Can anyone explain why \$\frac{\Delta V_x}{\Delta D} = V_{in}\$ here?
It seems to me that they are assuming that \$V_x = V_o\$ ( this is only correct for average voltage). However, if they use the assumption \$V_x = V_o\$ for average, then later in LC transfer function, they also should use this assumption to get \$\frac{\Delta V_o}{\Delta V_x} = 1\$. But from the transfer function of LC filter, we can see that they don't use that assumption.

I am really confused about this. It is not clear when we can use approximation and when not.

enter image description here


I think they assume that the average of the square wave generated by turning on and off the PMOS transistor is \$<V_x>=\text{D}V_{in}\$ and it doesn't look like they include the \$R_{on}\$ voltage drop associated with the PMOS transistor when it is on.

Now, \$V_{\text{o}}\$ isn't exactly equal to \$V_x\$ because of the voltage drop due to the series resistance associated with output inductor, but they should be really close.

So, I think of it as \$V_x\$ being a square wave without filtering with its average being portion of the input voltage (\$\text{D}V_{in}\$). And \$V_o\$ is the output voltage after filtering which should be the same as the average of the \$V_x\$ waveform minus the voltage drop caused by the output filter.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.