What is the name of this DC-DC converter topology?

Question

I spotted a circuit in the wild that generates symmetric (e.g. ±8 V) analog power rails from a single +5 V supply using a single boost converter and a single 10 μH inductor. This is a very common use case so I spent some time to reverse engineer it. The following is a schematics/simulation I made:

_{(source: imgh.us)}

U1 and V2 are my make-shift PWM power switch for simulation --- in the actual circuit the net is tied to the SW pin. Of course the positive output is regulated through feedback, but the negative output is not directly tied anywhere.

I found a post by Linear Technology that briefly mentions the negative branch of this circuit as a "inverting charge pump" (Fig. 4), but I cannot find other sources on this topology. The positive branch seems to be just flipping the direction of the two diodes, making it slightly different from the original boost topology...

So what is this really? And what are some advantages/disadvantages of this design compared to e.g.:

• Separate boost and inverting converters (e.g. TPS65131 which integrates both), which requires 2 inductors
• Boost converter followed by a fly-back-capacitor-type inverting charge pump

2017-03-19 EDIT:

After some digging I found this these links:

• Simplest possible 5V to +/- 10-15V converter (for op-amps)
• Inverting charge pump with boost converter loading characteristics
• http://www.ti.com/lit/an/slua288/slua288.pdf
• https://obsoletetechnology.wordpress.com/projects/studio-electronics/dc-dc-bipolar-power-supply-for-effect-pedals/

As I expected this is a very common use case. I think at this point I just need a textbook or equally authoritative source that documents the fundamentals of this topology. Any suggestions?

Answer 1

I think you're making a bit too much out of this, it is actually a very simple circuit.

The LMR62014 is also not required for this, almost any DCDC converter (buck or boost) can do this.

Have a look at these Voltage multiplier circuits, note how they all need a strong (it needs to provide power) AC signal (usually a square wave) at the input. Then the diodes and capacitors are used to make a high(er) DC output voltage.

If you flip the circuit "upside down" you would get a negative voltage.

The circuit you're looking at is basically only a one stage version of this. One for the positive voltage, one for the negative.

The Switcher IC, in combination with the inductor, is used to generate the squarewave signal that is needed.

Answer 2

At its output, the boost converter generates a waveform that varies between ground, when U1 is 'on', and whatever voltage the flyback spike reaches, when U1 turns off.

That waveform is rectified by C3, D1, D2, and the switching time of the boost converter is presumably fed back to regulate the output voltage. When everything has reached equilibrium, the boost inductor flyback voltage will be enough to match the required output voltage and the two diode drops.

For the negative rail, the same waveform is recitified by C5, D3, D4, but this time the +ve output is connected to ground rather than the negative.

The negative output voltage will generally track the positive output. However if differnt currents are drawn, their output impedance will result in slightly different output voltage drops.