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Dear Sagar,

welcome,

Your question is very interesting, it concerns the division of a DC voltage source into two equal parts. As we know voltage division can be accomplished by resistors. However this solution is lossy for power devices. So, the other solution is to use capacitors to divide the DC voltage source. What is required ideally?

One has a voltage source with a voltage VS and one would like to divide to to voltage sources with source voltage of VS/2.

Then one can use two equal capacitors with a capacitance say C. Noe it is required to get suitable value for C. To act as a voltage source the sub voltage source must have a very large capacitor. Very large capacitor are not feasible barbecue of the required operating voltage of the capacitor, it must safely withstand the maximum operation voltage. As the voltage increases the size of the capacitors for the same C becomes greater. The other thing is that as C gets larger charging it through voltage source may cause very large surge current that can damage the capacitor since i= CdV/dt. ,

Since the capacitor will act as a source of current for the circuit for interrupted period of time then practically the capacitor will be charged during the off period of the load from the main supply VS and it will discharge this charge in the load during switch on.

Since there is no ideal voltage source then VS it self has an internal source resistance which limits the recharging process and the highest rate of the current flow for the recharging process. IchxT= Idis Ton, Idis/Ich= T/Ton and the voltage drop during  discharge dV, = dQ/C=  (Idis-Ich) ton/C= Idis( 1- Ton/T)/C, where Ich is the charging current and Idis is the discharging current, T is the cycle time and Ton is the on period of the load on the capacitor.

From the above relation we have:

C= Idis( 1- Ton/T)/ dV for Ton/T =1/2,  Then C= Idis/2dV,

For a given ratings of Idis and allowed voltage droop dV we can calculate C.

Best wishes