Should I use a DC-DC converter or an LDO?

Question

In my project, I am designing a board which is supplied by a battery. The battery is a LiPo, 3.7 V, 1000 mAh, so its voltage range is 3 - 4.2 V.

All the components on the board need a fixed voltage of 3.3 V. In some configuration the board can draw 1 A, but most of the time it draws about 50 mA.

I need to convert the battery voltage to 3.3 V. I was thinking of 3 options:

• Option 1. Using a buck-boost converter
• Option 2. Using a buck converter and electronically cut the power supply when it is approaching 3.3 V (let's say 3.6 V)
• Option 3. Same as option 2 but with an LDO

Do you see better options?

My conclusions are as follows:

• Option 1 is the most expensive option, and price does matter.
• All the 3.3 V buck converters I have found only accepts input voltages starting at 4 V. Is it a normal behavior that bucks need a voltage gap to work?
• Option 3 seems to be the best option. But when the board draws 1 A, I would need to dissipate maximum (4.2 V - 3.3 V) * 1 A = 0.9 W, which is too much because the board is very small so it would badly dissipate heat, and heat does matter...

Which options would you use in such a case? Do you have advice?

Answer 1

It looks like LiPo batteries stay above 3.3V in normal charge levels, and drop to 3.0V when they're below ~80% capacity.

Image source: https://learn.adafruit.com/li-ion-and-lipoly-batteries/voltages

Based on this, you can stick with an LDO since it's the cheapest and easiest way to step down a voltage, assuming you don't mind the efficiency losses when your battery is >4.0V. LDOs will dissipate any unused power as heat; the greater the difference between Vin and Vout, the lower the efficiency. A benefit when compared with switching topologies is a much smaller ripple on Vout, which is good for sensitive electronics.

Answer 2

Consider using a boost converter chip in a SEPIC configuration. This can act as step up and step down in one circuit.