duffy said:
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.
duffy said:
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.
SneaKSz said:
at page 6 on the bottom at the right
Hello, SneaKSz
I don't see where, there are only 5 pages in the PDF.
SneaKSz said:
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.