A question about the loop gain and source contained port impedance

  • Thread starter genxium
  • Start date
  • #1
genxium
137
1

Homework Statement



I recently got a problem that confuses for days, the picture for this problem is in the attachment field, is there someone can tell me that, if an oscillator is topology is described using the negative impedance model in the picture, should I have R>r to ensure that the loop gain is larger than 1?

The second question is, is the current direction a significant element to consider ? Because to construct a negative impedance port, usually a voltage source or current source will be used, the source is directional.

I actually get different result by assuming different current directions in the second attached picture, I think the direction really matters, but if so, it'll be confusing that how I should choose the direction !


Homework Equations


I'm afraid there's no equation for this question.

The Attempt at a Solution


No meaningful attempt made. The port impedance I got for picture 2 is [itex]R_{eq}=\frac{U}{i_U}=\frac{Z_1 \cdot (Z_2+Z_3)}{(Z_1+Z_2+Z_3)+g_m \cdot Z_1 \cdot Z_2}[/itex], and for inverse direction current, the impedance I got is [itex]R_{eq}=\frac{Z_1 \cdot (Z_2+Z_3)}{(Z_1+Z_2+Z_3)-g_m \cdot Z_1 \cdot Z_2}[/itex]
 

Attachments

  • loop gain.png
    loop gain.png
    2.1 KB · Views: 327
  • impedance.png
    impedance.png
    3.5 KB · Views: 430
Last edited:

Answers and Replies

  • #2
gneill
Mentor
20,945
2,886
Which current(s) in particular are you referring to when you say "inverse current direction"? Is it the controlled source?
 
  • #3
genxium
137
1
Which current(s) in particular are you referring to when you say "inverse current direction"? Is it the controlled source?

Thanks for your reply ^_^

I'm sorry for the poor statement, when i said "INVERSE DIRECTION", I meant the direction of "current" in picture 1, and direction of "[itex]i_U[/itex]" in picture 2.
 
  • #4
gneill
Mentor
20,945
2,886
Hmm. If iU is reversed then would not this imply that U is reversed also, yielding the same U/iU for the impedance?
 
  • #5
genxium
137
1
Hmm. If iU is reversed then would not this imply that U is reversed also, yielding the same U/iU for the impedance?

Dear gneil, I'm afraid not, when I calculated [itex]R_{eq}[/itex] in the inverse current direction, I flipped the whole picture vertically, I think this reverses both the voltage and current :)
 
  • #6
gneill
Mentor
20,945
2,886
Dear gneil, I'm afraid not, when I calculated [itex]R_{eq}[/itex] in the inverse current direction, I flipped the whole picture vertically, I think this reverses both the voltage and current :)

I don't see how the orientation of the drawing on paper could possibly change the circuit's operation!
 
  • #7
genxium
137
1
My reasoning:

Use the transformation shown in the attachments.

[itex]Z_a=\frac{Z_1 \cdot Z_3}{Z_1+Z_2+Z_3}[/itex]
[itex]Z_b=\frac{Z_2 \cdot Z_3}{Z_1+Z_2+Z_3}[/itex]
[itex]Z_c=\frac{Z_1 \cdot Z_2}{Z_1+Z_2+Z_3}[/itex]

[itex]U=i_U \cdot Z_a+(i_U-g_m \cdot U) \cdot Z_c[/itex], the port impedance [itex]R_{eq}[/itex] is

[itex]R_{eq}=\frac{Z_1 \cdot (Z_2+Z_3)}{(Z_1+Z_2+Z_3)+g_m \cdot Z_1 \cdot Z_2}[/itex]
 

Attachments

  • Pi-Y-Trans.png
    Pi-Y-Trans.png
    14.7 KB · Views: 460
  • port-impedance.png
    port-impedance.png
    3.6 KB · Views: 402
Last edited:
  • #8
genxium
137
1
I don't see how the orientation of the drawing on paper could possibly change the circuit's operation!

I'm sorry gneill, you're right ~ Thanks a lot for your help ! But I'm still curious, is there any port impedance that would change subject to measuring direction (like flipping +/-)?
 
  • #9
gneill
Mentor
20,945
2,886
I'm sorry gneill, you're right ~ Thanks a lot for your help ! But I'm still curious, is there any port impedance that would change subject to measuring direction (like flipping +/-)?

Sure, if the network contained some non-linear component(s) such as diodes.
 

Suggested for: A question about the loop gain and source contained port impedance

Replies
1
Views
2K
Replies
2
Views
884
Replies
3
Views
2K
Replies
4
Views
6K
Replies
4
Views
3K
  • Last Post
Replies
8
Views
4K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
970
Replies
8
Views
7K
Top