# RMS voltage of a square wave

• posted

I was wondering if I take the RMS voltage of a messy looking square wave, a noisy square wave, with some measurement device, and I want to know the regular current of the thing can I take the RMS current and divide it by the square root of 2 in order to get my regular current?

I know you can do this for sinusoidal ac signals but don't know if the math still works out the same for other signals...

PS - I got my RMS current value by dividing the RMS voltage value by a known resistance I have in my circuit

Much Thanks for any responses I get

• posted

There is no formal definition of "regular current", so if you define it right, sure.

What do _you_ mean by "regular current"?

With sinusoidal waveforms, the peak quantity is the RMS quantity times the square root of two. With square waveforms, the peak and RMS quantities are equal.

• posted

Yes the RMS current is the RMS voltage devided by R.

No, the "regualr curent" What ever that is, is not the RMS current divided by the square root of two. With a sign wave, the peak current divided by the the square root of two is the RMS current, but ONLY for a sign wave.

With any other waveform, the the square root of two relation does NOT hold. It must be calculated from first principles or measured with a true RMS meter.

The equation for the RMS current is: Irms = ((int i(t)^2 dt )/ T)^1/2 Where i(t) is the current as a function of time, T is the time over which the integral is taken, ( 0 to T ), usually one period of the wave.

That is, in words, the RMS current is the square root of the average time integral of the current squared.

For a square wave of current Ip going plus and minus over a period of T the RMS value is:

Irms =( ( (Ip^2 *T/2 + (-Ip)^2 * T/2)) / T)^1/2 = ( (Ip ^2)/2 + (Ip^2)/2)^1/2 = (Ip ^2)^1/2

Irms = Ip, The RMS equals the peak value for a square wave. It very much more complicated if noise is present.

• posted

The RMS of an on and off wave is the average. Try it with 1V and 0V. Then you have to believe that it also works with 5v and 0v.

• posted

Thats right. In the example I gave the wave was plus and minus, not plus and zero.

• posted

Sample the waveform at whatever frequency you like, square each sample, average the squares, and take the square root (root of the mean squares). It doesn't matter what the waveform looks like, this algorithm works.

Absolutely. ...sorta by definition. ;-)

Yes. Think about it this way... RMS is used to measure power because power is proportional to the square of the voltage (or current), so to get the "effective" voltage of an arbitrary waveform you need to average the square of the voltages. This also has the benefit of deriving real power with negative voltages. The average of the voltages doesn't tell you much (the average of your wall outlet is zero).

No problem.

• posted

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D Not after sampling and squaring. The power is all positive, like the squared samples.

• posted

"BobG"

** Completely wrong !!

RMS current or voltage follow the square root of the duty cycle.

Eg: with a duty cycle of 0.5 ( square wave) the RMS value is 0.7071 of the peak.

..... Phil

• posted

Vrms = ((vp^2) *T/2 + 0) *1/T) ^1/2 = ((Vp ^2) / 2 )^1/2 = Vp/ (2 ^1/2)...

Yep, for 50% duty cycle, Vrms = Vp/ (sqr 2) if the other half of the wave is zero.

• posted

"Bob Eld" "Phil Allison"

** Keep it simple:

For a rectangular wave,

V (or I) rms = sq.rt Duty Cycle ( expressed as a decimal fraction )

For a bi-polar, rectangular wave - first rectify the wave.

..... Phil

• posted

Look at it this way: You have a 1KW space heater you run half the time (1 hour on and 1 hour off). The average power is 500 watts. Say the heater voltage is 100 VDC and the current is 10 amps, and the heater resistance is 10 ohms. So, the question is what RMS voltage will give you 500 watts at 10 ohms and 100 volts peak? Work that out and we find the voltage to be 70.7 and the RMS current to be 7.07 amps.

So, it looks like for a square wave, the RMS voltage or current is the peak divided by the square root of 2.

-Bill

• posted

IIRC, although it is easy to check, the rms voltage of a series of rectangular pulses will be the square root of the peak current times the average current.

Bill

• posted

============================================ Not after sampling and squaring. The power is all positive, like the squared samples.

----- But the average of the squared values is not the average value of the voltage and taking the root of the average of V*2 will not, in general, give you the average of V.

It will give the rms value which was devised originally as a way to get an "equal Power" to DC with the same load- as krw indicates.

• posted

Bob Eld and Phil Allison have dealt with this topic except for the question "how did you measure the rms voltage?". If you are using a true rms meter which samples at a high enough rate- then you are OK. If you are using a cheaper multimeter on the normal "rms" scale, then it simply assumes that the AC measurement is sinusoidal and scales from the DC average accordingly and will not give the correct value except for a sinusoid or for an On/Off cycle with equal on and off periods.

• posted

"Don Kelly"

** ?????????

The " true rms" function on a DMM is *not* obtained by sampling. It is normally obtained by the use of a " true rms to DC" converter IC - generally one made by Analog Devices like the AD736.

** Most DMMs use a precision rectifier circuit followed by RC averaging of the output and then DC scaling so that the final reading equals the rms value of a sine wave input ** within** the particular meter's specified frequency range.

Mostly, this range is very narrow - ie from about 30Hz to a mere 1kHz (with

+/- 1% accuracy).

So, unless you KNOW the energy spectrum of your wave falls inside this narrow band, the reading can be way wrong -ie well below the actual value.

Unfortunately, most " true rms " DMMs have the same frequency range issue - some of the more expensive ones work out to 20kHz or 100 kHz with good accuracy.

The only * low cost * solution is to examine the wave on a scope and then compute the rectified average and rms values - not possible with any accuracy if the wave is noise like.

..... Phil

• posted

*PROVIDED* you measured that voltage with a true RMS voltmeter, then yes, you have the true RMS current. But typical voltmeters are not true RMS devices. So beware, "Garbage In / Garbage Out"

daestrom

• posted

How about a precision resistor placed inside a calorimeter :-) The heat dissipated is a direct measure of RMS current squared.

Wasn't this done with some high-freq RF stuff at one time? Measured the temperature or wattage of a resistor or something like that?

daestrom

• posted

Calorimetry is notoriously difficult. Ask Pons and Fleishman. ;-)

Yes, I used an HP true RMS voltmeter when I was in college. It was a marvelously expensive widget and they didn't like mere students using it. ;-) It's likely the best way to measure true RMS voltage at high frequency. I suppose you could read current with the same meter. ;-)

• posted

Slick, though not perfect.

Normally is not

It *is* used, though perhaps not in handheld DVMs. IT is used in Kill-a-Watts.

How do you get better than 1% accuracy on a scope? Sampling works to any accuracy one desires and can go fairly high in frequency. Even higher if the waveform is repetitive.

• posted

"krw" Phil Allison "Don Kelly"

** Err - so just like you ?

** ?????

** Bollocks .

** You are not paying attention to the point at issue - fuckwit.

BTW:

Fix your stupid settings so the name of the poster you arr replying to is left visible.

...... Phil

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