So I watched this video to understand how DC-DC boost converter works.
Here's circuit diagram:
And I don't get, when switch is off, and polarity in the inductor is such that positive is on the right, negative is on the left.
Then in order for diode to conduct, anode should be at higher potential than cathode. In other words, voltage across inductor needs to be higher than voltage across capacitor.
Now video said, there's gonna be a spike in voltage in the inductor after switch is off due to collapsing magnetic field to keep current constant, but how big is this spike? And wouldn't charge, and therefore voltage across capacitor be so big, that the spike in inductor will no longer be higher, so diode would not conduct at all?
Correct?
Best Answer
The whole system starts when the FET is on, i.e. Vds = 0v in this situation. For a certain period called Duty Cycle the inductor is gonna charge up with the increasing current that flows through it, and like that it will store energy. As the FET is turned off, the current would normally decrease, but, as there´s a inductor it will try to force the current to same direction it was flowing due to Lenz´s Law. As a result of that, the voltage in the inductor will have changed its polarity in order to keep current flowing to the same direction and then the voltage will be greater than the input(9V) voltage. If the circuit is working under Steady State(SS) conditions then the current that was flowing through the inductor will have the same variation.
In discharge, as the voltage across the FET is now greater than input voltage, the diode will be successfully biased. The spike you talked about is going to be as big as your duty cycle and your load, because inductor tries to keep the current constant whatever resistance it sees ahead, then causing the voltage to go up. Regarding the duty cycle, in ideal models the appropriate formula is given by: Vo/Vi= 1/(1-D), which states that the greater your Duty Cycle, the greater your output voltage.