# How does a down converter work

## Forum: Analog electronics and circuit technology How does a buck converter work?

Hello everyone, I have problems understanding the circuit of the buck converter. Normally a coil and a capacitor are used there. I am now wondering what the purpose of the coil is exactly. Couldn't you just apply a PWM directly to the capacitor?

Put simply: Because you don't want to convert the difference between input and output voltage into heat. Let's first leave out parasitic influences, i.e. idealized. How much current flows when your output supplies 10V and the capacitor is at 5V? How does it look when you add the parasitic resistances of the output and the capacitor Not omits?

: Edited by user

In other words, the inductance stores the energy in the pulse pauses and thus ensures a relatively even flow of current.

Sandro wrote:> Normally a coil and a capacitor are used there. No. Actually only the coil is needed. Because that is the energy store. In addition, you need an input capacitor, an output capacitor, a switch (Mosfet) and a free-wheeling diode. Add a suitable controller and a good layout and the arbor is ready. There you will find the two important circuits: and you will see that the output is always involved, and therefore the current there is fairly "calm" or "continuous": http://www.lothar-miller.de/s9y/categories / 40 layout switching regulator

Sandro wrote:> Couldn't you just> apply a PWM directly to the capacitor? Example: In = 10V, out = 5V, PWM switch (Mosfet) = 0.1Ohm. deltaU / R_Mosfet = 50A With a very large capacitor and extremely short switch-on times, you're right. But it's damned impractical and brings a lot of problems with it. So you need the coil to temporarily store the energy that is not currently required in the magnetic field. The 'freewheeling' diode with anode to GND ensures that the collapsing magnetic field = induction voltage redirects the energy to the output capacitor in a 'rest cycle'.

Thanks for the answers. At what point does the capacitor have to deliver the current?

knoelke wrote:> It's damned impractical and brings a bunch of> problems with it. Unless the load is sluggish. Then this forms the rms value itself.

Martin Schwaikert wrote:> Unless the load is sluggish. Then this forms the rms value> itself. But then you don't need the output capacitor, in this case it is even counterproductive because of the high current peaks.

Sandro wrote:> At what point does the capacitor have to deliver the current? Actually only with rapid load changes. If the current of the load drops very quickly, the capacitor absorbs the energy still present in the coil and ensures that the output voltage does not rise in an impermissible manner. If the current of the load increases very quickly, it can briefly buffer the increase in current until sufficient current flows through the coil. It also smooths out voltage peaks that occur when switching and gives the control system time to react.

knoelke wrote:> With a very large capacitor and extremely short switch-on times> you are right. Even then you end up with the efficiency of a linear regulator and burn off the difference to the switching regulator in the internal resistance of the MOSFET. Just distributed differently in time.

: Edited by user

A. K. wrote:> in the internal resistance of the MOSFET. and the internal resistances of the capacitors

: Edited by user

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