Understanding project, flashing LED, with triangular wave.

[Thevenin]

Hello, this is my first post, i would like to understand better this project functioning.
I know good the functioning of the 555in the astable configuration, about the charge and cischarge of the capacitor, the comparators, the three 5k resistor voltage divider, the latch, all of it, but in this project I can't see the purpose of the zener diode, i Knowit will see in its catode the voltage at the capacitor is charging, if I use 12V=Vdd, the max voltage it will se is 8v and the minimun is 4V, but why is it there? how would affect not using it?
and what is the 47k resistor doing? just dividing the base current?

Thanks.

 

[Resqueline]

The zener is for dropping the voltage from the cap (that goes from 1/3rd to 2/3rd of the supply) so the LED light is modulated more deeply. The 47k is a pull-down (somewhat voltage dividing too) resistor that also ensures that the LED extinguishes fully when it's supposed to.
Using a zener instead of a resistor divider alone - has the effect that you retain the p-p amplitude whereas the divider would reduce the p-p amplitude as well as the bottom voltage.
Without one or both you'll see that the light from the LED won't appear to vary very much.
LED's (and your eyes too) are more efficient at low levels so therefore it's extra important to "kill" the last bit of current in this application.

 

[Thevenin]

Well, I made the circuit in my bread board and it didn't work, i mean the LED in the astable output was doing its job (oscillating) but the other one didn't do anything, the tried without the diode (direct connection form pin 2 to the base resistor) and it worked, it was flashing just how the charge and discharge wave of the capaciotor is, at least that's what I see. Then I tried with 3 1N4148, just 3 because I didn't have more, and it didn't work.

I'm confused now, I kind of understood what you said, but the capacitor will see a constant voltage that is Vz + Vin_10k_resistor + Vin_47k_resistor, right? what is the purpose of that? modulating the light more deeply? what does it mean?
By retaining the p-p amplitude you mean this wave on the picture attached?

Excuse me for the lot of questions.
Thanks in advance.

 

[Resqueline]

The optional LED is obviously the wrong way around in the diagram.
Are you sure you didn't point the 3 diodes the same way as the zener? That would be the wrong way around and explain the lack of light.
Cap voltage is not constant, but the p-p Voltage is a constant 1/3rd Vcc like in your attachment.
Lets say the 1/3rd point doesn't extinguish the LED, what do you do?
You could use a resistive divider and get it down to lets say 1/6th Vcc, but then the peak would go down to 1/3rd Vcc - and you are left with a signal of only 1/6th Vcc p-p. This method will in other words reduce the signal, - or the light intensity variation.
Better then to just subtract 1/6th of Vcc. That will retain the p-p amplitude of 1/3rd Vcc and so give a stronger signal to drive the LED with.
The 47k related to the 10k will just reduce the signal with 10/57th (17.5%) which is insignificant, but it would be even better if the 47k were connected to the other side of the 10k instead. That would make it not reduce the signal amplitude but still act as a pull-down.

 

[Thevenin]

The optional LED is obviously the wrong way around in the diagram.
Yes it's the wrong way, but i knew that and I'm using it right. It's actally working

Are you sure you didn't point the 3 diodes the same way as the zener? That would be the wrong way around and explain the lack of light.
Yes I'm sure I put them the opposite way of the zener, didn't work that way and didn't either with the zener.

Cap voltage is not constant, but the p-p Voltage is a constant 1/3rd Vcc like in your attachment.
Lets say the 1/3rd point doesn't extinguish the LED, what do you do?
You could use a resistive divider and get it down to lets say 1/6th Vcc,
A resistive divider how? in series with the capacitor, right?

but then the peak would go down to 1/3rd Vcc - and you are left with a signal of only 1/6th Vcc p-p.This method will in other words reduce the signal, - or the light intensity variation.
Better then to just subtract 1/6th of Vcc.
How do you substract 1/6 of Vcc exactly?

That will retain the p-p amplitude of 1/3rd Vcc and so give a stronger signal to drive the LED with.
So you're saying that substracting 1/6 of Vcc will make the peak go below 2/3Vcc but will keep the amplitud to 1/3 Vcc, if that's now what you're saying then tell me please.

The 47k related to the 10k will just reduce the signal with 10/57th (17.5%) which is insignificant, but it would be even better if the 47k were connected to the other side of the 10k instead.
You mean instead of being between the 10K resistor and the NPN base, between the 10k resistor and the zener or in my case the capacitor (taking count of it didn't work with it, but it did without it), I think that would make a difference in the cap voltage and not in the signal that is reaching the transistor base. Is that the point?

That would make it not reduce the signal amplitude but still act as a pull-down.

Thank you very much, hope help me you clear my doubts. I really want to understand the functioning of these circuit, and every one I make.

By the way, without the zener it extingishes very good. But please don't let those questions without
answering.