what an oscilloscope recommended for a woman passionate about electronics?

[rstofer]

Sorry for hijacking this thread but I did not want to open a new one for a quick question.

I will also be purchasing my first scope next week and cannot decide to go with a rigol or a keysight.

Which one of these models should I go for? 
Rigol DS1074Z Plus
Rigol DS1104Z Plus
Keysight eduX1052A

Have you read through this thread from the beginning?  Of the list you posted, I would probably say "none of the above".

The Rigol DS1054Z can be unlocked to 100 MHz/ 4 channels and it's been around a while longer than the newer Siglents
The Siglent SDS1202X-E is the subject of most of this thread, 2 channels, 200 MHz
The Siglent SDS1104X-E is 100 MHz 4 Channels and can be unlocked to 200 MHz.  Pretty expensive but some people NEED 4 channels

All of the possible comparisons are over on the Test Equipment forum and if 20 people respond, you will have 30 recommendations.

Look at the range of options considered in just this one thread!

[tautech]

Sorry for hijacking this thread but I did not want to open a new one for a quick question.

I will also be purchasing my first scope next week and cannot decide to go with a rigol or a keysight.

Which one of these models should I go for? 
Rigol DS1074Z Plus
Rigol DS1104Z Plus
Keysight eduX1052A

EDIT: I have now decided to go for the Siglent SDS1104X-E, thank you to everyone who have replied. 

You can find all the EU dealers here:
https://www.siglenteu.com/how-to-buy/

[CharlotteSwiss]

I was studying the siglent's compensation signal.
Square wave - 1Khz - Vp-p 3v (0v to + 3v)
This is what we see on the oscilloscope display (above in DC coupling, below in AC coupling):



As you can see I have marked the voltage area in white: in DC coupling the area is 1.5v, so we have an average value of 1.5Vdc.
Obviously in AV coupling the postive area is equal to the negative one, and we have an average value of zero.

For now everything seems easy, but what would I like to understand? 

If I measure this signal with my multimeter I have these results: (the multimeter is True RMS)
DC = 1.5vdc
AC = 1.5vac
Now, ok the measurement of the multimeter in DC corresponds to the calculation of the average value.
We come to the AC multimeter measurement: usually the multimeter reads the effective value of the alternating signal (which for a perfect sine wave is about VP / 1.4); here I see that the reading is 1.5vac, or the Vp if we look at the AC coupling signal.
Is there a simple explanation for all this?
But above all, in the presence of a signal like this, which setting of the multimeter should you have? (DC or AC?)
And how should a signal like this be defined? could be: unipolar periodic square wave / divided into: a DC component of 1.5vdc and an AC component of 1.5vac : could it be an ideal definition?

My mind is confused, but not much: I miss that clue to frame it all
(then I can move on to analyze a signal with offset)
thank you so much
 

[tggzzz]

Good for you for testing your understanding.

Some things you can twiddle to help to clear up your confusion.

Firstly, try a sine wave instead of a square wave. Does that change the DMM readings?

Secondly, with a square wave, measure it on the scope and DMM (AC and DC coupling), and reduce the offset to zero. How do the readings change?

Thirdly, insert a capacitor (>10nF) between the DMM and signal source, just as the scope does with AC coupling. How do the DMM readings change?

[CharlotteSwiss]

Good for you for testing your understanding.

Some things you can twiddle to help to clear up your confusion.

Firstly, try a sine wave instead of a square wave. Does that change the DMM readings?

Secondly, with a square wave, measure it on the scope and DMM (AC and DC coupling), and reduce the offset to zero. How do the readings change?

Thirdly, insert a capacitor (>10nF) between the DMM and signal source, just as the scope does with AC coupling. How do the DMM readings change?

thanks tggzzz 

first: to prove the difference between square wave and sine wave, I have to do it from the simulator. Then later I try.
(but I have already studied that with a sine wave, Vrms = Vp / 1,4. With different waves the factor is no longer valid)

second: but isn't it the proof I have already done just above? there is no offset in the siglent compensation signal!

third: this test too I have to do from the simulator though. I understand what you mean, in AC coupling the oscilloscope blocks the dc signal which in fact cannot be seen on the display, we only see the alternating AV wave. But my attention is not on the av coupling signal that the oscilloscope shows us, but on the real signal that comes out of the pins of the compensation signal, which is composed of both dc and ac: that I would like to understand how we should measure / define it

Note: my DMM it also has the AC + DC function, on the compensation pins I measure:
DC= 1.5 (ok Vm)
AC normal= 1,5vac
AC+DC function= 2,2v (I don't know what calculation it does)

I thought I had a less confused head, but perhaps it is more confused than I thought 

[tggzzz]

Note: my DMM it also has the AC + DC function, on the compensation pins I measure:
DC= 1.5 (ok Vm)
AC normal= 1,5vac
AC+DC function= 2,2v (I don't know what calculation it does)

RMS is "root of the means squared". So 2.25V = sqrt(Vac² + Vdc²)

[CharlotteSwiss]

Note: my DMM it also has the AC + DC function, on the compensation pins I measure:
DC= 1.5 (ok Vm)
AC normal= 1,5vac
AC+DC function= 2,2v (I don't know what calculation it does)

RMS is "root of the means squared". So 2.25V = sqrt(Vac² + Vdc²)

maybe I understood something
Ac+DC function is the true RMS value, and in the specific case it would be:
square root [(1,5x1,5 ac)+(1,5x1,5 dc)] = 2,12 vac RMS (ac+dc function DMM)

so we can say that the square wave compensation signal is composed as follows:
 -1,5vdc (Vm)
 -2,12 Vrms
(the 1,5vac value of the DMM should indicate the Vp of the square wave in AC coupling)
 

[Mortymore]

If the signal was sinusoidal, I suppose it would be something like this



EDIT: By mistake I wrote Vpp instead of Vpeak. Vpp is 4.24 not 2.12.
Vpeak is 2.12 (= sqrt2*(Vrms) = 1.41*1.5)
Vpp = Voltage peak-to-peak, 2*Vpeak = 2*2.12

[CharlotteSwiss]

Firstly, try a sine wave instead of a square wave. Does that change the DMM readings?

I have found a comparison between the mathematical formulas and the DMM reading, but I believe that I am still missing something to understand when I have a square wave (like this compensating wave) in front of me as I must rightly measure and define it.

I start with the first suggested point: for this I don't need to test with the software, the calculations are enough.
If we are going to measure a perfect sine wave alternating with a Vp of 1.5vac (Vp-p 3 vac), with the DMM I would read these values:
-DC = 0 (moreover also the Vm is 0)
-AC = 1.5 / sqrt2 = 1.06 Vrms

But this example, with the square wave example of my test I don't know how it can help me ..
 

[CharlotteSwiss]

If the signal was sinusoidal, I suppose it would be something like this

  opss
so perhaps tggzzz meant a sine wave but with a 1.5v center ...
I'm getting lost in a sea of electrons, I have to tidy up ideas ..
 

[rstofer]

One thing to keep in mind  The RMS voltage is the property of a waveform that would produce the same heating effect as an equal DC voltage.  The RMS voltage, say 120V_RMS would produce the same heating as 120V_DC for a given resistive (only) load.

It't the equivalent heating effect we want to measure.

So, what about that symmetric sine wave?  We already know the mean value is 0V, how does that work?  Well, conceptually, the resistor doesn't care which direction the current is flowing so we simply flip the negative excursion of the waveform to positive (the classic double hump waveform) and do the math on that.  The link below shows the AC derivation.

The case of the symmetric square wave is the easiest to see:  The part below 0V is flipped up and then we notice that V_RMS = V_PK.

https://www.rfcafe.com/references/electrical/square-wave-voltage-conversion.htm

The math for the sine wave is a bit more ugly but V_RMS = 1/sqrt(2) * V_PK

https://www.rfcafe.com/references/electrical/sinewave-voltage-conversion.htm

For more arbitrary waveforms, the math gets completely out of hand so we tend to talk about just square and sine waveforms.  At least in the early days.  Piecewise integration can wait for a later date.

V_RMS is all about the DC voltage that would have an equivalent heating effect as the test waveform.

[CharlotteSwiss]

One thing to keep in mind  The RMS voltage is the property of a waveform that would produce the same heating effect as an equal DC voltage.  The RMS voltage, say 120V_RMS would produce the same heating as 120V_DC for a given resistive (only) load.

It't the equivalent heating effect we want to measure.

So, what about that symmetric sine wave?  We already know the mean value is 0V, how does that work?  Well, conceptually, the resistor doesn't care which direction the current is flowing so we simply flip the negative excursion of the waveform to positive (the classic double hump waveform) and do the math on that.  The link below shows the AC derivation.

The case of the symmetric square wave is the easiest to see:  The part below 0V is flipped up and then we notice that V_RMS = V_PK.

https://www.rfcafe.com/references/electrical/square-wave-voltage-conversion.htm

The math for the sine wave is a bit more ugly but V_RMS = 1/sqrt(2) * V_PK

https://www.rfcafe.com/references/electrical/sinewave-voltage-conversion.htm

For more arbitrary waveforms, the math gets completely out of hand so we tend to talk about just square and sine waveforms.  At least in the early days.  Piecewise integration can wait for a later date.

V_RMS is all about the DC voltage that would have an equivalent heating effect as the test waveform.

thanks rstofer.
my study is only for the periodic square wave (sine wave already studied  )
The perfectly alternating square wave (zero value in the middle of the wave) is quite easy to understand.
The only thing that is not clear to me: we said that the average value Vm of an alternating periodic signal is always zero!
But if you look at your link:

https://www.rfcafe.com/references/electrical/square-wave-voltage-conversion.htm

Ok, I understand that in this wave the effective value is equal to the peak value.
But why do they say that the average value is also equal to the peak value? (and therefore not equal to zero?)
do they mean a different average value? (and not the sum of the areas of tension?)

---

My doubts, however, concern a square wave signal, for example unipolar (such as the compensation of the siglent, see my message 556).
So not alternating (and for now also without offset)
In the presence of such a signal and measuring with a DMM I understood that:
-the signal has a DC component of 1.5vdc (equal to the Vm of the signal by calculating the white area)
-the signal has an AC component (visible for example by setting the oscilloscope in ac coupling), which the DMM measures me in 1.5Vrms; I understood that the Vrms value of a signal like this is always the same as Vpk-pk / 2 (I did several tests with the simulator, changing the amplitude of the signal)
So I would like to understand, if in the presence of a signal like this (periodic square wave, not alternating, Vpk 3v, Vmin 0v), I can say this: the signal is a square wave, composed of a DC component of 1.5vdc and an AC component of 1.5Vrms
Can this statement of mine be correct and is it all that this sign can indicate?

I do not ask for more for now, just to understand this (then I switch to the same signal but with offset to study it ..)

thanks 

[tggzzz]

One thing to keep in mind  The RMS voltage is the property of a waveform that would produce the same heating effect as an equal DC voltage.  The RMS voltage, say 120V_RMS would produce the same heating as 120V_DC for a given resistive (only) load.

It't the equivalent heating effect we want to measure.

So, what about that symmetric sine wave?  We already know the mean value is 0V, how does that work?  Well, conceptually, the resistor doesn't care which direction the current is flowing so we simply flip the negative excursion of the waveform to positive (the classic double hump waveform) and do the math on that. 

There's no need to "flip the waveform"; just use the maths directly. The power dissipated in a resistor is V²/R, of course. If V=-20, then that is (-20)²/R -> +400/R

[CharlotteSwiss]

in my notes, I am trying to classify some types of periodic signals.
Here are 3 examples:
1 = periodic bipolar signal (0v is the center of the signal)
2 = periodic unipolar signal (obtained by adding an equal / higher offset of the Vp value)


I give up, you are unable to insert the image in this position, click on the image at the end of the message, thanks

Then I have signal 3, and here I have a doubt: in this case the offset value is less than the value of Vp.
The signal therefore has a positive part and a negative part.
Am I wrong or is it not right to call this signal with the same name as signal 1? (bipolar periodic signal)
Positive voltage values are higher than negative voltage values, not fair as in signal 1.

so you can't call both periodic signals bipolar, am I wrong?

Thanks
 

[CharlotteSwiss]

perhaps in my post above I asked for absurd things 

I think I understand the difference between the 1/2/3 signals in the image above:
-signal 1 = ALTERNATE bidirectional periodic signal
-signal 2 = unidirectional periodic signal (with offset)
-signal 3 = periodic bidirectional signal NOT ALTERNATE (with offset)
In conclusion we can say that a signal is ALTERNATE when its average value is zero (and therefore does not contain offset)
There are bidirectional signals NOT ALTERNATE (contain offsets)
I can move on now ...
 

[CharlotteSwiss]

slowly my study of the manual of the siglent 1202 continues..
I didn't understand what the EXT socket can do (it should be external trigger)

open attached image, direct insertion no longer works

the manual just says this:
The 2-channel oscilloscope’s trigger source includes analog channels, EXT, EXT/5 and AC Line

Analog channel input:
Signals input from analog channels can all be used as the trigger source.

External trigger input:
External trigger source can be used to connect external trigger signal to the EXT TRIG channel when all of the four channels are sampling data. The trigger signal (such as external clock and signal of the circuit to be tested) will be connected to EXT and EXT/5 trigger source via the [EXT TRIG] connector. EXT/5 trigger source attenuates the signal by a factor of 5. It extends the trigger level. You can set the trigger condition within the range of trigger level (-8 div to +8 div).

AC line:
The trigger signal is obtained from the AC power input of the oscilloscope. This kind of signals can be used to display the relationship between signal (such as illuminating device) and power (power supply device). For example, it is mainly used in related measurement of the power industry to stably trigger the waveform output from the transformer of a transformer substation.

Normally we use Ch1 or Ch2, I don't understand why I should use ext or AC line?
What are the advantages?
I haven't found any guides on the internet

thanks 

[rstofer]

AC Line is used for servicing televisions or, indeed, anything that is related to mains frequency.  You may not need to use it but it can be handy when you are looking at signals synched to line frequency.

The idea of not triggering on the two analog channels is that, sometimes you need to trigger on a signal but you don't need to see it.  More important, you have two signals you do need to see and the trigger isn't one of them.

The Rigol DS1054Z 4 channel scope doesn't have an external trigger input although it does have a Trigger Out for ganging other scopes or whatever.  A two channel scope will almost always have an external trigger input.

[rstofer]

If you try to scope the SPI protocol, you soon see why I bought a 4 channel scope.  There are 4 signals:


SS'  - Slave Select - a low going chip select signal that must remain low throughout the transaction.
SCK  - the clock signal generated by the master device
MOSI - Master Out, Slave In - data from the master to the slave
MISO - Master In, Slave Out - data from the slave to the master.


You would want to trigger on SS' even if you didn't display it and, with a two channel scope, you could only watch half of the conversation - perhaps SCK and MOSI to see what the master is telling the slave or SCK and MISO to see what is being sent back.  You wouldn't be able to see the clocking but it might also be useful to watch both sides of the transaction by probing MOSI and MISO already knowing that the clocking is correct.  In any event, the SS' signal is used as a trigger whether it is displayed or not.

Here is a long, yet boring, discussion of the SPI protocol:
https://en.wikipedia.org/wiki/Serial_Peripheral_Interface

It occurred to me that my 350 MHz Tektronix 485 with just 2 channels (plus external trigger) was going to be inadequate so I bought the 4 channel Rigol specifically for its ability to handle the SPI protocol.  I can see all 4 signals and this is important because the SS' signal is usually handled in code and it's important to know that the last bit has transferred before it goes inactive.  A lot of headscratching has started with raising the SS' signal too soon and the SPI buffer hasn't fully transferred.

Yes, I know about logic analyzers, I have some...  It's just that as long as I'm looking at signal timing, I might as well get the decode.

[CharlotteSwiss]

meanwhile, thanks rstofer, happy to hear from you: tonight I read calmly and write my thoughts
 
Charlotte

[Connecteur]

I don't know a lot about oscilloscopes, but why should gender matter?

[rstofer]

There are reasons for all the various controls and inputs/outputs and many will only come up when working on more specialized applications.  My experience is going to be primarily digital, I don't spend a lot of time on analog.  To me, logic analyzers and 4 channel scopes make sense.  The RF folks have entirely different needs.  They're going to care about 50 Ohm terminations and signal attenuators whereas they never come up for me.

Just keep asking questions, somebody around here will have an answer.

[CharlotteSwiss]

I don't know a lot about oscilloscopes, but why should gender matter?

if you mean man or woman, it was not to diversify the gender, but woman was referring only to me, as little expert, that's all 

[CharlotteSwiss]

thanks for the contributions rstofer: 
Ac line can therefore be used to trigger the trigger when we analyze signals with frequency of the home network.
Instead, I find it hard to understand why another probe should be connected to the ext trigger socket, I give an example:
- I connect the Ch1 probe to a point in the circuit.
- I connect ch2 to another point in the circuit
The two signals already trigger the trigger to show the two signals on the display.
Why should I connect a third probe to the circuit, and input a third signal from the ext socket? Maybe so I can see three signals on the display?
It may be that the siglent manual speaks little about it, it may be that I have not found any information on the net, but this thing does not enter my head .. uff 
thank you

[CharlotteSwiss]

maybe i figured something out about the external trigger.
If I am acquiring with the analog ch a signal with amplitude for example -2 + 2v, I can avoid that part of this signal passes into the oscilloscope trigger circuit. To achieve this I just set the ext trigger, and acquire the trigger signal from the ext-in socket. Obviously the signal must have an amplitude between -2 + 2v to be able to trigger the trigger in the signal acquired by the oscilloscope ch.
This procedure could have the advantage of having a more stable signal on the display, perhaps useful for signals of small amplitudes!
 

[rstofer]

Using the External Trigger input, I could have some logic determine that an event in a circuit has just occurred and I could probe some other logic signals on the channel inputs.  External Trigger is like having an extra channel that you can synch to but you can't display.

It's not uncommon in  FPGA projects to create logic just to provide a trigger on some event.  Since there is a pretrigger portion to the buffer, I can actually see what led up to the trigger event.  It's strange thinking about time before time t=0.

I could actually have several scopes all synched to an external trigger so they all start the display at the same time.


As I said earlier, AC Line trigger is useful for probing signals that are synchronized to the mains frequency.  Perhaps a phase shift lamp dimmer or the vertical ramp on an analog television set.  You can play with it, just select AC Line trigger and touch the probe tip with your finger to see the waveform is already stable.  Since you are looking at radiated mains voltage, it is already synched to AC Line.  The display will be stable regardless of the voltage level on the trace.

You may never find a use for either trigger scheme.  I certainly haven't used them very often but I do know that they're available when I come up with a need for either.

I think the User Manual does a pretty decent job of explaining the trigger capabilities starting on document page 55 or PDF page 79

https://www.siglent.eu/_downloads/51337474

Notice the 2d sentence where Siglent states that 4 channel scopes don't have an Ext Trigger input.  That's pretty common.  Two channel scopes will most always have the input.  Then they turn right around a couple of paragraphs later:

External trigger input:
External trigger source can be used to connect external trigger signal to the EXT TRIG
channel when all of the four channels are sampling data.

I'm pretty sure this is wrong but I don't have a 4 channel Siglent.  My DS1054Z 4 channel scope does NOT have an Ext Trigger input.  Of that, I am certain.