Output Capacitor for a Boost Converter The output capacitor is defined based on the maximum permissible voltage ripple and based on the maximum permissible voltage change (V droop) resulting from a load step. In our example we want to have a voltage ripple of maximum 1% (120 mV) at an output voltage of 12 V.
Let’s take a deeper look at the basics of how buck and boost converters work and the components, such as capacitors, inside them. The most common switching converter is the buck converter, which is used to down-convert a DC voltage to a lower DC voltage of the same polarity.
The boost converter is used to "step-up" an input voltage to some higher level, required by a load. This unique capability is achieved by storing energy in an inductor and releasing it to the load at a higher voltage. This brief note highlights some of the more common pitfalls when using boost regulators.
To successfully increase capacitor voltage beyond the driving voltage, we need to “pump” charge into the capacitor and prevent charge from flowing back into the source. The boost converter’s diode, which functions as a one-way valve for current, provides both of these actions:
The boost converter might still be able to output the desired current at that low input voltage because is the minimum switching current it can handle. But better to be safe than sorry. Here you can see the inductor will see a max of 0.94A at its lowest input voltage. Now we can chose the inductor for our design.
The voltage ripple is therefore For the boost converter and the buck-boost converter, the capacitor current waveforms are the same, since in both, the capacitor provides the output current when the switch is closed, and gets charged by the difference between the inductor current and the load current when the switch is open.