Switch between two DC sources using Schottky diodes

Question

I'm building a small power bank as a backup for my router due to frequent power outages. The router takes 12 V 1 A, even though, it can take between 10-15 V without an issue.

I'll be using CC/CV module to charge the 18650 batteries protected by 3S BMS board. I have two options for auto switching between power adapter (15 V 2 A) and 18650 battery power.

• 12 V relay.
• 2x Schottky diodes (1N5822).

Since max voltage of charged batteries will be 12.6 V, the adapter will power the router, since higher voltage will go through if I use diodes as switch.

Relay can also do the job, but it switches slowly, probably due to adapter dropping voltages slowly partly because of the capacitor in adapter slowly discharge (relay only switch after less than 9v). What I think of relay is the complete cut-off batteries when adapter power is On.

I want to understand will batteries be discharged slowly if only adapter is powering the router? Since the diodes are blocking (idling) batteries from supplying the power being still connected to the diode. If not, is there any better solution I can learn for the above?

Vf drop is not an issue in case of diodes.

Answer 1

The buck converter is not a lithium charger. It will not turn off the output to the batteries when they are fully charged. It likely does not have the precision of limiting the battery voltage to the allowed maximum within few millivolts of tolerance.

And no the batteries won't discharge if there is a buck converter supplying them. Batteries cannot charge and discharge at the same time, either current goes into them, out of them, or it is zero. The circuit will find an equilibrium.

Schottky diodes are leaky though. They are not very good in blocking, expect a few microamps to tens of microamps depending on voltage and temperature.

You have many other options too. Use the relay but design a circuit where it switched at 11V voltage instead of powering it diretly.

You can also use transistors for switching. You could also use a real lithium charger, so your buck converter does not damage the batteries by keeping them at float voltage indefinitely.

Answer 2

Lithium batteries are very fussy - they require very precise control of current & voltage during charging, and each cell must be individually managed so you need the BMS. I suggest consider using a lead-acid battery instead. Yes, it is old technology that has been around for decades, but it works.

Benefits:

1. The charger can be very simple: can be as simple as voltage source with series resistor for current limiting, although a simple analog current limited charger is better.
2. Simple series connection of cells for higher voltage, meaning: individual cell management is not required, so the is BMS not required.
3. Lower cost than lithium.

Down-side:

1. Lead-acid batteries are physically larger and heavier than lithium, but in your case the difference will be reduced since you don't need the BMS and the complex charger.

Example battery:
Here is a typical 12V 1.2AH SLA VRLA (sealed lead-acid, valve-regulated lead-acid) battery:
https://www.digikey.com.au/en/products/detail/b-b-battery/BP1-2-12-T1/653328

Battery chargers:
Have a look around this website for some ideas, here is a good place to start:

https://electronics.stackexchange.com/a/152641/341959