Equivalent buck-boost models

What happens if Vsw < VD In fact that is the situation in most commercial Flybacks. But note that to do a proper comparison, you have to reflect the diode drop to the primary side. And for that we have to multiply the diode drop by the turns ratio (see the equivalent Buck-Boost models of a Flyback section in my book, Switching Power Supply Design Optimization). So, for example, if the turns ratio is 20 and the diode drop is 0.6V, the effective VD we need to compare with Vsw for our time-sharing analysis is 0.6 x 20 = 12V. And that is usually greater than the (average) drop across the switch. Therefore, we tend to say that in a Flyback, decreasing D (increasing input) will worsen the total conduction loss and decrease the efficiency. But of course that never happens, because as we increase the... 

Figure 3-2 The Equivalent Buck-Boost Models of the Flyback...

Looking at the equivalent buck-boost models in Figure 3-2, the center of the ramp on the secondary side (average inductor current, II ) must be equal to Io/(l — D), as for a buck-boost (because the average diode current must equal the load current). This secondary-side inductor current gets reflected to the primary side, and so the center of the primary-side inductor current ramp is Ilr, where Ilr = Ii/n. Equivalently, it is... 

In a buck, there is a post-LC filter present. Therefore this filter stage can easily be treated as a cascaded stage following the switch. The overall transfer function is then very easy to compute as per the rules mentioned in the previous section. However, when we come to the boost and buck-boost, we don t have a post-LC filter — there is a switch/diode connected between the two reactive components that alters the dynamics. However, it can be shown, that even the boost and buck-boost can be manipulated into a canonical model in which an effective post-LC filter appears at the output — thus making them as easy to treat as a buck. The only difference is that the original inductance L (of the boost and buck-boost) gets replaced by an equivalent (or effective) inductance equal to L/(l—D)2. The C remains the same in the canonical model. 

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