Proper Usage of Oscilloscope (40 Marks)

[Ferrock]

Hey guys!

I have a bit of work I need to do, one of the questions is as follows (40 marks).

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Figure 1 below shows a two channel analog oscilloscope, which has a maximum input voltage of 10 volts. You are required to use the oscilloscope to measure:

i. a dc voltage which is approximately 4 volts in value.
ii. a signal comprising of a dc part (~ 4 volts) and an ac part (~ 100 mV peak to peak and 50kHz).

Describe in detail the proper measurement procedure to obtain accurate measurements of the two signals.
You should pay particular attention to the settings of the input, vertical and horizontal controls, and the timebase.

Link to figure 1: http://i59.tinypic.com/2mockdj.png
-----------------------------------------

I am not very good with oscilloscopes so far, but am keen to improve my understanding. If you could help me answer the question but also explain the 'why' behind your comments, that would be great - otherwise I'll never learn

Another thing that is throwing me off is that it is worth 40 marks, this seems a lot for just adjusting some controls so I am assuming there is more to it that I haven't heard of yet.

For the 4 volts DC signal:

You would need to first attach one channel (CH2) to ground and the other to the 4V supply voltage.
You would then set the volt/divisions to 1V (As this means each square will correspond to 1V, so you can easily count the amount of squares to measure the voltage)
The timebase would be 0.1mS?

I am not really sure on what else you would need to do but I NEED to expand on this as its worth so many marks.

For the ac+dv voltage, I don't really have much to go on yet, but hopefully you'll be able to help!

Thanks in advance

 

[davenn]

Hi Ferrock
welcome to EP

OK being a homework/study question
we are not going to give you the answers outright. but will help you through to a result

For the 4 volts DC signal:

You would need to first attach one channel (CH2) to ground and the other to the 4V supply voltage.
You would then set the volt/divisions to 1V (As this means each square will correspond to 1V, so you can easily count the amount of squares to measure the voltage)
The timebase would be 0.1mS?

That was sort of implying that you were going to use both channels ?

No, just use one channel

The timebase speed isn't too critical, what about the vertical amplitude setting ?

have a look at this 2 channel scope



see you have 2 vertical inputs CH1 VOLTS/DIV and CH2 VOLTS/DIV
so what are you going to set CH1 to and how are you going to connect the CH1 probe wires to the 4V supply ?

Dave

 

[Ferrock]

Thank you Dave!

Yeah thats perfect. Theres no point me just being given an answer anyway - I wouldn't learn anything!

Yeah sorry, we're only measuring one signal so would only need one channel.

CH1:
Would you set the volts/div to 1 volts/div? Since there are 8 divisions, that would make it easy to measure as you can count divisions. I did think 0.5 per division but that would be off the screen unless the vertical position was right at the bottom. You could also set the vertical position at 0 to 2 lines from the bottom. That way the 4 volt signal should be 2 from the top - so the whole thing is centered and clear.

This is all assuming the dc signal is connected positive to negative. If it was negative to positive and our vertical position was at line 2 (from the bottom) then our measurement would be 2 divisions off the bottom of the screen.
Should I therefore have the vertical position centered, and the volts/div greater than 1? Say 2, so that whichever way they connect the dc supply - it will get a reading either above or below?

As for CH2 - you wouldn't need to change this as it is not in use?

As for the time base, should I just say set it so any time as it is not important? Or maybe just 1Second as a general time?

To connect it to the supply, you'd connect one of CH1s wires (usually red) to the supply and the other (usually black) to ground?

Thanks!

 

[davenn]

CH1:
Would you set the volts/div to 1 volts/div? Since there are 8 divisions, that would make it easy to measure as you can count divisions. I did think 0.5 per division but that would be off the screen unless the vertical position was right at the bottom. You could also set the vertical position at 0 to 2 lines from the bottom. That way the 4 volt signal should be 2 from the top - so the whole thing is centered and clear.

Its always good to keep the line in the middle, cuz you never know if you may put the scope probe onto a negative or positive rail

This is all assuming the dc signal is connected positive to negative. If it was negative to positive and our vertical position was at line 2 (from the bottom) then our measurement would be 2 divisions off the bottom of the screen.
Should I therefore have the vertical position centered, and the volts/div greater than 1? Say 2, so that whichever way they connect the dc supply - it will get a reading either above or below?

Its even better to start with the V/Div set to be able to read a voltage much more than you think you will measure. You can always adj the V/Div down as required

As for CH2 - you wouldn't need to change this as it is not in use?

correct

As for the time base, should I just say set it so any time as it is not important? Or maybe just 1Second as a general time?

At 1 sec per division, its going to take 10 secs for the trace to sweep across the screen. I would set it at 0.1 sec/Div or a little faster

To connect it to the supply, you'd connect one of CH1s wires (usually red) to the supply and the other (usually black) to ground?

if it has a standard probe, you will see the centre end is a little hook that can extend out for hooking onto a wire
there is also a fly lead off the probe with a "crocodile" clip on it .... that will go to the negative/ground line of the battery/circuit

Dave

 

[Ferrock]

Yeah true, can't assume the user will know which way round to connect it. I can always say that they can adjust the setting once connected if its the right way around.

And ok - so I should say set the volts/div to something like 2 (with the vertical position centered), connect the signal - and once they have confirmed the reading is vaguely what they expect, lower the volts/div to get a more and more accurate reading?

Ahh okay, so the higher the volts/div the long it takes to produce a trace? So yeah, 0.1/sec or smaller is better as it allows a quick result and as it is DC - it doesn't really matter as the signal is not varying.

Those are the leads we had in our labs - with the hook. So hook on supply, then the other clip/hook to negative/ground

I think that is probably good enough for DC? As I can't really think of much else I would need to mention! I'm thinking that most of the 40 marks will be to do with the ac/dc combination signal, of which I don't really know where to start.
" a signal comprising of a dc part (~ 4 volts) and an ac part (~ 100 mV peak to peak and 50kHz)."

I would still only need to use one channel as its one signal?
The 100mV peak to peak would impose itself onto the DC signal giving it a slightly sinusoidal form so would the overall signal be 4v + 100mV in either direction? So a total of 4.2V?
The 50KHz.. Thats 1/50,000, so would need a very small time/div. 1 micro second is 1/100,000 so would that be a good time/div as it would allow.. 5 full wavelengths?

Sorry if I'm barking up the wrong tree with those thoughts. I don't feel right asking for help if I haven't had a think about it first myself!

Thanks again!

 

[davenn]

And ok - so I should say set the volts/div to something like 2 (with the vertical position centred), connect the signal - and once they have confirmed the reading is vaguely what they expect, lower the volts/div to get a more and more accurate reading?

yup, 2 or 5 etc ( this is more important if your are going in to test a circuit with unknown voltages)

Ahh okay, so the higher the volts/div the long it takes to produce a trace? So yeah, 0.1/sec or smaller is better as it allows a quick result and as it is DC - it doesn't really matter as the signal is not varying.

not for the timebase ( the horizontal sweep) its time/division seen the right centre large knob ... SEC/DIV

Those are the leads we had in our labs - with the hook. So hook on supply, then the other clip/hook to negative/ground

OK so presumably the leads have 2 crocodile clips and presumably they are red and black ?

I would still only need to use one channel as its one signal?

Yes

The 100mV peak to peak would impose itself onto the DC signal giving it a slightly sinusoidal form so would the overall signal be 4v + 100mV in either direction? So a total of 4.2V?

yes ... basically an AC signal with a +4V offset

Sorry if I'm barking up the wrong tree with those thoughts. I don't feel right asking for help if I haven't had a think about it first myself!

Thanks again!

no probs .... best learning comes from using the gear so every chance you get just play, play, play

good starting point is just clipping the probe lead onto the scope test calibration terminal ... on the scope in my pic its between the upper left 2 knobs ... see it ?

its usually a square waveform at a specific voltage and frequency

use that and play with the settings and see what effect the setting changes make to the displayed waveform

cheers
Dave

 

[KrisBlueNZ]

Ahh okay, so the higher the volts/div the long it takes to produce a trace? So yeah, 0.1/sec or smaller is better as it allows a quick result and as it is DC - it doesn't really matter as the signal is not varying.

As Dave pointed out, it's the time/div control that you're talking about here, not volts/div.

" a signal comprising of a dc part (~ 4 volts) and an ac part (~ 100 mV peak to peak and 50kHz)."

The 100mV peak to peak would impose itself onto the DC signal giving it a slightly sinusoidal form so would the overall signal be 4v + 100mV in either direction? So a total of 4.2V?

Not quite. The AC signal has an amplitude of 100 mV peak to peak. This means there is 100 mV between the highest peaks and the lowest valleys, and if it's a sinewave, the mean voltage is half way between those points. So the wave varies from 50 mV above the mean level to 50 mV below the mean level. When you add a constant DC offset of 4V, the mean votage of the sinewave is shifted up by +4V, so the resulting signal will range between (4V + 0.05V) = 4.05V and (4V - 0.05V) = 3.95V. And "a total of 4.2V" doesn't mean anything.

The 50KHz.. Thats 1/50,000, so would need a very small time/div. 1 micro second is 1/100,000

No, 1 µs is 1/1,000,000 second. If you choose a timebase setting of 1 µs per division, and there are ten divisions on the screen, the whole screen width corresponds to 10 µs, which is 1/100,000th of a second. Is that what you meant?

so would that be a good time/div as it would allow.. 5 full wavelengths?

You need to do these calculations more carefully. Do them step by step.

1. What is the duration of one cycle of a 50 kHz waveform, in microseconds?

2. If you want five cycles on the screen (which is reasonable), what duration do you need for the whole screen width?

3. How many divisions are there horizontally on the screen?

4. What time duration per division do you need to choose?

5. Double-check your calculations by working out how many divisions each cycle of the waveform will occupy.

Sorry if I'm barking up the wrong tree with those thoughts. I don't feel right asking for help if I haven't had a think about it first myself!

You're getting the idea. You just need to be careful with your quantities and units and calculations.

 

[Ferrock]

How do I go about quoting parts of the post? I can't seem to find an option/button where they'd normally be. Would be easier for you than trying to remember what I'm referring to each line!

Oops that was a typo on my part, the higher the sec/div the longer it takes the sweep.

Yeah we have the same leads/probes as you mentioned. One clip with a hook (red) and then one 'extra' (fly?) lead with a clip/hook for ground. The hooks are detachable.

Ah of course, so the 100mV AC signal would be shifted up by 4V (+4V DC offset). So in total it would only actually reach 4.05V? As it will be 50mV either side of the 4V mark as its 100mV peak to peak. So my 4.2V total was wrong.

Am I correct for my time/div estimate? (1 micro second/div).

I don't actually have access to an oscilloscope anymore.. this is an exercise we're doing after the lab (which was just reading a sine wave so.. didn't really get to try much), so we have to take what we've learnt and research it more - I guess so that we can grow our understand of oscilloscopes operation in general before the next lab, probably why it carries so many marks. Otherwise I'd definitely just have a play around with it - would be so much easier! I find I learn this kind of thing better by just doing it too, but... ah well, sadly can't!

Thanks

edit 1: Only just saw your post Kris, will edit this again once I've read it

edit 2:
Okay, I noticed my typo mistake on the time and my random rubbish calculation regarding the offset which was good. 4V±0.05 makes perfect sense.

As for the time/div, I know what I did wrong - I mistakenly look micro (10-6) to include the 1, so 1/100,000.. when it should have been 1/1,000,000 like you said.

1 full cycle of a 50kHz wave would be 20uS (1/50,000 divided by 1/1,000,000).

5 full cycles would therefore be 100uS

There are 10 divisions, so 100/10 = 10uS per division.

Hopefully that sounds more reasonable

Thanks!

 

[KrisBlueNZ]

To quote a post, click Reply at the bottom right corner of the post you want to reply to. The whole original post will be quoted, with [QUOTE] and [/QUOTE] around it.

Now if you want to reply to the quoted message in sections, delete parts of it that you don't want to quote, and when you find a place where you want to put a reply, type [/QUOTE] at the end of the quoted text, then type your reply, then type [QUOTE] before the continuation of the quoted text.

So each block of text from the original message has [QUOTE] and [/QUOTE] around it. Each block will appear in a separate quote box, and your replies will appear in between them.

Here's an example.

----- start of example
Here's a section I've quoted from your post.

Am I correct for my time/div estimate? (1 micro second/div).

No, that's wrong.
----- end of example

And here's what I put in my message to create the section between "----- start of example" and "----- end of example":

Here's a section I've quoted from your post.
[QUOTE]Am I correct for my time/div estimate? (1 micro second/div).[/QUOTE]
No, that's wrong.

 

[KrisBlueNZ]

There are 10 divisions, so 100/10 = 10uS per division.

That's right. And at 10 µs per division, a cycle (20 µs) will occupy two divisions, so there will be five cycles displayed on the whole screen.

BTW, the SI unit for seconds is a lower case 's', not a capital 'S'. You will see people writing "µS", "mS" etc all over the place - even experienced people, and I've even seen it on the display of an oscilloscope! But it's wrong. The capital 'S' is actually the unit of conductance, the siemens. Just to help you avoid embarrassment!

 

[Ferrock]

Ah ok thanks I was looking on the standard reply screen down the bottom for some sort of quote option but couldn't find any.

BTW, the SI unit for seconds is a lower case 's', not a capital 'S'. You will see people writing "µS", "mS" etc all over the place - even experienced people, and I've even seen it on the display of an oscilloscope! But it's wrong. The capital 'S' is actually the unit of conductance, the siemens. Just to help you avoid embarrassment!

- thank you, I often see it written both ways so have never been sure which is right or if it was just a case of whatever you prefer to write. I will stick to lower case from now on

The question asks for an accurate measurement of the signals. For AC - does that mean measuring the dc and ac components separately?

You can measure the frequency of the AC component easily by adjusting the timebase until you get one complete cycle on screen, and then work from there. Or 5/10 full cycles, not sure which would be better - depends on the accuracy of the timebase controls I suppose.
Measuring the DC, it would be from the vertical position to the center of the sinusoidal waveform. (How to make sure you have the correct center? change the volts/div until the waveform occupies 2 divisions then use the center line? Although that might send the vertical position baseline off screen)
Measuring the AC, would be from the bottom of the waveform to the top - so, counting the divisions between the two. I'd say measure the DC first, and then lower the vertical position and adjust the volts/div until the waveform takes up most of the screen and then count the divisions between top and bottom for a more accurate answer?

Not sure if this is possible with just an oscilloscope.. but can you set it to measure just AC or DC? If you changed the input to AC, would it only show the AC component? Same with DC? The oscilloscope Dave showed has options for AC, GND and DC. The one shown in the question though just has AC/DC (combined) and GND.

Does that mean the one Dave showed would display the components separately but the one shown in the question will show them combined so I would need to measure them in a way like I suggested?

Thanks

 

[davenn]

Now if you want to reply to the quoted message in sections, delete parts of it that you don't want to quote, and when you find a place where you want to put a reply, type /QUOTE in brackets at the end of the quoted text, then type your reply, then type

and there is an even quicker and easier way that doesn't require all that tedious deleting of unwanted text

just scroll mouse across the section of text to highlight it then click on the little reply tag at the end of the highlighted section

 

[davenn]

Not sure if this is possible with just an oscilloscope.. but can you set it to measure just AC or DC? If you changed the input to AC, would it only show the AC component? Same with DC? The oscilloscope Dave showed has options for AC, GND and DC. The one shown in the question though just has AC/DC (combined) and GND.

Does that mean the one Dave showed would display the components separately but the one shown in the question will show them combined so I would need to measure them in a way like I suggested?

you are going to see the combined voltages if the scope control is set to DC you will only see the AC voltage if the scope is set to AC, You will see nothing is the control is set to GND

 

[Ferrock]

Ah ok. So for the one pictured in my question - I will see both as that is the only option, so will have to separate them out via the methods I mentioned(?). Yet if it had a separate option for DC and AC like the one you linked, I could use them both to make measuring the separate components easier?

Is it ok for me to post my completed answer to the question on here for feedback? I am just about to write up what I can

 

[Neal]

Ah ok. So for the one pictured in my question - I will see both as that is the only option, so will have to separate them out via the methods I mentioned(?). Yet if it had a separate option for DC and AC like the one you linked, I could use them both to make measuring the separate components easier?

Is it ok for me to post my completed answer to the question on here for feedback? I am just about to write up what I can

The scope you have in your picture can switch between AC or DC, the switch is a push button switch. I would use one channel for the DC level and the other channel to display the AC component.

 

[KrisBlueNZ]

The question asks for an accurate measurement of the signals. For AC - does that mean measuring the dc and ac components separately?

Yes, I would say so.

You can measure the frequency of the AC component easily by adjusting the timebase until you get one complete cycle on screen, and then work from there. Or 5/10 full cycles, not sure which would be better - depends on the accuracy of the timebase controls I suppose.

Yes, either way.

Measuring the DC, it would be from the vertical position to the center of the sinusoidal waveform. (How to make sure you have the correct center? change the volts/div until the waveform occupies 2 divisions then use the center line? Although that might send the vertical position baseline off screen)

Yes you could try that. I would increase the time/div until the AC becomes a big thick blur, and estimate the centre of the blur.

Edit: This is only valid if the AC component has an equal enclosed area above and below the centre line. Any symmetrical waveform, such as a sinewave, triangle wave, sawtooth wave or square wave, does have equal area above and below the centre line, but pulse waveforms and rectangular waves that don't have 50% duty cycle do not, and the DC component of the signal will not be half way between the positive and negative peaks for those waveforms.

Measuring the AC, would be from the bottom of the waveform to the top - so, counting the divisions between the two. I'd say measure the DC first, and then lower the vertical position and adjust the volts/div until the waveform takes up most of the screen and then count the divisions between top and bottom for a more accurate answer?

If you switch the input coupling switch to AC coupling, the DC component will be removed and you can zoom in on the AC component. I think that's what Dave was suggesting too.

Not sure if this is possible with just an oscilloscope.. but can you set it to measure just AC or DC? If you changed the input to AC, would it only show the AC component? Same with DC? The oscilloscope Dave showed has options for AC, GND and DC. The one shown in the question though just has AC/DC (combined) and GND.

"AC/DC combined" doesn't make sense to me. The three normal options are:

• GND - the input is shorted to GND;
• DC - the input is passed through to the amplifier unmodified, so the amplifier sees the complete signal, including its DC component, if there is any;
• AC - the input is passed through a capacitor which "blocks" the DC component (if there is one), so the AC component is shifted so its mean level is 0V and you only see the AC component.

When you're interested in the DC component, you have to use DC coupling. When you're measuring the AC component, it's easier if you use AC coupling, because the AC component of the waveform is shifted down so its mean voltage is 0V and you can zoom in on it without the display jumping off the screen.

If the example oscilloscope really only has one input coupling type, I would expect it to be DC coupling (unless it's an antique scope); in that case, you can add an external capacitor (0.1 µF would be fine for this case) between the signal and the scope input, to block the DC so you can zoom in on the AC component.

 

[Ferrock]

Yes you could try that. I would increase the time/div until the AC becomes a big thick blur, and estimate the centre of the blur.

That would make it easier to see where the top and bottom are. Problem is it is still only an estimate. Although I suppose all measurements from the screen are estimates anyway so, they can't really use that point against me!

This is only valid if the AC component has an equal enclosed area above and below the centre line. Any symmetrical waveform, such as a sinewave, triangle wave, sawtooth wave or square wave, does have equal area above and below the centre line, but pulse waveforms and rectangular waves that don't have 50% duty cycle do not, and the DC component of the signal will not be half way between the positive and negative peaks for those waveforms.

I see what you're getting at. I think we can assume it is a sine wave for this so will be 50/50 however I will definitely mention that point as it is something to consider - thanks

If you switch the input coupling switch to AC coupling, the DC component will be removed and you can zoom in on the AC component.

Ok, that makes it a bit easier then than the method I suggested. So just set it to AC, and then you can do as we said above to find the center and the peak-peak amplitude / voltage.

"AC/DC combined" doesn't make sense to me. The three normal options are:

• GND - the input is shorted to GND;
• DC - the input is passed through to the amplifier unmodified, so the amplifier sees the complete signal, including its DC component, if there is any;
• AC - the input is passed through a capacitor which "blocks" the DC component (if there is one), so the AC component is shifted so its mean level is 0V and you only see the AC component.

I said AC/DC combined because thats what the switch looks like on the oscilloscope pictured on my question paper. It is this image:

It has AC/DC and GND. So I assumed it was a combined switch but.. that doesn't exist Although I guess DC could be classed as combined as it shows both - but thats just because it displays the signal unmodified?

So, is the AC/DC switch on this one actually just a DC input? Or would it cycle through AC and DC? So just press it and each press it will cycle through both the options over and over?

Hopefully it will cycle through DC and AC coupling as that would make it much easier to measure the different components as I don't think I can suggest using a capacitor as its asking about settings rather than using other physical components. Thats a handy tip to know though - very simple yet its the kind of thing I wouldn't have thought of.

Thanks

 

[KrisBlueNZ]

That would make it easier to see where the top and bottom are. Problem is it is still only an estimate. Although I suppose all measurements from the screen are estimates anyway so, they can't really use that point against me!

Right!

Ok, that makes it a bit easier then than the method I suggested. So just set it to AC, and then you can do as we said above to find the center and the peak-peak amplitude / voltage.

Yes. Set it to AC and zoom in until the AC component nearly fills the display, so your estimate of its amplitude will be as accurate as possible. The scope will probably display the amplitude as a number as well. But you don't need to find the centre of the waveform - you only do that when you're estimating the DC component.

You could at this point check that the waveform has equal enclosed area above and below the line half way between the positive and negative peaks, so later when you estimate the DC component, you can be sure that it is half way between the top and bottom of the blurred section.

I said AC/DC combined because thats what the switch looks like on the oscilloscope pictured on my question paper. It has AC/DC and GND. So I assumed it was a combined switch but.. that doesn't exist Although I guess DC could be classed as combined as it shows both - but thats just because it displays the signal unmodified?

In post #15 Neal says the switch does select between AC and DC coupling. It looks like it's a pushbutton to toggle between AC coupling and DC coupling. The second pushbutton chooses the third option, GND.

So, is the AC/DC switch on this one actually just a DC input? Or would it cycle through AC and DC? So just press it and each press it will cycle through both the options over and over?

It's marked AC/DC. So unless it makes the oscilloscope play Highway to Hell when you press it, it presumably toggles between AC coupling and DC coupling each time you press it.

Hopefully it will cycle through DC and AC coupling as that would make it much easier to measure the different components as I don't think I can suggest using a capacitor as its asking about settings rather than using other physical components. Thats a handy tip to know though - very simple yet its the kind of thing I wouldn't have thought of.

And I'm going to give myself a formal warning for giving direct answers to a homework question. :-(

 

[Ferrock]

In post #15 Neal says the switch does select between AC and DC coupling. It looks like it's a pushbutton to toggle between AC coupling and DC coupling. The second pushbutton chooses the third option, GND.

I wasn't sure if he was referring to my one or the one Dave linked since that one showed the AC, DC and GND separate. But then he did say 'push button' so I should have figured that out, oops.

It's marked AC/DC. So unless it makes the oscilloscope play Highway to Hell when you press it, it presumably toggles between AC coupling and DC coupling each time you press it.

That would be quite a surprise. However I didn't include those thoughts in my report

And I'm going to give myself a formal warning for giving direct answers to a homework question. :-(

No need! Its been an 18 post conversation discussing it and I have learnt a lot. Ended up writing just over 3000 words for the question.. so hopefully I'll get close to the 40 marks. I tried to cover all bases and therefore probably rambled on a bit, or a lot.

Thanks for your help everyone!

 

[KrisBlueNZ]

And I'm going to give myself a formal warning for giving direct answers to a homework question. :-(

No need!

It's Electronics Point policy to not give straight answers to questions in the homework help section. But the forum software doesn't let me give myself a warning, so I guess I got away with it this time