu}}/Q\end{aligned} \right. \tag{19}\]  
Afterwards, the numerator of Equation (18) is deduced as follows:  
\[\text{num}(CDR_{11}^{\prime})=nSoC_{1}^{n-1}SoC_{2}^{n-1}[i_{\text{sum}} \times SoC_{1}-i_{\text{satu}}(SoC_{1}+SoC_{2}+SoC_{3})]/Q \tag{20}\]  
Thus, the reachable restriction is represented as  
\[\frac{i_{\text{satu}}}{i_{\text{sum}}}>\frac{SoC_{1}}{SoC_{1}+SoC_{2}+SoC_{3}} \tag{21}\]  
Accordingly, \(CDR_{11}\) will decline. To actualize Inequation (21), the maximum of the right side needs to be found. In fact, it persistently drops until reaching \(1/3\), as the individual SoC levels come closer and closer under the balancing control. Thus, its maximum is just the initial value. Undoubtedly, Inequation (21) is guaranteed if only Inequation (16) is realized. When \(i_{\text{sum}}>2i_{\text{satu}}\), ESU#1 and ESU#2 are saturated in the beginning. With the SoC levels approaching each other, ESU#2 becomes unsaturated earlier, and the following condition is met:  
\[(i_{\text{sum}}-i_{\text{satu}})\times\frac{SoC_{2}^{n}}{SoC_{2}^{n}+SoC_{3}^{ n}}=i_{\text{satu}} \tag{22}\]  
By simplifying Equation (22), we have  
\[\frac{SoC_{3}}{SoC_{2}}=\sqrt[n]{\frac{i_{\text{sum}}}{i_{\text{satu}}}-2}=w \in(0,1] \tag{23}\]  
Then, \(i_{2}=i_{\text{satu}}\) and \(i_{3}=i_{\text{sum}}-2i_{\text{satu}}\). Combination with Equation (3) yields  
\[t_{0}=\frac{Q(wSoC_{20}-SoC_{30})}{(w+2)i_{\text{satu}}-i_{\text{sum}}}<\text {min}(\frac{QSoC_{30}}{i_{\text{sum}}-2i_{\text{satu}}},\frac{QSoC_{20}}{i_{ \text{satu}}}) \tag{24}\]Equation (23) implies that \(i_{\text{sum}}=(2+w^{n})\ i_{\text{stat}}<(2+w)\ i_{\text{statu}}\). Moreover, we determined that \(w=SoC_{3}/SoC_{2}>SoC_{30}/SoC_{20}\) as \(i_{2}>i_{3}\). Thus, \(t_{0}\) is positive. Another question is whether the moment \(t_{0}\) occurs before \(SoC_{2}\) and \(SoC_{3}\) decrease to zero as Inequation (24) states. This restriction is assumed correct, and then the balancing operation begins between ESU#2 and ESU#3. After the SoC levels of the two ESUs are identical, the situation is the same as that when \(i_{\text{sum}}<2i_{\text{statu}}\). Three balanced ESUs can be obtained once Inequation (16) is met. Based on Inequations (16) and (24), it is inferred that  
\[\frac{i_{\text{sum}}-2i_{\text{statu}}}{(w+2)i_{\text{statu}}-i_{\text{sum}}}< \frac{SoC_{20}+SoC_{30}-SoC_{10}}{(w+1)SoC_{10}-SoC_{20}-SoC_{30}}<\frac{SoC_{30 }}{wSoC_{20}-SoC_{30}} \tag{25}\]  
Therefore, it is ensured that \(t_{0}<Q*SoC_{30}/(i_{\text{sum}}-2i_{\text{statu}})\).