Hi everyone,

I am a bit confused about the operation of a boost converter, mainly with the changing current. In the attached image I replaced the diode with a switch. Now I keep reading that once the turn my W1 switch off and turn on my W4 switch the current will decrease and the voltage will drop. If the resistance of the W4 switch is the same as the W1 switch wouldn't that mean that I just simply switched paths of the current flow, but kept the resistance the same and therefore the current would be kept constant?

 

When you turn on W1 the current will rise in the inductor. When you turn W1 off the current still wants to flow in the inductor so the voltage will rise. Because W4 is now closed the capacitor will be charged to a higher voltage. The big capacitor in your drawing goes on the other side of W4.

 

With W1 closed and W4 open the current will increase in the inductor, causing inductive energy to be stored.

With W1 open and and W4 closed, the inductor current will continue to flow (as inductor current is prone to do) and add current (charge) to the output capacitor. This causes the inductor to lose energy and the inductor current to decrease.

The reason it can increase voltage is because the inductor voltage across the inductor when charging can be lower than the reverse polarity voltage generated when it is discharging (the inductor voltage will go as high as needed to keep the current flowing until the inductor current stops).