A few questions about TRUE RMS

[PA4TIM]

Because that is what it is.
A 5V squarewave with a 2,5V DC offset, flip the tops down and you have 2,5V DC signal

[tszaboo]

Why does my Brymen BM257 multimeter read 2.5 V when I measure a 5V PWM signal at 50% duty cycle?
According to theory and simulation it should read 3.54 V

sqrt(0.5 * (5^2 + 0^2)) = 3.54 V

My meter reads 2.5 V both when dialed to AC and DC

You are right, that is how you calculate RMS.
 The meter will measure RMS AC, which means it is AC coupled, so the measured signal is plus and minus 2.5V.

[eKretz]

So RMS AC would be 2.5V, RMS AC + DC (offset) would be 3.54V - that's the confusion here?

[oldway]

When dealing with household electrical projects, is true RMS really required or an average measurement is sufficient? (what can be the error between them in % in common house hold scenarios?)

The mains voltage is sinusoidal and for this reason it is not necessary to use a true rms multimeter.

But with regard to current measurements, this is different because there are many non-linear loads and the currents therefore do not have a sinusoidal waveform.

For example, the simple magnetization current of a transformer is strongly distorted and its measurement with a non-true rms multimeter will give an erroneous value.

But we must avoid using the multimeter to measure currents on mains.

For safety and convenience reasons, use a true rms current clamp.

[cs.dk]

You are right, that is how you calculate RMS.
 The meter will measure RMS AC, which means it is AC coupled, so the measured signal is plus and minus 2.5V.

sqrt(0.5 * (2,5^2 + -2,5^2)) = 0,000 V

What is RMS then? I can see the logic behind, avarage of 0-5V DC is 2,5V when the high and low times are equal. ie. 50% duty cycle.

I don't get it

[oldway]

sqrt(0.5 * (2,5^2 + -2,5^2)) = 0,000 V

Back to school 
(-1)^2= +1
sqrt(0.5 * (2,5^2 + -2,5^2)) = 2.5V

[cs.dk]

My TI-89 gives me 3,06 on that.

[eKretz]

Why does my Brymen BM257 multimeter read 2.5 V when I measure a 5V PWM signal at 50% duty cycle?
According to theory and simulation it should read 3.54 V

sqrt(0.5 * (5^2 + 0^2)) = 3.54 V

My meter reads 2.5 V both when dialed to AC and DC

Because it's reading the 2.5V DC (50% pulse width 5V peak signal) in DC mode, and it's reading 2.5V AC in AC mode because the meter is measuring the voltage as an AC signal - the center of the signal becomes zero and there's 2.5V above and 2.5V below that artificial "zero." If you use your oscilloscope to measure the same PWM signal, it will show the same result - as long as you set it correctly! When measuring DC you need to DC couple the vertical channel. When measuring AC you need to AC couple the vertical channel. Both will show 2.5V. An RMS measurement (depending on your scope) will usually show RMS AC + DC - which will give you your 3.54V.

sqrt[(2.5²(this is RMS AC)+2.5² (this is DC)]=3.54V

RMS AC is good for measuring irregular waveforms - RMS is used to calculate the area of these irregular shapes. When we talk about DC there shouldn't need to be an RMS because DC is defined as a steady voltage. There is a third option which is a DC signal with an AC component (like pulsed DC) - and this is what is measured by meters with a TRMS AC + DC setting. On my scope this is just labeled as RMS - but there's a twist - if the scope is DC coupled you'll get the RMS AC + DC voltage (DC voltage with AC component) and if it's AC coupled you'll get RMS AC voltage only.

This is pretty easy to demonstrate/view on any scope with probe compensation and it's 50% pulse width 0.5V (or whatever) pulsed DC squarewave signal even if you don't have a function generator.

[perkabrod]

Happy new year!
Thanks eKretz for a great writeup, amazing how many replies I've gotten, I signed up to ask this question and seems to be an amazing group of people

Yes I used a DIY scope and swapped between DC and AC coupling and noticed how in AC it looks like a 2.5 V amplitude square wave centered around 0 V.

If I have a 1 volt sine wave that has a 1 volt DC offset, type y=1+sin(x) in google search to see a graph of what signal I mean
Will the multimeter in DC-mode give me the DC equivalent, and in AC mode the meter would read .707 V, from the 1V/sqrt(2) equation, because the 1 V DC offset can't pass the AC ”capacitor coupling”

I don't have a signal generator to try this but am interested in finding out what the meter would show in DC mode for a 1 V DC offset 1 V sine signal Would clear things up in terms of measuring techniques.

[perkabrod]

My TI-89 gives me 3,06 on that.

(0.5*2.5^2+2.5^2)^.5 = 3.06

Problem is the bold part needs to be within parentesis because we're using HALF of 2.5^2 + -2.5^2

(0.5*(2.5^2 + -2.5^2))^.5 = 2.5

[cs.dk]

If I have a 1 volt sine wave that has a 1 volt DC offset, type y=1+sin(x) in google search to see a graph of what signal I mean
Will the multimeter in DC-mode give me the DC equivalent, and in AC mode the meter would read .707 V, from the 1V/sqrt(2) equation, because the 1 V DC offset can't pass the AC ”capacitor coupling”

I don't have a signal generator to try this but am interested in finding out what the meter would show in DC mode for a 1 V DC offset 1 V sine signal Would clear things up in terms of measuring techniques.

Have a look in this thread - It's more complicated then so;https://www.eevblog.com/forum/testgear/brymen-multimeters-fault/

[cs.dk]

My TI-89 gives me 3,06 on that.

(0.5*2.5^2+2.5^2)^.5 = 3.06

Problem is the bold part needs to be within parentesis because we're using HALF of 2.5^2 + -2.5^2

(0.5*(2.5^2 + -2.5^2))^.5 = 2.5

Thanks..
Well, my math is not used everyday - But something seems wrong, have a look at the attached photos. (Sorry for crappy quality)

EDIT: Welcome to the community btw.

[slicendice]

LOL! Sorry for laughing but that is what happens when relying on calculators too much.

Try typing in this:

sqrt(0.5*(2.5^2+(-2.5)^2))

and you get your 2.5V


if you type:
sqrt(0.5*(2.5^2+ -2.5^2))

you get sqrt(0.5*(2.5^2+ -(2.5)^2))
which is 0

Please see attached screenshot

Edit: added a second calculation where you actually divide by 2 instead of multiplying by 0.5

Edit2: something was wrong with the second image, uploaded again

[perkabrod]

Thanks for the Brymen link, interesting info there about autoranging etc.
Will update here with actual readings on a DC offset sine wave when I get time.

Attached a photo from the Casio fx9860, been very happy with it, coming from a TI82.
Some features I miss (or don't know how to do) but overall a very nice calculator, with BACKLIGHT and nice input/output in terms of fractions, etc.

Here's a graphing speed test of the Casio vs TI82 I did
https://youtu.be/lYE8YmZ4yoc

[cs.dk]

LOL! Sorry for laughing but that is what happens when relying on calculators too much.

Try typing in this:

sqrt(0.5*(2.5^2+(-2.5)^2))

and you get your 2.5V

You are absolutely correct - Just tested it. It gave me 2,5

Laugh all you want to, i take it as a "lesson learned".. Its many years since i used calculus for anything serious (In school as marine engineer), and haven't touched it since. 

[HKJ]

My meter reads 2.5 V both when dialed to AC and DC

Even true rms meters uses average on DC

[eKretz]

Happy new year!
Thanks eKretz for a great writeup, amazing how many replies I've gotten, I signed up to ask this question and seems to be an amazing group of people

Yes I used a DIY scope and swapped between DC and AC coupling and noticed how in AC it looks like a 2.5 V amplitude square wave centered around 0 V.

If I have a 1 volt sine wave that has a 1 volt DC offset, type y=1+sin(x) in google search to see a graph of what signal I mean
Will the multimeter in DC-mode give me the DC equivalent, and in AC mode the meter would read .707 V, from the 1V/sqrt(2) equation, because the 1 V DC offset can't pass the AC ”capacitor coupling”

I don't have a signal generator to try this but am interested in finding out what the meter would show in DC mode for a 1 V DC offset 1 V sine signal Would clear things up in terms of measuring techniques.

Yes there are a bunch of great guys here, it is a good place. Here is a link that might help you further understand:

http://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-sine-wave-with-a-dc-offset/

[eKretz]

Oh, and for the y=1+sin(x) question, I believe the meter on RMS AC would read .707V; on DC would read 1.0V (the offset - the AC "ripple" cancels out); and on RMS AC + DC would read 1.225V.

[perkabrod]

Oh, and for the y=1+sin(x) question, I believe the meter on RMS AC would read .707V; on DC would read 1.0V (the offset - the AC "ripple" cancels out); and on RMS AC + DC would read 1.225V.

 

Interesting. Curious, why not 1.707 V for AC+DC? Not that I have ever come across a multimeter that had a setting for AC+DC.

I noticed something on the brymen bm257 but it may be applicable to other DMM, just in case anyone has wondered about why the Hz not always working. I ”discovered” I had to be in AC when entering Hz mode, at least for a pwm signal. Maybe certain signals will trigger also in DC mode when trying to read the frequency.

[eKretz]

Oh, and for the y=1+sin(x) question, I believe the meter on RMS AC would read .707V; on DC would read 1.0V (the offset - the AC "ripple" cancels out); and on RMS AC + DC would read 1.225V.

Interesting. Curious, why not 1.707 V for AC+DC? Not that I have ever come across a multimeter that had a setting for AC+DC.

Did you forget your earlier lesson already? I will answer a question with a question: why on a 50% pulse width 5VDC signal did your meter read 2.5V RMS AC and on DC read 2.5VDC but measured 3.54V RMS AC + DC? Why shouldn't it have read 5V?

BTW for the fun of it and to check myself I just went ahead and set up a 2V peak to peak sinusoidal wave with +1VDC offset - DC coupled, my scope showed 1.22V RMS (AC + DC) and AC coupled measured 0.707V. My multimeter read 0.707V on AC (TRMS) and 1.0V on DC.

[slicendice]

The RMS for a AC voltage (sinewave) of 1V with an offset of 1V we put in this formula:

SQRT([OFFSET]^2 + [VOLTAGEAC]^2 / 2)

which becomes:

SQRT(1^2 + 1^2 / 2)

that equals 1.2247.....

If we have a voltage of 2V and an offset of 1V it would become this:

SQRT(1^2 + 2^2 / 2) = 1.732...

With a voltage of 3VAC and offset of 2VDC we get:

SQRT(2^2 + 3^2 / 2)
--> SQRT(4 + 9 / 2 )
--> SQRT(4 + 4.5)
--> SQRT(9.5) = 2.915...

[rrinker]

LOL! Sorry for laughing but that is what happens when relying on calculators too much.

Try typing in this:

sqrt(0.5*(2.5^2+(-2.5)^2))

and you get your 2.5V

You are absolutely correct - Just tested it. It gave me 2,5

Laugh all you want to, i take it as a "lesson learned".. Its many years since i used calculus for anything serious (In school as marine engineer), and haven't touched it since. 

 "Knowing" -2.5^2 is +6.25 (also known as, I haven't used this calculator in ages and couldn't find the negate button), I got the right answer with my TI-83. Once I figured out the negate key though - the answer it returns if typed in as originally specified is 0, because it applies the negation AFTER the exponent. So the extra set of parentheses is necessary and typing it in as (-2.5)^2 results in the same final answer of 2.5
 Potential trap for young players as these sort of calculators allow you to type in equations as written and will apply correct order of operations and result in the correct answer. MOST OF THE TIME!
 I usually prefer Casios, but I sort of inherited the TI plus it reminds me of my old TI-57 I wish I could find.

[perkabrod]

Did you forget your earlier lesson already? I will answer a question with a question: why on a 50% pulse width 5VDC signal did your meter read 2.5V RMS AC and on DC read 2.5VDC but measured 3.54V RMS AC + DC? Why shouldn't it have read 5V?

On a 50% 5VDC signal I measured 2.5V in AC and 2.5V in DC on the multimeter. I never measured 3.54V (only in ltspice simulation), though I did the math and got confused as to why in DC mode my multimeter didn't read 3.54V. You said When we talk about DC there shouldn't need to be an RMS because DC is defined as a steady voltage. which makes sense.

Why I was thinking 1.707 for 1+sin(x) was because if a 1 VAC sine reads 0.707 and I offset that with 1 V I thought it would just become 1+0.707 but maybe not that easy Do you know the math for AC+DC situation? For the example with 1 V offset and 1 VAC sine

[eKretz]

Yes I posted it in my answer to you earlier in this thread. Your meter won't read the 3.54V unless it's RMS AC + DC. It will read 2.5V RMS AC and 2.5V DC (averaged). RMS AC+ DC would read 3.54V. The equation to figure it is RMS AC + DC = sqrt[(RMS AC)² + (DC V)²].

So for the y=1+sin(x) example:

sqrt [.707VAC² + 1.0VDC²] = 1.225V

Your pulsed waveform example is not straight DC - it is DC with an AC component - so to correctly measure it you need to use RMS AC + DC. Your scope simulation is measuring RMS AC + DC because you had it set to DC coupling. When the scope is DC coupled it passes all signals, so you get RMS AC + DC when you measure RMS. When the scope is AC coupled it will only pass the AC component of the signal - so you get RMS AC when you measure RMS.

[slicendice]

All these calculations are actually integrals (calculating areas below any curve within a certain range) with a lot of variables, but if you read the text in the link, posted by eKretz, that talk about how to derive from the original formula, you can clearly see that most components cancel out. This is why the formulas become what they are.

About graphing calculators...I bought a TI-85 when I was in high-school. Many of my friends had HP or Casio, except one who also had TI-85. We were the only ones who could calculate any problem that was given to us, using the calculator only. All others had issues because there was something always missing. :-) TI series calculators are really good, and they have pretty much everything, but the formulas and constants for certain things can be found under unexpected menus. But once you understand how the calculations are done, one can find everything needed to get the right answer from one single equation. :-)

Love TI, and don't like the other brands that much. Not bad calculators, but for me they just don't make any sense at all. Casio and HP had a lot of features that TI did not have, but for pure Math, Chemistry and Physics problems those features were of no help to solving the problem.