I want to use a 4D Systems OLED display with an ATmega32u4 based PCB. I'm still working on designing the PCB and have hit a problem. In the data sheet it says that one of the power supply pins for the display requires 13.5-14.5 volts (which is VDDh). The problem is I will be powering the ATmega32u4 from a lipo battery with a 3.3V regulator. How do I convert 3.3V to 14V? Would I need to use a boost converter?
Electronic – Convert 3.3v to 14v
3.3vatmegaboostlcdoled
Related Solutions
Do I need to limit the current progressively as the charge progresses? Yes. My understanding is that the battery will 'accept' lower and lower current as it nears the end of charge. You will need to be able to recognize that and then choose a value at which the battery is full. For example when charging current becomes 10% of original charging current then battery is full. For a faster charge, some people might choose a larger percentage as a cutoff.
Do I cutoff charging when voltage reaches 25.2 volts? As you can tell from the answer to number 1, no. To charge the lithium chemistry you start with a constant CURRENT until the battery reaches final voltage (25.2) then you charge at constant VOLTAGE allowing the current to slow down as the battery fills up, terminating charge as described in question 1.
Any IC to build charger around? Yes, lots. You do need to know how to build a charger around the IC but look here for an example. http://www.ti.com/product/bq24616. I am sending this via my phone so please excuse me if this link doesn't work. Just go to the TI.com website and search under power management and you will find around 200 to choose from.
Finally, the BMS that you are going to use could technically control the termination of the charging of the battery and it will in fact provide the all important balance charging that your battery will require, but most people recommend using it only as a backup to a proper battery charger and not to rely on the BMS as a battery charger.
Good luck.
Your LCD's LED typically accepts 4.2~4.6V. It should be self regulating, so no further resistor needed. You have 11V or 5V available. Below are the three options available. First is use a Adjustable regulator from 11V. The Second is use a fixed 5V regulator and then a regular silicon Diode (1n4001 or similar) to bring it down another ~0.7V. In either mode, the regulator wastes 1 Watt of heat, a heatsink might be recommended (attach the regulator to the module's metal casing is one method).
Or as mention, tap off the already existing 5V regulator, and just throw on the diode. 0.160A is not exactly a large load, it should be fine.
simulate this circuit – Schematic created using CircuitLab
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Best Answer
You will need to use a boost convertor. For best efficiency, you'll want to use a dedicated boost IC, rather than spin your own. It's not the best in performance, but I like the LM3578 for use by inexperienced people. It's adjustable, and the output voltage is set through a pair of resistors. As long as you put a reasonable LC network (see the datasheet) on the output and use a Schottky diode, it will regulate. It's also available in a DIP package for easy prototyping.