# Buck-Boost small signal

I read the paper "Simplified Analysis of PWM Converters Using The Model of The PWM Switch", and I have some questions about the model of DCM (Discontinuous Conduction Mode) flyback.

This is in the DCM condition. and I would like to use PWM switch to model it. 1.) In this paper, about equation(3) I don't know why vac equal to VL, why it doesn't consider the Vin, it looks like neglect the Vin. • Group the 4 equations in a 2-by-2 way, by considering that two of them are a function of d, and the other of d2. It should make sense then. Apr 21, 2021 at 9:29
• I don't get it. I am confused why the equation (3) and (4) can ignore the Vin and R. if Vac = L*ipk/dt which mean the input voltage is short. Apr 21, 2021 at 9:48

You are in DCM, so the switch is neither in the a position, nor in p, but floating in the middle. Therefore there are two voltages, V(a,c) and V(c,p), which are dependent on the previous states because of the inductor and capacitor -- reactive elements, or elements with memory.

So, if coming from the CCM side, the switch was in the a position, thus the inductor was energized, and now that the switch is opened, you get the same voltage, with a negative sign. But since the voltage is expressed as V(a,c), not V(c,a), there is no need for the minus. All this happens on the ON time, thus for $$\D\$$.

On the p side, the capacitor would have been charged by this voltage, and the load is supplied only by the capacitor. The voltage that appears is the same as the one transferred by the inductor, hence the equation as a function of the OFF time, or $$\D'\$$, or $$\D_2\$$.

I have to stress out the DCM part -- Discontinuous Conduction Mode. In this mode, you no longer talk about $$\D\$$ and $$\D_2\$$, only, you are inherently talking about $$\D_3\$$, which is the moment the switch is not in the a position, or in p, it's "floating" in the x position: The waveform is also no longer a repeated triangle, it has the $$\D_3\$$ portion which is flat. See fig. 3 in that document, it clearly shows that after $$\D\$$ and $$\D_2\$$ there is a portion where there is no current flowing: And those formulas stem from the memory effect of the reactive components. Looking at the schematic, when the switch was in the a position, Vin would fall on L for the duration of $$\D\$$, giving the formula (3) (but with a negative sign, I explained before). Then it would switch to p for the duration of $$\D_2\$$, where the charged inductor discharges to the capacitor + load. Finally, it switches to the x position and stays there for the duration of $$\D_3\$$, when nothing happens because the inductor is discharged and the only volage on the load is that of the discharging capacitor -- the only one holding the line.

So, yes, all those 4 equations are from the CCM part, but now there is another part, the $$\D_3\$$ part, where nothing happens in the inductor, there is no current flowing. The 5th and 6th equations would look like this:

\begin{align} i_x&=0\tag{5} \\ v_{xc}&=0\tag{6} \end{align}

• Hi But the equation(1) is in the turn-on, and equation(2) is in the turn-off. Apr 21, 2021 at 12:21
• can you draw a figure for me, I still consfused Apr 21, 2021 at 12:21
• @Jitter456 I've updated the answer, not sure how else to explain. Apr 21, 2021 at 16:38
• Thanks for your answer, I know what's you said. in the paper it show the relationship between ia and ip, so I can dertive the equation(1) and equation(2). but how do I know the relationship between vac and vap, that is my confused point. Apr 21, 2021 at 16:53
• @Jitter456 I explained that both in the first part, and in the second. At his point, I don't what else to say. Best wait for another answer. Apr 21, 2021 at 17:19