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Tricky load calculation when using boost converters.

Discussion in 'General Electronics Discussion' started by Richard22, Feb 5, 2014.

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  1. Richard22


    Feb 5, 2014
    I wonder if anyone can work out an equation that describes the voltage seen on the input of a DC-DC boost converter when the output of the converter is providing current at a regulated/fixed voltage and the voltage source has an internal resistance that is not negligible.

    This is important when you consider a small 1.5V battery having its voltage boosted up and driving a load (to, say, 2.5V). The battery has a few 100milliOhm resistance. When a load is put on the converter, the battery supplies current. That current produces a voltage drop across the internal resistance of the battery, and so the voltage at the converter drops.

    Because there is less voltage, more current is required to maintain the power, so more current is drawn and the voltage drops further.

    Basically, after a point, the battery voltage collapses rapidly and disproportionaly to the increase in load.

    Its doing my head in trying to work out an equation for this relationship. Anyone able to help me out here?

    Observed voltage on Battery = func of (Vbatt, Rbat, Iload, Vload)
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    There is no simple answer, but you can determine from the battery voltage and the internal resistance where the peak power output is.

    From that you can (assuming some fixed efficiency) determine the output power.

    From that you can determine the maximum current that can be delivered to the load (i.e. at which the regulator can maintain output voltage). This is all assuming that the regulator can work at the reduced input voltage.

    As the load increases further, the power available will fall and because the regulator is not designed to pass the maximum power to the load (as an MPPT regulator might) the output voltage is likely to fall precipitously.

    How this happens depends on a lot of things including: The amount of input capacitance, the actual efficiency of the regulator at lower voltages, the ability of the regulator to even operate at lower voltages, the characteristics of the battery, the characteristics of the load, etc., etc.

    It's something you could test. You might substitute for the battery a regulated power supply (close to an ideal voltage source) in series with a fixed resistor, and then apply a changing load (constant resistance, current, power -- all would be different) and then graph the results.

    I'd be interested to see what you get.

    Maybe you could derive a formula that closely described one particular regulator (in a particular configuration) being driven by a particular source into a particular type of load.

    Whether you could generalise that depends on how predictive you want it to be. It's going to break down once you get into the "interesting" part, and possibly even before that. (The interesting part is when the effective load resistance begins to approach the series resistance)
  3. Richard22


    Feb 5, 2014
    Thank you for considering this for me Steve.

    You have summed up my dilemma. I feel that I should be able to construct an equation describing this behaviour, but I end up with sheets of equations and not being able to pull them together into a single model. I am suspecting that it may indeed be a quadratic equation and there could be two solutions to each state. I will have to go back to my maths books and swot up on solving quadratic equations!

    Incidentaly, this is not just an academic problem. I have a circuit that behaves like this. I have substituted the battery for a bench power supply with a series resistor, and this produces head scratching results.

    To extract the most energy from the battery, I have to remove the power peaks that could flip the circuit into the collapsed state. It would be nice to be able to work out the shape of this hysteresis.

    Ah well, back to tacking on resistors and staring at a ‘scope I guess!

    I will graph the results, that will be interesting.
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