The Op-Amp in Commutation

[Jugurtha]

Hello,


This second lab assignment is about the Operational Amplifier In Commutation, multivibrators (Monostable and Astable specifically).

It is the theoretical work that you do before getting to the lab, and when you go to the lab there is the "practical part" where you manipulate and read values on the scope and tinker with things.

I found that there is a terrible lack of covering the maths behind the multivibrators.

I hope this basic paper will help unveil some aspects of it, and help understand a little better.

Any feedback would be appreciated and incorporated to the following drafts. Why not do a small detailed guide.

 

[duffy]

You might add waveforms to the schematics - take up half or an entire page with this, it's an important visual aid. Your use of the term "in vigor" is brief, but very awkward - it might be better to say "the output swings to it's maximum positive (minimum negative) value" instead.

 

[SneaKSz]

Hello,

at page 6 on the bottom at the right, the C will indeed discharge very quickly but it won't reach 0 because the diode is forward biased so it voltage drop is ~ 0,7V., the current will decrease (Rdiode increases) so Vdiode and also Vc so slowly it will decrease.

 

[Jugurtha]

You might add waveforms to the schematics - take up half or an entire page with this, it's an important visual aid.

Hello, Duffy

100% with you. I always draw the signals trace, it tells more things than words. The reason I didn't put waveform is that the schematics are done with Multisim, but I'll draw the waveforms with LT Spice.. Will include them soon.

Your use of the term "in vigor" is brief, but very awkward - it might be better to say "the output swings to it's maximum positive (minimum negative) value" instead.

It might seem awkward, but I didn't find any term that serves the context better than this one. It's like saying that an order is effective until midnight. It's in vigor.

I think there is a misunderstanding here, V+ and V- aren't the maximum positive nad minimum negative; V+ and V- are the voltages at the non inverting terminal when the op-amp swings to its maximum positive value VSAT+ and VSAT-.

I didn't want to emphasize the output, I wanted to emphasize the voltage at the non-inverting terminal by saying V+ or V- is in charge now .. You know, something that controls the circuit.

Thanks for the feedback, I'll add those traces as soon as I finish them.

at page 6 on the bottom at the right

Hello, SneaKSz

I don't see where, there are only 5 pages in the PDF.

the C will indeed discharge very quickly but it won't reach 0 because the diode is forward biased so it voltage drop is ~ 0,7V., the current will decrease (Rdiode increases) so Vdiode and also Vc so slowly it will decrease.

I don't understand what you just wrote, but I think you are talking about the entry-circuit (C1, R1, D1)

Well, once Ve = 0, you have a closed circuit formed by C1, R1 (in parallel with D1). For one moment, the voltage across R1 is (-E) which biases the diode D1 directly and it conducts.

It is true that it will have a certain voltage drop, though a diode voltage drop isn't a generator.

In other words, the capacitor will discharge like there is no tomorrow.

 

[SneaKSz]

I meant page 2, well in the calculations you search the time that is generated due to input change, therefore you set the threshold value for V+ to 0V, all I wanted to say it that it will be a little bit less because V- will have a slightly negative voltage due to the voltage drop across the diode.

 

[Jugurtha]

I meant page 2, well in the calculations you search the time that is generated due to input change, therefore you set the threshold value for V+ to 0V, all I wanted to say it that it will be a little bit less because V- will have a slightly negative voltage due to the voltage drop across the diode.

I frankly don't understand. Page 2, Okay. V+ and V- are the thresholds, the diode does have nothing to do with them. As I said above, a diode isn't a battery. The voltage across the diode will be 0 after it shorts the resistance and the capacitor discharges. Why ? Because there will be no current, if there is no current then VR=R*I=0=VD. No voltage drop across the resistor, which means no voltage drop across the diode. There is no 0.7volts anymore. That's only the case when it conducts, and it's not.

I am sorry if I misunderstood what you wanted to say again..

 

[SneaKSz]

Well I've plotted what I'm trying to tell you :

https://i39.tinypic.com/10e0k1y.jpg

 

[duffy]

Permit me to rephrase: Nobody knows what the hell "in vigor" means here.

 

[JimB]

Jugurtha

Your French expression "en vigueur" would be better translated to English as "in effect" or "effective"

So the expression
V+ is in vigor when ...

Would be better as
V+ is effective when...


JimB

 

[Jugurtha]

Duffy

JimB, when I wrote it, the word was on my tongue but couldn't get out.. I used it in post #4 to make it clearer for Duffy. (Although "vigor" sounds more Hard-core ) (I'll use "in effect" or "effective")



SneaKSz .. I'm sorry I didn't understand you earlier. That plot is awesome, I didn't figure out how to do that in LT Spice just yet..

Maybe the trace looks like that in a simulation environment, but wire it and look at it in a real scope.. Vminus isn't that. It isn't that at all. It's a really tight impulse, a positive decaying exponential and not "squary" at all.

That's why simulation environments drive me nuts sometimes. The math is obvious, you can trace it with your hand, but you can't do it with a computer (at least, not as intuitively as you do it with your hand).

Here's how it should look something like this (see attachment)

I could've used an alternative circuit (like the one used in the first lab assignment) see attachment Alternative.JPG

Thanks,

 

[Jugurtha]

Here's the revision 5

Hey again,

I have added the waveforms for many parts. I didn't put them last time because I couldn't get LT Spice to do square waves to excite the Monostable, so I wired the Astable in it.

I think it's pretty much detailed now. What do you think ?

PS:I'll upload it in Articles.

 

[duffy]

Beautiful! Nice work.

 

[Jugurtha]

All right then,

Thanks for your help, guys...