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

1. Oct 27, 2011

### genxium

1. The problem statement, all variables and given/known data

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 !

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

3. 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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• ###### impedance.png
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Last edited: Oct 27, 2011
2. Oct 27, 2011

### Staff: Mentor

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

3. Oct 28, 2011

### genxium

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.

4. Oct 28, 2011

### Staff: Mentor

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

5. Oct 29, 2011

### genxium

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 :)

6. Oct 29, 2011

### Staff: Mentor

I don't see how the orientation of the drawing on paper could possibly change the circuit's operation!

7. Nov 2, 2011

### genxium

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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• ###### port-impedance.png
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Last edited: Nov 2, 2011
8. Nov 2, 2011

### genxium

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. Nov 2, 2011

### Staff: Mentor

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