Cuk converters, so help me
Really, it has an accent over the C, but I cannot remember what type and I am going to leave it out, with apologies to Slobodan Cuk.
Anyway, the Cuk converter is a very neat thing.
It has continuous current flow on both input and output – so low noise and little filtering required.
And can make output voltages higher and lower than the input with only one switch.
However, the output voltage has to be the opposite polarity to the input, which is not so neat for two reasons:
1, for easy feedback it is nice to have everything on the same side of the 0V rail.
2, I struggle to think upside-down and the right way up in the same circuit, especially when inductors (with all their already difficult polarity-reversing-while-current-stays-in-the-same-direction behaviour) are involved. Too many negated negatives.
Having never got past step A – how they work at all – let alone getting onto coupled-inductor variants, I had a really good read around and discovered a lot of explanations.
An excellent one can be found on a website called SimonBramble, and it is well worth reading the Cuk section because of its refreshing approach.
With its Ls and Cs, the Cuk looks very much like it should be some sort of resonant converter (even harder to visualise), but it is hard-switched just like a buck or a boost.
Then, it helps to know that once it is running steadily the voltage across the power transfer capacitor is equal to Vin minus Vout – so if it is 8Vin and -4Vout, the Vcapacitor is 8-(-4)=12V (I said there would be double negatives).
Ignoring the ripple on it, this voltage is going up and down like a sewing machine, with the switch clamping it to 0V or, when the switch is off, the diode clamping it to 0V.
This makes the output an LC-filtered version of that square-wave voltage train.
I realise this is a back-to-front way of thinking about the operation, as the Vin-Vout only really comes out at the end of analysis, but at least it gave me a fighting chance of seeing how the waveforms within the Cuk work.
The transfer function is Vout/Vin = (-D)/(1-D), where D is duty cycle.
And knowing that D = Ton/(Ton+Toff),
and that Vout/Vin = Iin/Iout for switching converters
this collapses to Vout/Vin = Ton/Toff = Iin/Iout
Linear Tech and National make Cuk controller chips.
If you wanted a positive output, and didn’t mind an extra switch or two, this (right) form of boost-buck converter will do it for you. The switches are operated in-phase, and a p-fet is best for the second switch (I read).
And while you are at it you might as well put an n-fet across that Schottky, driven in anti-phase, and remove that 0.4V drop.