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So I'm building a simple 12v router UPS and I have trouble with one concept - Nominal Voltage. So the build is very simple, some common 18650 cells, DC-DC voltage step up board, and a BMS (battery management system) board to charge and discharge (safely) the 18650 cells. Connecting them all is super easy too. Issue is, I need to determine how many cells I want to use. The build guide I followed used two batteries, but the only BMS board I could get my hands on needs 3 cells. Easy enough to adjust my build, but my question about the cells being fully charged and the effect of the higher voltage.

What I could research is that 18650 cells have a nominal voltage of 3.6v, so with some easy maths we know that 3 cells (in series) will produce 10.3v nominal voltage for the whole circuit. Easy enough, I can still use the step-up board to boost the voltage to the desired 12v. thing is, if the cells are fully charged they will be at 4.2 v, which will result them all combined to provided 12.6v (too much). So what does nominal voltage really mean? and will the step up board still work when the batteries are fully charged because then it technically wont be stepping up the voltage, its already past 12v.

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    \$\begingroup\$ So what does nominal voltage really mean? For most circuits a higher voltage isn't an issue so very often the circuit design just needs to be designed with the lowest battery voltage. This is usually 3.6 V for Li-Ion cells. However, that 3.6 V is a choice resulting from compromise between battery lifetime and usable capacity. If we discharge down to say 3.5 V then we can extract more energy from the cell but also stress it more and that impacts its lifetime (it will wear out sooner). \$\endgroup\$ – Bimpelrekkie May 7 at 8:31
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    \$\begingroup\$ Depending on the design of the step-up board it might work OK with an input voltage that is above 12 V. Do realize that then the output voltage will also be above 12 V. Your load (the router I guess) needs to be able to handle that. 9 out of 10 routers I have seen use a voltage regulator at their input so they can handle a higher voltage just fine. \$\endgroup\$ – Bimpelrekkie May 7 at 8:36
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    \$\begingroup\$ An alternative is to not use 3 cells in series but only 2 so that the battery voltage is always lower than the output voltage. Then you would always get 12 V at the output. Consider using 2 x 2 cells in series-parallel to get 7.2 V but with more capacity than 3 cells if the energy stored in 2 cells is too low. \$\endgroup\$ – Bimpelrekkie May 7 at 8:36
  • \$\begingroup\$ Thanks @Bimpelrekkie! That makes a lot of sense. I think a 2X2 cell design is a better/safer solution all round. \$\endgroup\$ – Marnus Steyn May 7 at 8:41
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The typical step-up converter (see the 2nd circuit, "Boost", on top right in: https://en.wikipedia.org/wiki/Boost_converter) actually has a permanent series diode so even at 12.6V input the load will still see ~ 12V (maybe 12.1 if it's a very good Schottky diode).

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As a compilation of the Bimpelrekkie's comments, heres a good explantion of nominal voltage w.r.t 18650 cells, as well as an solution to my problem:

So what does nominal voltage really mean?

For most circuits a higher voltage isn't an issue so very often the circuit design just needs to be designed with the lowest battery voltage. This is usually 3.6 V for Li-Ion cells. However, that 3.6 V is a choice resulting from compromise between battery lifetime and usable capacity. If we discharge down to say 3.5 V then we can extract more energy from the cell but also stress it more and that impacts its lifetime (it will wear out sooner).

...will the step up board still work when the batteries are fully charged because then it technically wont be stepping up the voltage.

Depending on the design of the step-up board it might work OK with an input voltage that is above 12 V. Do realize that then the output voltage will also be above 12 V. Your load (the router I guess) needs to be able to handle that. 9 out of 10 routers I have seen use a voltage regulator at their input so they can handle a higher voltage just fine.

As for the amount of cells:

An alternative is to not use 3 cells in series but only 2 so that the battery voltage is always lower than the output voltage. Then you would always get 12 V at the output. Consider using 2 x 2 cells in series-parallel to get 7.2 V but with more capacity than 3 cells if the energy stored in 2 cells is too low

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