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There are two battery operated devices connected to the garage door in my house. One is a bought solution which opens/closes the garage door, the other is a self-made module which detects whether or not the garage door is open, and communicates this with a receiver in the house.

Both are running on batteries, and both eat through batteries like crazy, so I want to power them both with a wall adapter.

I wanted to use a single wall adapter and two buck/boost modules to power them both, but was wondering which was the best practice. One device needs a coin cell (3.2V), the other an MM21 12V battery. The way I see it, there are three options:

  1. Use a 3.2V wall adapter, connect device one directly, and use a buck/boost module to power the 12V device
  2. Use a 12V wall adapter, connect device one via a buck/boost module, and power device two directly
  3. Use a wall adapter with a voltage in between (say 6V), and use a buck/boost module for each device

Which option is best practice, and why?

Thanks!

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  • \$\begingroup\$ A coin cell and an MM21 12 V battery (which consists of many very small coin cells in series) do not use a lot of power. The fact that your batteries drain quickly is because these batteries simply have a small energy capacity. Also the devices could be poorly designed. Anyway, the point I want to make is that you do not need to use buck/boost converters, you can, but there is no need. I would get an adapter with a proper, regulated 12 V DC output (measure that it is actually 12 V when there is no load). Use that 12 V for the 12 V device.... \$\endgroup\$ Apr 6, 2021 at 9:11
  • \$\begingroup\$ Then use a 3 V regulator (module), a 3.3 V regulator would be OK as well as a fresh 3 V coin cell is close to 3.3 V and use it to make 3 V (or 3.3 V) from that 12 V. Only if you need to use a 5 V (USB) adapter would I use a boost converter for the 12 V. The 3 V can still be made directly from the 5 V. \$\endgroup\$ Apr 6, 2021 at 9:14

2 Answers 2

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Use a 12 volt adapter and check that it works exclusively on the device that needs 12 volts before trying to work a dual solution. Once you are happy with that, get an isolating DC-to-DC converter such as 12 volts to 12 volts or 12 volts to 9 volts and connect that to the output of the 12 volt adapter. Make sure the system that needs 12 volts still works.

Then, I'd use a linear regulator to drop the output of the DC-to-DC converter to 3.2 volts. There are plenty to choose from and, many cheap ones you can buy have potentiometers on them that can be used to adjust the output to what you want. Add that regulator module to the output of the DC-to-DC converter and make sure the original system that needs 12 volts still works fine.

Then you should be OK to connect the output of the linear regulator to the 2nd system. Good luck. If you have a multimeter, use it to double check voltages at the stages listed.

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It's easier to step voltage down than to step it up. So a good starting point would be an off-the-shelf 12V power supply.

Then to give the 3.2V supply, use an off-the-shelf DC/DC converter, or make up a simple linear voltage regulator. Anything that runs off a coin cell only needs a tiny current, so there's no need to design a more complex buck regulator.

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  • \$\begingroup\$ Unfortunately I cannot accept two answers,so I accepted the answer from Andy aka because it was the more complete one (and because he also hinted at testing along the way). But thank you for your answer! \$\endgroup\$
    – Joe
    Apr 7, 2021 at 10:11

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