Can somebody explain me about the working and waveforms of Buck converter with LC input filter? I can't figure out the output voltage expression using Vg, R and D(duty cycle) andenter image description here the dc voltage and current of inductor L2 ,L1,C1 and C2. Please explain in details.

  • \$\begingroup\$ Everything to the right of, and including, Q1 work the same as normal. L1 simply decouples the power supply from the high frequency switching currents, leaving C1 to do the bulk of that work. \$\endgroup\$ – Unimportant Nov 2 '20 at 7:38
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    \$\begingroup\$ There are plenty of resources online for buck converter, with more details that you'd ever need. \$\endgroup\$ – a concerned citizen Nov 2 '20 at 8:06
  • \$\begingroup\$ But I can't understand how to express the output voltage V0 in terms of Vg,D and R \$\endgroup\$ – Camila's voice Nov 2 '20 at 8:15

The buck converter operations has been the object of many descriptions in the literature and its operations can be explained in a variety of ways. I have taught an entire APEC seminar on the buck converter and an excerpt is shown below:

enter image description here

The easiest way is to first consider the continuous conduction mode (CCM) and see the converter as a low-impedance square-wave generator whose output is filtered by the \$LC\$ network. The output voltage of the buck converter is thus in a first approximation the average voltage developed across diode \$D_1\$ which theoretically swings between 0 V and \$V_g\$. Considering power switch \$Q_1\$ turned on during \$DT_{sw}\$, the output voltage of a perfect buck converter is simply \$V_{in}=DV_g\$. In reality, there are losses incurred to the power switch, the diode forward drop and its recovery losses, the inductor series resistance and so on which make the actual output voltage deviate from this formula. Then, when the load current reduces, the converter enters a discontinuous conduction mode (DCM) or operation and the formula changes to a more complicated expression in which time constants and switching frequency now matter.

From your circuit - looks like it is coming from Colorado Edu. - there is a front-end filter made of \$L_1C_1\$. This filter will ensure that high-frequency pulses are delivered by \$C_1\$ while an almost continuous non-pulsating current is delivered by \$V_g\$. The effects of inserting this filter are well beyond this short text but, in a nutshell, the converter is now a 4th-order system in CCM and instability can happen if you close the loop. A detailed analysis is provided here.

As a conclusion, why not running a quick simulation with the program of your choice to visualize the variables you want? SIMPLIS can do that in a few seconds as shown below and the demo version Elements is good for this:

enter image description here


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