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LP5907: Efficiency of LP5907 in low output current

Bardia Barai
Prodigy 70 points
Part Number: LP5907

Hello, I'm designing a low power device, as I know LDO regulators are less efficient than Buck converters. I want to know about the efficiency of LP5907 when load current is about 100uA to 10mA.

  • 0
    TI__Genius 10490 points

    Hi Bardia,

    Any regulator's efficiency is given by

    Pout/Pin = (Iout * Vout)/(Iin * Vin)

    For LDO's, Iin is the sum of the output current and Ignd, the current required by the device's internal circuitry for operation. Iin = Iout + Ignd. 

    Pout/Pin = Iout/(Iout + Ignd) * Vout/Vin

    At very light loads the ground current of an LDO will be comparable to the load current and degrade efficiency. At higher loads the ground current of the LDO is negligible compared to the output current and the efficiency equation simplifies to Vout/Vin. Most LDO datasheets will include a graph of Ground Current vs. Load Current to help with efficiency estimations. The one for the LP5907 is shown below:

    In general the LDO's input and output voltage will have the largest impact on efficiency. Mark explains how reducing headroom (Vin - Vout) can significantly improve efficiency in this blog post:

    How LILO LDOs increase system efficiency
    High efficiency for power supplies has historically been attributed to switching controllers or converters, whereas designers assume that linear regulators (LDOs) have bad efficiency. But linear regulator…
    By in Technical articles > Power management

    Thanks,

    Gerard