Buck boost converter + LDO or 2 Buck boost converter

Question

I am currently designing a circuit where we have a lipo battery powering two rails:

• Rail 1: some RGB LED with 5V power supply
• Rail 2: a bluetooth chip with 3.3V power supply

---

• Option 1: use 2 buck boost converter to go up from [3V-4.2V] to 5V one side and to 3.3V on the other side

• Option 2: use 1 buck boost converter to go up to 5V creating rail 1, then use rail 1 + a LDO or DC-DC to go from 5V to 3.3V to create rail 2

What is it better to do and why?

Answer 1

Certainly the second option is the more cost effective. An LDO does not require as much external component as a buck-boost converter. This soultion saves some money by reducing the PCB size and the componenet costs.

On the other hand an LDO won't be as efficient as a buck-boost converter, but it all depends on the current demand of the bluetooth chip. I suppose we are talking about max 30 - 40 mA peak consumption here. That will be \$ \text{1.7 V} \times \text{40 mA} = \text{68 mW} \$ wasted energy. And it is the peak, most of the time the bluetooth chip will draw much less. So I would definietly go with the second option.

In general the switching converters efficiency is worse at light loads than at high (though you could find one with light-load-efficiency feature) so an LDO could be even better here.

Answer 2

You also have an Option 3:

• Rail 1: Use 1 boost only converter (cheaper than buck-boost) to go from battery to 5V.
• Rail 2: Use a low-dropout linear regulator (LDO) to power 3.3V rail directly from the battery (3V to 4.2V). This would be the cheapest, simplest option I think.

Most BT chips can run at 3V or even lower and current consumption is so low that a decent low dropout linear regulator output can get very, very close to input voltage in dropout mode (3V to 3.3V input voltage). Hope that helps.