Physically, how does an increase in duty cycle lead to an increase in the output voltage of a buck converter?

Question

From my understanding, to achieve a certain voltage at the output, we can use a capacitor to store on it a certain amount of charge \$CV_0 = q_0\$. The moment it loses some charge \$\Delta q\$, we charge it again the same amount, so the charge of the output capacitor is \$q(t) = q_0 + \Delta q\$. We can do this using a resistor, however, due to losses, an inductor is more efficient and it offers an easier way to control the current. Now, I don't understand how the average output voltage is linked to the duty cycle, given how the duty cycle only controls the amplitude of the current through the inductor \$ L\frac{di}{dt}\$, but this \$\Delta i_L\$ is only used to compensate for the lost charge on the capacitor, not to charge \$q_0\$ needed to get the desired output voltage. So, any increase in duty cycle only leads to an increase in the output voltage \$\Delta v_{out}\$ not the average voltage, because an increase in the duty cycle means the current increases to a higher value through the inductor, which means more charge gets stored on the capacitor. However, the capacitor will lose the same amount of charge the next cycle, so this new charge doesn't contribute to the average output voltage.

Clearly my understading is wrong, so my question is, how exactly does the capacitor gets the \$q_0\$ charge needed, and how an increase in duty cycle increase this \$q_0\$ charge since the variation only contributes to the compensation of the loss of charge? I'm not looking for equations; I can derivate the conversation ratio \$V_{out} = DV_g\$.

Answer 1

Clearly my understanding is wrong, so my question is, how exactly does the capacitor gets the q0 charge needed, and how an increase in duty cycle increase this q0 charge since the variation only contributes to the compensation of the loss of charge?

Actually, you are nearly there; you just need to join things up.

An increase in the loss of charge is due to more load current on the output and, if this isn't countered, it will gradually lead (over several switching cycles) to a lower average output voltage so, we increase the duty cycle to put more charge into the capacitor so that the increased charge removed by the heavier load maintains the average output voltage at the correct value.

Of course, this means more output ripple voltage and, that's the price we pay when we use a buck converter.

Answer 2

You're ignoring many important details, and assuming values are fixed when they are not. For example, a loss of charge Δq on the capacitor causes a proportional change in voltage across the capacitor (Q = CV, therefore Δv = Δq/C). And the current through the inductor is proportional to the voltage across it, which is a function of both the input duty cycle (and voltage) and the capacitor voltage. So even if you didn't change the duty cycle, the current through the inductor would increase because of the reduced voltage on the capacitor.