# A question about the loop gain and source contained port impedance

## 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 $R_{eq}=\frac{U}{i_U}=\frac{Z_1 \cdot (Z_2+Z_3)}{(Z_1+Z_2+Z_3)+g_m \cdot Z_1 \cdot Z_2}$, and for inverse direction current, the impedance I got is $R_{eq}=\frac{Z_1 \cdot (Z_2+Z_3)}{(Z_1+Z_2+Z_3)-g_m \cdot Z_1 \cdot Z_2}$

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gneill
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Which current(s) in particular are you referring to when you say "inverse current direction"? Is it the controlled source?

Which current(s) in particular are you referring to when you say "inverse current direction"? Is it the controlled source?

I'm sorry for the poor statement, when i said "INVERSE DIRECTION", I meant the direction of "current" in picture 1, and direction of "$i_U$" in picture 2.

gneill
Mentor
Hmm. If iU is reversed then would not this imply that U is reversed also, yielding the same U/iU for the impedance?

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 $R_{eq}$ in the inverse current direction, I flipped the whole picture vertically, I think this reverses both the voltage and current :)

gneill
Mentor
Dear gneil, I'm afraid not, when I calculated $R_{eq}$ 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!

My reasoning:

Use the transformation shown in the attachments.

$Z_a=\frac{Z_1 \cdot Z_3}{Z_1+Z_2+Z_3}$
$Z_b=\frac{Z_2 \cdot Z_3}{Z_1+Z_2+Z_3}$
$Z_c=\frac{Z_1 \cdot Z_2}{Z_1+Z_2+Z_3}$

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

$R_{eq}=\frac{Z_1 \cdot (Z_2+Z_3)}{(Z_1+Z_2+Z_3)+g_m \cdot Z_1 \cdot Z_2}$

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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 +/-)?

gneill
Mentor
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.