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LM3481: Using a boost converter with a battery input

Matt Lang
Intellectual 560 points
Part Number: LM3481

I have the LM3481 SEPIC dev board. I had a question about using a battery as an input to a boost converter.

Could you explain why the following logic is wrong? Or if it's right, what is the solution.

In a boost converter, if you want to maintain a 12V output at 2amps or 24W, you must draw at least that much power on the input. More due to efficiency. 

So the lower the input voltage goes, the more input current must be drawn to maintain that 24W on the output.

Batteries have internal resistance. So once current starts to flow, the input voltage to the boost lowers due to the voltage drop on the battery resistance, which means more input current is needed to maintain the output. Drawing more current drops more voltage on the internal resistance even more, which makes for an even higher input current. This cycle just keeps going like a "runaway" system until there is not input voltage left on the battery or something fails.

Is this scenario incorrect? 

  • 0
    TI__Guru* 78190 points
    Hi Matt,

    You are right that the input voltage may "run away" if your battery internal impedance is excessive which does not support that full power. If this is the case with your battery, it is not the right battery to start with, and you should choose a better one.

    A more scenario is not the batter capacitor issue. More often the inrush current during start up can cause problem, because extra current is needed to charge up the output capacitor. In such case, you need to slow down the soft start to limit the inrush current. Try to avoid employing the excessive output capacitors.

    Thanks,
    Youhao