In a buck converter, the switch controls energy storage in the inductor. The average of the square wave applied to the filter will be the DC output level (12V @ 41.6% duty cycle = 5V average). The inductor acts as a current source to keep the output capacitor charged.
Depending on the load, switching frequency and inductor size, a fixed-frequency buck converter can operate in one of two modes. If the DC output current level is greater than half of the inductor current ramp, the converter is said to be in CCM (continuous conduction mode); if not, it's said to be in DCM (discontinuous conduction mode).
\$ I_o > \dfrac{1}{2} \cdot \dfrac{[V_i-V_o]\cdot T_{on}}{L} \$ or
\$ I_o > \dfrac{1}{2} \cdot \dfrac{-V_o\cdot T_{off}}{L} \$
In CCM, the inductor is sourcing current the entire time. The inductor current never goes to zero. The capacitor is always being charged by the inductor, so it never has to support the load by itself. The duty cycle will remain essentially fixed regardless of output current once in CCM.
In DCM, because the output current is less than half the inductor ramp current, the only way to regulate properly is to decrease duty cycle. This decrease in duty cycle leads to a third mode of operation, where the switch is off and the inductor has completely discharged.
(Some controllers will operate in what's called critical conduction mode, where instead of operating with fixed frequency and variable duty cycle, it operates with fixed duty cycle and adjusts the frequency to keep the converter exactly at the DCM/CCM threshold.)
But a Low Pass filter is the one that attenuates frequencies higher
than a desired frequency and passes the lower frequencies. So How does
this definition fit in this case?
The important thing you haven't grasped is that the on-off square wave has an average level that is totally dependent on the duty cycle. This mathematically average level gets converted to a real dc level by the L and C low pass filter action.
Here's a square wave that rises to 3V from 0V (gold trace): -
The blue trace has a little bit of low pass filtering applied - can you see that if a lot more low pass filtering were applied, the blue trace would be almost a constant DC level of 1.5 volts?
Best Answer
The catch diode insures a proper return path for the inductors current. Without the diode there is risk of damage to the MOSFET switch and a greatly reduced output.
There is also a possibility of incorrect polarity at the output, possibly causing damage or drawing excessive current from the source. The diode solves many problems on both the ON and OFF cycle of the MOSFET.
To answer your question,yes, the current would keep flowing from source to load if the MOSFET stayed in an ON state. However Vout would equal Vin with no voltage or current control provided by the SMPS IC.
If the diode is missing, there can be NO current flow if the MOSFET is OFF, as the path has been broken. The capacitor will quickly drain to zero volts.
If the diode is missing and the switch is OFF after being ON, then the stored inductors current will pass through the MOSFET switch, likely destroying it. MOSFET's fail as a short, so it would have to be replaced.