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Help in deriving Zin small-signal expression

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promach:
Could anyone guide me in deriving the Zin expression below ? How do I model the two transconductance block in small-signal model including their output impedance ?

The Electrician:
A model for a single transconductance would be like this:



This is a two-port: https://en.wikipedia.org/wiki/Two-port_network

Its two-port Y parameters can be written like this:



But you have another transconductance going in the reverse direction.  Its admittance matrix looks like this:



A gyrator is two transconductances going in opposite directions like you have.  See: https://en.wikipedia.org/wiki/Gyrator

The Y matrix (admittance matrix) for a gyrator with infinite output impedances for your circuit would be:

The Electrician:
If you add a capacitor CL at the right side of the gyrator, that load can be added to the admittance matrix of the gyrator in the 2,2 element of the admittance matrix:



Now to find the input impedance of the gyrator with that CL load on the output, we need only invert our admittance matrix:



The 1,1 element of the inverse matrix is your input impedance.  Compare that to what your book gives for the first case.

To add output impedances to the two transconductances we need to add a couple of terms to the admittance matrix:



Now we need to add the capacitor  CL to the output:



Finally, to get the input impedance, we invert the admittance matrix:



The desired input impedance is the red 1,1 element of the inverse matrix.

promach:
You made things too complicated. Please see the derivation process below:




--- Quote ---The first gm stage plus CL forms an integrator. What is it integrating? The voltage at node X. The second gm stage completes a feedback loop which tries to keep the voltage at node X held constant at the point where you get no output current from the first gm stage. If you inject a current into node X, the feedback loop will respond to adjust the voltage on CL until the 2nd gm stage absorbs all of that current. This gives you a low impedance at low frequencies. How low? Just 1/(gm1rout1gm2). I.e. 1/total_gm. What happens at high frequencies? CL becomes low impedance effectively removing the effect of the feedback loop. Now the impedance you see is high. How high? Ignoring capacitances associated with the gm stages, Rout2. So you have a low impedance at low frequencies that becomes high impedance at high frequencies. That is an inductor. For the two break frequencies think about it this way. At very low frequencies, Rout1 has a lower impedance than CL and it isn't until Rout1 = |1/(jwCL)| that you start to see appreciable effects from CL. That is just w = 1/(Rout1CL). At the high frequency end, you can solve for when the inductive part equals Rout2, or you can disconnect node X from all but the gm stages and realize that CL sees a resistance of 1/(gm2Rout2gm1) and so the pole frequency is 1/(gm2Rout2*gm1). Note that this analysis and the plot you show both ignore the effects of capacitances associated with the transistors in the gm stages. Those will be important at some point.
--- End quote ---

Regarding the quote above, How do I derive the equation "Just 1/(gm1rout1gm2). I.e. 1/total_gm" ?

The Electrician:
I don't think I made it too complicated.  The image you show has everything all together on one page.  I showed everything in linear fashion, one step at a time.

If you want my solution to look less complicated here's the version without all the steps shown in excruciating detail:



As to your question, the large quote you show apparently was posted on another forum I haven't seen, and I believe there are some errors in it.

Look at the Bode plot associated with the original problem statement.  It shows an upper corner frequency at (Gm1 Gm2 Rout2)/CL due to the pole in the expression for Zin.  It is not exactly correct, but it is a good approximation if you can neglect the (+1) term in the denominator of the Zin expression.

The denominator of the Zin expression is (Rout1 CL s + Gm1 Rout1 Gm2 Rout2 + 1).  If this is solved for s, the result is:

(1+Gm1 Rout1 Gm2 Rout2)/(Rout1 CL) which is the exactly correct pole frequency.

In order to get the pole frequency shown in the Bode plot we have to neglect the (+1) term
and solve (Rout1 CL s + Gm1 Rout1 Gm2 Rout2) for s.  Then we get a pole frequency of (Gm1 Gm2 Rout2)/CL

The large comment says "If you inject a current into node X...This gives you a low impedance at low frequencies. How low? Just 1/(gm1rout1gm2)"

That is not the result I get for the resistance at node X.  To get the resistance at node X, simply remove the +s CL term from the admittance matrix and invert it.  The resistance seen at node X is the red element of the inverse:



In order to get the expression in the large comment (from another forum) you show, it's necessary to ignore the (1+) term in the resistance expression I show in red.  I believe the expression in the large comment is in error.

The expression in the large comment for the resistance seen by CL suffers from the same neglect of the (1+) term.  The commenter simply asserts: "...you can disconnect node X from all but the gm stages and realize that CL sees a resistance of 1/(gm2Rout2gm1) and so the pole frequency is 1/(gm2Rout2*gm1)"

He doesn't show how he calculated that "CL sees a resistance of 1/(gm2Rout2gm1)", and then he makes a mistake when he says "the pole frequency is 1/(gm2Rout2*gm1)"; the pole frequency would be gm2 Rout2 gm1/CL if the expression for the resistance seen by CL were correct.  Perhaps he didn't actually calculate the resistance seen by CL, but just asserted that it is 1/(gm2Rout2gm1) because that would match up with what is shown on the Bode plot.

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