I'm designing a circuit that has a built-in single-cell LiPo battery with its charger for a GSM circuit. The board feeds from a 12V power supply and uses the battery as a backup, and when the power supply goes out, the battery must power the load, and when the supply goes back battery should begin charging, and the load should be fed from the power supply. I used two Schottky diodes to select between the two. But I'm not sure what happens when the battery is being charged and powering the board at the same time. Another problem is that both regulator circuits supply 5 volts, but how can I make it use the step-down regulator (power supply) when both regulators are working? My load uses 5V and 1~2A, and the maximum dropout should not get over 600 millivolts.


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


If you set or design the step-down regulator to 5.6 V but the step-up converter (after the battery) slightly lower, for example to 5.3 V then as long as the 12 V power is "up" the step-down regulator will always "win" (be used, supply the current) as it delivers a slightly higher voltage.

When the 12 V power is off, then the step-down regulator cannot supply a voltage so the step-down regulator will will be used.

I recommend adding a battery protection circuit between the battery and the rest of the circuit so that when the 12 V power is off for a long time, the step-up converter cannot fully deplete the battery. It should stop at about Vbat = 3.0 V to prevent over discharging of the battery which damages the battery.

A battery protection circuit will simply disconnect the battery when Vbat gets too low (and too high as well if the charger circuit breaks).

But allow me to suggest an alternative solution: many power banks come with an input and an output socket and allow for supplying 5 V and charging the battery at the same time. Then you would have a complete finished, working solution. If the powerbank needs 5 V in put then you only need a step-down regulator module.

  • \$\begingroup\$ Thanks. Can you check this, It uses a feedback from 5v to en pin of the step-up chip. Can you suggest any battery protection circuit that suits this circuit. \$\endgroup\$ Commented Jul 20, 2018 at 14:28

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