I have a doubt regarding average PWM voltage. Let's take a buck convertor with Vin=10V, Vout=5V. With this we can determine the duty cycle of PWM needs to be 50%. Thus means the MOSFET of the buck is switched on for half of the time and switched off for remaining time.

How can we say that output voltage will be 5V. It needs to be a varying voltage from 0 to 12V, since the voltage is being turned on and off by PWM. Why are we using the average voltage?

  • \$\begingroup\$ So this must be a perpetual motion PWM? 10V goes to 12V… \$\endgroup\$
    – Solar Mike
    Jun 24 at 21:45
  • 5
    \$\begingroup\$ Are you aware of how the inductor and freewheel diode works? The switcher provides bursts of current to charge the inductor and the output filtering capacitor holds the output end at constant voltage. Post a circuit for analysis if you want. \$\endgroup\$
    – Transistor
    Jun 24 at 21:54
  • 4
    \$\begingroup\$ A buck converter is basically a pwm switch followed by a filter to remove the pwm and leave us with the average dc voltage. Think of it like a water tap going into a pipe. The mass of water serves to filter or average the water flow. Whilst the tap way be turned on/off at various duty cycles, the result at the end of the pipe is a continuous flow of water - it might fast or slower depending on the duty cycle. \$\endgroup\$
    – Kartman
    Jun 24 at 21:55
  • 3
    \$\begingroup\$ "PWM" is not necessarily the same as "buck converter." PWM without an inductor (and possibly a capacitor) is NOT a buck converter. \$\endgroup\$
    – mkeith
    Jun 24 at 21:56
  • 1
    \$\begingroup\$ electronics.stackexchange.com/questions/367574/… \$\endgroup\$
    – Andy aka
    Jun 25 at 11:40

2 Answers 2


In every DC-DC (buck) converter you will find an inductor that is connected in series to your PWM output and a parallel capacitor that is connected to the inductor output and ground. Together, these two passive elements will form a second order lowpass filter that attenuates every frequency content that is bigger than:
$$f_c = \frac{1}{2 \pi \sqrt{L C}}$$

If you now can guarantee that your PWM switching frequency is much higher than your cut-off frequency \$f_c\$, you will essentially have an almost DC output.

The point of using a PWM modulation in such converters is that they allow you to have a very efficient voltage conversion, where the ratio between output power and input power is usually in the range of 85-95%.
A mosfet can be operated nearly lossless when it is turned fully on and it is also really good in blocking any voltages when it is turned off. Everything in between is inefficient because a lot of power is dissipated over the mosfet itself and will thus not be available on the output.
The PWM signal is favored because it mainly consists of these two operating points (mosfet fully on or fully of) where the mosfet is efficient and some very short (and inefficient) transitions in between.


If you have a simple first order RC filter which removes PWM switching frequency but leaves the DC voltage, it has the correct output voltage, but due to the resistor, the output impedance at DC is the RC filter resistance so it can't provide much current or support a high load.

An LC filter is better, because it is a second order filter, and because the inductor of the LC filter has very low impedance at DC it can provide current into the load.


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