I'm building a sort of homemade tablet out of a 10" touchscreen (designed for RPi-like devices) and an Asus TinkerBoard SBC. The SBC takes the standard 5V 2.5A power source, but the screen requires 12V 1.5A (it can power the SBC from a built-in driver board). I've come to the part where I need to design the battery, but I'm stuck wondering how to get 12V out of a battery that will fit in a tablet.
Do normal 10" tablets have similar 12V power requirements for their screens? That seems unlikely, given that I can't figure out how a 4C LiPo could fit and still provide hours of use. But I'm not sure how else to approach the power problem.
Best Answer
You're limiting yourself with the choice of components.
A tablet manufacturer would probably not choose the display you're using. 12 V 1.5 A is 18 Watts of power so this is a very power hungry display.
A tablet display should use LEDs for the backlight (perhaps yours uses CCFL tubes) and use only a couple of Watts. LEDs used in displays need about 3 to 4 Volts each, for efficiency many are connected in series and for this a high voltage (20 - 40 V DC for example) is needed. So the tablet manufacturer would design a boost converter to make that 20 - 40 V directly from the battery voltage. That is the most efficient way.
You could also use a boost converter to make 12 V 1.5 A from the 5 V however that means at least 3.6 A at 5 V, in practice it might be 4 A due to losses in the boost converter.
Boost converters can be used to increase the voltage of a single cell battery but it will cost effciency losses especially with your power hungry display.
Fitting 3 or 4 cells in series in a tablet can be done but it is inconvenient because then buck converters are needed to make the 5 V or lower for the electronics. Also charging from 5 V USB would then require a boost converter. Also the charge-managing of cells in series is more complex than a single cell design. In practice only laptops use this scheme as they require more power than most tablets and laptops usually do not charge form 5 V USB as that would take too long due to the large battery capacity.