From the formula of charge relating capacitance and voltage, Q = CV, capacitance is inversely proportional to voltage.
This is how I got the idea that possibly when increasing the capacitance, the output voltage would decrease for both buck and boost converters.
But I think that idea is invalid because I'm just grasping at straws for the formula because it has a voltage. I have a feeling that the behavior of increasing capacitance might result differently for buck converters and boost converter.
I would like to ask what the answer to this is with an explanation of how it works. I just can't understand what would happen in the given situation of the question.
Best Answer
A larger capacitor will decrease the output ripple for a given fixed load.
First, on your equation: your logic isn't right because Q, the charge on the capacitor, isn't fixed. What the equation in fact says instead is that for a given output voltage, a capacitor with more capacitance C will store more charge on its plates. This is related to the reason why more capacitance will decrease the ripple, which we'll talk about next.
Second, (for example) a boost converter ultimately works by quickly charging something up (say, an ordinary inductor), then connecting the charged thing in series with the power supply in order to increase the output voltage above the supply voltage. Then, after a short time, the converter disconnects the charged thing in order to charge it up again because it will inevitably quickly begin to discharge when it is supplying power to the load.
Putting a smoothing capacitor across the output (i.e. with the other side of the capacitor connected to ground) of such a converter will cause the capacitor itself to charge to the output voltage. Thus, in the periods where the converter cannot itself supply current (i.e. when the converter is charging up the e.g. inductor inside it), the capacitor can supply current to the load instead. Current is charge per unit time, so more current means the smoothing capacitor will discharge faster, and so fail to smooth out the converter output for as long. (Note, your equation says this: as Q decreases for fixed capacitance C, the voltage V must decrease proportionally.) To compensate, a capacitor with greater capacitance can be used, since it stores more charge for the same voltage, and thus it will take longer for the output voltage to droop.
Attendum: Of course, eventually after the capacitor has been discharging for some time the switching converter will reconnect and begin to supply to power to charge the capacitor back up as well as supply power to the load. Thus, we don't need an infinitely large capacitor to get acceptable performance: instead, for a given load we are designing the converter for, we only need to use a capacitor which will reduce the output ripple by an acceptable amount for however long the switching converter needs to charge itself back up again.
Also, if you think about it the capacitor will never charge all the way up to the output voltage of the converter when under load, and this is what happens in real life.