EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: SG-1 on March 21, 2018, 11:19:42 am
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I have a resistor & measure 240 VAC RMS across it & then I measure 300 VDC across it. Both signals are present at the same time. Am I thinking correctly that the voltages add directly on the real number line ?
The resistor is actually seeing 540 volts RMS across the terminals.
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You could be measuring 240V AC offset by 300V. Which is not common.
But you should explain your measurement methods more detailed (with schematic for example) to say for sure.
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It depends on how you're defining DC and RMS.
Strictly speaking, DC is the average voltage and RMS is the voltage, with equivalent heating power to the same DC voltage, so no it's not possible. The RMS voltage will always be equal to or greater, than the average DC voltage. The RMS voltage is always the DC+AC voltage.
In reality multimeters are often AC coupled on the AC setting, so the RMS measurement only works down to a certain lower cut-off frequency.
Some instruments may also measure the peak voltage, when set to DC, rather than the average voltage.
It's true, a 240VRMS sine wave (the voltage you'd measure if it were AC coupled), with 300VDC offset, will be 540VRMS in total.
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Am I thinking correctly that the voltages add directly on the real number line ?
Yes, as long as you're talking about a resistively loaded signal or a completely unloaded signal. As you kind of imply, things get complex (pun intended) if you have a reactive load.
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It depends on how you're defining DC and RMS.
What other definition of RMS do you think is possible but the "square Root of the Mean of the Square of the signal"?
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It depends on how you're defining DC and RMS.
What other definition of RMS do you think is possible but the "square Root of the Mean of the Square of the signal"?
The square Root of the Mean of the Square of AC part of the signal, which is what most meters give, when set to read AC volts. I know this is not the mathematically correct definition of RMS, but in practise is more often than not, what most true RMS meters actually measure, unless set to AC+DC.
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Both DC and RMS voltage only make sense with a given averaging time. Also, RMS on multimeters often means "AC RMS" which is RMS after a high-pass filter with some cutoff frequency. RMS always means "root mean square of X", but X can be anything you want.
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It depends on how you're defining DC and RMS.
What other definition of RMS do you think is possible but the "square Root of the Mean of the Square of the signal"?
The square Root of the Mean of the Square of AC part of the signal, which is what most meters give, when set to read AC volts. I know this is not the mathematically correct definition of RMS, but in practise is more often than not, what most true RMS meters actually measure, unless set to AC+DC.
If you say "defining" when answering questions in the beginners' topic, when you mean something else, some of them, being beginners, and believing you to be more experienced and knowledgable, will believe you. You go on to begin a sentence "Strictly speaking" and then offer another loose definition of RMS that isn't actually a definition.
You can be as sloppy as you like elsewhere and leave people to second guess what you really mean, but I think the poor old beginners deserve a bit more precision and clarity, they're already confused or they wouldn't be asking a question. If you say something is a definition, or is a strict interpretation in instruction to a beginner I don't think it's too much to ask that you then deliver an actual definition or a strict interpretation. C'mon, with 9770 contributed messages to this forum under your belt you must be able to see the sense of that.
I'm not taking issue with 'It depends on how your meter measures RMS' - which you didn't actually say; you didn't actually mention metering or make it clear that was what you were really talking about. The thing is that, forced to think about what you were really trying to say, you've got it right above, not omitted your assumption that you're talking about what a meter sees.
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I think that for voltage or current, using the RMS of the AC, it is: (https://www.eevblog.com/forum/beginners/ac-dc/?action=dlattach;attach=405698)
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It depends on how you're defining DC and RMS.
What other definition of RMS do you think is possible but the "square Root of the Mean of the Square of the signal"?
The square Root of the Mean of the Square of AC part of the signal, which is what most meters give, when set to read AC volts. I know this is not the mathematically correct definition of RMS, but in practise is more often than not, what most true RMS meters actually measure, unless set to AC+DC.
If you say "defining" when answering questions in the beginners' topic, when you mean something else, some of them, being beginners, and believing you to be more experienced and knowledgable, will believe you. You go on to begin a sentence "Strictly speaking" and then offer another loose definition of RMS that isn't actually a definition.
You can be as sloppy as you like elsewhere and leave people to second guess what you really mean, but I think the poor old beginners deserve a bit more precision and clarity, they're already confused or they wouldn't be asking a question. If you say something is a definition, or is a strict interpretation in instruction to a beginner I don't think it's too much to ask that you then deliver an actual definition or a strict interpretation. C'mon, with 9770 contributed messages to this forum under your belt you must be able to see the sense of that.
I'm not taking issue with 'It depends on how your meter measures RMS' - which you didn't actually say; you didn't actually mention metering or make it clear that was what you were really talking about. The thing is that, forced to think about what you were really trying to say, you've got it right above, not omitted your assumption that you're talking about what a meter sees.
Yes, I should have deleted that first line. If you look closely, you'll see I edited the post about 1h:30 later. You're lucky you didn't see the first version: it was even worse.
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I have a resistor & measure 240 VAC RMS across it & then I measure 300 VDC across it. Both signals are present at the same time. Am I thinking correctly that the voltages add directly on the real number line ?
The resistor is actually seeing 540 volts RMS across the terminals.
Can you please provide a sketch of your circuit (or schematic) so that we can evaluate your question? Also, please give either a picture of your meter reading both conditions, or a meter model number and a picture of your test setup (or both).
What you might be reading on your DC setting on your meter is the p-p voltage of the circuit. The peak-to-peak (p-p) voltage of the AC should be ~340 VAC, which is calculated by multiplying the RMS reading X 1.414 (or the square root of 2, your choice). However, on some meters, if you measure using the pulsed DC setting on your meter, it reads a little less than the p-p (depends on the meter). The meter model number would also be helpful.
If the latter is the case (which I suspect to be true), then there is no DC on the circuit. If you were getting 540 VAC/DC then your pulsed DC setting on the meter would show ~540 volts.
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Sorry for the delay I had to leave for work four hours early today & operating on about fours of sleep. We were using two meters a Fluke 27II & a Flir 284. Both meters were reading within a volt or two of each other. The Flir was set in LoZ, the Fluke had a SSV225 (LoZ accessory). Neither meter can measure AC + DC, the mode(Flir) or dial(Fluke) has to be changed between AC & DC .
The test specimen has flown the coop, I am trying to make sense of what we measured, because something is nagging at me about it.
I can supply a schematic with notes in a few hours, but not t this moment.
I am currently in the process of making a mock up to reproduce the original circuit in my private lab. It will be operating at 1/10 voltage.
I appreciate the overwhelming replies so far. All are greatly appreciated.
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Some notes:
You are looking at the schematic for a MV 3 phase Circuit Breaker. The EDR 5000 is the brains. It can be set to provide nearly any type of protection required like over-current, over-voltage, under-voltage, sync check, negative sequence voltage,...
The breaker is shown open & closing springs not charged. The entire circuit is shown de-energized.
M is a small universal motor, like the ones found in hand held drills. It charges a pair of closing springs.
Y is a logic level relay coil. Stops the breaker from pumping. (repeated closing onto a fault)
SR is the spring release coil (close coil).
TC is the trip coil.
CS C is the Control Switch Close
CS T is the control switch Trip
CAP Trip is a stored energy device in the event the AC is lost. Stores around 20 joules. A PTC prevents it from being used as a power supply.
The circles with G & R are Green & Red LED indicator lights.
The diode below R05 is 400 PIV 40A. Some overkill on the current.
The origional question pertains to the voltage measured across DI1 (Digital Input 1) on the far right between X1-6 & X1-5. An ohm meter reads about 600K ohms across those points. It has a 300volt max AC or DC limit. The Diode was added because when the Trip Coil was energized DI1 was destroyed.
Normally the "a" contacts will be closed, because the breaker will be closed. (all "b" contacts will be open)
DI1 is monitoring the health of the trip coil.
The 240V source is a variable auto-transformer. The Control Power Transformer is being simulated.
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No, you don't sum voltages like that.
If you are trying to get RMS (meaningful to account for power in a resistor for example) you do Sqrt(a+b) a and b could be DC and AC or different type of AC, different frequency sine for example. So in your case you have 384.2V. This assumes your AC measurement was RMS to start with.
If you are concerned about peak voltage you need to know the AC peak and sum it with the DC. AC could be asymmetrical, so you want the peak with the same sign of the DC. Let's say you have a pure sine and you have the RMS reading. In that case you do DC+AC*1.414, in your case the answer is 639.4V.
If you have a resistor across this voltage, it must be rated to withstand 639.4V and a power of 384.2^2/R. Other components ratings are different, for example, caps have a voltage rating for their dielectric rupture, this must be higher than the 639.4V, but they also must handle the current from the AC, that's a function of the voltage and the frequency. Inductors shouldn't have DC voltage across them at all, you might see some as they are a bit resistive but in theory not a µV.
JS
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I said a couple of incorrect things in one of my previous posts, which I've crossed out, rather than deleted, as people have since replied to them.
RMS is always the root mean square and is calculated the same, whether the signal has a net DC bias or not.
Meters often measure the AC component of the signal. They have a high pass filter which eliminates the DC offset, before the RMS calculation.
I forgot that you simply don't add the voltages together. Putting a 300VDC source in series with a 240VAC sine wave gives 384.2VRMS, not 540VRMS.
a = 240V
b = 300V
VRMS = (a2+b2)0.5 = (3002+2402)0.5 = (90000+57600)0.5 = 1476000.5 = 384.2VRMS
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Thanks to all who replied. I will probably be back after I get the mock up running. ( Maybe this weekend )
One other interesting thing cropped up after the diode was added. 600VDC was measured between X2-3 & X1-5 on the EDR5000. I think the diode connected to X1-5 is acting as a second power supply & is in series with the cap trip.
Again, I thank everybody for the tutorial. :-+
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Just coming back to this thread. There's one thing which I don't think has been explained: why the RMS voltage isn't simply the sum of the sinusoidal and DC components.
The reason is, because the AC voltage is bipolar, so the positive cycles add to the DC voltage, but the negative cycles cancel out part of the DC voltage. If the AC waveform was full wave rectified, then added to the DC, then the RMS voltage, would simply be the sum of the two voltages.
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Just coming back to this thread. There's one thing which I don't think has been explained: why the RMS voltage isn't simply the sum of the sinusoidal and DC components.
The reason is, because the AC voltage is bipolar, so the positive cycles add to the DC voltage, but the negative cycles cancel out part of the DC voltage. If the AC waveform was full wave rectified, then added to the DC, then the RMS voltage, would simply be the sum of the two voltages.
It is a simple sum of the sinusoidal and DC components, it's the algebraic sum of the sinusoidal and DC components. That's the magic word that people usually miss out. So it's not Adc + Aac, it's Adc + Aac sin(omega t).
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Just coming back to this thread. There's one thing which I don't think has been explained: why the RMS voltage isn't simply the sum of the sinusoidal and DC components.
The reason is, because the AC voltage is bipolar, so the positive cycles add to the DC voltage, but the negative cycles cancel out part of the DC voltage. If the AC waveform was full wave rectified, then added to the DC, then the RMS voltage, would simply be the sum of the two voltages.
It is a simple sum of the sinusoidal and DC components, it's the algebraic sum of the sinusoidal and DC components. That's the magic word that people usually miss out. So it's not Adc + Aac, it's Adc + Aac sin(omega t).
That is correct, but the confusing thing for a beginner is, even a true RMS meter doesn't display the components of the signal like that. Switch it to AC mode and you get the RMS value, of a band pass filtered signal, change it to DC and you get the average. It's important to note that simply algebraically adding the two displayed values together, doesn't give the total RMS value.
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An attempt to avoid misconceptions, confusion and wrong information:
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I can confirm the same result as JS above (384.1Vrms) by using a bit more involving equation based on what I had already on my cellphone calculator and the definition below:
http://mathworld.wolfram.com/Root-Mean-Square.html (http://mathworld.wolfram.com/Root-Mean-Square.html)
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Follow up question.
If the DC value is measured with a negative sign, does one use the absolute value of the measured DC or does it subtract from the AC RMS value ?
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Follow up question.
If the DC value is measured with a negative sign, does one use the absolute value of the measured DC or does it subtract from the AC RMS value ?
It doesn't matter. If you look at the formula, trying both positive and negative values, you'll see why.
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Follow up question.
If the DC value is measured with a negative sign, does one use the absolute value of the measured DC or does it subtract from the AC RMS value ?
It doesn't matter. If you look at the formula, trying both positive and negative values, you'll see why.
Yes. Given that RMS uses the quadratic (power of 2) of the voltage components, the negative sign will be cancelled.
BTW, the equation I am using on the calculator is:
\$V_{RMS} = \int_{0}^{freq} (V_{DC}+V_{AMP}*\sin(2*\pi*freq*x))^2dx\$
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Follow up question.
If the DC value is measured with a negative sign, does one use the absolute value of the measured DC or does it subtract from the AC RMS value ?
It doesn't matter. If you look at the formula, trying both positive and negative values, you'll see why.
Yes. Given that RMS uses the quadratic (power of 2) of the voltage components, the negative sign will be cancelled.
BTW, the equation I am using on the calculator is:
\$V_{RMS} = \int_{0}^{freq} (V_{DC}+V_{AMP}*\sin(2*\pi*freq*x))^2dx\$
Yes, the RMS voltage is always a positive number.
If all you have is the RMS value of the AC component of the signal and the DC value, then that formula can be simplified. You don't need to know the frequency or the time. It just becomes VRMS = (VAC2+VDC2)0.5
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So you can see the fruits of your labor, here is what I have done so far...
Please, none of this :-DD
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Hah, I love your CAD tool, but I am much more in love with your scopemeter. What a beauty!
(...)
If all you have is the RMS value of the AC component of the signal and the DC value, then that formula can be simplified. You don't need to know the frequency or the time. It just becomes VRMS = (VAC2+VDC2)0.5
Yep, frequency is not part of the RMS at all - I was carried away back then. Fancy tools at your disposal tend to do that to you. :-DD
I also have the RMS equations for triangular and square with DC levels, which are also useful to me.
(I love my 48GX! Both the physical and the soft versions)
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Here are my CAD tools. The only thing missing is the eraser dust bag! :-DD I was one of the last drafters Westinghouse had in medium voltage switchgear before it began moving to CAD. I moved back to the Test department.
Notice the different size erasers for different size screw ups. Culminating on the right with the electric eraser for screw ups of Biblical Proportions.
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Here are my CAD tools. The only thing missing is the eraser dust bag! :-DD I was one of the last drafters Westinghouse had in medium voltage switchgear before it began moving to CAD. I moved back to the Test department.
Notice the different size erasers for different size screw ups. Culminating on the right with the electric eraser for screw ups of Biblical Proportions.
Hehehe... I have never heard about an electric eraser before. You americans and your super cool toys.
I used to have one of these (http://www.oprojetista.com.br/produto/221_Gabarito-diagramas-eletronicos-TRIDENT-E-25.html) when I wanted to be fancy with my electronics drawing. It worked really well. In university years I started using one of these (https://www.oprojetista.com.br/busca/subsubcat_5_13_14_Materiais-para-Desenho-Reguas-Normografos.html) as my writing was subpar, but I then quickly improved it as it was too slow to write longer texts with it.
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I seriously doubt that only a few people here would know about an electric eraser unless they were a pre-CAD drafter.
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Today I learned about electric erasers.
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Lets go one more step. You would think the electric eraser would be rather blunt. It is quite capable of very fine touch ups. Just as good or better than the manual erasers. It can even erase the fused ink that a copier lays on paper. One can erase a hole through the paper in a heartbeat too ! |O Works best on vellum, bond paper needs a light touch.
For very small areas one uses an erasing shield. It is the silver template looking thingy next to the blue drafting pencil. Place the appropriate size hole over the mistake & it protects the surrounding area from the eraser, manual or electric.