I have a circuit that is usually powered by +9V and -9V (uses two 9 volt batteries), but I am trying to power this from a rechargeable 3.7 V battery. The circuit does not draw much power (can be powered down to about +2V and -2V), but it has op-amps that require both a negative, and positive swing (LF412).

I was wondering if it would be possible to create a dual supply from the battery, and if so, what the most efficient method would be to do this. I have a couple of ideas that i might try:

1.) Use a boost converter to step the voltage up, generate a pulse (555 timer), send it to a transformer, and use a center-tap to create the separate voltages, which can then be filtered.

2.) Follow the steps for idea 1, but instead of a 555, use a micro controller to generate a sin wave (DAC lookup table method), which can be amplified, send this to the transformer, as this would be more efficient for transferring the power.

I would really appreciate the help!!

  • \$\begingroup\$ Would simply boosting the voltage to 12V and using a 7809 and 7909 be sufficient? It's not a great set-up, but it's simple and works. \$\endgroup\$ – Mast Dec 17 '17 at 13:48

There are off the shelf chips that do what you want. Also, you don't need any transformers since there is no need for isolation in this case.

For the positive supply, you need a boost converter. This is assuming you connect the negative side of your 3.7 V battery to ground. There are also switcher chips that are intended for making a negative supply from a positive one. If your negative current demand is low enough, a charge pump might be all you need.

These things are available off the shelf. For your low voltages and currents, you can find chips that contain the switching element. Generally you supply the inductor, a few caps, and sometimes a external diode or two. Look around at offerings from Microchip, TI, LT, Linear, and many others. Or use a distributor web site to drill down parametrically across the vendors carried by that distributor.


For low currents and analog circuits, I prefer charge pumps to inductor-based boost and boost-buck circuits.

If the current required is relatively low, then a charge pump inverter will turn some of the +9 V into -9 V. Maxim, Linear Tech, TI.

For the 3.7 V source, you can use two charge pump parts, one to double the 3.7 V to 7.4 V, and one to invert that to -7.4 V. IIRC, someone makes a single chip that does everything, double plus invert. Sounds like something Maxim would do.

Found it: https://pdfserv.maximintegrated.com/en/ds/MAX680-MAX681.pdf


Why not just use two rechargable 9V batteries? That way you can still recharge it when it runs out of power, and you won't have to tinker with its electronics and risk breaking it.


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