Ah ok... Sure... Yes I can understand! (After looking 3 times at the circuit I find my self really stupid! ...)

 

OK. Psy
So with the understanding what effect V1 and D3 will have when a Vin of -15V is input to the circuit, you should now be able to state what Vout is and also the operation of the diodes.
E
BTW: did the Prof specify the -15V input.?

 

OK. Psy
So with the understanding what effect V1 and D3 will have when a Vin of -15V is input to the circuit, you should now be able to state what Vout is and also the operation of the diodes.
E
BTW: did the Prof specify the -15V input.?

Ok, I can try...

Well, the problem statement is kinda split in 2. The first part says to find Vout in function of Vin. The second part says to draw Vout if Vin is 15cos (2*pi*1000t).

So I was trying to work out the first part but as I was struggling I tried with a discrete value for Vin to try to figure something out!

 

So, for the original circuit, I think I can say that we will have (Vin - Vzener) at zener's anode and (Vin - Vzener) - Vd1 + V1 at Vout node. Am I correct?

 

So, for the original circuit, I think I can say that we will have (Vin - Vzener) at zener's anode and (Vin - Vzener) - Vd1 + V1 at Vout node. Am I correct?

As with your previous circuits, there is a level of -Vin at which the Zener will not be conducting in the zener mode,
eg: when -Vi is < than -(Vz +Vd)

Also you should consider the case where Vin is positive, as this will require a different set of limits for Vout.

Try to solve part 1 of your question

EDIT:
You should remember its the transfer function of the circuit as a whole for any voltage the user may connect as Vi, not a specific solution for a particular Vi.

 

Oh God... I'm realising that I'm not going to make through this!

There are too many variables to take into account at the same time and I'm not being able to tackle it!

Is there any way that I can analyse this in separate parts, like breaking the circuit into 2 or 3 parts and analyse each one separately and then gather them all at the end (mathematically)?

 

Don't get into flap.
You have virtually done the -Vin part of the problem.
From -Vi = 0V to -(Vz+Vd) the zener/diode path is not conducting, so -Vo = -Vi
When -Vi >= -(Vz+Vd) then the zener/diode path conducts, so -Vo = -(Vi - (Vz+Vd))/2
Remember V1 and D3 are not in circuit for negative Vi.
Check this and see if you agree.

 

You have virtually done the -Vin part of the problem.
From -Vi = 0V to -(Vz+Vd) the zener/diode path is not conducting, so -Vo = -Vi

Was this your circuit from post #10?

If so, wouldn't we have a more negative voltage at D1 cathode and a more positive voltage at D1 anode, making it to conduct and a more negative voltage at zener's anode and a more positive voltage at zener's cathode making it also to conduct? I'm confused where you say I already virtually solved that part!

 

I do not follow why you are saying you are confused?

Post #10 shows the current path and voltages when the Vi = -15V, so the diode and zener are both conducting, the Vout is close to -10V.
You said you understood that diagram.?

The Ground reference on the circuit is at 0V, the zener's Cathode is at 0V and its Anode is -5V.

ALL the voltages on that circuit are referenced to 0V.

EDIT:

From -Vi = 0V to -(Vz+Vd) the zener/diode path is not conducting, so -Vo = -Vi

Is it this part thats causing you a problem.?
If yes, consider what happens when Vi is less than -(5V+Vd), will the zener/diode be conducting.?

 

I do not follow why you are saying you are confused?

Post #10 shows the current path and voltages when the Vi = -15V, so the diode and zener are both conducting, the Vout is close to -10V.
You said you understood that diagram.?

The Ground reference on the circuit is at 0V, the zener's Cathode is at 0V and its Anode is -5V.

ALL the voltages on that circuit are referenced to 0V.

EDIT:


Is it this part thats causing you a problem.?
If yes, consider what happens when Vi is less than -(5V+Vd), will the zener/diode be conducting.?

I think so... Let me just try to say the condition that makes things work.

As we have tried before, if we have 0V at the Zener's cathode, we will have -5V at it's anode and we will have -5.7V at D1 cathode, right? This way we are saying that D2 is working as a zener and D1 is working as a regular diode!
So, now I need to find the Vin range that makes this possible. This is my struggle, I think.

Replying to your post, it's that and not only. My struggle, I think, is to find the Vin ranges that makes a specific circuit condition to be possible!

I would like to think about some approach that would allow me to apply to any circuit and find these ranges and equations.

What I have just told about the voltage at D1 cathode, would be a key voltage, right? the -5.7V would be a voltage that would make the circuit to change behaviour by means of diodes changing states!

 

I wrote it as Vout = (Vin - Vd1 - Vz)/2. It's the same

Lets set Vin to -4v in order to test your equation.

Vout= (-4 - 0.7 - 5) / 2 = -4.85v ; [actual is -4v].

You have to consider the two conditions for -Vin, first when -Vin is > -(Vz+Vd) and second when -Vin < -(Vz+Vd)

 

Lets set Vin to -4v in order to test your equation.

Vout= (-4 - 0.7 - 5) / 2 = -4.85v ; [actual is -4v].

You have to consider the two conditions for -Vin, first when -Vin is > -(Vz+Vd) and second when -Vin < -(Vz+Vd)

So, the total voltage drop at this branch will be a key voltage, right? At this voltage, -5.7V, the circuit will change somehow by means of diodes changing their states, right?

But in the meantime, we already stated that the other branch, the one with the V1 = 10V and D3 is out because D3 is OFF, correct? Or were we analysing the circuit ignoring this branch at the right side of the circuit?

 

So, the total voltage drop at this branch will be a key voltage, right? At this voltage, -5.7V, the circuit will change somehow by means of diodes changing their states, right?

Yes, this is the point I have been trying to get you to understand in the last 10 posts.

will change somehow by means of diodes changing their states,

The change of states is either the diodes/zener are conducting or not.

So write the transfer function that covers that.

Then do the same for when Vin is 0V to say +15V

You are making this hard work Psy.!


edit corrected '=' to '+'

 

Yes, this is the point I have been trying to get you to understand in the last 10 posts.



The change of states is either the diodes/zener are conducting or not.

So write the transfer function that covers that.

Then do the same for when Vin is 0V to say +15V

You are making this hard work Psy.!


edit corrected '=' to '+'

I'm really sorry... It's being really hard to me to understand this, I don't know why! I'm really disappointed by now because this shouldn't be so hard to understand.

By transfer function you mean the Vout equation?
And this value of -5.7V is a key voltage but when I write down an equation for Vout, I also need to find a range in within this Vout equaiton is valid. At this point I only have a single value, which is -5.7V.

So for this situation to happen, D1 -> ON and D2 -> zener, Vout would be, I think, (Vin - Vd1 - Vzener)/2. This would be a voltage divider for either of R1 or R2 as they have the same resistance!

But I still need to find the other end of the Vin range that makes this situation possible. At this point I only know that for some Vin to -5.7V, D1 is ON and D2 is a zener. I don't know yet if it's from "- some Vin" up to -5.7V or if it's from -5.7V up to "some other Vin".

I know you already said that it's from -5.7V up to 0V, but I still haven't figured out why the other end of the Vin range is 0V!

 

This will have to be last post of today.

Lets look at -Vin in a different way.

Lets start with Vin = -1V
We know that Vz+Vd =-5.7V, this means as Vin is only -1V, as the zener/diode requires -5,7V to conduct no current flows thru the zener, so Vout will be -1V

Let Vin= -4v , thats still too low to make the zener/diode conduct, so Vout = -4v.

Let Vin= -5.7v , its equal Vz+vd [ 5.7v] so the zener is at the point of conducting, but Vout still equals Vin ie: -5.7v

Let Vin = -6.7v, which is greater than Vz+Vd [5.7v] so the zener and diode will be conducting.
The current flowing now flowing thru R1 D1 Z1 will cause the Vout to drop lower than Vin.
The drop in Vout will be due to voltage drop across R1.

So -5.7v is the voltage at which the change in Vout occurs.

 

Ok, so, that means that for this situation to occur, Vin must be lower than -5.7V, right? -6.7V is lower than -5.7V. So the range would be Vin < 5.7V.

So Vout = (Vin - Vd1 - Vzener)/2 for Vin < -5.7V. Hope this is correct because I think I'm finding a small inconsistency with what you wrote at your post #27

When -Vi >= -(Vz+Vd) then the zener/diode path conducts, so -Vo = -(Vi - (Vz+Vd))/2

-Vi >= -(Vz+Vd) <=> Vin <= Vz + Vd

which is not exactly the same as I have -- Vin <= -(Vz + Vd)

 

So Vout = (Vin - Vd1 - Vzener)/2 for Vin < -5.7V

How can that be correct.?

Consider if Vin < -5.7V , how can the zener/diode path be conducting and be causing a drop due to current thru R1.?

Vin < -5.7V , means that Vin is between 0v and -5.7v
[ so -Vo = -Vin. zener/diode Off]

Vi > -5.7v means that Vi is between -5.7v and say -15v. [zener/diode On]

 

Morning...

Are you considering -5.7V a lower value than -15V???
If I draw the real line and mark there the -5.7V, if I say lower than 5.7V I'll draw an arrow to the left going to values lower than -5.7V, like -6.7, -7.7, -8.7... If I say greater than -5.7V, I'll draw an arrow to the right towards 0V, so -4.7 > -5.7 and -3.7 > -5.7, and so on!

 

Are you considering -5.7V a lower value than -15V???

What do you think.?
Consider that you had a car with a 12V battery and the positive terminal was connected to the car chassis.

If you measured the battery with a DMM from chassis to the negative battery terminal when the battery was full charged and it read say -12v
and if the battery battery was partially discharged and you measured say -11v, which is the lower voltage.???

You seem to be having a problem understanding polarities and referencing to a common point of measurement.

 

What do you think.?
Consider that you had a car with a 12V battery and the positive terminal was connected to the car chassis.

If you measured the battery with a DMM from chassis to the negative battery terminal when the battery was full charged and it read say -12v
and if the battery battery was partially discharged and you measured say -11v, which is the lower voltage.???

You seem to be having a problem understanding polarities and referencing to a common point of measurement.

Ok, I get your point. I was thinking mathematically!
Anyway, this is turning to hard to understand and also to explain via forums and via web. I think it would be better to me to move forward to another problem and next week I'm going to schedule a day to meet my teacher so that he can help me, face-to-face, understanding how to tackle these diodes problems!

Eric, I appreciate all your efforts and I sincerely hope that you can keep helping me out with another problems... I have started to solve a new problem. I would appreciate if you could give a help there!

Thanks
Psy