Can you help with operational amplifier homework?

[Pablo3]

Homework Statement

Good morning everybody,I'm french and my english isn't good but I need help for this exercice please:

I tried to do this exercice,however I'm not sure of my answers,can you tell me if they are good , or what mistakes I committed ?
Here the exercice :

2. Revelant equations
The questions are:
a)Calculate Vs1 and Vs2 according to Ve

Calculate Ie according to Is and I1.
Calculate Ie according to Ve.
b) Show that this dipole is equivalent to a coil L,calculate L according R1,R2,and C.

Deduce the impedance of the assembly(mounting)

3. The Attempt at a Solution

And here is what I have said:

Question a)We use the mesh(or stitch) 1 so Ve-Vs1=

Ve-Z1.ie=0=>Ve=Z1.ie= (R1+R1).ie=2R1.ie(the resistors are in series).

Then for the mesh(or stitch) 2,on a Ve=Z2.ie=R2.ie.

Then to express ie depending on I1 and Is, it can be said according to the law for the nodes ie=i1+is.

After we know that Ze.ie=Ve ,so ,ie=Ve/Ze that's why Ze=Ve/ie=2R1.
b) I don't know

 

[Hesch]

From your second attached figure, it seems that you try to solve the problem by means of KVL. Instead I would suggest that you regard the circuit as a closed loop, included two inner loops, namely the left amplifier circuit ( A1+R1+R1+R1 ) and the right amplifier circuit ( R2+C+A2 ).

The outer loop has a feed back ( Is ) through the upmost R1 ( current feedback ).

As for the inner loops, you must calculate/setup transfer functions. The rightmost is the most easy, so I will choose that to show you what I mean
( I hope you know about Laplace transforms, Laplace was from France, I think ):

The sum of currents into A2_inp- = 0 →
V_s1/R2 + V_s2*sC = 0 →
V_s1/V_s2 = -R2⋅Cs →
V_s2/V_s1 = -1/(R2⋅C)s

So you simply sketch this rightmost transfer function as a "block" inside the outer loop.

Likewise you sketch the transfer function for the leftmost inner block. You should ( by intuition ) see that
V_s1/V_e = 2
because otherwise the amplifier circuit will not be "in balance". ( A1_inp- = A1_inp+ ).

Having sketched the outer loop, apply "Masons rule" to calculate the transfer function, and calculate
V_e/I_e
which is the input impedance of the circuit, and it should be possible to write it in the form:
V_e/I_e = sL
which is the answer to:

b) Show that this dipole is equivalent to a coil L,calculate L according R1,R2,and C.

 

[Pablo3]

From your second attached figure, it seems that you try to solve the problem by means of KVL. Instead I would suggest that you regard the circuit as a closed loop, included two inner loops, namely the left amplifier circuit ( A1+R1+R1+R1 ) and the right amplifier circuit ( R2+C+A2 ).

The outer loop has a feed back ( Is ) through the upmost R1 ( current feedback ).

As for the inner loops, you must calculate/setup transfer functions. The rightmost is the most easy, so I will choose that to show you what I mean
( I hope you know about Laplace transforms, Laplace was from France, I think ):

The sum of currents into A2_inp- = 0 →
V_s1/R2 + V_s2*sC = 0 →
V_s1/V_s2 = -R2⋅Cs →
V_s2/V_s1 = -1/(R2⋅C)s

So you simply sketch this rightmost transfer function as a "block" inside the outer loop.

Likewise you sketch the transfer function for the leftmost inner block. You should ( by intuition ) see that
V_s1/V_e = 2
because otherwise the amplifier circuit will not be "in balance". ( A1_inp- = A1_inp+ ).

Having sketched the outer loop, apply "Masons rule" to calculate the transfer function, and calculate
V_e/I_e
which is the input impedance of the circuit, and it should be possible to write it in the form:
V_e/I_e = sL
which is the answer to:

b) Show that this dipole is equivalent to a coil L,calculate L according R1,R2,and C.

From your second attached figure, it seems that you try to solve the problem by means of KVL. Instead I would suggest that you regard the circuit as a closed loop, included two inner loops, namely the left amplifier circuit ( A1+R1+R1+R1 ) and the right amplifier circuit ( R2+C+A2 ).

The outer loop has a feed back ( Is ) through the upmost R1 ( current feedback ).

As for the inner loops, you must calculate/setup transfer functions. The rightmost is the most easy, so I will choose that to show you what I mean
( I hope you know about Laplace transforms, Laplace was from France, I think ):

The sum of currents into A2_inp- = 0 →
V_s1/R2 + V_s2*sC = 0 →
V_s1/V_s2 = -R2⋅Cs →
V_s2/V_s1 = -1/(R2⋅C)s

So you simply sketch this rightmost transfer function as a "block" inside the outer loop.

Likewise you sketch the transfer function for the leftmost inner block. You should ( by intuition ) see that
V_s1/V_e = 2
because otherwise the amplifier circuit will not be "in balance". ( A1_inp- = A1_inp+ ).

Having sketched the outer loop, apply "Masons rule" to calculate the transfer function, and calculate
V_e/I_e
which is the input impedance of the circuit, and it should be possible to write it in the form:
V_e/I_e = sL
which is the answer to:

b) Show that this dipole is equivalent to a coil L,calculate L according R1,R2,and C.

Thank you for your help,yes Laplace was from France effectively.
I will review my course to completely understand what you say,but I have a question,what does mean "inp-" and what it is "s"?

 

[Hesch]

what does mean "inp-" and what it is "s"?

For example the integrated amplifier, A1 ( leftmost ), has one output, and two inputs ( a negative and a positive ). I have just named the negative input A1_inp- , and thus the positive input A1_inp+ .

"s" is a complex variabel, used in the Laplacian domain, like "t" in the time domain.

For example you can describe a sinusoidal signal in the time domain by: F(t) = V*sin ωt.
The Laplace transform of this signal will be: F(s) = V*ω / ( s² + ω² ).

Now, if you want to determine dF(t)/dt, you do this in the Laplace domain by multiplying F(s) by s, so
s*F(s) = V*ω*s / ( s² + ω² )
which is the Laplace transform of G(t) = V*ω*cos ωt.

In the frequency domain, the impedance of a capacitor is described by: Z_c(ω) = 1 / jωC.
In the Laplacian domain it is described by: Z_c(s) = 1 / sC.

L(dF(t)/dt) = s*F(s)
L( ∫(F(t)⋅dt ) = F(s) / s

That's the clever thing about Laplace transforms. It is very comfortable to do calculations by Laplace. You will see that when you review your course.

 

[Pablo3]

Thank you ,I'm trying to improve my English to post on your forum more often
it's a shame that most people only post on their forums countries ,but it is true that the language barrier is not easy.

 

[Hesch]

Well, in Denmark ( where I'm staying ), about 70% of the books used for studying are in english due to the small size of DK.

"Then we can learn it", as we say.

 

[Pablo3]

70% percent it's very big!
for us it is only scientific book in higher study.
Yes ,you dont't have choice,if only it was the same thing in France,french will be better in english ahahah.