Ok, I want to thank you all for the insights of each one but I need to keep focus on the basics and on the academic context. I'm not sure if I'm correct, but looks like some of us here in the thread are not quite sure about what each block means. I'm not saying that you guys don't know what the blocks are, but I think this might help narrowing the options:

All already noticed this is a 3 block circuit.
The AO means only OpAmp in Portuguese, which stands for AmpOp. We use this acronym to refer to OpAmps in general without specifying if it's on amplifier mode or on comparator mode!

So, I already got the meaning of the 3 blocks.
Block 1 - bistable waveform generator
Block 2 - integrator that will create a triangular waveform shape from the Block 1 coming waveform signal
Block 3 - somehow a Duty Cycle modulator/controller with Half-Wave rectification.

Tomorrow I'll get into the maths that our teacher is asking, namely, (d.)) calculate the oscillation frequency for the circuit composed by Blocks 1 & 2, (e.))draw the waveform for that circuit on key points/nodes, (f.)) then draw the waveforms when the rpot is at 1/2, 1/3 and 1/4 positions at the AO3 output and at the circuit output, meaning before and after the diode and lastly to repeat d.) and e.) replacing R3 by Block 4... I still have a lot to do!

 

And is the resistor next to the diode, a pull-down resistor or what is it doing there?

Yes, we can call it a pulldown resistor

What about my attempt to set the Vin voltage as I tried? Why doesn't it work?

Because by mistake you set Tdealy as a 50s.

So, I already got the meaning of the 3 blocks.
Block 1 - bistable waveform generator
Block 2 - integrator that will create a triangular waveform shape from the Block 1 coming waveform signal
Block 3 - somehow a Duty Cycle modulator/controller with Half-Wave rectification.

But block 1 alone is nothing more than a Schmitt trigger.
Only when we connect Block 1 + Block 2 together via a feedback we get waveform gen.
http://www.piclist.com/images/www/hobby_elec/e_ckt16_2.htm
http://www.piclist.com/images/www/hobby_elec/e_ckt16.htm
"Duty Cycle modulator" - We never use this name, PWM is the right term.

Tomorrow I'll get into the maths that our teacher is asking, namely, (d.)) calculate the oscillation frequency for the circuit composed by Blocks 1 & 2, (e.))draw the waveform for that circuit on key points/nodes, (f.)) then draw the waveforms when the rpot is at 1/2, 1/3 and 1/4 positions at the AO3 output and at the circuit output, meaning before and after the diode and lastly to repeat d.) and e.) replacing R3 by Block 4... I still have a lot to do!

The task is not that hard but good luck with that.

 

I have heard of PWM but I've never heard it called PXX. Kind of strange to make a three letter acronym for a three letter acronym.

Hi,

That's funny

I guess we could have something like PxM though, which would stand for any kind of pulse modulation like PWM, PPM. PSM, whatever. I would rather see it written out though. Being more clear is always better than being less clear

 

Yes, we can call it a pulldown resistor

Because by mistake you set Tdealy as a 50s.


But block 1 alone is nothing more than a Schmitt trigger.
Only when we connect Block 1 + Block 2 together via a feedback we get waveform gen.
http://www.piclist.com/images/www/hobby_elec/e_ckt16_2.htm
http://www.piclist.com/images/www/hobby_elec/e_ckt16.htm
"Duty Cycle modulator" - We never use this name, PWM is the right term.

The task is not that hard but good luck with that.

Ok, thanks...

Yet about that pull-down resistor... The role of that resistor is to make sure that a lower voltage potential is at the diode's anode and make sure the diode is forward biased??


One other question:
Teacher is asking now to analyse the circuit with Block 1 and Block 2. In this case I already don't need a specific Vin as for when I analysed Block 1 alone where I used a voltage supply at node A????

Hum, I think I need a Vin to make that circuit to work!

 

Ok, thanks...

Yet about that pull-down resistor... The role of that resistor is to make sure that a lower voltage potential is at the diode's anode and make sure the diode is forward biased??


One other question:
Teacher is asking now to analyse the circuit with Block 1 and Block 2. In this case I already don't need a specific Vin as for when I analysed Block 1 alone where I used a voltage supply at node A????

Hum, I think I need a Vin to make that circuit to work!

No Vin is needed, it will oscillate all by itself. The comparitor will output a square wave op amp, the integrator op amp will give you the triangle wave.

 

Ok, thanks...

Yet about that pull-down resistor... The role of that resistor is to make sure that a lower voltage potential is at the diode's anode and make sure the diode is forward biased??


One other question:
Teacher is asking now to analyse the circuit with Block 1 and Block 2. In this case I already don't need a specific Vin as for when I analysed Block 1 alone where I used a voltage supply at node A????

Hum, I think I need a Vin to make that circuit to work!

I PMed a nice analysis to you.

 

Yet about that pull-down resistor... The role of that resistor is to make sure that a lower voltage potential is at the diode's anode and make sure the diode is forward biased??

This resistor is here to "set" the voltage at point D at 0V when diode is OFF.

Teacher is asking now to analyse the circuit with Block 1 and Block 2. In this case I already don't need a specific Vin as for when I analysed Block 1 alone where I used a voltage supply at node A????
Hum, I think I need a Vin to make that circuit to work!

If you are gonna to analyze Block 1 alone then yes, node A is the input terminal.
https://en.wikipedia.org/wiki/Schmitt_trigger#Non-inverting_Schmitt_trigger

 

Ok, I want to thank you all for the insights of each one but I need to keep focus on the basics and on the academic context. I'm not sure if I'm correct, but looks like some of us here in the thread are not quite sure about what each block means. I'm not saying that you guys don't know what the blocks are, but I think this might help narrowing the options:

All already noticed this is a 3 block circuit.
The AO means only OpAmp in Portuguese, which stands for AmpOp. We use this acronym to refer to OpAmps in general without specifying if it's on amplifier mode or on comparator mode!

So, I already got the meaning of the 3 blocks.
Block 1 - bistable waveform generator
Block 2 - integrator that will create a triangular waveform shape from the Block 1 coming waveform signal
Block 3 - somehow a Duty Cycle modulator/controller with Half-Wave rectification.

Tomorrow I'll get into the maths that our teacher is asking, namely, (d.)) calculate the oscillation frequency for the circuit composed by Blocks 1 & 2, (e.))draw the waveform for that circuit on key points/nodes, (f.)) then draw the waveforms when the rpot is at 1/2, 1/3 and 1/4 positions at the AO3 output and at the circuit output, meaning before and after the diode and lastly to repeat d.) and e.) replacing R3 by Block 4... I still have a lot to do!

Hi again,

I dont know if i would look at that last stage as a "rectifier" circuit. I tend to think of it more as a logic level translator. If the power supply for the circuit is plus and minus 5v but we need a 0 to 5v logic signal at the output for the next stage which is purely logic, then that last stage could be called a "logic level translator" or "logic level shifter" because it takes a -5v to +5v signal and converts it to 0v to +5v only, and that is what a logic level converter does. The comparator itself however generates the square (rectangular) wave in the first place, but the diode and resistor convert the logic level to an acceptable level for the next stage (which is not shown yet).
So "rectifier" although acceptable may be too general whereas "logic level converter" is more specific to what it actually is being used for. Maybe call it, "rectifier that does the logic level translation".

 

Ok, guys... Thanks once more for the help given...

I'll use the term "logic level converter" for the 3rd block.

Now I'm trying to simulate the 2 first blocks in LTSpice and as this is kind of a trick stuff to simulate because something weird happens. The triangle waveform is clipping either in the upper side or in the lower side depending on which waveform is being integrated!

My simulation is attached. How can I improve it?

Note: the asc file contains the setup for a triangular wave in a text box, if needed, so that no one need to create it from scratch every time it is needed.

 

You are seeing the offset due to the startup conditions.
Adding the directive .ic V(out) =5V to set the initial output voltage will help.

Otherwise just connect R1 to the Output signal as shown in your original diagram and it will oscillate to give you a proper triangle wave [select .uic (Skip initial operating point solution) to get it to start].

 

You are seeing the offset due to the startup conditions.
Adding the directive .ic V(out) =5V to set the initial output voltage will help.

Otherwise just connect R1 to the Output signal as shown in your original diagram and it will oscillate to give you a proper triangle wave [select .uic (Skip initial operating point solution) to get it to start].

Like the new attached picture?

Weird start, though!

Also I didn't understood the 2nd option. What is .uic? A directive?

 

Like the new attached picture?
View attachment 94921
Weird start, though!

Also I didn't understood the 2nd option. What is .uic? A directive?

Yes, but you need to increase the value of R3 or the input frequency to prevent clipping of the integrator output.

The start is not "wierd". The integrator is initially at zero voltage output so the integration starts from there.

Yes, .uic is a directive, as I stated in my previous post.

I don't understand what "2nd option" you don't understand.

 

You are seeing the offset due to the startup conditions.
Adding the directive .ic V(out) =5V to set the initial output voltage will help.

Wasn't this one option to get rid of the offset?

Otherwise just connect R1 to the Output signal as shown in your original diagram and it will oscillate to give you a proper triangle wave [select .uic (Skip initial operating point solution) to get it to start].

Wasn't this a second option?

I tried both options and this one

Yes, .uic is a directive, as I stated in my previous post.

says that

Error in line 15 : .uic
Unkown control card

The first one, I'm not sure of it's correct...

Pictures attached!

 

Check your connection, the left side of a R1 resistor must be connected to integrator output.
And next time upload asc file also.

 

.uic (Skip initial operating point solution) is located in the Simulation Command window.

 

ok, does this looks correct now or can it yet be improved?

 

Hi again,

I dont know if i would look at that last stage as a "rectifier" circuit. I tend to think of it more as a logic level translator. If the power supply for the circuit is plus and minus 5v but we need a 0 to 5v logic signal at the output for the next stage which is purely logic, then that last stage could be called a "logic level translator" or "logic level shifter" because it takes a -5v to +5v signal and converts it to 0v to +5v only, and that is what a logic level converter does. The comparator itself however generates the square (rectangular) wave in the first place, but the diode and resistor convert the logic level to an acceptable level for the next stage (which is not shown yet).
So "rectifier" although acceptable may be too general whereas "logic level converter" is more specific to what it actually is being used for. Maybe call it, "rectifier that does the logic level translation".

Ok, I tried to verify that what you said but I couldn't get the expected effect...
At the Output, I still have negative voltages... What is wrong here? I also stepped the wiper value to watch the Duty Cycle modulation!

 

ok, does this looks correct now or can it yet be improved?
View attachment 94967

Looks okay to me.

 

Looks okay to me.

But how come we have negative output voltages if it was supposed the diode to rectify that?

 

But how come we have negative output voltages if it was supposed the diode to rectify that?

It's the output of the comparator that's rectified, not the output of the integrator.