the teacher said nooo questions ..

That explains a lot.

OK, here's the solution:

Get a teacher who isn't too lazy to do his job. The one you have is useless.

I spent a lot of years preparing and teaching electronics material. Your teacher needs to get the boot.

If he thinks giving problems that can't be solved is clever, he needs to be clever in the unemployment office.

 

If and when the teacher provides a "solution", please post it so we can "learn" something.

I agree this could be a good learning experience for you: it's good to learn that sometimes people who should know something don't.

You have people on an electronics forum with probably a combined length of experience exceeding 60 years or more saying the question is nonsense. Make your own decision.

Maybe the "purpose" of the problem is to see if you can recognize that it is gibberish.

 

If and when the teacher provides a "solution", please post it so we can "learn" something.

Maybe the "purpose" of the problem is to see if you can recognize that it is gibberish.

Of course I will post the solution when I get it : )

He is asking for "mathematical derivation" ..
Can you prove mathematically that this question is "gibberish"?

 

Can you prove mathematically that this question is "gibberish"?

Sure, assume an ideal Op Amp:

Post the equation for input and output impedance. (can be done with a few assumptions, Any assumptions are generally examples to prove a negative or trend, not as data, therefore, not valid.)

Post the equation for gain. (infinity, hence comparator like output)

Amplification = PowerOut > PowerIn Is that satisfied?

 

I'm sorry .. Please bear with me .. As I told you, My info is so limited ..
I can't understand a lot of things you said : (

Now, if it was with negative feedback .. This is how it would be solved ..
right?
http://i.imgur.com/5bifS.jpg

With the positive feedback .. How is it gonna be different?
Pleeeease simple language ..

 

I hope you can understand that it is literally impossible for us to answer your question. The question is nonsensical. I can not postulate or extrapolate any direction it is supposed to go that makes any sense whatsoever.

At the risk of repeating myself for the fifth time, I can not help but point out that you said:

I'm sorry .. Please bear with me .. As I told you, My info is so limited. I can't understand a lot of things you said

Take those words with you and say them to your teacher. That should lead to a positive outcome.

 

I hope you can understand that it is literally impossible for us to answer your question. The question is nonsensical. I can not postulate or extrapolate any direction it is supposed to go that makes any sense whatsoever.

At the risk of repeating myself for the fifth time, I can not help but point out that you said:



Take those words with you and say them to your teacher. That should lead to a positive outcome.

Then what was thatoneguy talking about?

I mean proving mathematically that the questions doesn't make sense ..

 

Then what was thatoneguy talking about?

I mean proving mathematically that the questions doesn't make sense ..

That is why the question is gibberish. The equations cannot be filled in without making a huge number of assumptions.

Due to the fact that no equations can be written (besides the positive feedback resulting in comparator like output), the question is gibberish, as has been stated many times.

I am eager to see what the prof has to explain about this, myself. If he suggests there is an answer with only the info given, you'll have a terrible time learning anything useful.

 

check out this tutorial. It shows very nicely how to derive equations when solving op-amp problems. It even has your last question in there - you will see how it works as there is a circuit called voltage follower, but this isn't quite it.

 

I asked this question in another forum ..
Here is one of the replies ...

"""""
The output voltage is:

Vo = Vi / ( 1- 1/A)
where A is the open loop gain of the op amp. so A>>1
you can disregard 1/A then you have.
Vo=Vi
That is a voltage follower. In theory the same as the standard one with negative feedback.

But in real life the output will smack against one rail or the other and stay there. Any difference between the two inputs is amplified and increases the difference rather than reducing it.

""""""


So what he is saying ..
If the Op-amp is ideal .. ( I+ = I = 0 ) and ( V+ = V- ) .. Then , Vin = Vout in this circuit ..

and if it is not ideal ,, the op-amp will be saturated and Vout will be equal to the value of the external power supply ..

What do you think of this???

 

But in real life the output will smack against one rail or the other and stay there. Any difference between the two inputs is amplified and increases the difference rather than reducing it.

He is saying the same thing I said previously. Go back and read my posts. Since this is the case, the term "amplification" does not apply to this circuit in any way or form. That being true, this proves the question is gibberish. It is something like:

You have five ducks. You add three more ducks. Now, tell me how many elephants do you have?

 

He is saying the same thing I said previously. Go back and read my posts. Since this is the case, the term "amplification" does not apply to this circuit in any way or form. That being true, this proves the question is gibberish. It is something like:

You have five ducks. You add three more ducks. Now, tell me how many elephants do you have?

Loool ,, exactly ..
But what do you think about what he said about the ideal op-amp?

 

...
where A is the open loop gain of the op amp. so A>>1
you can disregard 1/A then you have.
Vo=Vi
That is a voltage follower.
...

But in real life the output will smack against one rail or the other and stay there. Any difference between the two inputs is amplified and increases the difference rather than reducing it.
...

Because the circuit clearly has infinity gain (as myself and others have said). Any discussion of Vo=Vi and a gain of 1 is just silly.

 

Loool ,, exactly ..
But what do you think about what he said about the ideal op-amp?

I don't think it matters.

I agree with:

Because the circuit clearly has infinity gain (as myself and others have said). Any discussion of Vo=Vi and a gain of 1 is just silly.

 

The circuit may or may not have any practical utility, but it can most certainly physically be constructed, its expected behavior described mathematically, and that mathematical behavior comparied to the behevior expected of a linear amplifier.

You can choose to consider that the opamp has infinite gain or just a very large finite gain. You can assume that the output voltage is unbounded are not. You can assume that there the input bias current is zero or just that it is a small finite value. The conclusion iunder any of these assumptions are, for all intents and purposes, the same.

But you cannot assume that V+ = V- because there is no feedback mechanism for making this happen in this circuit.

When trying to work with any kind of a mathematical derivation, itis generally best to avoid working with any kind of "infinity" as long as possible. Instead, use symbolic quantities that can be allowed to go to infinity in the limit at the appropriate point. So use Av for the open-loop voltage gain of the op-amp and use something like Vmax and Vmin for the greatest and least voltage outputs attainable by the op-amp. Then draw Vout versus Vin. Do this for Vin starting below Vmin and going to above Vmax and then for Vin starting above Vmax and going to below Vmin.

 

I went to the teacher's office today and asked him about the question ..
I told him what I learned from you here ..

He was like you are just saying what's in the books or on the internet ..

He said that he wants a mathematical proof that the positive feedback makes the amplifier saturated! ..

 

here is a very good lecture on positive feedback from MIT. I don't know if you have done models of op-amps yet, but he explains it well.

 

Answer the following questions:

The input voltage is Vin.

Q1) Assuming that the assumption V+ = V- is valid, what does Vout need to be?

Now, assume that Vout moves to Vout + dV (perhaps due to noise).

Q2) What are the conditions at V+ and V- now?

Q3) Will the output tend to return to its original value or be driven further away from it?

 

I went to the teacher's office today and asked him about the question ..
I told him what I learned from you here ..

He was like you are just saying what's in the books or on the internet ..

He said that he wants a mathematical proof that the positive feedback makes the amplifier saturated! ..

You know, that takes me back 35 years to when I was in university. We had two teachers who were idiots. One had tenure, which meant he could not be fired, so they just took all his classes away from him and let him do nothing. The second had no tenure, so he was fired.

To your "teacher", I don't see what "mathematical" proof is necessary or even applicable:

You simply look at the input voltages. Whenever the + input voltage of an op is higher than the - input, the output will move positive. If the output is tied back to the + input, it pulls the + input more positive and it keeps going.

If you want to impress the idiot teacher, tell him the effect is REGENERATIVE and see if he knows what that means.

The point is, it keeps going until the output is pinned to the most positive voltage it can reach.

If the + input is pulled negative with respect to the - input, the output swings negative and the output will end up at the negative limit of it's range by the same action.

This is how op amps work. Any fool with basic knowledge of op amps understands this, and someone who demands "mathematical proof" of how it works is, in my opinion, not just any fool but a really special kind of fool.

I have reviewed this "problem" a number of times, and I don't know what your teacher wants to obtain from this process other than a sense of superiority or the sociopathic delight of tormenting others because he can.