A question about the loop gain and source contained port impedance

In summary, the conversation discusses a problem with an oscillator and its description using the negative impedance model. The main questions are whether R>r is necessary for the loop gain to be larger than 1 and if the current direction is significant in constructing a negative impedance port. The conversation also touches on the different results obtained by assuming different current directions. However, it is determined that the orientation of the drawing on paper does not affect the circuit's operation. There is also a mention of the possibility of the port impedance changing when measuring in different directions, but it is noted that this would only occur with non-linear components such as diodes.
  • #1
genxium
141
2

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
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  • impedance.png
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  • #2
Which current(s) in particular are you referring to when you say "inverse current direction"? Is it the controlled source?
 
  • #3
gneill said:
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
Hmm. If iU is reversed then would not this imply that U is reversed also, yielding the same U/iU for the impedance?
 
  • #5
gneill said:
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
genxium said:
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
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
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  • port-impedance.png
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  • #8
gneill said:
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
genxium said:
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.
 

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

How do you calculate the loop gain in a circuit?

The loop gain in a circuit can be calculated by finding the open-loop gain of the circuit and multiplying it by the feedback factor. The open-loop gain is the gain of the circuit without any feedback, while the feedback factor is the ratio of feedback signal to the input signal.

What is the significance of the loop gain in a circuit?

The loop gain is an important parameter in a circuit as it determines the stability and performance of the system. A high loop gain indicates a stable system, while a low loop gain can lead to oscillations and instability.

How does the source contained port impedance affect the loop gain?

The source contained port impedance refers to the impedance seen by the source in a feedback circuit. It can affect the loop gain by altering the open-loop gain and feedback factor of the circuit. A higher source contained port impedance can result in a lower loop gain and vice versa.

Can the loop gain be negative?

Yes, the loop gain can be negative in certain circuits. This indicates that the feedback signal is out of phase with the input signal. In such cases, the circuit can become unstable and produce oscillations.

How can the loop gain be optimized for a circuit?

The loop gain can be optimized by designing the circuit with appropriate component values. This involves adjusting the open-loop gain and feedback factor to achieve the desired loop gain. Additionally, reducing the source contained port impedance can also improve the loop gain.

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