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This is the circuit, where some of the components values where given and the others, I found myself. (Sorry I have no link to the design, I needed membership there)

enter image description here

The following part says to draw the corresponding AC-circuit and find the magnitude of the transfer function U2/U1 at ω= 0 , ω = infinite , ω =1/sqrt(LC)

This is the AC drawing that I have done enter image description here

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I am asking this question for the purpose of hoping that I get some feedback about the way I solved this problem and the fact that I am not sure if it’s done correctly. By no means this is homework. I am just trying to learn transistors and not knowing what the correct answer is causes doubts if I did it right or not

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  • \$\begingroup\$ How can the transfer function be unity? You haven’t explained why. \$\endgroup\$
    – Andy aka
    Jan 25 '20 at 13:52
  • \$\begingroup\$ the transfer function was given, saying that is approx to 1 so we just use it as one \$\endgroup\$
    – E199504
    Jan 25 '20 at 16:09
  • \$\begingroup\$ Then why is R2 down at 100 ohms. If it were more like 10 kohm I’d agree with you. \$\endgroup\$
    – Andy aka
    Jan 25 '20 at 16:55
  • \$\begingroup\$ @Andyaka It's probably because the problem creators want those solving the problem to assume that the base is biased for the load. \$\endgroup\$
    – jonk
    Jan 25 '20 at 20:59
  • \$\begingroup\$ @Andy aka, actually Urb = 7,3V and Ib= 20μ so Rb= Urb/Ib \$\endgroup\$
    – E199504
    Jan 25 '20 at 23:23
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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When analyzing a transistor we need to use the following relations:

  • $$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$
  • Transistor gain \$\beta\$: $$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$

Using KCL, we can write:

$$ \begin{cases} \text{I}_\text{y}=\text{I}_\text{c}+\text{I}_4\\ \\ \text{I}_\text{t}=\text{I}_3+\text{I}_4\\ \\ \text{I}_\text{x}=\text{I}_2+\text{I}_3\\ \\ \text{I}_8=\text{I}_5+\text{I}_7\\ \\ \text{I}_7=\text{I}_6+\text{I}_\text{y}\\ \\ \text{I}_1=\text{I}_5+\text{I}_6\\ \\ \text{I}_\text{x}=\text{I}_1+\text{I}_2\\ \\ \text{I}_8=\text{I}_\text{t}+\text{I}_\text{c}\\ \\ \beta=\frac{\text{I}_\text{c}}{\text{I}_\text{t}} \end{cases}\tag3 $$

Using KVL, we can write:

$$ \begin{cases} \text{I}_\text{x}=\frac{\text{V}_\text{x}-\text{V}_5}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_5}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_5-\text{V}_4}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{y}-\text{V}_4}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_1}{\text{R}_5}\\ \\ \text{I}_7=\frac{\text{V}_1-\text{V}_2}{\text{R}_6}\\ \\ \text{I}_7=\frac{\text{V}_2}{\text{R}_7}\\ \\ \text{V}_4-\text{V}_1=\alpha \end{cases}\tag4 $$

Now, you can solve your problem using:

FullSimplify[
 Solve[{Iy == Ic + I4, It == I3 + I4, Ix == I2 + I3, I8 == I5 + I7, 
   I7 == I6 + Iy, I1 == I5 + I6, Ix == I1 + I2, \[Beta] == Ic/It, 
   I8 == It + Ic, Ix == (Vx - V5)/R1, I2 == (V5)/R2, 
   I3 == (V5 - V4)/R3, I4 == (Vy - V4)/R4, I5 == (V1)/R5, 
   I7 == (V1 - V2)/R6, I7 == V2/R7, V4 - V1 == \[Alpha]}, {Ix, I2, I3,
    It, I1, I4, Ic, I8, I5, I7, I6, Iy, V5, V4, V3, V1, V2}]]

EDIT

Let's assume \$\alpha=\frac{7}{10}\$, \$\beta=100\$ and \$\text{V}_\text{x}=10\$. In the (complex) s-domain we need to solve the following system of equations (I used Mathematica 12.0 to solve it):

FullSimplify[
 Solve[{Iy == Ic + I4, It == I3 + I4, Ix == I2 + I3, I8 == I5 + I7, 
   I7 == I6 + Iy, I1 == I5 + I6, Ix == I1 + I2, I8 == It + Ic, 
   100 == Ic/It, Ix == ((10/s) - V5)/600, 
   I2 == V5/(((((s*20*10^(-6)))*((1/(s*2*10^(-9)))))/(((1/(s*2*10^(-9)\
))) + ((s*20*10^(-6)))))), I3 == (V5 - V4)/(1/(s*1*10^(-9))), 
   I4 == ((18/s) - V4)/100, I5 == V1/2000, 
   I7 == (V1 - V2)/((1/(s*100*10^(-9)))), I7 == V2/100, 
   V4 - V1 == (7/10)/s}, {V1, V2, V4, V5, Ic, It, Ix, Iy, I1, I2, I3, 
   I4, I5, I6, I7, I8}]]

Now, I used inverse Laplace transform to plot the voltage \$\text{V}_2\left(t\right)\$:

enter image description here

Where:

  • $$\lim_{t\to0}\text{V}_2\left(t\right)=-\frac{7}{10}\tag5$$
  • $$\lim_{t\to\infty}\text{V}_2\left(t\right)=0\tag6$$

The maximum occurs when \$t\approx1.20098\space\mu\text{s}\$ and is then equal to:

$$\hat{\text{V}}_2\left(1.20098\right)\approx16.6022\space\text{V}\tag7$$

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  • \$\begingroup\$ if I may ask why the circuit is different? when I said I had to find the components for Dc. I just had to find what Ib using your formula from above and R2. My problem is that I don't know if the AC drawing is good and the magnitude. If you could help me with that I would appreciate it a lot \$\endgroup\$
    – E199504
    Jan 25 '20 at 16:13
  • \$\begingroup\$ can you please have a look at the comment above. I would really appreciate it \$\endgroup\$
    – E199504
    Jan 26 '20 at 20:38
  • \$\begingroup\$ @be1995 I do not understand your question \$\endgroup\$
    – Jan
    Jan 26 '20 at 20:39

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