For this design, a backup time duration of 75ms with a maximum backup power requirement of 28W is required. Here, use the 2S supercapacitors with total internal resistance to be around 45mΩ with VCH = 5.4V. Therefore, calculate the needed minimum capacitance as:

1. Equation 3. $V_{CL} = min\left(\right(V_{out}{I}_{out})/(\eta I_{lim})+I_{lim}Rin), V_{BUBUV}+\left(V_{out}{I}_{out}{R}_{in}\right)/\left(\eta V_{BUBUCV}\right))$
   Equation 4. $$\begin{gathered} V_{CL} = 28W / \left(\right(0.95 \times 15A)+(15A \times 45m\Omega \left)\right) \\ = 2.64V \end{gathered}$$
2. Calculate VCL:
3. Calculate parameter b and k:
   Equation 5. $b=\left(4V_{out}{I}_{out}{R}_{in}\right)/\eta =5.305W.\Omega$$$\begin{gathered} k=[{V_{CH}}^2-{V_{CL}}^2+V_{CL}\surd ({V_{CL}}^2-b)-V_{CH}\surd ({V_{CH}}^2-b)+b(ln(V_{CH}+\surd ({V_{CH}}^2-b\left)\right)-ln(V_{CL}+\surd ({V_{CL}}^2-b\left)\right)\left)\right]/4 \\ =7.445V^2 \end{gathered}$$
4. Calculate C_min:
   Equation 6. $C_{min}=\left(2V_{out}{I}_{out}T\right)/\left(\eta \right({V_{CH}}^2-{V_{CL}}^2)-2k)=0.68F$
5. Verifying Results:

   According to the previous calculation results, when a 0.68F super capacitor is selected. So the maximum energy that can be discharged is:

   Equation 7. $E_{scap}=[C_{scap}{V_{CH}}^2-C_{scap}{V_{CL}}^2] /2=7.544J$

   The energy consumption for load is:

   Equation 8. $E_{load}=V_{out}{I}_{out}T=2.1J$

   The loss on TPS61381 is:

   Equation 9. $los{s}_{BST}=E_{load}\left(\right(1/\eta )-1)=110.52mJ$

   The total energy consumption is:

   Equation 10. $E_{total}=E_{load}+los{s}_{BST}+los{s}_{Rin}=7.273J$

   The energy margin is:

   Equation 11. $E_{margin}=E_{scap}-E_{total}=271mJ$

   Here, select a supercapacitor of capacitance 5F with a V_CH of 5.4V and an internal resistance of 45mΩ. While selecting a supercapacitor, it is critical to make sure that the internal resistance of the selected supercapacitor is as low as possible to minimize any power loss, to maximize the energy being transferred to the output.