Consider the following Tension_Current Converter.

 

Consider the following Tension_Current Converter.Explain the behaviour of the circuit if:

a)R1 gets interrupted

b)R2 gets interrupted

c)Zener short-circuits

d)The load short-circuits


a)If R1 gets interrupted (i suppose that means R1 is like an open circuit) and thefore V+=0 V,thefore Vout increases ,so ib will increase and so will Ie and the output of the OpAmp will saturate for -VCC

You mention Vout, yet Vout is not labeled on your diagram anywhere, so are we just supposed to throw a dart at the diagram and guess which node you are talking about?

If you are talking about the output of the opamp, if it is increasing, how would it saturate at -VCC?

But why would it increase in the first place? If V+ goes to 0V, then why wouldn't the opamp try to find an output that makes V- also equal to 0V?

b)If R2 is open circuit the i think that the transistor will possibly blow up,because there is not a ground reference in the mesh,so the load will stop being feed

Why would it blow up? What is the damage mechanism?

C)If the zener shorts then the load current will no longer the value that was intended

Duh! If any of these happen you don't expect the load current to remain what was intended.

How is this situation significantly different from (A)?

d)V- increases,and Vout increases

Why would V- increase?

If V- did increase, why would Vout increase?

 

You mention Vout, yet Vout is not labeled on your diagram anywhere, so are we just supposed to throw a dart at the diagram and guess which node you are talking about?

If you are talking about the output of the opamp, if it is increasing, how would it saturate at -VCC?

Yes i am talking about the output of the opamp.It would increase and so as there is not anymore feedback i thought it would saturate at any point..

But why would it increase in the first place? If V+ goes to 0V, then why wouldn't the opamp try to find an output that makes V- also equal to 0V?

Yes,so we get what 0 V at the output of the opamp?

Why would it blow up? What is the damage mechanism?

The damage mechanism is the transistor because there is not any ground reference point and the current that passes through the colector of the transistor is much larger without R2..

Duh! If any of these happen you don't expect the load current to remain what was intended.

In some of them it could be the same ,but that in fact it is not true here.

How is this situation significantly different from (A)?

It isn't.We would have 0 V at V+?The problem is that there is not anymore a reference in that mesh..

Why would V- increase?

If V- did increase, why would Vout increase?

The load shorts IC increases,IE increases therefore V- increases, thefore as V->V+ that implies that vout increases

 

Yes i am talking about the output of the opamp.It would increase and so as there is not anymore feedback i thought it would saturate at any point..

Why do you say there is no feedback? The feedback is still there, you simply have an unintended, but completely well defined, input signal.

But even if Vout increased, as you said, why would it saturate at -Vcc (i.e., the lowest of the voltages)? How does it go from increasing in value to ending up at
at the lowest voltage?

Yes,so we get what 0 V at the output of the opamp?

Is that the voltage at Vout that results in the voltage at V- (the two are NOT the same) being 0V? That might work. Are there any other values for Vout that would probably also be consistent with V- being 0V?

The damage mechanism is the transistor because there is not any ground reference point and the current that passes through the colector of the transistor is much larger without R2..

How is a component ("the transistor") a damage mechanism? By "damage mechanism", we mean the physical process by which the transistor is being damaged, such as excessive voltage, current, or power.

If R2 is open circuited, where is the current that is passing through the collector going? Remember KCL.

It isn't.We would have 0 V at V+?The problem is that there is not anymore a reference in that mesh..

You keep throwing around the phrase "reference in the mesh". What does that mean?

The load shorts IC increases,IE increases therefore V- increases, thefore as V->V+ that implies that vout increases

Why does IC increase if the load short?

How does V- being greater than V+ imply that Vout is increasing? If V- is greater than V-, then is the differential input voltage positive or negative?

As I've said before, most of your problem stem from extremely weak circuit (and physics) fundamentals. I think I've seen enough decent reasoning skills from you that, if you go back and overcome those weak fundamentals, you will be fine. But until you do, you will continue to flounder becuase your reasoning is being based on fundamentally flawed basics.

 

Why do you say there is no feedback? The feedback is still there, you simply have an unintended, but completely well defined, input signal.

But even if Vout increased, as you said, why would it saturate at -Vcc (i.e., the lowest of the voltages)? How does it go from increasing in value to ending up at at the lowest voltage?

Is that the voltage at Vout that results in the voltage at V- (the two are NOT the same) being 0V? That might work. Are there any other values for Vout that would probably also be consistent with V- being 0V?

If V-=0V and V+=0V also,i am not seing what other value for the output since the diferential tension in the input is equal to 0V

How is a component ("the transistor") a damage mechanism? By "damage mechanism", we mean the physical process by which the transistor is being damaged, such as excessive voltage, current, or power.

If R2 is open circuited, where is the current that is passing through the collector going? Remember KCL.

It is going into the inverting input of the opamp

You keep throwing around the phrase "reference in the mesh". What does that mean?

It means that there is not ground reference along a certain mesh in the circuit and therefore the components in that mesh are going to be subject to high values of tension

Why does IC increase if the load short?

The load is attached to the colector of the transistor.So if the load disapears then Ic rises

How does V- being greater than V+ imply that Vout is increasing? If V- is greater than V-, then is the differential input voltage positive or negative?

It is called feedback and in an opamp i have learned that with VOUT linked to V-, the effect produced at the output of amplifier fedaback negatively (VOUT) is in the opposite direction to cause the produces, ie:

Vout will diminuish if V->V+ and increase if V+>V-

As I've said before, most of your problem stem from extremely weak circuit (and physics) fundamentals. I think I've seen enough decent reasoning skills from you that, if you go back and overcome those weak fundamentals, you will be fine. But until you do, you will continue to flounder becuase your reasoning is being based on fundamentally flawed basics.

 

If V-=0V and V+=0V also,i am not seing what other value for the output since the diferential tension in the input is equal to 0V

In analyzing opamp circuits we always assume that, if the opamp is in the active regime, that V+ = V- and that the differential voltage is 0V. Regardless of what the output is. In fact, there is a very small differential input of a few microvolts to a few millivots in order to sustain the actual output.

It is going into the inverting input of the opamp

Really? And what is the input resistance of the inverting input of an opamp?

It means that there is not ground reference along a certain mesh in the circuit and therefore the components in that mesh are going to be subject to high values of tension

This is basically gibberish. You are basically saying stuff like this when you run out of other ways to describe something. Again, it is a reflection of very poor fundamentals that you really need to correct.

The load is attached to the colector of the transistor.So if the load disapears then Ic rises

Which only underscores that you are not yet equipped to analyze circuits with opamps in them because of the poor fundamentals.

It is called feedback and in an opamp i have learned that with VOUT linked to V-, the effect produced at the output of amplifier fedaback negatively (VOUT) is in the opposite direction to cause the produces, ie:

Vout will diminuish if V->V+ and increase if V+>V-

I don't think you have really "learned" much of anything. You are really just making poor immitations of partial concepts and passing those off as analysis.

Do yourself a favor and start working through the fundamentals. The E-books available on this site are a reasonable place to start. Build up your fundamental skills. You simply are not yet equiped to be dinking around with opamp circuits or any of the other topics you have started threads on. If you want to start working through those I will be more than happy to work with you on question you have as you do so, but I am done trying to help you with concepts that you just aren't ready to deal with.

 

In analyzing opamp circuits we always assume that, if the opamp is in the active regime, that V+ = V- and that the differential voltage is 0V. Regardless of what the output is. In fact, there is a very small differential input of a few microvolts to a few millivots in order to sustain the actual output.



Really? And what is the input resistance of the inverting input of an opamp?

It is infinite i forgot about that.On the other hand Ro of the opamp equals to zero so the current can only go to the output of the opamp unless is magicaly dissapears but Kirchhoff would not like that very much would he?

This is basically gibberish. You are basically saying stuff like this when you run out of other ways to describe something. Again, it is a reflection of very poor fundamentals that you really need to correct.

So this means what that having a real ground is irrelevant??So i suppose the circuit should work because there is a potential difference and electrons are driven through the circuit. So the ground is only something that provides a zero reference voltage and this does not matter to anything else?

I don't think you have really "learned" much of anything. You are really just making poor immitations of partial concepts and passing those off as analysis.

You can assume whatever you want, i am not interested in what you assume or not,if you got something to say that would me help,i appreciate otherwise just do not answer because no one forces you to do so.
But you must have some serious skill for analysing so quickly at person do or does not know. when you are sitting there at your desktop somewhere near the rainbow .Besides i did not even mention what i am studying so i can even be someone who has got a degree in some other area and is trying to learn something about electronics because i need it for a specfic project.But i guess you did not thought of that beucase you were just to busy doing some other stuff and replying at the same time

And i am sure what i said is correct i say it again
It is called feedback and in an opamp i have learned that with VOUT linked to V-, the effect produced at the output of amplifier fedaback negatively (VOUT) is in the opposite direction to cause the produces, ie:

Vout will diminuish if V->V+ and increase if V+>V-