How to obtain a dual power supply ±2.5V from a battery for a portable system?

I’m working on a portable system that uses a microcontroller and an analog/digital converter. I want use this ADC in dual power supply mode so I need a ±2.5V .

In my previous project, the system was wire-powered by USB and made use of an isolated dual output DC/DC converter which provides a ±3.3V, so with two LDO positive and negative regulator I obtained the ±2.5V.

Now I want to transfer the project in a portable contest and I need to make some changes to reduce the system size. For this reason, I can’t use the previous DC/DC converter type yet cause they are too big for my intent.

I tried to find on the web distributors some IC that provide a dual ±2.5V power supply from a single positive source, but I found one or two solutions that don’t satisfy me at all. Furthermore, these IC use a charge pump that create on the output a very high ripple, but I think that if I use two LDO regulator with high PSRR the problem is in part solved.

So, my questions for you are:

• Do you have any suggestion to obtain a dual power supply ±2.5V from a battery, with the noise as low as possible?

• Do you know some good IC that made this?

• What battery? If 5V, just take a 2.5V LDO – Gregory Kornblum Jan 24 '17 at 20:28
• @GregoryKornblum The battery is a 3.7V Li-Po. If it was a 5V battery, I would have to get the -2.5V from the ground to the output of the LDO and the 2.5V from the battery to the output of the LDO (virtual ground on the LDO output)? Is this a good method without any consequences on the power supply quality? – thoraz Jan 24 '17 at 20:49
• It is a good method. So you can create 5V from this battery by using a boost converter. Another option is to use two buck converters, one for 2.5V and another one for - 2.5V. If you need extremely silent rails- LDO to 2.5V, buck to - 4V or so, and a negative LDO. Those LDOs come as complimentary positive and negative regulators, for RF applications. – Gregory Kornblum Jan 24 '17 at 20:51
• @GregoryKornblum If I use a boost converter, I do not add unnecessary noise? Can you give me some part example for the buck converters? – thoraz Jan 24 '17 at 20:57
• Noise- yes, definitely, but you should filter it out and use those LDOs, although i don't believe you actually need something that silent. Example- sorry, it's late and i am tired. Remind me tomorrow, i will recommend specific components. But in the meantime look at ti.com, they have everything – Gregory Kornblum Jan 24 '17 at 20:59

I came across the same problem when I looked into a similar ADC with +/- 2.5V supply. So I checked the evaluation kit they were providing: ADS131A04 Evaluation kit (power shematic p.29)

Apparently, they are doing the following:

• +2.5V generated by LDO from the main power (5V from USB I think): U34
• Charge pump used to generate negative voltage from the main power: U36
• -2.5V generated by another LDO from that negative voltage: U37

• Do you know some IC That make whole this job? – thoraz Jan 26 '17 at 12:46
• No, I don't. I think it is a common practice to use several ICs. – Edesign Jan 26 '17 at 13:48

Creating a virtual ground using LDO is possible if you have a 5 V source, as mentioned. here is one such reference:
http://cds.linear.com/docs/en/design-note/dn115f.pdf

In order to have a stable 5 V, i would go for any boost converter. Below is one such example considering 3.7 V input supply available.
http://www.linear.com/product/LT1307

Points:

1. Since the negative ref. voltage is only needed for ADC, the current requirement will be too low. A resistor divided may also be sufficient instead of dedicated linear 2.5 V regulator for virtual ground.
2. Also, since 5 V is from a regulator with low noise and small quiescent current, it is suitable for battery powered application.
3. using second LDO for positive 2.5 V reference, keeps the supplies free form noise (less noise)

Everything assuming a current need of less than 200 mA at 5V.