what an oscilloscope recommended for a woman passionate about electronics?

[CharlotteSwiss]

Perfect!  Until we come up with a real application, I'm willing to move on from the integral.  It's nice that the math and the scope agree.

the integral, I can archive it; tomorrow I write a few lines about it and then I move on to math-square ...
And I try to write something on the sinusoid too, if my ideas in the previous post are reliable.
Well it's bedtime for me, good evening over there in the united states
 

[alsetalokin4017]

If you monitor Voltage on one channel and Current on another channel, multiply the two traces to get Instantaneous Power in watts then Integrate the result over time ... you have Energy in joules.

[CharlotteSwiss]

If you monitor Voltage on one channel and Current on another channel, multiply the two traces to get Instantaneous Power in watts then Integrate the result over time ... you have Energy in joules.

certainly, voltage * current = power.
But the question is: how can an oscilloscope channel tell me the current strength? The oscilloscope shows the voltage over time, not the current ... for that I should use the multimeter
 

[tautech]

If you monitor Voltage on one channel and Current on another channel, multiply the two traces to get Instantaneous Power in watts then Integrate the result over time ... you have Energy in joules.

certainly, voltage * current = power.
But the question is: how can an oscilloscope channel tell me the current strength?

2 common ways.
Voltage across a current shunt with units selected as Amps in the channel menu.
Current probe.....which provides the same thing, a voltage.

Current probes while expensive provide insulation while care must be used when using a current shunt so to not make ground loops with how you might connect probe reference leads when measuring voltage and current in the same circuit.
There are other things to consider like is the current is DC or AC but that is for another lesson. 

[CharlotteSwiss]

however, an additional device / or probes for the current are required; for the current I easily use one of my two multimeters, or my father's current clamp
 

[alsetalokin4017]

Those devices do not give you the shape of the current waveform, thus cannot be useful to find instantaneous power of anything but steady DC. However if you had enough channels on your oscilloscope you could do a floating differential measurement using two channels A and B (measuring the voltage drop across, say, a one-ohm resistor) and a voltage measurement on a third channel C. Then the scope's Math can perform the operation INTegrate((A-B)*C) to arrive at the energy trace.  For example say you need or want to know the energy per pulse of a PWM motor driver, or you'd like to determine if your perpetual motion machine is using less energy than you are supplying to it.
 
Well, somebody did ask for a practical application of the integration function. Sorry that the application requires a current probe and/or more than two channels.

Remember Steorn?

[rstofer]

In Dave's video (linked above), the CPU runs for a very brief period so there is a definite current spike.  Then the CPU sleeps for a relatively long time but still consuming power.  No much power but if integrated over an even longer time it is possible to see the total consumption over that period.

There are clearly other ways to get this integral and some kind of datalogger might be one of them.  The problem it, the spike is narrow and needs a high sample rate.  For a datalogger, this means a lot of storage used up during sleep just to accommodate the high sample rate.

I think there is a place for the differential probe and integration technique but I would try to find a better way to do it.  I'm not sure what that would be.

The Analog Discovery 2 has 2 fully differential scope channels so making the measurement is easy.  It doesn't seem to have an integration function but it does allow for scripting.  It will do the product and display the instantaneous value.

One place where differential measurements are required is the simple RC circuit where we want the phase shift across the capacitor, we want to see where the current leads the voltage by 90 degrees.  Measuring the capacitor voltage is no problem but measuring the current is best done with a differential probe across the resistor.  The AD2 is ideal for this kind of thing.

This is the ICE in "ELI the ICE man" - I leads E in a Capacitive circuit and E leads I in an inductive circuit.  Do they still teach it that way?

[CharlotteSwiss]

Well, somebody did ask for a practical application of the integration function. Sorry that the application requires a current probe and/or more than two channels.

why do you say this? my siglent has only two channels, i don't have any additional probe .. and the integral gives me the V * s value at any point of the signal i want .. it could be useful, don't you think?
 

[alsetalokin4017]

Absolutely useful. And there are some relatively inexpensive current probes that will be a good match for your intended use cases, once you get to that point. You don't need superhigh bandwidth for your audio work.

Here I must mention the "old way" of integrating a scope trace: You photograph or even sketch the trace using tracing paper, then cut out the waveform as solid shapes. Then the shapes can be weighed on an analytical balance. The area under the curve of the trace is proportional to the mass of the paper or photograph. The scale factor is easily determined by simple experiments. Then to integrate all you need to do is trace the trace, cut out the shapes, weigh them, apply the scaling factor and Bob's your uncle.
 

(I have also applied a more modern version: counting (manually!) pixels in a digital photo of a trace displayed on an analog oscilloscope.)

[CharlotteSwiss]

In Dave's video (linked above), the CPU runs for a very brief period so there is a definite current spike.  Then the CPU sleeps for a relatively long time but still consuming power.  No much power but if integrated over an even longer time it is possible to see the total consumption over that period.

There are clearly other ways to get this integral and some kind of datalogger might be one of them.  The problem it, the spike is narrow and needs a high sample rate.  For a datalogger, this means a lot of storage used up during sleep just to accommodate the high sample rate.

I think there is a place for the differential probe and integration technique but I would try to find a better way to do it.  I'm not sure what that would be.

The Analog Discovery 2 has 2 fully differential scope channels so making the measurement is easy.  It doesn't seem to have an integration function but it does allow for scripting.  It will do the product and display the instantaneous value.

One place where differential measurements are required is the simple RC circuit where we want the phase shift across the capacitor, we want to see where the current leads the voltage by 90 degrees.  Measuring the capacitor voltage is no problem but measuring the current is best done with a differential probe across the resistor.  The AD2 is ideal for this kind of thing.

This is the ICE in "ELI the ICE man" - I leads E in a Capacitive circuit and E leads I in an inductive circuit.  Do they still teach it that way?

I have to see dave's video, I have it listed; interesting this graph, very informative.
Of course for my needs the basic things that my siglent can do are enough, I'm sure they could certainly be enough.
The important thing is to know the instrument thoroughly and what it can do, then when you are in front of a circuit to be repaired the possibility of placing the probes in the right places and understanding what the displat teaches us, everything will be easier.
 

[CharlotteSwiss]

Absolutely useful. And there are some relatively inexpensive current probes that will be a good match for your intended use cases, once you get to that point. You don't need superhigh bandwidth for your audio work.

Here I must mention the "old way" of integrating a scope trace: You photograph or even sketch the trace using tracing paper, then cut out the waveform as solid shapes. Then the shapes can be weighed on an analytical balance. The area under the curve of the trace is proportional to the mass of the paper or photograph. The scale factor is easily determined by simple experiments. Then to integrate all you need to do is trace the trace, cut out the shapes, weigh them, apply the scaling factor and Bob's your uncle.
 

(I have also applied a more modern version: counting (manually!) pixels in a digital photo of a trace displayed on an analog oscilloscope.)

however you must understand that I entered the oscilloscope world from July; I'm starting with the basics, then there will be time to think about more complicated things or accessories.
The size / weight exercise could be instructive, but I challenge which technician would go that far
 

[CharlotteSwiss]

just to play for a moment: for my siglent, to get the values of the integral of a certain point of the wave, the interested part (ramp, descent ...) must be isolated with gate and the cursors will tell us the V * s value of that point.
In the image I isolated the negative half-wave of the wave: we can see that the cursors give me the integral value of the two points indicated (yellow and green); one -17uV * s and the other -112uV * s.
I would say that this is enough
 

[CharlotteSwiss]

Lacking a good source of sine waves, I don't think you can replicate this final experiment so let's walk through the math because I found it interesting.

We know that the integral of sin(x) over the range 0 to 2 pi is exactly 0 because the upper lobe has the same area as the lower lobe (which is negative) and they cancel.  But what about over the range 0 to pi - just the upper lobe.  We know that the integral of sin(x) over that range is 2 (kind of a strange answer).  Just ordinary 2 - no pi's, no weird functions, just 2.  Symbolab.com will do that definite integral.

Suppose we draw a rectangle surrounding that lobe with a sine wave with an amplitude of 1 (Vpeak if you wish) and extending along the X axis from 0 to pi (.3.14).  The area of that rectangle is 3.14 and fully encompasses the sine wave lobe with an area of 2.  So, the sine wave fills up 0.637 of the rectangle and that is esactly the integral of that lobe: 0.637 units (V ms, or whatever).  2 / 3.14159 -> 0.637.  2 is the area of the lobe, 3.14159 is the area of the rectangle enclosing the lobe.

It took me a while to tumble to this solution but, voila', the scope gets the same answer.  The settings aren't exact but here is a scope image using the integral function over that limited range.  The measurement we're looking for is AY = 0.636 V.  Actually, it took more time to tumble to the math than it did to get the scope display.

To make the numbers neat, I used 2 Vp-p and 500 Hz.  That way the upper lobe is 1 ms wide and 1V high.

I have never used the integration function on a scope.  Dave has a video that shows a splendid use:

i saw dave's integration video - some doubts returned to me about what i thought i learned!
Looking at your image attached to message 713, let's see if I understand what it explains in the video:
You have isolated the integral of the positive half-period: the integral trace represents in practice the area included under the signal and included within zero volts; the total value of this area is represented by the highest peak of the integral, in your case 636.0
So we are only interested in the highest peak value of the integral, an intermediate value makes no sense, only the highest value represents the energy that releases this positive half-wave of the signal, am I right? (with this I want to emphasize that perhaps detecting the intermediate values of the integral as in my previous post, perhaps it does not make much sense ..)

Second doubt: I thought I understood what was meant by the measurement unit of the integral, you also explained it to me but I think I misunderstood: let's take your half-period again, we have a total value of that integral of 636uV * s; from what i understood, this meant that the signal delivers 1v in a time of 636us, but after dave's video i think i didn't understand anything about it ..; the correct answer is perhaps that 636uV*s indicates that that half-period is capable of delivering 636uV in the time of 1 second?

We hope to understand definitively ...
thank you
 

side note: I liked to see how with the oscilloscope you can capture those small peaks that every 2 seconds activate the display of the microcontroller .. instructive.

[rstofer]

We know that, in my integral, the peak voltage is 1V and I'm integrating for 1 ms (that's why I picked 500 Hz).  If it was a square wave, I would get 1V ms or 1000 uV s as a result of integrating the rectangle.  But the sine wave only occupies 63.7% of that area so 637 uV s.

Dave is using One Shot mode.  That is the MOST IMPORTANT feature of the modern DSO - in my view.  Since his integral increases without bound (the device consumes current forever), he has to start the integration at some point and call that t=0.  Assuming a constant voltage, integrating current (which is a function of time), we get Watt seconds (which will be converted to Watt hours or mW hr).    Assuming that all cycles are identical, he can integrate just one cycle - like us.  But if there are other possible waveforms, he has to account for those as well.  All of our signals have been repetitive and only integrated over a short period of time.

As I look at the definition of a volt second, I keep getting referred to magnetic flux.  I'm not sure the unit, by itself, is very useful.  Ampere second is charge and useful when dealing with capacitors.  Or just assume a constant voltage like Dave's power example.  But scopes measure volts and time.  I think there is some mental gymnastics involved in assigning the actual units.

[CharlotteSwiss]

We know that, in my integral, the peak voltage is 1V and I'm integrating for 1 ms (that's why I picked 500 Hz).  If it was a square wave, I would get 1V ms or 1000 uV s as a result of integrating the rectangle.  But the sine wave only occupies 63.7% of that area so 637 uV s.

Dave is using One Shot mode.  That is the MOST IMPORTANT feature of the modern DSO - in my view.  Since his integral increases without bound (the device consumes current forever), he has to start the integration at some point and call that t=0.  Assuming a constant voltage, integrating current (which is a function of time), we get Watt seconds (which will be converted to Watt hours or mW hr).    Assuming that all cycles are identical, he can integrate just one cycle - like us.  But if there are other possible waveforms, he has to account for those as well.  All of our signals have been repetitive and only integrated over a short period of time.

As I look at the definition of a volt second, I keep getting referred to magnetic flux.  I'm not sure the unit, by itself, is very useful.  Ampere second is charge and useful when dealing with capacitors.  Or just assume a constant voltage like Dave's power example.  But scopes measure volts and time.  I think there is some mental gymnastics involved in assigning the actual units.

certainly the more time passes, the more the microcontroller consumes, but it seems to me that even making the integral of many signal periods, the maximum peak value of the integral is always the same, each ramp will have approximately the same consumption as the other ramps; but I don't concentrate on how much the summative value of several periods could be, but it was enough for me to understand what the max value of the integral represents, for example relative to a single half-period; in fact the peak of the integral of your half period represents the value of that area (636).
For the unit of measure, Dave seems to be telling us that the integral value represents that value in mV or uV I believe in the time of 1 second .. he explains it about 11 to 13 minutes of the video, but with subtitles maybe yes understands wrong.
It would be good to know, in your example, if 636 is uV per second, or is it us in 1 volt ....

Maybe I better go to bed, the neurons are dancing in my head
 

[rstofer]

certainly the more time passes, the more the microcontroller consumes, but it seems to me that even making the integral of many signal periods, the maximum peak value of the integral is always the same, each ramp will have approximately the same consumption as the other ramps; but I don't concentrate on how much the summative value of several periods could be, but it was enough for me to understand what the max value of the integral represents, for example relative to a single half-period; in fact the peak of the integral of your half period represents the value of that area (636).
For the unit of measure, Dave seems to be telling us that the integral value represents that value in mV or uV I believe in the time of 1 second .. he explains it about 11 to 13 minutes of the video, but with subtitles maybe yes understands wrong.
It would be good to know, in your example, if 636 is uV per second, or is it us in 1 volt ....

Maybe I better go to bed, the neurons are dancing in my head
 

The unit is <whatever> seconds.  Microvolts, millivolts, megavolts, <whatever> seconds.  Always seconds.  Yes, we can play with the math to get everything aligned but AFAIK, we always want to wind up with seconds.

Maybe other units of time are appropriate.  Small batteries are rated in mA hours and this is exactly where Dave wants to wind up.  He may need to convert the integral from mA seconds to mA hours...

There is is no condition under which the integral of battery current in Dave's example will diminish.  The CPU may go to sleep and the integral not change very much for some time (a nearly straight line on the integral graph) but it never reverses.  The only way the integral of current can diminish is if there is a battery charger recharging the battery.  In Dave's case, pumping current back into the battery which the circuit clearly doesn't do.

When you think of the sine wave being integrated over 0 to 2pi, the last half of the waveform is at a negative voltage and subtracts from the previously accumulated value.  At 2pi, the result is 0 uV s.  We integrated over pi radians to keep the integral increasing until it became asymptotic at pi radians.  In the very next delta t, the value would start to decline because sin(pi+delta t) is negative.  Try integrating from 0 to 3*pi/2 -- 270 degrees.  It won't be back down to 0 but the value is clearly decreasing.

In my example, the AREA is in terms of uV seconds.  Microvolts on the Y axis and time on the X axis.  It is NOT the same as uV/second.  Y is uV, X is time (some variation of seconds) and the result is an AREA of Y * X or uV * time or uV second (implicit multiply symbol omitted by convention).

We expect to get an AREA when we integrate.  Converting that to the real world is a math exercise.

In Dave's example, he is continually draining water out of the previously full bucket.  Sooner or later the bucket is empty.  Unless somebody shows up with a hose...

[atmfjstc]

This conversation reminds me of a girl I used to know, who bought herself a Yamaha keyboard so she could learn how to play the piano. She behaved exactly like Charlotte, carefully perusing every single section of the manual, particularly the details of advanced features such as the metronome, auto accompaniment, MIDI interface etc. She became an expert in the manual and the product's capabilities, sure, but several months in, she still couldn't play a coherent melody.

Point being, all those advanced features are irrelevant until you've achieved some skill with the device's core functionality - for the piano, it's being able to play a tune; for the oscilloscope, it's getting some experience with poking around in a few devices checking signals and troubleshooting. The manual can't help you with that. You can't learn tennis by reading a book. You have to put in the time and "do", instead of reading.

[borjam]

certainly the more time passes, the more the microcontroller consumes, but it seems to me that even making the integral of many signal periods, the maximum peak value of the integral is always the same, each ramp will have approximately the same consumption as the other ramps; but I don't concentrate on how much the summative value of several periods could be, but it was enough for me to understand what the max value of the integral represents, for example relative to a single half-period; in fact the peak of the integral of your half period represents the value of that area (636).

There is a clear use case for such a feature. Imagine you are developing a small gadget that must run on batteries. It uses a modern microcontroller and some other devices, notably some radio transmitter in order to upload data.

I mentioned a modern microcontroller because they have important power optimization features. So, turns out that the way you write your software can affect useable battery life a lot.

Using an oscilloscope with a current probe, monitoring supply voltage and current, you can evaluate, for example, which versions of your program are more efficient by determining power. As current can vary a lot depending on which devices you turn on or off and, as I said, how your program works, integration will help you make a much more accurate measurement.

You can't do that with a multimeter, for instance.

[CharlotteSwiss]

This conversation reminds me of a girl I used to know, who bought herself a Yamaha keyboard so she could learn how to play the piano. She behaved exactly like Charlotte, carefully perusing every single section of the manual, particularly the details of advanced features such as the metronome, auto accompaniment, MIDI interface etc. She became an expert in the manual and the product's capabilities, sure, but several months in, she still couldn't play a coherent melody.

Point being, all those advanced features are irrelevant until you've achieved some skill with the device's core functionality - for the piano, it's being able to play a tune; for the oscilloscope, it's getting some experience with poking around in a few devices checking signals and troubleshooting. The manual can't help you with that. You can't learn tennis by reading a book. You have to put in the time and "do", instead of reading.

thanks and for the suggestion, but I don't agree: I believe that if you don't study the manual with all its functions well, 90% of users will only observe lines on the display, without ever being interested in some functions that could simplify the work.
Let's relate the speech to the young students: according to your reasoning should young people learn in the field and not in books? practice on the front line is important, but it should only be done after a good basis of study.
 

[CharlotteSwiss]

The unit is <whatever> seconds.  Microvolts, millivolts, megavolts, <whatever> seconds.  Always seconds.  Yes, we can play with the math to get everything aligned but AFAIK, we always want to wind up with seconds.

Maybe other units of time are appropriate.  Small batteries are rated in mA hours and this is exactly where Dave wants to wind up.  He may need to convert the integral from mA seconds to mA hours...

There is is no condition under which the integral of battery current in Dave's example will diminish.  The CPU may go to sleep and the integral not change very much for some time (a nearly straight line on the integral graph) but it never reverses.  The only way the integral of current can diminish is if there is a battery charger recharging the battery.  In Dave's case, pumping current back into the battery which the circuit clearly doesn't do.

When you think of the sine wave being integrated over 0 to 2pi, the last half of the waveform is at a negative voltage and subtracts from the previously accumulated value.  At 2pi, the result is 0 uV s.  We integrated over pi radians to keep the integral increasing until it became asymptotic at pi radians.  In the very next delta t, the value would start to decline because sin(pi+delta t) is negative.  Try integrating from 0 to 3*pi/2 -- 270 degrees.  It won't be back down to 0 but the value is clearly decreasing.

In my example, the AREA is in terms of uV seconds.  Microvolts on the Y axis and time on the X axis.  It is NOT the same as uV/second.  Y is uV, X is time (some variation of seconds) and the result is an AREA of Y * X or uV * time or uV second (implicit multiply symbol omitted by convention).

We expect to get an AREA when we integrate.  Converting that to the real world is a math exercise.

In Dave's example, he is continually draining water out of the previously full bucket.  Sooner or later the bucket is empty.  Unless somebody shows up with a hose...

so it is already clearer: certainly the integral of your positive half-period (let's call that area +636 anything), is equal but of opposite sign of the integral of the next negative half-wave (let's call that area -636 anything); it is obvious that the sum of the two integrals is zero.
Unit of measurement: yes, the unit of measurement of the integral is a product (not division) between voltage and time; let's assume a square wave, so the area of the half-period is easy to calculate: we have a Vp of 2V and a half-period of 500us: the area will be 2V * 500us! the result will therefore be 1000uV * s (i.e. the sum of two squares with sides 1V * 500us = 1 square with side 1V * 1000us)

I believe that all my confusion about the unit of measurement of the integral function arose from the fact that they indicate uVs or mVs, that is u and m attached to V.
If we take the above result that is 2V * 500us = 1000uV * s, and we change it by adjusting the time in seconds we will have: 2V * 0.0005 = 0.001V * s; here in my opinion V * s is less confusing than uV * s, even if the product is the same!
This made my poor neurons dance ..
 

[CharlotteSwiss]

certainly the more time passes, the more the microcontroller consumes, but it seems to me that even making the integral of many signal periods, the maximum peak value of the integral is always the same, each ramp will have approximately the same consumption as the other ramps; but I don't concentrate on how much the summative value of several periods could be, but it was enough for me to understand what the max value of the integral represents, for example relative to a single half-period; in fact the peak of the integral of your half period represents the value of that area (636).

There is a clear use case for such a feature. Imagine you are developing a small gadget that must run on batteries. It uses a modern microcontroller and some other devices, notably some radio transmitter in order to upload data.

I mentioned a modern microcontroller because they have important power optimization features. So, turns out that the way you write your software can affect useable battery life a lot.

Using an oscilloscope with a current probe, monitoring supply voltage and current, you can evaluate, for example, which versions of your program are more efficient by determining power. As current can vary a lot depending on which devices you turn on or off and, as I said, how your program works, integration will help you make a much more accurate measurement.

You can't do that with a multimeter, for instance.

ok thanks for the example; This is already an advanced use of the instrument, but it gives a perfect idea! In the meantime, I just need to understand the simple operation with standard voltage probes, but in the future you will never know, the roads are endless ...
 

[CharlotteSwiss]

I finally got to the last Math function: square root. 

The siglent manual is very helpful    to me and says:
Square Root
Square root () calculates the square root of the selected source.
Where the transform is undefined for a particular input, holes (zero values) appear in the function output.
 

For now I have understood that the unit of measurement is V raised to 1/2, and if I look at the square wave (coupling AC) with Vp1.5V, it makes me an identical square graph, with half-period amplitudes of about 1.22 V raised to 1/2
I have to study it ... I hope it doesn't take me a week like the integral
 
 

[rstofer]

An AC coupled sine wave is nore interesting.  Use your Gate setting to capture the square root of a slice somewhere in the positive lobe and you will get something that resembles a sine wave, more or less.  The math function will track the sine function.  But let the Gate get into a portion where the value of the sin(x) is negative and, on my scope, it just displays NaN (Not a Number) and gives up in disgust.  The square root of a negative number is, at best, an imaginary number.  Lacing an imagination, the scope just decides it is undefined.

[rstofer]

You may find it interesting to plot some sqrt(sin(x)) equations at Desmos.

On the attached plot, the waveform around 0 is just sin(x) and sqrt(sin(x)).  Notice that the square root is always greater than or equal to the function.  That's because the square root of a number less than 1 is greater than the number itself.  Sqrt(0.5) = 0.707 and .707 is greater than 0.5

The second pair is offset by +3 and this complicates things (but saves on attachments and paper) so draw an X axis at 3 if you print this.  Here the square root is both larger and smaller than the function itself.

In both cases, the square root of a negative number is undefined.

Did I mention I love playing with Desmos?

[rstofer]

It actually works - the scope image is a lot like the Desmos graph:

Edit:  I changed the scope image