buck/boost converters currents

Question

I have seen that the buck and the boost converters have capacitors in them and a switching circuit that switches the circuit on/off. I want to ask whether the buck and boost converters increase the current since due to switching the current is very high. This will help me reduce the coil thickness since the current at input could be low and after switching it could be made high. Am I right or wrong??

Answer 1

A buck converter will have greater average output current than average input current. A boost converter will have less average output current than average input current.

The extra output current in the buck converter is drawn from the ground through the switching diode or synchronous transistor.

The peak current through the coil will be greater than the average current because of the switching. Peak current matters for things like saturation of the inductor coil. However, heating of the coil wire does not depend on the peak current, but the average.

The difference between peak current and average current depends upon the design of the converter and the operating conditions. This ripple current may be large or small. When a converter is operated in CCM, the ratio of peak to average current will be smaller than if the converter is operated in DCM.

So whats the reason behind this increase in the current in the buck converter. Like if I am using a voltage regulator I won't be able to use the relation that as the voltage decreases the current increases.

The output current of a "linear" voltage regulator cannot be greater than the input current. This is because a linear regulator simply passes (most of) the input current to the output while dropping the voltage. The voltage voltage drop in such a regulator is due to the resistance of the pass transistor. In a switching buck converter, on the other hand, current passes through the switching transistor into the inductor for part of the cycle, and then current passes from ground through the diode and through the inductor for the rest (or a part of the rest) of the cycle. The inductor does not want to see a change in current, so when the switching transistor turns off, the inductor, figuratively "sucks" current from the ground through the diode, in order to maintain the current in the inductor.

^{simulate this circuit – Schematic created using CircuitLab}