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This is a voltage-current feedback circuit. If I break the loop at the gate of the transistor then excite a voltage at gate and measue the return signal at the other end I got the loop gain T = gm*ro where gm, ro are transconductance and output resistance of the transistor, respectively.

However, if I break the loop by removing the feedback component Zf and account for the loading effect, the loop gain is calculated T = gm*(ro|| Zf) as in the figure.
As you can see they're different. In the second method you actually assumed unilateral for the derivation so the result here is not completely correct.

  1. Is the loop gain T = gm*ro in the first one correct?
  2. Is there any method to solve for loop gain (expressed in analytic form) without having to break a loop?

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  • \$\begingroup\$ @AJN why do you think that the loop should should include Zf? That is normally true but is it a must? The current source is idea and to make the loop possibe let assume that there is also Rs from the input voltage source. \$\endgroup\$
    – emnha
    Jun 4 at 5:06
  • \$\begingroup\$ For transistor is only ro is considered. Other than that they are ideal. \$\endgroup\$
    – emnha
    Jun 4 at 5:08
  • \$\begingroup\$ @AJN I added the figure with small signal model. For other questions I'll try to address after work. \$\endgroup\$
    – emnha
    Jun 4 at 5:52
  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$
    – AJN
    Jun 4 at 6:16
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    \$\begingroup\$ It seems like a good job for the extra-element theorem or EET: determine the transfer function \$H_0\$ without \$Z_f\$ then determine the resistance \$R_d\$ "seen" from \$Z_f\$'s terminals with a zeroed excitation and, finally, determine \$R_n\$ when the output is nulled. Assemble and you have \$H(s)=H_0\frac{1+\frac{R_n}{Z_f}}{1+\frac{R_d}{Z_f}}\$. \$\endgroup\$ Jun 4 at 6:35
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Here is a way to calculate the return ratio T (or modern loop gain) analytically.

I use your small signal circuit with the Vs and Rs.

Delta is the network determinant.

Delta0 is the network determinant with gm=0 (the active element).

The return difference

F = Delta / Delta0

and

T = F - 1

See e.g. Wai-Kai Chen "Active network analysis" page 226.

Now how to get the network determinant.

The network determinant is independent of voltage/current sources. It depend only on the topology of the circuit with voltage sources shorted and current sources opened.

The network determinant is the denominator of a transfer function which does not change the topology.

E.g. for the circuit above.

H1 = Vo/Vs = (Ro* (1 - GmRf)) / (Rf + Ro + Rs + GmRo*Rs)

and

Zout = Vo / Iinj = (Ro * (Rf + Rs)) / (Rf + Ro + Rs + GmRoRs)

where Iinj is an injected current at the output.

The denominator of H1 and Zout is the same. It is the network determinant.

So we get:

Delta = (Rf + Ro + Rs + GmRoRs)

Delta0 = (Rf + Ro + Rs)

F = Delta / Delta0 = (Rf + Ro + Rs + GmRoRs) / (Rf + Ro + Rs)

T = F - 1 = (GmRoRs) / (Rf + Ro + Rs)

Sure its all algebraic.

But if you open the loop at the gate and inject 1 V you get a current of Gm which is split between Ro and the feedback path and produce a feedback voltage above Rs.

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  • \$\begingroup\$ Could you add the detail on how you get the H1 = Vo/Vs? Also do you need to break loop here? If not what is the advantage of this method compared to the conventional method of breaking loop? \$\endgroup\$
    – emnha
    Jun 6 at 19:58
  • \$\begingroup\$ H1 (and Zout) are here closed loop transfer functions, which you get by normal pen and paper calculation or a symbolic solver. You can also build the indefinite admittance matrix. But in this case you have already the network determinant. What is the advantage? Is is less work to calculate a transfer function or the loop gain? I don't know. You asked for a analytical method without breaking the loop. \$\endgroup\$
    – JosefC
    Jun 7 at 5:37
  • \$\begingroup\$ One advantage is that you can handle multiple feedback loops. To your 1. question. T=gm*Ro is correct if your input is a pure current source. I don't know how you calculate the loading if you remove the feedback Zf. \$\endgroup\$
    – JosefC
    Jun 7 at 5:59
  • \$\begingroup\$ I don't mind calculation. I need a method that can calculate the loop gain without breaking the loop as it's not always easy to break a loop properly. I'll read more about the method you mentioned. \$\endgroup\$
    – emnha
    Jun 7 at 7:08
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    \$\begingroup\$ Yes sure. I will answer this in the thread electronics.stackexchange.com/questions/570844/… . \$\endgroup\$
    – JosefC
    Jun 13 at 16:37
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properly breaking the loop requires proper terminations AND evaluation in both directions. The passive feedback element, Zf, is bi-directional (as is the transistor through Cgd which you are not modeling). For the model you are using, Cgd=0, Cgs=0, the loop gain is LG~(yf-gm)*Rf||ro, where yf=1/Rf, Cgs=gate-source capacitance, Cgd is gate to drain capacitance, and the 'reverse loop gain' is zero. yf is normally lost in the (yf-gm) factor when opening the loop and not taking into account the forward signal path through yf from input to output.

Loop gain depends on what you are leaving out and how you are terminating open loops. Including Cgs for example but still leaving out Cgd, you get:

LG~(yf-gm)*Rf||ro * yf/(yf+(Cgs)*s), reverse loop still zero. Note that for Cgs=0 this defaults to the approximation above.

Adding Cds, drain-source cap coupling,

LG=(yf+Cds*s-gm)Rf||ro * (yf+Cdss)/(yf+(Cgs+Cds)*s). Again, letting Cgs-Cds=0 returns the first approximation.

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