# How does a boost converter allow for a gain in output voltage by just switching the positions of the inductor and switch?

I get that in a buck converter, a low pass LC circuit was attached and this effectively filters out high frequencies. If the bandwidth is configured so that it is lower than the switching frequency, a pure DC output is produced. But what I don't get is that why in a boost converter, the positions of the switch and inductor are interchanged so that the the output voltage increases. How does this work?

I guess I could also ask for how the buck boost converter works through putting the switch and inductor in series.

• I've always thought it's easier to understand these if you think of the switch as two switches, the way it actually is in real circuits. Commented Dec 22, 2020 at 0:01
• The key is to learn Lenz's Law, and follow the inductor current. (and start by assuming the load, R, is high resistance)
– user16324
Commented Dec 22, 2020 at 0:03
• @BrianDrummond has it, but you can also think in terms of the energy the inductor gains when the switch is on has to be transferred to the output cap when the switch is off if the output is to remain at the desired voltage. Commented Dec 22, 2020 at 0:08
• I agree with @BrianDrummond. You have to follow the small ripple approximation and keep and eye on inductor volt-sec balance. Commented Dec 22, 2020 at 11:21

Consider this simplified schematic of a boost converter: -

In the first half of the switching cycle the MOSFET is activated and the inductor is placed directly across the incoming supply $$\V_{IN}\$$: -

That action causes current to rise linearly in the inductor at a rate of $$\\frac{V_{IN}}{L}\$$ amps per second. As current increases, so does the stored energy. $$\\text{Energy} = ½ L\cdot I^2\$$

Then, in the second half of the switching cycle, that magnetic energy is released when the MOSFET disconnects (or deactivates): -

Released energy can only pass to the output capacitor and load. That energy makes the output voltage higher than the input voltage because, the inductor is attached to the input voltage source and, that means that the output voltage receives the input voltage AND the extra magnetic energy that was stored in the inductor during the first half of the switching cycle.

This makes the output voltage larger than the input voltage.

• Where did you get those visuals? I love your explanation, and slightly sad that my question did not ask about buck converters because I would have loved to see your take on the buck converter with matching visuals to get a sort of parallelism. Your explanation makes things really clear. Commented Dec 22, 2020 at 11:25
• @AndroidV11 my website has a tear-down of a boost converter and it also has a boost converter calculator that shows the inductor currents like an oscilloscope. I guess I should also get round to doing one for the buck converter (one day). Commented Dec 22, 2020 at 11:35

Think about the current flowing through the inductor; in the buck configuration, it's limited by the resistance of the load (OK maybe you get more if the capacitor is discharged but in steady state the capacitor is - or should be at - something close to your desired output voltage). Essentially you're seeing the inductor storing a little bit of energy and using it to power the load when the switch is in position 2. In the boost configuration, it's shorted straight across the source when the switch is in position 1 so it's storing a much larger relative amount of energy and pushing it into the load when the switch is in position 2.