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The output of buck regulator is the average value of the output across the regulator. So below are my questions.

1) How the average value is coming across the load resistor practically ?

2) Whether the switching action (ON/OFF) leads to create the average value at the output ? If so, how ?

3) The average value is coming across the resistor because of what force or impact (practical perspective)?

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Your question seems to be "How does a buck converter work?"

To summarize:

  1. The switch is turned on
  2. Vin rises fast but the inductor delays the rise of Vout and it expands it's magnetic field
  3. The switch is turned off when Vout is near/above the desired voltage
  4. The inductor's magnetic field collapses and powers the circuit
  5. When Vout is near/below the desired voltage, goto #1

The fluctuation of Vout due to switching is called ripple. A capacitor is typically added in parallel with the load to reduce ripple.

Specifically:
http://en.wikipedia.org/wiki/Buck_converter

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  • \$\begingroup\$ Hi, Noted your above answers.But the function of induction and capacitor is smoothing (or filtering) the output. induction and capacitor will not regulate the output. My question is who is proving the average value of volatge to across the load resistor ? We can make buck regulator Without inductor or capacitor. That time the output will be rippled form. Note that inductor or capacitor for smoothing the ripples to one level. \$\endgroup\$ – Mohammed Azlum Feb 23 '15 at 10:02
  • \$\begingroup\$ @MohammedAzlum: A buck without an inductor? Thats not a buck regulator. The inductor isn't for smoothing, it is for regulating the output voltage. Read through the points 3 and 5 that Jon mentioned specifically. \$\endgroup\$ – PlasmaHH Feb 23 '15 at 10:18
  • \$\begingroup\$ TON and TOFF time is for regulating the output voltage. Considering I have 39V input. How do you going to buck the output to 13V ? Please explain in terms of inductor ? But I think I can explain in the terms of switch ON and OFF time; If the duty cycle D of the switch is 0.33, then the average voltage is 39 ● 0.33 = 13Vdc. \$\endgroup\$ – Mohammed Azlum Feb 23 '15 at 10:32
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    \$\begingroup\$ Without the Inductor & capacitor, the output would be a 39V square wave. The inductor and capacitor form a low-pass filter, whose output is roughly the average of the 39V square wave at its input. There is usually feedback from the output back to the converter so that the duty cycle can be adjusted as the load changes. \$\endgroup\$ – sbell Feb 23 '15 at 12:26
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    \$\begingroup\$ Ignoring losses, Vout = Vin * duty cycle. But this is only for a steady load. If the load changes, then Vout will also change, temporarily. In order to deal with changing loads, the buck regulator senses output voltage and modulates duty cycle to maintain nearly constant voltage when the load changes. \$\endgroup\$ – mkeith Feb 23 '15 at 16:03
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Without the Inductor & capacitor, The circuit output will be average value of inout siganl with high ripples. Once you place L (Series to the switch) and C (parallel to the load), then you will get smooth output wave form. The smooth output wave form you can all as " Bucked voltage"

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