I'm doing a project measuring ambient light with 'photodiode + op-amp + ADC + cortex-M0+.' For a while, I used stable DC source for providing 3.3v to every components.

The next milestone is making the module portable. What I found was 18650 Li-ion batteries, which have voltage range from 4v to 2.7v through a discharging cycle.

As I use photodiodes, op-amps, and ADCs, I think the voltage provided to those analog parts should be stabilized. So, I'm worried that is it okay to use 3.3v LDO for them, even the Li-ion battery has voltage under 3.3v.

Will it be okay? Or, what can be other solutions?

  • \$\begingroup\$ If the battery voltage falls below 3.3 V the LDO won't be able to produce a 3.3 V output. \$\endgroup\$ Jun 14, 2016 at 10:12
  • \$\begingroup\$ Buck-boost DC/DC converter. There are plenty of it on the market, specialized for battery powered devices. \$\endgroup\$ Jun 14, 2016 at 10:13
  • \$\begingroup\$ I would not discharge a Li-ion battery down to 2.7 V as that decreases it's lifetime. I would stop discharging at about 3.5 V, and that happens to be just enough for a decent LDO to make 3.3 V. Depending on the current consumed at 3.3 V it might be more efficient to use a switched regulator. Does your circuit stop working immediately below 3.3 V or can it still operate at 3.0 V ? \$\endgroup\$ Jun 14, 2016 at 10:19

3 Answers 3


If you are not drawing high current then an LDO regulator should be fine. At low current the discharge voltage will be higher, and stay high until the battery is almost empty. In the example below, at 0.2C (5 hour discharge rate) the voltage stays above 3.6V right until the end, providing at least 0.3V headroom (plenty enough for a true LDO regulator).

Modern LDO regulators often have extremely low quiescent current, much lower than a typical switching regulator. If your device spends significant time 'sleeping' to save power then this could be an advantage.

A switching regulator could be significantly more efficient at the start when voltage is higher, but will probably have higher ripple and may induce noise into sensitive analog circuits.

enter image description here

  • 1
    \$\begingroup\$ What's up with the Y axis values on that graph? 3.0, 3.8, 4.0, 4.8?? \$\endgroup\$
    – jpa
    Jun 14, 2016 at 13:38
  • \$\begingroup\$ @jpa I think they meant 3.5/4.5, there is no way a Li-ion cell is near empty at 3.8V \$\endgroup\$ Jun 14, 2016 at 15:54
  • 1
    \$\begingroup\$ @Dmitry Grigoryev I think you are right, where it says 3.8V it should be 3.5V (just my luck to choose a faulty datasheet!). I will revise my answer. \$\endgroup\$ Jun 14, 2016 at 19:54

If you really want to discharge the battery to voltages below 3.5 V, you won't be able to use an LDO regulator, since it needs to be powered with voltage higher than it outputs. However, considering how little charge remains in the cell below 3.5V, I would consider using the LDO anyway and stop the discharge earlier.

enter image description here


Granted, your system will run about 20% less time on battery, but it will be much safer, since a Li-ion cell discharged to 2.7V is very vulnerable: leaving it uncharged for a couple of weeks will drop the voltage even lower, resulting in permanent damage.

  • \$\begingroup\$ Note that most Li-Ion cells are considered discharged at 3.3V because below this value they cannot provide sufficient current to power-up the target devices they were designed for. \$\endgroup\$
    – Overmind
    Jun 14, 2016 at 12:01

You could do that, and it would indeed be reliable, but you would not be able to get the full charge out of the battery. Whether or not that would be okay for you is up to your requirements for battery life time.

A linear regulator (often called LDO) can only deliver a lower voltage than its input. Think of it as akin to a resistive voltage divider and you get the picture. The 'LDO' part means that it is low-dropout, so the input voltage does not need to be much higher than the output voltage (but always higher). Maybe 100 mV higher. Depends on the LDO, load current and other factors. So when the battery voltage drops below 3.4V or so, the LDO will not be able to sustain a 3.3V output voltage.

Another downside of using a linear regulator is that there is a power loss equal to the voltage difference from input to output, multiplied by the load current. Additionally, this power loss will be dissipated as heat in the linear regulator, so if the voltage difference and/or current is too great, the part will become too hot and desolder or destroy itself. For the circuit you have described, heat is very unlikely to become a problem though.

Enter the buck-boost switching regulator, which can deliver a stable output voltage with either higher, equal and lower input voltage, AND does not have the resistive power loss associated with a linear regulator. Efficiency will still not be 100% though, it never is. You should be able to achieve something like 90% though. That figure varies a lot, both in datasheets and in real applications, as many factors play in. More importantly though, you will be able to run on the battery until it is empty of energy. Finding a buck-boost with 2.7-4V input range and 3.3V output will be no problem at all. Plenty of those exist for exactly the type of application you are describing, and exactly which one you will need depends on a host of other requirements.

Finally, you should not dischage the battery too far if you intend to recharge it. Choose a buck-boost with UVLO (under voltage lock out) and set the resistor divider to disable the regulator around 3V. There is little to energy to be used anyway once the voltage has dropped that far. Take a look at a discharge curve and for the battery to help you choose the exact voltage.


In your place, I would either: 1. Use an LDO if there is not really a need to optimize for battery life time. 2. Find a buck-boost regulator already soldered to a small PCB with all the external components. You can get something like that on eBay. Just be sure to check the various requirement, like voltages and output current.


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