Summarised Solution:
Over 1 kV out from say 5 to 12V in if optimised for this purpose and your switches and output capacitor can handle it.
In real use, a few hundred volts would be doable but more than was sensible.
Long:
The split Pi is more complex than you need for what you say you want.
This is essentially a synchronous input, synchronous output full bridge boost buck converter used in many modern designs. It is fully bidirectional - ie it will convert any voltage at left to any sensible voltage at right OR can convert equally well in the reverse direction.
All you need to start is the parts up to C3. This is a standard boost converter. In non-synchronous versions S2 is replaced with an "upwards pointing" diode. (Parts to right of C3 will down convert the voltage on C3).
The available output voltage is "heaps" but in practice if maximum voltage was an aim you would place a second winding on L1 to provide a stepup ratio.
In real world use you could convert from say 12V to hundreds of volts as shown if necessary. The usual limits are the voltage ratings of L1, S1, S2, C3 (ignoring components beyond that).
In extreme cases the interwinding breakdown voltage if L1 becomes important (ask me how I know :-) ) with the voltage at the start of one winding and the end of the next being two layer voltages apart when windings are layed successively right to left then left to right. In some cases the voltage towards the winding ends can exceed the breakdown voltage of the wire insulation - making interwinding insulation important and how the wires are handled mechanically at the ends of each layer important. In most cases this is not an issue.
The absolute maximum output voltage occurs when the inductive energy in L1 is transferred into its stray capacitance. Consider:
Open all switches.
Close S1 until I in L1 is at a max.
Open S1 and DO NOT close any switch.
Voltage on L1 will now "ring" positively until the 0.5 x L1 x I^2 energy is stored in stray capacitance of L1 as 0.5 x Cstray x Vring^2.
So
- Vring_max = sqrt ( L1 x I^2 / Cstray)
With 12V in (or almost an Vin as Iin_max is what counts) this could be more than 1 kV in a carefully optimised case.
S1 and S2 are liable to object in most cases.
Tha diodes are there to absorb over-voltage (resulting in reversed current flow) shoved back into this power supply from the motors, I'm thinking. Any inductive load can "return energy back to the source" and those diodes are there to absorb such energy. Think of the reverse polarity diodes across relay coils, same thing.
Those input and output filter caps cover an impressive set of orders of magnitude. I wouldn't change them at all. This is an atomic force microscope, yes? That's an amazingly precise instrument. Those different caps are there to soak up noise and avoid self-resonance at different frequencies.
If your AFM works now, good enough!! Look at some atoms!!!
Best Answer
In boost convertors, for part of the cycle, all the load current comes from the capacitor.
Buck converters can deliver current continuously through the inductor to the load (alternately from the voltage source and from ground), so only a fraction of the load current ever comes from the capacitor (in continuous current mode).
Therefore, for the same ripple voltage, the capacitor can be that fraction of the size of that in the boost convertor.
(This is not the whole answer; a more accurate answer would take the different duty cycles for different voltage conversion ratios into account too)