This might seem like a stupid question, but nevertheless I'm uncertain. To get an accurate measurement of the RMS voltage value of a modulated DC signal, which setting of the voltage meter is correct, for AC voltage or for DC voltage?

-(puzzled) Pete

Edit: In other words, I mean the voltage drop across a load of fixed resistance from a pulsed direct current, i.e., a periodic current (in my case, not sinusoidal) changing in value, but not direction through the load.

In the case of what I want to measure, there is no constant DC current component. Each pulse starts and stops at 0 Amps.

 

You need to consult the manual for your meter. Many meters have a DC blocking capacitor for AC measurements. If your manual doesn't say, just check a DC source on the AC range - if it reads zero (or close, depends on noise) then there is a blocking cap.

If there is no blocking cap, then the reading will reflect the RMS value. If there is a blocking capacitor, measure the DC value on DC range and AC value on AC range and calculate the RMS (square root of (DC squared + AC squared)). This can be a bit problematic if the meter sampling rate is such that you can't get a stable reading due to the AC component.

These should be helpful:
http://literature.cdn.keysight.com/litweb/pdf/5988-6916EN.pdf
https://www.keysight.com/main/edito...g&ckey=458358&nid=-536902435.626391&id=458358

 

Your meter needs to be a true RMS type to measure the RMS voltage accurately.

 

You need to consult the manual for your meter. Many meters have a DC blocking capacitor for AC measurements. If your manual doesn't say, just check a DC source on the AC range - if it reads zero (or close, depends on noise) then there is a blocking cap.

If there is no blocking cap, then the reading will reflect the RMS value. If there is a blocking capacitor, measure the DC value on DC range and AC value on AC range and calculate the RMS (square root of (DC squared + AC squared)). This can be a bit problematic if the meter sampling rate is such that you can't get a stable reading due to the AC component.

These should be helpful:
http://literature.cdn.keysight.com/litweb/pdf/5988-6916EN.pdf
https://www.keysight.com/main/edito...g&ckey=458358&nid=-536902435.626391&id=458358

Thank you. At the moment I'm doing computer simulation but I would think that what you are saying is relevant to the simulation as well as taking an actual voltage reading.

 

Your meter needs to be a true RMS type to measure the RMS voltage accurately.

Yes, thanks, I'm not using it at the moment, but my DMM is termed a true RMS-type measuring AC voltage accurately from 40 Hz- 1 kHz. Right now I have been measuring voltage in a computer simulation (Electronics Workbench).

 

@ebp

Thanks for the Application Note (pdf) discussing AC rms readings. The explanation is very good.
On page 6 of the note, I believe the formula for AC rms value of a pulse train is stated incorrectly. Corrected I believe it should be

Vrms = (Vpk / CF) * sqrt [ 1- (1 / CF)^2]

Calculating with the corrected formula I get 280 mV for the waveform of Fig. 5 of the note, this being what the note states should be the result of using the formula.

 

Does anyone care to comment whether or not the formula in my previous post #6 above is correct?

If it is correct, then for example a DC square wave pulse with Vpk= 2V and duty cycle of 50% should have an RMS voltage equal to about 1V. 1 Vrms makes sense. In a simulation with Electronic Workbench, the DC voltage reading of this source was 2 V.

Because I have only had LTspice for a day now, I could not correctly configure a simulation in that program to get a RMS voltage reading. That would be fantastic if anyone who knows how to do this simulation in LTspice would give it a shot and let me know the RMS voltage reading.

 

That would be fantastic if anyone who knows how to do this simulation in LTspice would give it a shot and let me know the RMS voltage reading

Go to LTspice XVII / Help / Index / Waveform Arithmetic.

 

Does anyone care to comment whether or not the formula in my previous post #6 above is correct?

If it is correct, then for example a DC square wave pulse with Vpk= 2V and duty cycle of 50% should have an RMS voltage equal to about 1V. 1 Vrms makes sense. In a simulation with Electronic Workbench, the DC voltage reading of this source was 2 V.

Because I have only had LTspice for a day now, I could not correctly configure a simulation in that program to get a RMS voltage reading. That would be fantastic if anyone who knows how to do this simulation in LTspice would give it a shot and let me know the RMS voltage reading.

Hello there,

The RMS value of ANY waveform is defined precisely as "the Root of the Mean of the Square" which of course is "RMS" spelled out. Mathematically, this means square the wave, calculate the average, then take the square root.
The square is easy:
V^2
but then we have to find the 'mean' which is the average:
Avg=(1/T)integral(v,t)
where v is the squared wave and T is the total period.
Then all we have left to do is take the square root:
sqrt(Avg)
and that gives us the RMS value of the original waveform.

Since pulses are flat topped, they are particularly easy to calculate the RMS value for.
So we start with a constant value V, then square it and get:
V^2
and since the wave is 'on' for a given time say T1 and the period is T, we integrate:
integrate(V^2,t,0,T1)

then multiply by 1/T:
(1/T)*integrate(V^2,t,0,T1)

and the result of this is simply:
T1*V^2/T

and now we take the square root of that and get:
V*sqrt(T1/T)

and that is the RMS value of V, the pulse that went from t=0 to t=T1 with total period T and amplitude V.

As to your example of a 2v pulse that has 50 percent duty cycle, the RMS value is:
sqrt(2)=1.4142

Note that this makes more sense than the incorrect result of '1' because a 1.4142vrms voltage across a 1 ohm resistor yields a current of 1.4142 amps, and power is volts times amps which here is sqrt(2)*sqrt(2)=2 watts, and 2 volts across 1 ohm creates 2 amps,and 2 amps times 2 volts is 4 watts, but 'on' for only 1/2 the time is 2 watts. So the two views produce the same results and that's the definition of RMS in reference to the production of power. So the integral definition and the power definition both produce the same power in a resistor, but the result of "1vrms" will not produce 2 watts in a 1 ohm resistor, it will only produce 1 watt, which is not enough.

 

Go to LTspice XVII / Help / Index / Waveform Arithmetic.

LTspice- the sophisticated and powerful simulator.

 

@MrAl
In typical fashion, I tend to open my mouth with my brain out of gear. At least I recognize that I tend to do this which is half-way there to not doing it so often. Thank you for your very clear explanation.

 

@MrAl
In typical fashion, I tend to open my mouth with my brain out of gear. At least I recognize that I tend to do this which is half-way there to not doing it so often. Thank you for your very clear explanation.

Hello again,

You are welcome, and one more little thing i should mention.

That is, i noticed your thread title was "Measure RMS voltage of modulated DC" which could have a couple interpretations.
The first interpretation is of course what we already talked about, where we have pulses that are all the same width and spacing and we want to know the RMS equivalent.
The second interpretation is that the DC is being modulated into another type of wave, such as a sine wave. This means it would be good to know the average value of the DC pulses too, as they average over time to create a sine wave after some minimal filtering. Since a sine wave has RMS value Vpeak/sqrt(2), you could use that to calculate the RMS value of the resulting wave. Alternately, you'd have to calculate the square root of 1/n times the sum of squares of all the individual RMS values of each section.

Just for the heck of it i did a PWM sine wave modulated with pulses, starting with only 3 pulses (including 2 at zero for a total of 5) per half cycle up to 400000 per half cycle.
The results are (4 pulses is actually 3 plus two zero values, 40 is actually 39 plus two zero values, etc., etc.):
0.816496581 [4]
0.716114874 [40]
0.707992325 [400]
0.707195186 [4000]
0.707115620 [40000]
0.707107665 [400000]

and the true value as a float is:
0.707106781

so we can see that as we increase the number of pulses in the synth sine wave, we get an RMS value that is closer and closer to the RMS value of a real sine wave.

This uses the formula of the square root of the sum of squares of individual RMS values divided by n:
Total_RMS=sqrt((v1^2+v2^2+v3^2+...+vn^2)/n)

 

Hello again,

You are welcome, and one more little thing i should mention.

That is, i noticed your thread title was "Measure RMS voltage of modulated DC" which could have a couple interpretations.
The first interpretation is of course what we already talked about, where we have pulses that are all the same width and spacing and we want to know the RMS equivalent.

The waveform that I wanted to calculate the RMS voltage of is an AC triangle wave with Vpk = 5V, the duration of each half-wave is 3 mS, and the period T = 10 mS. Here is how I thought to solve it.

For the triangle wave,

Vrms = 0.58 * Vpk = 0.58 * 5V = 2.9V

So in terms of solving for RMSV, I can think of the triangular half-wave as being equivalent to a square wave with Vpk = 2.9V with the same duration, 3 mS.

Then next mentally full-wave rectify the AC waveform and combine the two half-waves per period into one square wave with a duration of 6 ms.

Finally, based on what you very helpfully provided in your post # 9,

Vrms = Vpk * sqrt( t / T)

where t= on time and T = period

Vrms = 2.9V * 0.77
Vrms = 2.25V

Correct?

 

The waveform that I wanted to calculate the RMS voltage of is an AC triangle wave with Vpk = 5V, the duration of each half-wave is 3 mS, and the period T = 10 mS. Here is how I thought to solve it.

For the triangle wave,

Vrms = 0.58 * Vpk = 0.58 * 5V = 2.9V

So in terms of solving for RMSV, I can think of the triangular half-wave as being equivalent to a square wave with Vpk = 2.9V with the same duration, 3 mS.

Then next mentally full-wave rectify the AC waveform and combine the two half-waves per period into one square wave with a duration of 6 ms.

Finally, based on what you very helpfully provided in your post # 9,

Vrms = Vpk * sqrt( t / T)

where t= on time and T = period

Vrms = 2.9V * 0.77
Vrms = 2.25V

Correct?

Hello again,

Well whatever method you come up with you have to match the 'known' results for a triangle wave which is:
Vrms=A/sqrt(3)

where A is the peak amplitude.

If your triangle is not 'on' for the entire time, then you have to factor in the square root of a ratio that corresponds to the duty cycle, but this points out the fact that we need to know the entire wave exactly as it will occur because there are different forms of a triangle wave that may go above and below zero and also as mentioned not be 'on' the entire time.
So it would be better if you make a drawing of your waveform and then we can be sure we get the right result.

A standard triangle as mentioned is A/sqrt(3) and that is for a triangle that repeats without any spacing. So if the peak was 2.9v then the RMS value would be 2.9/sqrt(3) which is about 1.674vrms.
Just for example, if the wave above was 'on' for only 50 percent of the time, then the RMS value would fall to 2.9/(sqrt(3)*sqrt(2)) or about 1.184vrms.
The factor is:
K=sqrt(ontime/(ontime+offtime))
so we get:
Vrms=A*K/sqrt(3)

But again, there are variations so seeing the waveform is best to ensure we get the correct result.

 

@MrAl

Thanks for the response. The waveform is a positive triangle with Vpk = 5V with a "base" lasting for 3 mS. This is followed by a space of 2 mS (0 Volts), then a negative triangle with again Vpk = 5V and lasting for 3 mS. Finally there is a space of 2 mS (0 Volts) after the negative triangle and preceding the next wave.

So I would suspect (based on how RMSV is calculated as you have pointed out) to be really accurate I would need to derive a special factor defining the ratio of peak to rms voltage of each triangular half-wave. Or LTspice could tell me.

Edit: After giving it more thought, given that the above waveform is symmetrically centered on ground (0 Volts), I don't see why the calculated RMS voltage of a single half-wave wouldn't produce the same ratio of RMS to peak voltage as that of a continuous full-wave.

 

@MrAl

Thanks for the response. The waveform is a positive triangle with Vpk = 5V with a "base" lasting for 3 mS. This is followed by a space of 2 mS (0 Volts), then a negative triangle with again Vpk = 5V and lasting for 3 mS. Finally there is a space of 2 mS (0 Volts) after the negative triangle and preceding the next wave.

So I would suspect (based on how RMSV is calculated as you have pointed out) to be really accurate I would need to derive a special factor defining the ratio of peak to rms voltage of each triangular half-wave. Or LTspice could tell me.

Edit: After giving it more thought, given that the above waveform is symmetrically centered on ground (0 Volts), I don't see why the calculated RMS voltage of a single half-wave wouldn't produce the same ratio of RMS to peak voltage as that of a continuous full-wave.

Hello,

I drew the waveform based on your description. And yes, the rectified wave will have RMS value the same as the actual wave, and not only that but then we can integrate over one half cycle.
In the attachment the formula is shown along with the clear markings of where we get the timing and amplitude from. The triangle has duration T, the period (or half period if you will) is Tp, and the amplitude is A. Note the triangle can be any shape as long as it is still a triangle, and that includes a regular triangle as well as a sawtooth. The sawtooth form is shown.
This should be easy to understand now. The RMS value is simply the RMS value of a pure triangle A/sqrt(3) times the square root of the ratio of times: sqrt(T/Tp).

 

Hello again,

Here is the RMS value for a set of triangles with various spacing and different 'on' times and different amplitudes.

From the drawing it looks evident that we can take this farther:
Vrms=sqrt(A1^2*T1+A2^2*T2+A3^2*T3+...+An^2*Tn)/sqrt(3*Tp)

where the An are the amplitudes and the Tn are the respective times for each An and Tp is the total period of the entire waveform.

 

Your meter needs to be a true RMS type to measure the RMS voltage accurately.

As of yesterday, my meter is a HP3478A. This meter reads RMS voltage of any waveform in the frequency range of 20- 200 kHz not lower than 100 mV with 0% error! My multimeter is used and about 40 years old, so accuracy may not be quite that good, but this meter is probably plenty accurate for my needs.