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

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\$$.

• Oh, you are right, it makes sense now. So, as we increase the output current, it takes out more charge from the capacitor, which means now the average voltage decreases. So to increase it back, we must pump more charge into it, and we achieve this by increasing the duty cycle. My problem was thinking that the average voltage and the output $\Delta v$ are 2 independent components, but the average is just the average of the $\Delta v$, right? Commented Aug 13 at 15:34