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I am trying to implement a secondary battery in one of my projects. When the main battery (24V) dies, the 3.3V battery takes over and the system operates in limp mode (some emergency functions). I am planning to connect the batteries like this, as my system runs on 3.3V and the secondary battery I'm going to use is a LiFePO4 cell (3.2V nominal). This way, my secondary also gets charged (to a safe 3.3V) when main battery is available. My load is hardly 500 mA with full functioning, and 200mA in limp mode. Is there anything I'm missing here? Will the LDO be fine when mains battery is not present? I'm attaching LDO's functional diagram as well.

I checked the oring diode methods and LT's powerpath ICs, but I want the circuit to be as simple as possible. LT has supply issues as well in my country, so they have to be imported.

Thanks!

My Circuit Diagram

My LDO Block Diagram

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  • \$\begingroup\$ Is the 1 ohm resistor for limiting the charging current in case of depleated battery? \$\endgroup\$
    – winny
    Apr 5 '18 at 13:46
  • \$\begingroup\$ Yup, just to limit the charging current \$\endgroup\$
    – harshr2
    Apr 5 '18 at 16:12
  • \$\begingroup\$ unless you already have a 7V buck converter and don't want to buy/build a new converter, you will have greater efficiency if you use a 5V or even 4V converter before your 3.3V LDO. // When drawing 200mA in limp mode, your 1 Ohm resistor will drop 200mV, leaving you with 3.2-.2 = 3.0 V. Is that sufficient? \$\endgroup\$ Feb 7 at 18:20
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I'm not very familiar with LiFe technology, but a simple resistor in series with the LiFe battery might not be ideal. While charging, depending on the current drawn by the LiFe battery, the series resistor will generate a voltage drop that may fall below the nominal voltage of your battery thus reducing its usable capacity and life span. Same matter when the main battery is removed, you should double check the low voltage supply limit of your load, for the voltage drop across the series resistance may be too high.

Perhaps you could use an integrated LiFe IC charger which will go through the constant current / constant voltage cycle. If you reckon the circuitry simple enough it will ensure proper charging of your battery.

As for the LDO, I am not sure what will be its behavior regarding the LiFe powering up the output pin. But I guess a diode wouldn't do the trick. You could put a N type MOSFET whose gate would be driven by the LDO. With a pull-down resistor to ground, when the LDO is not powered by the main battery, the MOSFET will turn OFF and prevent the LiFe from damaging the LDO.

schematic

simulate this circuit – Schematic created using CircuitLab

Side note about the buck converter : why choose a 7V output converter instead of a 5V one? Does your LDO need a voltage drop larger than 1.7 V? Again, just a minor question that does not help with your question.

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  • \$\begingroup\$ Well my circuit needs a 5V supply as well, so I have another 5V LDO after the buck stage. That's why its 7V. \$\endgroup\$
    – harshr2
    Apr 6 '18 at 6:49
  • \$\begingroup\$ At 200mA, the series resistor would drop 0.2V, the output being 3.1V which is okay for the limp mode (since most critical operations are stopped). While charging, the resistor would allow limit the max charge current to a safe level, and will ultimately (but slowly) become equal to 3.3V. Not being able to utilise the full capacity is also not a problem for me, and LiFe charge >80% at 3.3V which is enough for my application. \$\endgroup\$
    – harshr2
    Apr 6 '18 at 6:59

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