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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?

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It could be possible to use a boost converter but depending on how much current the OLED draws, you could run into problems. You might be better off using a higher voltage supply then use a buck converter to drop it down to the voltage for the Atmega board. –  Nick Williams Jun 1 at 23:03
Please note that you don't want to convert 3.3V to 14V. You want to upconvert to 14V directly from the battery voltage. Using the output of the 3.3V regulator as input to the boost converter unnecessarily loads the regulator and wastes energy as heat. –  Ben Voigt Jun 2 at 0:04

3 Answers 3

up vote 4 down vote accepted

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.

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What you will need is a Boost converter, to convert the voltage from 3.3V to 14V. There are ways of doing this with discrete components (which I'm not really qualified to comment on), but another alternative would be an all-in-one package, such as MEE1S0315DC, which can convert 2.9-3.6V to 15V (but only ~1W, ~60mA).

Looking at the datasheet for the OLED, that should be enough current, and the voltage can be dropped to ~14V with an inline diode (ie a 1N4001).

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You've mentioned on my answer that the Vout on an op amp can't be greater than the supply voltage. So I would like to know why that's the case. As far as I know an op amp is to boost a voltage higher than the supply voltage when it's connected as non-inverted. And that the Vout will be lower when connected as inverted. –  Handoko Jun 1 at 23:06
@Handoko It's a detail of how an op amp is implemented; if you look at this diagram, the output is generated by driving transistors sourcing/sinking to V+ and V-. So, it can amplify a small signal (0.1V to 10V) but only if V+ can source the voltage. –  CoderTao Jun 1 at 23:10
@Handoko, it's pretty simple. Op amps have push-pull outputs. The highest possible voltage achievable is the positive rail. –  Matt Young Jun 1 at 23:11
@Handoko there are a few other details to it though; depending on the op amp you may only be able to get within a diode drop (~1V) of either rail, or with a rail-to-rail op amp you should be able to hit either rail. And then there's open-drain op amps that require Output be pulled high (again, at/above the max signal level) –  CoderTao Jun 1 at 23:13
Ah ok. So an op amp isn't only getting a Vin and Vout but also VCC and VSS. And the Vout can't be bigger than the VCC. Thanks for helping me to understand the op amp! –  Handoko Jun 2 at 7:56

As others have said, you need a boost converter. For the best efficiency you want to go directly off of your battery voltage, not off of the 3.3V. Also, the 3.3V regulator might not be able to supply enough current. You could spin your own design with an existing chip like Matt suggested, but I've heard that even the circuit board layout is difficult for switched mode power supplies. It would probably be worth it if you are making your own board already and going into production, but for prototypes there are numerous boards available with adjustable output voltages. Mostly they're on sites like Ebay or Alibaba. None of them are "name brands" or anything, you're buying direct from the factory, but I've had pretty good experiences with them. Here's a few examples. You obviously want to make sure they cover your whole battery voltage range, can output the voltage you need, and have enough current capacity.

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