+5V Rail - What is the best way to implement?

[Vulpecula]

Hello all!

I would like to supply a circuit with voltage, however, I am still somewhat inexperienced and hope that someone here may share his knowledge with me.

Background: My circuit is powered by a 2S LiPo battery via a battery charge controller (Texas Instruments BQ25792). So the output voltage of the charge controller ("VSYS") is in the range of 6 to 9 volts, with a nominal voltage of 7.4 volts.

I now need two voltage rails: One rail with +3.3 volts and one rail with +5 volts. The 3.3V rail will need about 1 amp and for this I have provided a buck converter as effective as possible. The 5V rail will draw much less current with a maximum of 200mA (generously sized).

I am unsure of the best way to implement the +5V rail now. One option would certainly be to add another buck converter. Is this something that would normally be done? What alternatives do I have (especially in terms of space requirements and number of components)? A simple linear regulator would be nice, but would most likely run at the temperature limit quickly, right?

Thanks a lot,
Vulpecula

[fchk]

Suggestion:
https://www.richtek.com/assets/product_file/RT6212A=RT6212B/DS6212AB-08.pdf
https://www.digikey.de/de/products/detail/richtek-usa-inc/RT6212BHRGJ6F/16376743

This has a fairly low component count and low pcb space requirements. I'd use two of them on the board, one for each rail.

Using a LDO for 5V you would have to design your system to remove ((9V-5V)*0.2A)=0.8W to the outside world.
Look at page 1 of this datasheet:
https://assets.maxlinear.com/web/documents/spx3819.pdf

SOT23-5 has a thermal resistance of 191°C/W. Your 0.8W will cause the chip temperature to rise by 0.8*191°C=152.8°C. Aua!
DFN8 has a thermal resistance of 59°C/W. Your 0.8W will cause the chip temperature to rise by 0.8*59°C=47.2°C. Still a lot.
You could choose TO263 or TO220, but thats more space and more volume.

The best way is to reduce or eliminate the 5V rail. This would be the most efficient solution.

[mariush]

There are switching regulators with 2 outputs but most are more expensive than 2 single ones, so the only benefit (in the case of a simple application like this) is less pcb used.

For example:

TPS542951PWPR - 2A and 3A outputs : https://www.digikey.com/en/products/detail/texas-instruments/TPS542951PWPR/3548340

Datasheet : https://www.ti.com/lit/ds/symlink/tps542951.pdf

RT7273GQW - 3 outputs : 3A , 2A , 2A  max output voltage = input voltage -1v :  https://www.digikey.com/en/products/detail/richtek-usa-inc/RT7273GQW/3986811
 
Datasheet : click on button on https://www.richtek.com/Products/Switching%20Regulators/DC_DC%20StepDown%20Convertor/RT7273?sc_lang=en&specid=RT7273

If you have enough space, you can use two separate switching regulators.

You should also consider if it would be better to just parallel your two batteries to have 3.7v..4.2v input
A switching regulator will be much more efficient converting <4.2v to 3.3v and you have the benefit of cheaper charging ICs that can be powered from 5v (linear charging ICs).
Then, you can use a cheap boost regulator IC to boost that single cell voltage to 5v and that will also be done with higher efficiency.

example of fixed 5v output boost regulator : TPS613222ADBVR https://www.digikey.com/en/products/detail/texas-instruments/TPS613222ADBVR/8638376
or  RT9266BGE (adjustable output) : https://www.digikey.com/en/products/detail/richtek-usa-inc/RT9266BGE/2547122


You don't even have to worry that much about cell balancing, lots of products just put two batteries in parallel and don't care about it.

[CountChocula]

Mornsun makes these buck regulators that are drop-in replacements for a standard 7805. They work great, and are both pretty efficient and reasonably inexpensive.

I don't think that a linear regulator is the right solution here; besides the heat dissipation challenges you're bound to face, it would probably end up eating through your battery capacity without delivering a lot of value.


—CC