Author Topic: what an oscilloscope recommended for a woman passionate about electronics?  (Read 144558 times)

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Offline rstofer

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if I understand correctly, the most frequent use of the signal generator is for those who design and create electronic circuits, to do the signal tests during the project finishing phase

I think the signal source comes in a lot earlier in the design.  It will clearly be useful for prototyping.

Most designs are done with a process known as 'stepwise refinement' (from Niklaus Wirth) where we model something, test how it works, figure out a better design and iterate until we get as much as we can out of the circuit.

https://www.cs.cornell.edu/courses/JavaAndDS/stepwise/stepwise.html

If I am designing a transistor amplifier, I might start with a simple common-emitter design and then add an emitter resistor and some kind of capacitor across the emitter resistor to control the AC gain.

Here's a video on the subject:


The point is, there is a lot of experimentation in design.  Real parts have tolerances.
 

Offline CharlotteSwissTopic starter

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I think the signal source comes in a lot earlier in the design.  It will clearly be useful for prototyping.


nice video, it must be fantastic to go from the project on paper to the final realization, a lot of satisfaction
 ;) ;)
 

Offline CharlotteSwissTopic starter

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ETA:  I used a 220 uF capacitor and a 150 Ohm resistor.  The generator output is 1Vp-p riding on a constant offset of 1VDC.  I added a little schematic using LTspice.  I really don't like creating schematics this way but I don't have anything any better.
We're having fun now!

I also did the experiment with the software to understand the work of the coupling capacitor: AV 2vp-p - 2Vdc offset.
It is clearly seen that in the output with capacitor (purple signal) the dc has disappeared and the signal has zero volts in the center of the wave.
While in the output without condenser (green signal) we have the center of the 2v wave above the zero display (2v offset)





 ;)
« Last Edit: July 09, 2020, 10:04:23 am by CharlotteSwiss »
 

Offline tggzzz

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I also did the experiment with the software to understand the work of the decoupling capacitor: AV 2vp-p - 2Vdc offset.
It is clearly seen that in the output with capacitor (purple signal) the dc has disappeared and the signal has zero volts in the center of the wave.
While in the output without condenser (green signal) we have the center of the 2v wave above the zero display (2v offset)

Another experiment: replace the sine wave with a square wave, and see what happens. Then change the duty cycle to 10% and 90%, and explain the what you see to yourself.

FYI "condenser" is a very old term now. Youngsters probably won't understand it :)
There are lies, damned lies, statistics - and ADC/DAC specs.
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Offline CharlotteSwissTopic starter

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Another experiment: replace the sine wave with a square wave, and see what happens. Then change the duty cycle to 10% and 90%, and explain the what you see to yourself.

FYI "condenser" is a very old term now. Youngsters probably won't understand it :)

as soon as I can (maybe tonight) I try with the square wave
Young people prefer cap?
 ;)
 

Offline rstofer

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as soon as I can (maybe tonight) I try with the square wave
Young people prefer cap?


Even old people prefer cap or capacitor.  I haven't heard the term condenser since the early '50s.  Back then we had micro-microfarad, none of this picofarad stuff.  Nobody used nanofarads either,  The schematic would should something like 100 mmf instead of 100 pf
 

Offline CharlotteSwissTopic starter

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Even old people prefer cap or capacitor.  I haven't heard the term condenser since the early '50s

I learned the term condenser from my father, old school (not electrotechnical, but he was an electricity worker = electrician)
Sometimes I help him to fix the electrical network at home and I never fail to warn him that I'm wasted in that field  ;D
 

Offline CharlotteSwissTopic starter

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Another experiment: replace the sine wave with a square wave, and see what happens. Then change the duty cycle to 10% and 90%, and explain the what you see to yourself.

measurement only on the output with capacitor.
What do I know? nothing understandable!
It seems to me that the curve is the discharge of the capacitor, then the difference between 10 and 90 is that one signal is opposite to the other.
It feels too difficult for me now



 

Offline rstofer

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A suggestion:  Always plot both the input and output.  That's why scopes have at least 2 channels.  There is some relationship between the input and output and that information is not apparent without having the other trace.

Sometimes it is even better if both signals share the same 0V reference (they are overlaid except for the DC offset).  Other times we shift off the DC value such that only signal changes are overlapping.  Whichever provides the most intuition.
 

Offline CharlotteSwissTopic starter

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A suggestion:  Always plot both the input and output.  That's why scopes have at least 2 channels.  There is some relationship between the input and output and that information is not apparent without having the other trace.

Sometimes it is even better if both signals share the same 0V reference (they are overlaid except for the DC offset).  Other times we shift off the DC value such that only signal changes are overlapping.  Whichever provides the most intuition.

I'm still immature in the trade, in fact with the track of in and out you have more information to compare, thanks
 ;)
 

Offline alsetalokin4017

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Here's a classic capacitor charge-discharge curve:



It should be possible to achieve something like this with parts from the scrap pile.
The easiest person to fool is yourself. -- Richard Feynman
 

Offline CharlotteSwissTopic starter

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Here's a classic capacitor charge-discharge curve:
It should be possible to achieve something like this with parts from the scrap pile.
I therefore imagine that with the oscilloscope it is easy to identify a capacitor that no longer works well, observing its charge - discharge curves. Great, thanks  ;)
 

Offline CharlotteSwissTopic starter

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there is one thing that does not fit well in my head: here I learned that normally 10x probes are used, ok.
I would like to give an example and understand if I have learned well: we have a Vp-p voltage of 30v. I measure with the 10x probe, the probe resistance will reduce the voltage input in the oscilloscope to 30/10 = 3vp-p. The oscilloscope set correctly with a 10x probe, on the screen will indicate the right amplitude of 30Vp-p (and not 3v)
I hope I understood..
 :phew:
 

Online tautech

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there is one thing that does not fit well in my head: here I learned that normally 10x probes are used, ok.
I would like to give an example and understand if I have learned well: we have a Vp-p voltage of 30v. I measure with the 10x probe, the probe resistance will reduce the voltage input in the oscilloscope to 30/10 = 3vp-p. The oscilloscope set correctly with a 10x probe, on the screen will indicate the right amplitude of 30Vp-p (and not 3v)
I hope I understood..
 :phew:
:-+
Nailed it !  8)
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Offline tggzzz

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Another experiment: replace the sine wave with a square wave, and see what happens. Then change the duty cycle to 10% and 90%, and explain the what you see to yourself.

measurement only on the output with capacitor.
What do I know? nothing understandable!
It seems to me that the curve is the discharge of the capacitor, then the difference between 10 and 90 is that one signal is opposite to the other.
It feels too difficult for me now

It is difficult to start with. So is riding a bicycle, or skiing, or learning English German/French/Italian/Romansh :) All are worthwhile.

For the particular parameters you have chosen, yes you are observing the capacitor charge curve. The key is in the ratio of the period (166ms, 6Hz) to the RC time constant (150*220e-6 = 33ms).

Change the relative times so they are the other way around. You could decrease the period 1000 times (i.e. 6000Hz), or increase the resistor 100* to 150000, or increase the capacitor 1000* times to 220e-3.

Theory: your current circuit is operating below the cutoff frequency of the high pass filter. Making the changes will move it to above the cutoff frequency. Don't worry about the details, but in the diagrams at https://www.electronics-tutorials.ws/filter/filter_3.html you are currently operating on the 20dB/decade slope. The changes will move you onto the flat part.

And yes, showing both input and output can be revealing.
There are lies, damned lies, statistics - and ADC/DAC specs.
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Offline rstofer

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Another way to view that charge/discharge example above is to enlarge the square wave such that its V/div is the same as the charge waveform.  Then shift it down so that both channel reference 0V lines are in the same location.  Now it is apparent that the upper and lower voltage levels of the charge/discharge waveform are related to the high and low voltages of the square wave.  It might also help in visualizing how many time periods (Tau) are being displayed for the charge or discharge portion of the waveform.

The circuit for the example below is a 0.1 ufd capacitor and a 10K ohm resistor which will result in a 1 ms time constant (Tau=R*C).  We want to show 6*Tau high and 6*Tau low (to reach 99.75% of the input voltage) so 2 ms per division times 3 divisions is just right.

See Reply 54 here:
https://www.eevblog.com/forum/testgear/starter-scope/50/

The traces look ike this:
« Last Edit: July 10, 2020, 09:54:53 am by rstofer »
 

Offline CharlotteSwissTopic starter

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:-+
Nailed it !  8)

sometimes I write the right things too  :palm:  ;)

It is difficult to start with. So is riding a bicycle, or skiing, or learning English German/French/Italian/Romansh :) All are worthwhile.

For the particular parameters you have chosen, yes you are observing the capacitor charge curve. The key is in the ratio of the period (166ms, 6Hz) to the RC time constant (150*220e-6 = 33ms).

Change the relative times so they are the other way around. You could decrease the period 1000 times (i.e. 6000Hz), or increase the resistor 100* to 150000, or increase the capacitor 1000* times to 220e-3.

Theory: your current circuit is operating below the cutoff frequency of the high pass filter. Making the changes will move it to above the cutoff frequency. Don't worry about the details, but in the diagrams at https://www.electronics-tutorials.ws/filter/filter_3.html you are currently operating on the 20dB/decade slope. The changes will move you onto the flat part.

And yes, showing both input and output can be revealing.


something I understand, thanks. But maybe I should get there step by step, taking too long a leap could be dangerous for my poor head
 ;)

Another way to view that charge/discharge example above is to enlarge the square wave such that its V/div is the same as the charge waveform.  Then shift it down so that both channel reference 0V lines are in the same location.  Now it is apparent that the upper and lower voltage levels of the charge/discharge waveform are related to the high and low voltages of the square wave.  It might also help in visualizing how many time periods (Tau) are being displayed for the charge or discharge portion of the waveform.

this image helps a lot to understand when charging and discharging takes place, very educational
thanks  ;)
 

Offline rstofer

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Regarding 1x versus 10x probes:

When you create a Tau=1 ms RC circuit, you can use a wide range of values:  10k and 0.1 ufd or 100k and 0.01 ufd or 1M and 0.001 ufd as examples.

But, when the probe is on 1x the scope impedance is 1M ohm.  This doesn't form much of a voltage divider with the 10k resistor but it divides the charge voltage in half for the 1M circuit.  You have 1M feeding the capacitor and another 1M across the capacitor to ground dividing the voltage in half.  The scope loading is significant in 1x mode on a 1M circuit.

It is only 1/10th as significant if the probe is 10x.  Then you might have a 1M resistor in series with a 10M input impedance which is in parallel with the capacitor.  The capacitor sees 10/11 of input voltage.

That's one reason for choosing the 10k 0.1 ufd components.  Even on 1x, the scope isn't a really significant load and on 10x it is negligible.

Always think about how your scope probe is upsetting the circuit.  Use 10x almost always and there is probably a reason for 100x probes if the voltage is high enough to display on the scope after dividing by 100.  A 5v circuit would measure fine with 100x probes because the scope would be displaying 0.05 (50 mV) and most scopes can handle much lower values.

I didn't work through the numbers, I just grabbed the closest 0.1 ufd capacitor laying on my bench.  Then I chose the 10k resistor to come up with Tau=1 ms.  I just lucked out because I really didn't include the scope impedance when I began the experiment.  I was using the Analog Discovery 2 with the 1x flyleads.  Because 10k ohm is much lower than 1 M ohm, the output voltage reaches very close to the input voltage (within 1%).

If you want to reallly understand what I wrote, build the 1M/0.001ufd circuit and measure it in 1x and 10x mode.  Then build the 10k/0.1ufd circuit and compare the results.  Or model it!  I don't prefer the models, they're not real components with a real scope with real input impedance.  But the ideas will be the same.

The input frequency should be related to 12 Tau (6 to charge, 6 to discharge) so 12 * 1ms = 12 ms and 1/0.012= 83.3 Hz

There's a lot of learning going on with these simple circuits.
« Last Edit: July 10, 2020, 01:54:38 pm by rstofer »
 

Offline CharlotteSwissTopic starter

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Regarding 1x versus 10x probes:

When you create a Tau=1 ms circuit, you can use a wide range of values:  10k and 0.1 ufd or 100k and 0.01 ufd or 1M and 0.001 ufd as examples.

But, when the probe is on 1x the scope impedance is 1M ohm.  This doesn't form much of a voltage divider with the 10k resistor but it divides the charge voltage in half for the 1M circuit.  You have 1M feeding the capacitor and another 1M across the capacitor to ground dividing the voltage in half.  The scope loading is significant in 1x mode on a 1M circuit.

It is only 1/10th as significant if the probe is 10x.  Then you might have a 1M resistor in series with a 10M input impedance which is in parallel with the capacitor.  The capacitor sees 10/11 of input voltage.

That's one reason for choosing the 10k 0.1 ufd components.  Even on 1x, the scope isn't a really significant load and on 10x it is negligible.

Always think about how your scope probe is upsetting the circuit.  Use 10x almost always and there is probably a reason for 100x probes if the voltage is high enough to display on the scope after dividing by 100.  A 5v circuit would measure fine with 100x probes because the scope would be displaying 0.05 (50 mV) and most scopes can handle much lower values.

I didn't work through the numbers, I just grabbed the closest 0.1 ufd capacitor laying on my bench.  Then I chose the 10k resistor to come up with Tau=1 ms.  I just lucked out because I really didn't include the scope impedance when I began the experiment.  I was using the Analog Discovery 2 with the 1x flyleads.  Because 10k ohm is much lower than 1 M ohm, the output voltage reaches very close to the input voltage (within 1%).

If you want to reallly understand what I wrote, build the 1M/0.001ufd circuit and measure it in 1x and 10x mode.  Then build the 10k/0.1ufd circuit and compare the results.  Or model it!  I don't prefer the models, they're not real components with a real scope with real input impedance.  But the ideas will be the same.

The input frequency should be related to 12 Tau (6 to charge, 6 to discharge) so 12 * 1ms = 12 ms and 1/0.012= 83.3 Hz

There's a lot of learning going on with these simple circuits.

rstofer we are already in difficulty for me, but nothing is impossible. However, I did not understand how the circuit to be made on broadboard is composed (I have a choice of mixed components). Are you referring to some scheme that you have previously attached?
Thanks for the valuable advice, now I run to work, I need me out there ...
 ;)
 

Offline tggzzz

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But maybe I should get there step by step, taking too long a leap could be dangerous for my poor head

That's the way to do it!

Taking a leap is best done when you understand where you are going, and that all intermediate steps can be jumped over. That is only possible after you have experience, or with boring things.

One skill you might like to cultivate is to learn how to twiddle things and see what happens :) The most exciting words in science isn't "eureka" but "that's funny".
There are lies, damned lies, statistics - and ADC/DAC specs.
Glider pilot's aphorism: "there is no substitute for span". Retort: "There is a substitute: skill+imagination. But you can buy span".
Having fun doing more, with less
 
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Offline rstofer

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The circuit goes all the way back to that link I provided a couple of posts back

https://www.eevblog.com/forum/testgear/starter-scope/50/  Starting at Reply 52

Basically, there are just two components: a 10k resistor and a 0.1 ufd capacitor.  I do 4 experiments in that thread starting at Reply 52

The signal goes into a 10k resistor in series with a 0.1 ufd capacitor to ground.  The output is measured across the capacitor.

The point of the 3 sets of values is that the charge/discharge would look like my plot above only in the case of the 10k/0.1 ufd setup because, as the resistor gets larger, the voltage divider effect of the scope probe in series with the resistor drops the voltage to the capacitor.

Draw 2 1M resistors in series between, say, 1V and ground.  What is the voltage at the junction where the capacitor would be connected?  Well, it would be exactly 1/2 of 1V because the first resistor drops 0.5V and the 2d resistor (emulating the scope impedance) drops the other half.

Now, replace the first resistor with 10k but leave the 1M to ground.  Now the voltage at the junction will be 0.99V and that is within 1% of the applied voltage.

Here is a better explanation of the voltage divider than I can provide.  Using the early examples, R1 is 10k and R2 is 1M, calculate Vout.  Then make R1 = 1M with R2 (the scope impedance) still 1M and see what happens.  The voltage will be cut in half.

Then change R2 (the scope impedance) to 10M and recalculate the two values as above.  In all cases, use 1V for Vin.

https://www.elprocus.com/voltage-divider-rule-with-examples/

Nothing, even using a scope, is as easy as it seems.  Ohm's Law jumps up and bites us from time to time.  BTW, Ohm's Law really is a Law, not simply a suggestion like speed limits.

The point of this long-winded explanation is just this:  Don't believe everything you see on your scope.  If the trace shows 0.5V, is it because the point being probed is low impedance and the voltage actually IS 0.5V or is it because your 1x probe is upsetting the measurement by creating a voltage divider?  In 10x you still have the voltage divider, it's just not as noticeable.  100x just might be necessary for some high impedance circuits.

Where this comes up, for me, is with my analog computers.  I want Tau=1 second and I use 1 ufd capacitors.  Therefore, the resistor is 1M and it's important to remember what is being probed.

« Last Edit: July 10, 2020, 01:45:19 pm by rstofer »
 

Offline Mr. Scram

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The Japanese still call them condenser (コンデンサー), hence the names Rubycon, Nichicon and Chemicon.
Other languages do too. I'm not sure from which part of Switzerland Charlotte is from but the German, French and Italian translations are Kondensator, condensateur and condensatore respectively. A lot if not most of Europe uses a word related to condenser.

Note, wurdz hard.
« Last Edit: July 10, 2020, 09:27:11 pm by Mr. Scram »
 

Online tautech

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I'm not sure from which part of Swiss Charlotte is from ......
Switzerland.
Best you change your forum settings so you can see everyones profile then while you're in there set a country flag to your own like most other members do.
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Offline Mr. Scram

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Switzerland.
Best you change your forum settings so you can see everyones profile then while you're in there set a country flag to your own like most other members do.
Autocomplete.  :palm: I feel you missed the point though. I can see Charlotte's flag the same I can see everyone else's. Switzerland has three major official languages, plus a load of in-betweens and offshoots I'm not sure qualify as human language at all. Hence German, French and Italian. Depending on the area she's from she may speak any or all of them. You don't really need to worry about my own profile as it's set appropriately. :)
 

Offline CharlotteSwissTopic starter

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That's the way to do it!

Taking a leap is best done when you understand where you are going, and that all intermediate steps can be jumped over. That is only possible after you have experience, or with boring things.

One skill you might like to cultivate is to learn how to twiddle things and see what happens :) The most exciting words in science isn't "eureka" but "that's funny".

certainly, but to make the jumps you need to have experience in the sector, I still lack it  ;)

The circuit goes all the way back to that link I provided a couple of posts back
https://www.eevblog.com/forum/testgear/starter-scope/50/  Starting at Reply 52
Basically, there are just two components: a 10k resistor and a 0.1 ufd capacitor.  I do 4 experiments in that thread starting at Reply 52

thanks, as soon as i have time i try to run these circuits, let's see if i will understand something  ;)

Other languages do too. I'm not sure from which part of Switzerland Charlotte is from but the German, French and Italian translations are Kondensator, condensateur and condensatore respectively. A lot if not most of Europe uses a word related to condenser.

where I live, German / Romansh prevails, but part of my family comes from Italian Switzerland  ;)
« Last Edit: July 11, 2020, 09:53:26 am by CharlotteSwiss »
 


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