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Blog How to Compute the Average Versus the RMS Value of a PWM Waveform
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EMI-Reduction-Techniques
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  • Author Author: kkazem
  • Date Created: 7 Jan 2022 3:19 AM Date Created
  • Views 11651 views
  • Likes 9 likes
  • Comments 2 comments
  • pwm
  • Design
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How to Compute the Average Versus the RMS Value of a PWM Waveform

kkazem
kkazem
7 Jan 2022

A PWM waveform is, of course, a modulated rectangular wave with the bottom at zero volts (typically). This differs from a square wave where the center of the waveform is at zero and the bottom is at a negative voltage. This blog will discuss how to compute the RMS value of such a waveform.

We'll start at first principles and use a voltage waveform, but a current waveform would have the same results. For a simple average of a rectangular waveform, we have: V(t) = A*ton/T, where A is the peak DC voltage of the rectangular wave, ton is the on-time, and T is the period or equivalently, ton+toff. We can also write this in terms of duty cycle, d, where d = ton/(ton+toff). We can then rewrite the average value of a rectangular waveform as V(t) = A*d. And intuitively, this makes sense since there's really only a usable value of voltage during the on-time since during the off-time, the voltage is zero.

Although this average value is useful in DC-DC Converter and other PWM design, for many calculations like the power dissipated in a FET switch over one cycle, we need to use the RMS value and it's quite different from the average value. I'll give the answer first, then the theory behind it. The RMS value of a rectangular wave is A*SQRT(d). We'll start at time=0 with the rising edge of the waveform. 

VRMS = SQRT(Average Value(V^2)), with the average value being: (1/T)*Integral(0 to T)[V(t)*dt]. The RMS value is: SQRT((1/T)*Integral(0 to T)[A^2*ton*dt]), which equals: A*SQRT(d) or equivalently, A*SQRT(ton/T).

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  • kkazem
    kkazem over 3 years ago in reply to phoenixcomm

    Thanks, I appreciate the comment.

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  • phoenixcomm
    phoenixcomm over 3 years ago

    love it. !!

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