Switching Regulator basics

I have a basic query on the switching regulator.Basically on the usage of the inductor at the "SW" pin(**broken link removed**).

Is it only for ripple suppression? Will i get proper voltage if i dont use the inductor?
What will be the output Voltage and current waveform at this SW pin and after the inductor?

The inductor is vital for the operation of the circuit, and any SMPS. In a buck converter, you could see it as being part of a lowpass filter that reduces ripple, but its real purpose is storing energy and cycling it between the input and output. Without the inductor, it wouldn't function properly at all.

If you use a resistor instead of an inductor, and if you're lucky, then it's possible that you would get the proper voltage at the output. However, the efficiency would be terrible (completely defeating the purpose of using a switching regulator in the first place). If you really don't want to use an inductor, you should use a linear regulator (LDO) instead, being mindful that your efficiency will not be good. If you want high efficiency, you must use a switching regulator with an appropriate inductor.

The IC will act as a fast repeating on/off switch, whose "SW" output alternately connects one end of the coil to the input voltage, or disconnects it. So you might expect a square wave signal (of sorts) on this pin.

When connected to the input voltage, the coil's current gradually increases (since Vin > Vout), and Vout rises a little while the coil stores power in its magnetic field. When disconnected, the coil 'pulls' current through the diode, Vout drops a little, and the current gradually decreases (since Vin < Vout) while the coil retrieves power from its magnetic field. The IC will vary duty cycle according to Vin/Vout ratio, output load, internal timing elements etc (using Vfb pin to sense output voltage on other side of the coil). Average current through the coil will be the same as what's consumed on the output. Average voltage seen on SW pin will be a little higher than the output voltage, since the coil has some DC resistance too.

In other words: the basic principle of a buck converter (aka step-down DC-DC converter).

An inductor (coil) is the most essential part: no coil, no efficient power conversion. :sad:

About buck, boost, and buck-boost regulators (aka converters)...

I have a Youtube video consisting of animated simulations of these circuits.

It portrays how the coil is the center of action, how it charges and discharges, along with its changing emf and flux field.

www.youtube.com/watch?v=FT_sLF5Etm4

Dear VJKris
Hi
In fact as we know Vave of each square function will given by 1/T integral over Vs dt from zero up to ton . if we solve it , we will have Vave= VS*D.C ( D.C is duty cycle ) . so that inductor and capacitor will provide a 2nd order integrator circuit to take average from square wave , thus if you change the D.C the out put voltage will change .
there are many ways available to justify behavior of that circuit . such as things that mtwieg told in his post .
Best Wishes
Goldsmith