Ocsilliscope - Doing something wrong?

[mikefromsac]

I'm a newbie and learning to use an oscilloscope. When hooked up to a function generator (DC), all looks perfect, but when I hooked up to two different toy-train transformers (both AC), the waves read high (see attachment). This image was with the transformer set at 8 volts (according to the voltmeter on the transformer and confirmed by a stand-alone DMM). Why is the scope showing around 11.2 volts? Same thing happens with the other toy-train transformer. Do I have something set incorrectly?

[nugglix]

Hi!

You just discovered the difference between peak-to-peak and rms voltage.
The 8V are in the measurements, look at VrmsP.

For a start you might want to look here:
https://en.wikipedia.org/wiki/Root_mean_square

Cheers

[Joule Thief]

Consider instruments which read AC voltages in RMS versus peak values. Do you think this difference may apply to your situation?

[alsetalokin4017]

Mike, you may also want to review the issue of "AC vs DC input coupling".

In this case since you are measuring the unloaded output of a simple transformer, it's not going to make much difference whether you use AC or DC coupled input, but in general one should use DC-coupled input unless there is some specific reason to use AC coupling. Just because a signal might technically be "AC" this does not mean that AC-coupling is appropriate for measuring it! The AC-coupling feature puts a capacitor _in series_ with your probe's input to the oscilloscope's front-end; this has the effect of removing DC offset from the signal and moving the _average_ of the signal down (or up) to the channel baseline. Sometimes you want this... and sometimes it doesn't matter, like in the present case. But whenever one uses AC-coupled inputs on a scope there should be some clear understanding of the whys and wherefores of doing so.

I think that this issue of AC vs. DC coupled inputs is perhaps the most often misunderstood feature of oscilloscopes, whether digital or analog. "AC coupling" isn't just for measuring "AC" signals! If the control were labelled "Direct Coupling" and "Capacitor Coupling" some of the confusion might be avoided, but we are stuck with what we've got. Probably the most common appropriate use of AC-coupled inputs is for removing a large DC offset in order to be able to resolve a smaller AC ripple sitting on top of the DC offset.

Again, in this case it's not going to make much difference if any. It's just a particular bugbear of mine, since I often see beginners (and even some experienced scoposcopists) try to use AC-coupled signals in power computations, completely masking any DC contribution to the measured power that might be there.

[marinedalek]

I'm curious as to why your model train controller is outputting 60 Hz AC. Usually they will produce either continuously-variable DC (or at least recified AC), or ~10 kHz square wave AC for DCC systems. 12V RMS AC at 50 or 60 Hz is often provided for peripherals such as lighting, point motors etc., but that would be around +- 16V peak voltage. Just nosy!

[John_ITIC]

Hi!

You just discovered the difference between peak-to-peak and rms voltage.
The 8V are in the measurements, look at VrmsP.

For a start you might want to look here:
https://en.wikipedia.org/wiki/Root_mean_square

Cheers

What the Wikipedia article explains, is that the RMS ("Root Mean Square") voltage is a scaling of the peak-peak voltage. This is done such that the "usable" power calculations into resistive loads can directly be comparable with DC voltages. Since your sine wave is only at its peak voltage part of the time (part of the time even zero) then the voltage has to be "scaled down" by a factor 1/sqrt(2), i.e. Vrms = Vpp/sqrt(2), to have the same total power as the same load used with a DC voltage source.

Where does the 1/sqrt(2) come from? When looking closer at one cycle of your sine wave, it can be seen that the "average" voltage is the same as the area between the x-axis and the sine wave's voltage over one period. This is the same as the integral over one cycle time. This is done in the Wikipedia article for the current (I) but can just as well directly be done for the voltage since resistive load (current and voltage in phase). When one does the math, the scaling factor comes out to 1/sqrt(2).

[mikefromsac]

Wow, you guys are awesome. More learning to do ... Thanks so much!

[SaabFAN]

I'm curious as to why your model train controller is outputting 60 Hz AC. Usually they will produce either continuously-variable DC (or at least recified AC), or ~10 kHz square wave AC for DCC systems. 12V RMS AC at 50 or 60 Hz is often provided for peripherals such as lighting, point motors etc., but that would be around +- 16V peak voltage. Just nosy!

Its most likely an AC-System like the one used by the german Model Train-Company Märklin. They use either 50Hz Variable AC Voltage on the rails, or constant 16V with control-signals for the Decoders inside the locomotives on a higher carrier-wave that is sitting on top of the 50Hz AC (Makes a distinct grasshopper-like sound when a train derails and a short-circuit occurs^^).

About Vrms: A 50% DutyCycle Square-Wave should put out 1/2*Vpp, right? I'm asking because my Uni-T Multimeter shows 0.7Vrms at the cal-output of my PM3320A Oscilloscope when it is set right on spec (1Vpp).

[mikefromsac]

I'm curious as to why your model train controller is outputting 60 Hz AC. Usually they will produce either continuously-variable DC (or at least recified AC), or ~10 kHz square wave AC for DCC systems. 12V RMS AC at 50 or 60 Hz is often provided for peripherals such as lighting, point motors etc., but that would be around +- 16V peak voltage. Just nosy!

They're both for AC toy trains. One is an MTH Z-4000 and the one from the screen-shot is a 1950's era Lionel Type LW.

[mikefromsac]

Consider instruments which read AC voltages in RMS versus peak values. Do you think this difference may apply to your situation?

I don't have a situation or need. I'm, just a newbie trying to lean how to use, read, and interpret an oscilloscope.

[mikefromsac]

Mike, you may also want to review the issue of "AC vs DC input coupling".

In this case since you are measuring the unloaded output of a simple transformer, it's not going to make much difference whether you use AC or DC coupled input, but in general one should use DC-coupled input unless there is some specific reason to use AC coupling. Just because a signal might technically be "AC" this does not mean that AC-coupling is appropriate for measuring it! The AC-coupling feature puts a capacitor _in series_ with your probe's input to the oscilloscope's front-end; this has the effect of removing DC offset from the signal and moving the _average_ of the signal down (or up) to the channel baseline. Sometimes you want this... and sometimes it doesn't matter, like in the present case. But whenever one uses AC-coupled inputs on a scope there should be some clear understanding of the whys and wherefores of doing so.

I think that this issue of AC vs. DC coupled inputs is perhaps the most often misunderstood feature of oscilloscopes, whether digital or analog. "AC coupling" isn't just for measuring "AC" signals! If the control were labelled "Direct Coupling" and "Capacitor Coupling" some of the confusion might be avoided, but we are stuck with what we've got. Probably the most common appropriate use of AC-coupled inputs is for removing a large DC offset in order to be able to resolve a smaller AC ripple sitting on top of the DC offset.

Again, in this case it's not going to make much difference if any. It's just a particular bugbear of mine, since I often see beginners (and even some experienced scoposcopists) try to use AC-coupled signals in power computations, completely masking any DC contribution to the measured power that might be there.

Thanks for this explanation. I choose AC Coupling for exact reasons you stated, I assumed this was for measuring AC. Another thing learned ...

[mikefromsac]

Hi!

You just discovered the difference between peak-to-peak and rms voltage.
The 8V are in the measurements, look at VrmsP.

For a start you might want to look here:
https://en.wikipedia.org/wiki/Root_mean_square

Cheers

What the Wikipedia article explains, is that the RMS ("Root Mean Square") voltage is a scaling of the peak-peak voltage. This is done such that the "usable" power calculations into resistive loads can directly be comparable with DC voltages. Since your sine wave is only at its peak voltage part of the time (part of the time even zero) then the voltage has to be "scaled down" by a factor 1/sqrt(2), i.e. Vrms = Vpp/sqrt(2), to have the same total power as the same load used with a DC voltage source.

Where does the 1/sqrt(2) come from? When looking closer at one cycle of your sine wave, it can be seen that the "average" voltage is the same as the area between the x-axis and the sine wave's voltage over one period. This is the same as the integral over one cycle time. This is done in the Wikipedia article for the current (I) but can just as well directly be done for the voltage since resistive load (current and voltage in phase). When one does the math, the scaling factor comes out to 1/sqrt(2).

Thanks for the explanation as the Wiki article was a bit over my head at the moment. As you described, I calculated using 1/sqrt(2) and it worked (as you knew it would) and was 8 volts. Since the scope can do the math for me, in the future can I just look at the VrmsP measurement and assume this is the voltage output or is there more to it than this? I'm not opening a particle accelerator lab, just trying to learn general electronics and maybe restore a few old AM radios ...

[alsetalokin4017]

My advice, fwiw, is to rely as little as possible, at first anyway,  on the "numbers in boxes" that your scope spits out and try to learn how to interpret the actual waveform trace on the screen, using the graticule markers and timebase settings and the waveshape. That is, force yourself to use the scope as if it were an _analog_ scope until you have the basics of waveform interpretation down pat. One should always view the DSO's "numbers in boxes" measurements with a grain of salt and try to confirm them as much as possible by the actual visual interpretation of the waveforms.

Also, many DSOs will only use the portion of the waveform that is actually displayed on the screen when giving you the "measurements". So, for some signals, if you are looking at the waveform in some horizontal timescales, you may see very different "numbers in boxes" than if you use other scales on the same signal.

[Electro Fan]

Someone here might say this is an unwise shortcut, but I think for a sine wave you can multiply the peak by .707 and get the approximate RMS value.  Sometimes it can be handy for quickly getting your bearings.  (Or RMS multiplied by 1.414 gives the approximate peak value.)

[obiwanjacobi]

Someone here might say this is an unwise shortcut, but I think for a sine wave you can multiply the peak by .707 and get the approximate RMS value.  Sometimes it can be handy for quickly getting your bearings.  (Or RMS multiplied by 1.414 gives the approximate peak value.)

+1 Thats how I always do it 

[mikefromsac]

Someone here might say this is an unwise shortcut, but I think for a sine wave you can multiply the peak by .707 and get the approximate RMS value.  Sometimes it can be handy for quickly getting your bearings.  (Or RMS multiplied by 1.414 gives the approximate peak value.)

That makes sense if it's a constant for sine waves. Does it change with different wave types?. So why would it be bad practice?

[Len]

That makes sense if it's a constant for sine waves. Does it change with different wave types?. So why would it be bad practice?

Yes, it's different for different waveforms. The wikipedia article linked previously has a table of formulas for various waveforms. The .707 rule of thumb is fine as long as you know that you're dealing with a sine wave.

[tggzzz]

That makes sense if it's a constant for sine waves. Does it change with different wave types?. So why would it be bad practice?

Yes, it's different for different waveforms. The wikipedia article linked previously has a table of formulas for various waveforms. The .707 rule of thumb is fine as long as you know that you're dealing with a sine wave.

And you'll note the waveform shown on the scope isn't a sine wave: it is a slightly truncated sine wave. I've seen the same thing in my house and at the local hackspace, and it isn't my imagination: an FFT does indeed show many harmonics.

So, what causes that? Certainly it wasn't there when I previously did the experiment in the 70s. My best guess is that nowadays there are far more small AC/DC converters that mainly charge up the PSU capacitors at peak AC voltage - and that the peak current draw causes the sinewave peak to "droop".

I'm far from convinced by that explanation, and would welcome other suggestions.

[Zero999]

That makes sense if it's a constant for sine waves. Does it change with different wave types?. So why would it be bad practice?

Yes, it's different for different waveforms. The wikipedia article linked previously has a table of formulas for various waveforms. The .707 rule of thumb is fine as long as you know that you're dealing with a sine wave.

And you'll note the waveform shown on the scope isn't a sine wave: it is a slightly truncated sine wave. I've seen the same thing in my house and at the local hackspace, and it isn't my imagination: an FFT does indeed show many harmonics.

So, what causes that? Certainly it wasn't there when I previously did the experiment in the 70s. My best guess is that nowadays there are far more small AC/DC converters that mainly charge up the PSU capacitors at peak AC voltage - and that the peak current draw causes the sinewave peak to "droop".

I'm far from convinced by that explanation, and would welcome other suggestions.

I believe you're correct. Non-linear loads such as rectifiers draw the peak current at the high point of the sine wave, resulting in harmonic distortion.

Unfortunately too much harmonic distortion can increase heating in transformers and motors. The neutral current can also be higher than normal as the currents from x3 harmonics combine in the neutral conductor.

[tautech]

That makes sense if it's a constant for sine waves. Does it change with different wave types?. So why would it be bad practice?

Yes, it's different for different waveforms. The wikipedia article linked previously has a table of formulas for various waveforms. The .707 rule of thumb is fine as long as you know that you're dealing with a sine wave.

And you'll note the waveform shown on the scope isn't a sine wave: it is a slightly truncated sine wave. I've seen the same thing in my house and at the local hackspace, and it isn't my imagination: an FFT does indeed show many harmonics.

So, what causes that? Certainly it wasn't there when I previously did the experiment in the 70s. My best guess is that nowadays there are far more small AC/DC converters that mainly charge up the PSU capacitors at peak AC voltage - and that the peak current draw causes the sinewave peak to "droop".

I'm far from convinced by that explanation, and would welcome other suggestions.

It'a interesting you make this assumption based on the OP's timebase setting, I'd want to see it expanded somewhat before I reach a similar conclusion.

But you could be quite correct.
As for mains waveforms being not clean sinewaves, this is not uncommon, especially if the connection is some distance from the Elco's supply transformer.
It is predominantly caused by the prevelance of SMPS in many of our appliances.

But no need for ya'll to rush out and connect your scopes to mains to check  just take my word for it. 

[tggzzz]

That makes sense if it's a constant for sine waves. Does it change with different wave types?. So why would it be bad practice?

Yes, it's different for different waveforms. The wikipedia article linked previously has a table of formulas for various waveforms. The .707 rule of thumb is fine as long as you know that you're dealing with a sine wave.

And you'll note the waveform shown on the scope isn't a sine wave: it is a slightly truncated sine wave. I've seen the same thing in my house and at the local hackspace, and it isn't my imagination: an FFT does indeed show many harmonics.

So, what causes that? Certainly it wasn't there when I previously did the experiment in the 70s. My best guess is that nowadays there are far more small AC/DC converters that mainly charge up the PSU capacitors at peak AC voltage - and that the peak current draw causes the sinewave peak to "droop".

I'm far from convinced by that explanation, and would welcome other suggestions.

It'a interesting you make this assumption based on the OP's timebase setting, I'd want to see it expanded somewhat before I reach a similar conclusion.

Yes, but I'd make a small wager I'm right. I certainly wouldn't have jumped to that conclusion without having seen - and been puzzled by it - myself.

But you could be quite correct. As for mains waveforms being not clean sinewaves, this is not uncommon, especially if the connection is some distance from the Elco's supply transformer.
It is predominantly caused by the prevelance of SMPS in many of our appliances.
But no need for ya'll to rush out and connect your scopes to mains to check  just take my word for it. 

I have no direct evidence, but I don't believe what I saw is simply a distance-from-transformer issue.

[tautech]

That makes sense if it's a constant for sine waves. Does it change with different wave types?. So why would it be bad practice?

Yes, it's different for different waveforms. The wikipedia article linked previously has a table of formulas for various waveforms. The .707 rule of thumb is fine as long as you know that you're dealing with a sine wave.

And you'll note the waveform shown on the scope isn't a sine wave: it is a slightly truncated sine wave. I've seen the same thing in my house and at the local hackspace, and it isn't my imagination: an FFT does indeed show many harmonics.

So, what causes that? Certainly it wasn't there when I previously did the experiment in the 70s. My best guess is that nowadays there are far more small AC/DC converters that mainly charge up the PSU capacitors at peak AC voltage - and that the peak current draw causes the sinewave peak to "droop".

I'm far from convinced by that explanation, and would welcome other suggestions.

It'a interesting you make this assumption based on the OP's timebase setting, I'd want to see it expanded somewhat before I reach a similar conclusion.

Yes, but I'd make a small wager I'm right. I certainly wouldn't have jumped to that conclusion without having seen - and been puzzled by it - myself.

But you could be quite correct. As for mains waveforms being not clean sinewaves, this is not uncommon, especially if the connection is some distance from the Elco's supply transformer.
It is predominantly caused by the prevelance of SMPS in many of our appliances.
But no need for ya'll to rush out and connect your scopes to mains to check  just take my word for it. 

I have no direct evidence, but I don't believe what I saw is simply a distance-from-transformer issue.

You should, it is because of the higher impedence of the supply as a result of the distance from the mains grid (low impedence).
As Hero999 points out non-linear loads are the cause, often from SMPS.

A Uni Dr EE mate had severe problems with mains sinewave purity and it was traced to classrooms full of computer suites.....all powered by SMPS.

[tggzzz]

That makes sense if it's a constant for sine waves. Does it change with different wave types?. So why would it be bad practice?

Yes, it's different for different waveforms. The wikipedia article linked previously has a table of formulas for various waveforms. The .707 rule of thumb is fine as long as you know that you're dealing with a sine wave.

And you'll note the waveform shown on the scope isn't a sine wave: it is a slightly truncated sine wave. I've seen the same thing in my house and at the local hackspace, and it isn't my imagination: an FFT does indeed show many harmonics.

So, what causes that? Certainly it wasn't there when I previously did the experiment in the 70s. My best guess is that nowadays there are far more small AC/DC converters that mainly charge up the PSU capacitors at peak AC voltage - and that the peak current draw causes the sinewave peak to "droop".

I'm far from convinced by that explanation, and would welcome other suggestions.

It'a interesting you make this assumption based on the OP's timebase setting, I'd want to see it expanded somewhat before I reach a similar conclusion.

Yes, but I'd make a small wager I'm right. I certainly wouldn't have jumped to that conclusion without having seen - and been puzzled by it - myself.

But you could be quite correct. As for mains waveforms being not clean sinewaves, this is not uncommon, especially if the connection is some distance from the Elco's supply transformer.
It is predominantly caused by the prevelance of SMPS in many of our appliances.
But no need for ya'll to rush out and connect your scopes to mains to check  just take my word for it. 

I have no direct evidence, but I don't believe what I saw is simply a distance-from-transformer issue.

You should, it is because of the higher impedence of the supply as a result of the distance from the mains grid (low impedence).

Why does increased impedance on its own cause distortion? Impedance is linear, we are looking at a nonlinear effect.

As Hero999 points out non-linear loads are the cause, often from SMPS.
A Uni Dr EE mate had severe problems with mains sinewave purity and it was traced to classrooms full of computer suites.....all powered by SMPS.

I don't have any problem with that; it matches my presumptions

[Zero999]

Why does increased impedance on its own cause distortion? Impedance is linear, we are looking at a nonlinear effect.

Increased impedance on its own will not cause any distortion. Non-linear loads + high impedance = distortion. On the other hand, if the impedance is extremely low, there will be hardly any distortion, even in the presence of non-linear loads.

[tggzzz]

Why does increased impedance on its own cause distortion? Impedance is linear, we are looking at a nonlinear effect.

Increased impedance on its own will not cause any distortion. Non-linear loads + high impedance = distortion. On the other hand, if the impedance is extremely low, there will be hardly any distortion, even in the presence of non-linear loads.

Precisely. It is the electronic PSU, SMPS or traditional, that is probably the cause.

There is no reason to believe that the two locations in which I observed the same waveform have abnormal impedance. Indeed, the hackspace is in an light industrial "artists colony" in the middle of a city, and is likely to have a slighly lower than normal impedance. I made that measurement at night, so there was unlikely to be significant load from elsewhere in the unit.