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Which among the circuit is the most efficient one for a battery backed up solution. Would be helpful if you can state the reasons as well.

schematic

simulate this circuit – Schematic created using CircuitLab

Notes

  • Power Supply - 15V, 1.4Amps (SMPS)
  • Battery - 12V, 11.2Ah (SLA)
  • Load - 3.3V, 250mA
  • Charger - Float Charging (13.6V Max)
  • Voltage Regulator - DC to DC buck (4.5 ~ 50V input)
  • Some components are not shown since they are not relevant
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Efficient depends on battery life and reliabilty.

Version 1: charges the battery and the battery gives the power to the circuit. Zero transition time in case main power fails, cause load is always served from the battery. High strain on battery.

Version 2: charges the battery but also gives power from main line to load. Battery also charges, but only when it leaks or was drawn on main power fail. Normal strain on battery. Fast take over in case of main power fail.

Version 3: Battery only charged if not full. Main power feeds load directly. Bit slower take over, due to relay needs to decharge (If its a normal coil based relay). Close to no strain on battery.

At least for starters this should be adequate.

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  • \$\begingroup\$ Can you please explain on the statement 'Efficient depends on battery life and reliability' ? How can the battery life affect the options listed? Also for Version 1: The charger has sufficient capacity to charge and supply current to the load simultaneously, hence the strain on the battery should be minimal I guess. \$\endgroup\$ – Zac Mar 22 '20 at 20:19
  • \$\begingroup\$ The question was about efficiency of the given versions. Efficieny also has to take into the account how long the battery will stay alive. A battery that is constantly discharged and charged again will have shorter life time. Thus I would but -1 for efficiency on this kind of version. For Version 1: Sure the charger has enough "power" but the load always drains the battery. Then the charger puts that back into the battery. We have a constent ebb and flod in the battery. Thus the wear. \$\endgroup\$ – ownedcore Mar 23 '20 at 7:42

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