Electronics Forums > Re: RMS voltage of a pulse width mod. waveform.

# Re: RMS voltage of a pulse width mod. waveform.

Dave Martindale
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Posts: n/a

 04-05-2008, 08:28 PM
Steve <(E-Mail Removed)> writes:
>How do I calculate the RMS voltage of a pulse width?

For any waveform, the RMS voltage is the square root of the mean square
voltage. In concept, you take the voltage as a function of time, square
that, integrate it over one cycle of the waveform, divide by the period
of one cycle, and then take the square root.

In the case where the waveform in question is a square wave whose
voltage is 0 for part of the period and Vp for the rest of the period,
evaluating the above is particularly simple. You ought to work it out
for yourself to see just how simple, but the answer is

Vp * sqrt(Ton / (Ton + Toff))

>I have a Ton 5.4us, Toff 10.2us, frequency 64kHz, and the peak (measured on
>a scope) is 14 volts.

So the true RMS voltage is 8.24 V.

is out of calibration. On the other hand, most meters do *not* read RMS
voltage, they read either peak voltage or average voltage, and then
scale that by a correction factor so they read RMS voltage *for a
sine wave*. Their reading isn't valid for non-sine waveforms.

>How do I calculate the pulse to get my 5.6v?

Do you really want 5.6 V RMS output? If so, you need to adjust Ton and
Toff, or possibly Vp.

Dave

Dave Martindale
Guest
Posts: n/a

 04-06-2008, 08:00 AM
Steve <(E-Mail Removed)> writes:

>Hmmm what does a Fluke 87 measure in? I must have lost certain brain
>cells and begining to confuse myself.

It's a true RMS meter, isn't it?

>The peak of (US) AC outlets is 170v peak. The value the Fluke measures
>is 120 or (0.707) * 170.

That's correct. Although integrating a sin^2 function is somewhat more
difficult than integrating a square wave (I recommend looking it up in a
table of integrals, or (these days) checking online), it turns out that
the integral of sin^(t) over one cycle is t/2, or the mean square of a
sine function with peak 1 is 0.5. That means the RMS voltage of a sine
wave is 1/sqrt(2) times the peak.

Interestingly, that's *also* true for a square wave with a low voltage
of 0 V, high voltage of 1 V and a 50% duty cycle. So RMS is 1/sqrt(2)
times peak for this waveform too. But the *average* voltage is not the
same - the mean absolute value for the sine wave is 108V, while the
square wave's mean value will be 85. So an average-reading meter that
measures 108 V and "converts" it to 120 V for display will be fooled by
the 85 V mean of the square wave and display a bogus voltage, not 120
V.

Dave

Dave Martindale
Guest
Posts: n/a

 04-06-2008, 05:30 PM
Steve <(E-Mail Removed)> writes:

>>>I have a Ton 5.4us, Toff 10.2us, frequency 64kHz, and the peak
>>>(measured on a scope) is 14 volts.

>> So the true RMS voltage is 8.24 V.

>>>How do I calculate the pulse to get my 5.6v?

>Hmmm what does a Fluke 87 measure in? I must have lost certain brain
>cells and begining to confuse myself.

Are you saying that the Fluke 87 reads 5.6 VDC on *this* waveform?

>The peak of (US) AC outlets is 170v peak. The value the Fluke measures
>is 120 or (0.707) * 170.
>So I should be measuring the RMS of the pulse.

The Fluke displays the correct RMS voltage when measuring a 60 Hz sine
wave with 170 V peak. But virtually *any* meter will do that, even if
it's average-reading, doesn't work above 100 Hz, and is accurately
calibrated on only the 200 V scale.

Now, the Fluke probably is accurately calibrated on all scales, and it
is a RMS-reading meter. But as someone else pointed out, it can only
read the RMS voltage of a waveform accurately under limited conditions:
the peak-to-average ratio of the waveform has to be within some maximum,
and all the significant harmonics of the waveform have to be within the
meter's measuring bandwidth. The *fundamental* of your waveform is 64
kHz, and the meter bandwidth is only 20 kHz, so you shouldn't expect it
to provide an accurate measurement of even a sine wave at that
frequency.

It looks like you need something with a bandwidth of 1 MHz or so to
accurately measure the RMS voltage of this waveform. For example,
digital storage scopes can capture the waveform, and some of them have
built-in functions that display the RMS voltage measured over one cycle
of a waveform.

Dave

Dave Martindale
Guest
Posts: n/a

 04-06-2008, 06:05 PM
(E-Mail Removed) (Dave Martindale) writes:

>Although integrating a sin^2 function is somewhat more
>difficult than integrating a square wave (I recommend looking it up in a
>table of integrals, or (these days) checking online), it turns out that
>the integral of sin^(t) over one cycle is t/2, or the mean square of a
>sine function with peak 1 is 0.5.

This is getting somewhat off topic, but I saw a particularly elegant
explanation on the Web about why the integral of sin^2(t) over one
cycle is t/2:

* Because of Pythagoras, and the definition of sin and cos,

sin^2(x) + cos^2(x) = 1, for any x

* So, the integral of (sin^2(t) + cos^2(t)) dt is just t

* So the definite integral of this sum over any interval 2*pi in length
is just 2*pi.

* But cos(x) is really the same function as sin(x) with a phase shift.
The same is true of cos^2 and sin^2. If you integrate over one
period of the waveform, the integral is always the same no matter what
start and end points you pick.

* So the sin^2 and cos^2 terms contribute equally to the integral (as
long as you integrate over one period), and each one is half the
total.

* So the integral of sin^2(t) dt over a 2*pi period is just 1*pi, and
the mean value of the amplitude is 1/2.

Dave

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