Q: What are the design oriented forms of the DC equations for the stress on the switch in the three Cuk Converters?
A: On both the transistor Q, and the diode P, the stress is:

OFF Voltage: Vs = Vg/D'
ON Current: Is = I/D'

Vg is the source voltage, I is the load current.

Note that if the converter is isolated N:1, where N = Np/Ns, the transistor current, and the diode voltage are 1/N the values given above.

Q: What is the stress on the switches in the Boost-Buck cascade?
A: The OFF Volatage for both switches: Q1 & P1, Q2 & P2 is just Vg/D'.

The ON Current in the Boost's switch is the input current Ig. The ON Current in the Buck's switch is the load current I.

Q: What is the sum of the stresses in the two switches?


A: It is just twice (Vg + V) x (Ig + I).

You'll see that this is just the same as in the Cuk topologies if you do the math!
Note that in both cases, the transistor stress equals the diode stress.

Q: Is this true even in the step up/down case, with N not equal to one?
A: Yes.

Q: Why?
A: Stress is in units of Power [Watts = Volts x Amps.] Since Energy is a real physical quantity, which can be neither created nor destroyed, and since time never stops ("tempis fugit!" the Romans said,) Power, which is Energy per second, is the same on either side of a 1:N transformer.

Q: What does happen?
A: Suppose we step down, say 2:1. The transistor current will be half the diode current, but the diode voltage will be half the transistor voltage. The stress on each will be identical!