I’m using this 3.3V LDO (https://www.sparkfun.com/products/526) to power a device which normally takes 2 AAs (3V), using a 3.7V LiPo battery. The device won’t power on without the regulator (it can withstand variance in AA voltages but only until about 3.7V total).

I realized that the regulator I’m using has a 1V dropout voltage, yet the device powers on and works just fine until the battery drains. My question is this: why does it power on when 3.7V is well within the 1V dropout range? It is supposedly not regulated at this point, correct? Is there something else going on that happens to drop the voltage just enough that the device allows itself to power on? (For reference, it’s a Gameboy Advance, and it protects itself from powering on outside its expected range).


1 Answer 1


It's an emitter follower with voltage and current feedback, so it just follows the input by the same dropout voltage if at the same current.

Due to saturation of Vce at dropout threshold, Rce resistance rises a bit and current gain has reached a minimum ~10 so voltage dropout tends to increase with more current.

But a 1V dropout is unsuitable for your application, so search for a <50mV dropout LDO using FET technology rather than BJT's. There are many.

  • \$\begingroup\$ Did you mean <500mV? The parts you linked have a 0.26V dropout. \$\endgroup\$
    – Andrew M.
    Commented Apr 11, 2018 at 1:35
  • \$\begingroup\$ Also, there’s quite a bit of jargon in your answer. Could you explain in layman’s terms? \$\endgroup\$
    – Andrew M.
    Commented Apr 11, 2018 at 1:36
  • \$\begingroup\$ have you tried looking anything up? MOSFET's being Vgs controlled Ron resistors, follow V=I*Ron for the drop voltage, but bipolar BJT's have Darlington higher Vin-out drops, If you know the lod current , then look for a suitable LDO with < 50mV drop or <500mv if you care less but dont run < 3V when discharged \$\endgroup\$ Commented Apr 11, 2018 at 3:35

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