Switch Mode Alternative to LM7805/LM7905

Question

I am attempting to design a dual rail power supply PCB for a university project. There are a number of restrictions - namely that it maintains > 80% efficiency regardless of the current draw. I was originally just going to use a single battery (Maybe LiPo) with a voltage divider to provide a virtual ground connected to the common pins of both a 7805 and a 7905 to provide +-5V (For a microcontroller + numerous op amps), however, they have a relatively horrible efficiency.

I should also note that I need to supply enough current for 4 x 10 DIL BAR LED's, up to 10 op amps, an Arduino Mini. I've roughly estimated this at 0.5A to be on the safe side.

My current solution involves a TL2575-05 switch mode regulator for the +5V @ 1A, hooked up as per the example in the datasheet. My -5V is currently problematic. I've looked at the LT1074HV, but I can't find its efficiency, and it's ~$22 a unit.

Surely there must be a simple way to produce a +-5V power supply from a single battery, that is both efficient and cost effective?

EDIT: Current schematic

Answer 1

There are specific converters for this purpose, named Split Rail Converters. These ICs consits of two converters in one package and have two output, +Vout, -Vout for each. Adventage is that the design will be much simpler and BOM size will be smaller too.

Example, TPS65133 :

This one is arround 2.6$. Note that this particular example could not supply as much current as you need, but I am sure that you can find a suitable one now. I just wanted to point out to search for one converter instead of two.

Answer 2

Efficiency isn't a property of the converter IC alone. It's a function of the inductor copper losses (resistance), the forward voltage drop and reverse recovery charge of the diode, the switching time and parasitic capacitance of the transistor, the switching frequency, and the duty cycle, among other things. The efficiency numbers in the TL2575 datasheet are typical values, and therefore not guaranteed. Proper selection of the diode, inductor, and capacitor are extremely important. Section 10 of the datasheet goes into some detail about this.

Answer 3

Best calculate currents for positive and negative rails separately. Usually your negative rail is much less demanding. A simple charge pump IC may do the trick.

Maxim's 4434 app note describes a cute way of generating a negative rail from a simple boost converter.

Answer 4

There is some advantage to designing it yourself (and it may be a requirement for the particular school project), but if it is not actually a core requirement to design yourself, you might consider using a SMPS module. Something like this, which doesn't seem to have the required current for your application. Alternatively, you can keep your current +5V solution and use a lower current module to generate the -5V rail.