So, I have this circuit which needs <50mA and can be supplied with a voltage between 2V and 3.3V.

I want to supply it with a Li-Ion cell, which has 2.7-4.2V. I don't really want/need the complexity of a buck/boost converter. I know I could just regulate it down to anything below 2.5V, but it still would be great if my circuit would get 3.3V for Input-Voltages above 3.3V and the input voltage for anything <=3.3V.

So basically I want a voltage regulator that only regulates for voltages >3.3V and which doesn't (and ideally doesn't use any power) for voltages below.

What is this thing/circuit called?


3 Answers 3


You need a voltage regulator that regulates to 3.3V and as the input voltage drops to 3.3V or below, the output remains close to the input voltage despite it not being able to regulate any more - in other words it acts like a <0.25ohm resistor when unable to regulate.

The LP3964 has a drop-out voltage of 24mV at 80mA and its output will follow the input voltage if the input voltage is too low for regulation at 3.3V. Here is the pdf file for it.

Figure 11 speaks volumes - this is for the 2.5V fixed version but the adjustable-version (set for 3.3V operation) will work just as well: -

enter image description here

I think this nails it really with one exception - it still draws about 3 mA when the voltage is not regulating. Can you live with this? If you can live with the losses of a linear regulator when the battery is at 4.2V and the regulator is producing 3.3V at 50mA (0.9V x 0.05A = 45mW) then 10mW (3.3V x 0.003A) doesn't seem a problem really.

  • \$\begingroup\$ This is very similar to the answer I was about to post. Most fixed voltage regulators will simply supply the input voltage when it falls below the fixed output level. \$\endgroup\$ May 4, 2013 at 18:38
  • \$\begingroup\$ Yes - this is what I suggested. As noted, many regulators do this. But, many misbehave. The good ones are probably in the minority. | \$\endgroup\$
    – Russell McMahon
    May 4, 2013 at 19:37
  • \$\begingroup\$ @RussellMcMahon - you're quite correct - I looked at a few and this was the only one that delved into the sub-regulation properties. \$\endgroup\$
    – Andy aka
    May 4, 2013 at 19:46

What is your load current?
This influences the best answer.

A minority of regulators shut down gracefully and have little Vin to Vout voltage drop.
However, if it doesn't say so in the spec sheet it is quite likely not to do it in practice.

Yould could achieve close to what you want by using a bypass MOSFET across the regulator and switching it on with a voltage detector IC or similar. A TLV431 clamp regulator plus a bipolar pnp or p channel MOSFET (to invert TL431 signal) would do this. Extra current for the regulator is about 100 uA. You can get around 10-20 microamps using an LM385.

If you do not mind using "select on test" and some imprecision of changeover point you could use just resistors and 2 MOSFETs and get extremely low quiescent and operate currents - under 1 uA if necessary.

There are other possible solutions, but knowing load current will help.

Using MOSFETs to bypass regulator at low Vin.

See circuit below. This is my 1st use of the workbench cct drawer - please to excuse the several mistakes and cobby look. By all means fix it - I'm going to bed (7am plus, up all night, need to be up by 9:15). Connect M1 source to Vin. Regulator = LDO. MOSFETs are what works for you.

This circuit critically depends on the Vgs threshold voltage of M1 and the sharpness of the Vds - Ids turn-on knee. As the load has a substantial voltage range over which it will operate this "softness" of turn on may be acceptable.

For a sharp transition replace M1 with a TLV431 shunt regulator on the ground rails to M2 gate and move non gate end of R3 to Vin. Use TLV431 and not TL431 due to Vin_min being below TL431 range and also TLV431 gives lower Vclamp_min so FET driven harder.

When Vin too high R1-R2 turn on FET M1 which turns off FET M2 and regulator functions as normal.

When Vin falls too low M1 is turned off. M2 is turned on by R3 and the regulator is bypassed. It will still draw quiescent current via Vin to ground.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ He said his circuit draws <50mA \$\endgroup\$
    – user17592
    May 4, 2013 at 14:09
  • \$\begingroup\$ "resistors and 2 MOSFETs"??? Could you be a little more specific? \$\endgroup\$
    – Dave Tweed
    May 4, 2013 at 14:34
  • 1
    \$\begingroup\$ The downvoters never fail to amaze me with their ignorance of the meaning of the term "useful", or, it seems, their knowledge of electronics. \$\endgroup\$
    – Russell McMahon
    May 4, 2013 at 19:10
  • \$\begingroup\$ @DaveTweed - That was meant to imply ... as well as the regulator - see cct. Can be done with resistors and MOSFETS only IF Vgsth well enough defined and as load has a good voltage range that would be achievable. When Vout too high turn off series in FET. When Vout > Vmax turn on series in FET. Very soft BUT acceptable here, I think. A 5c of components solution in volume :-). \$\endgroup\$
    – Russell McMahon
    May 4, 2013 at 19:31

You want a Low-dropout regulator. They're a commonly available IC.

  • 2
    \$\begingroup\$ Can you please add more detail, why do we want a LDO? What does an LDO do? \$\endgroup\$
    – Kortuk
    May 4, 2013 at 14:14

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