# Low frequency gain and extra element theorem

I was reading 'Fast Analytical Circuit Techniques' from Vatché Vorperian.

Problem 3.7 states:

This is the circuit we wish to analyze:

This is the equation we must obtain:

I fear I misunderstand something important. From what I understand, the theorem never states that $$\A_o\$$ is a DC gain.

My questions:

1. Why was $$\C_E\$$ left out of the expression of $$\A_o\$$ and $$\A_o\$$ referred to as a low-frequency gain!? Shouldn't $$\A_o\$$ simply be $$\v_o(s)/v_{in}(s)\$$ with $$\R_f \rightarrow \infty\$$, thus contain $$\1/sC_E\$$?
2. This brings me to my second question. Suppose you would have a circuit with two reactive elements (e.g. capacitors) and some resistive elements. Suppose you apply the EET to a resistor. From what I understand, you would have to calculate 3 expressions, namely the gain with the extra element nulled ($$\A_o\$$), the null driving point impedance $$\\mathcal{L}^{(1)}\$$ and the ordinary driving point impedance $$\Z_{(1)}\$$. From what I understand those 3 expressions should contain the expressions of the two reactive elements (since they aren't designated as extra elements).
3. There is the EET and the NEET. Is it technically possible to not use the NEET and only use the EET? For example say we have a resistive circuit from which we know the transfer function H(s). Say now we want to add a reactive element, a capacitor. We could technically apply the EET to have the new transfer function $$\H_{C1}(s)\$$ to add the new capacitor C1. Then, if we want to add a second capacitor C2, we could, starting from $$\H_{C1}(s)\$$, apply again the EET and get the new transfer function $$\H_{C1,C2}(s)\$$. Thus, the NEET in this case would be optional, no?

When writing a transfer function, you try to approach a so-called low-entropy form in which a leading term is associated with a fraction made of a numerator $$\N(s)\$$ and a denominator $$\D(s)\$$. The leading term carries the unit while the fraction is unitless. For instance, if you consider the below circuit, you can write that $$\H(s)=H_0\frac{1}{1+\frac{s}{\omega_p}}\$$ in which $$\H_0\$$ is the dc gain obtained when $$\s=0\$$:

$$\s=0\$$ is like in a SPICE simulation: you open all the capacitors and short all the inductors then redraw the circuit and determine the gain. In the quick example, you immediately see that the dc gain is $$\H_0=\frac{R_2}{R_2+R_1}\$$. Then, to determine the pole, you reduce the stimulus to 0 V and replace the source by a short circuit. Then you temporarily disconnect the capacitor and "look" through its terminals the resistance $$\R\$$ you see. In this example, you see that $$\R=R_1||R_2\$$ which leads to $$\\tau_1=C_1(R_1||R_2)\$$ and immediately gives a pole located at $$\\omega_p=\frac{1}{\tau_1}\$$. And this is it, no need for equations, just inspection.

Now coming back to your questions:

1. if you want the low-frequency gain of this circuit analyzed for $$\s=0\$$, you have to remove the capacitor since, in dc, a cap. is an open-circuit (no current at steady-state). In this case, you analyze the circuit without $$\C_E\$$ but with $$\R_f\$$ still there. $$\R_f\$$ is your extra element and complicates the analysis. If you apply the extra-element theorem or EET, you start by letting $$\R_f\$$ go infinite (or reduce it to zero, you choose which one leads to a simpler intermediate circuit) and determine the gain in this mode. This is what Vatché did in Figure 2.13 (without $$\C_E\$$) and resulted in (2.41a) where $$\A_0\$$ is determined. Then you apply the EET involving $$\R_f\$$ and you have (2.47) where you see that the first term to the left is your $$\A_0\$$ of (2.45a) and the following fraction is $$\\frac{1+\frac{R_n}{R_f}}{1+\frac{R_d}{R_f}}\$$. However, the result is rearranged in a friendly way and Vatché truly excels in the exercise :)

2. If you apply the 2EET in the formal way, you certainly consider the two energy-storing elements in the intermediate steps but the final result is far from being a low-entropy version of the final transfer function. The best is to determine the time constants associated with each energy-storing elements and write $$\D(s)=1+s(\tau_1+\tau_2)+s^2(\tau_1\tau^1_2)\$$ if you adopt a slightly different formalism used in my book on FACTs. I have actually shown in Chapter 4 of my book that dealing with two energy-storing elements and applying the EET twice gives 4 possible expressions with different conditions for the intermediate reference calculations. I have adopted the simplest option which, like SPICE, considers open caps. and shorted inductors. However, you could adopt any combination you like but then need to juggle with the rest of the combinations.

3. As I said, you can apply the EET several times but the obtained expression becomes difficult to deal with. The best is use the general form which lets you determine transfer function up to the order $$\n\$$ by determining time constants associated with each energy-storing elements.

I have published many examples on the subject and you can have a look at my APEC 2016 seminar which introduces the subject. Good luck with the FACTs: the topic can seem difficult at first but when you master it, there is no going back to the classical method.

• Are the 2 figures of the circuit top above ... "equivalent" ?
– user288518
Jun 20 at 20:43
• Which figures, what do you mean? Can you please be more specific in your question, thank you. Jun 20 at 21:16
• @VerbalKint I see now. Vatché used the techniques presented in chapter 2 where we can determine a time constant by looking at the equivalent resistance seen from the capacitor. He then calculates the RC pole. He calculates the zero by solving s for the null condition vo=0 (thus ic=0). And he calculates Ao as a DC gain. Confused me thought he applied the EET while somewhat ignoring CE to the whole vo/vin. Thus I thought it was implied the gain calculated in the EET (part of the 3 calculations) was by definition a DC gain. Glad it isn't. Thank you very much for your detailed answer. Jun 21 at 22:03
• Regarding question 2 I wanted to know if the theorem was still valid if it contained s-terms in Ho, Zn and Zd. I get your point that it would however make the transfer function confusing (the poles/zero would be difficult to guess I suppose). I will have a look at your book, eventually, to reinforce my EET knowledge. I must say it is a bit because you keep raving about FACTs here, especially the EET, that I decided to read Vatché's book :) Jun 21 at 22:08
• Glad to read these comments. Yes, the EET applied several times does not lead to low-entropy formulas in which poles and zeroes appear clearly. That's what I show in my book before going to the other options including a generalized formula which saves from running a NDI. My book can be seen as a stepping stone to Vatché"s but also as a complement since I used SPICE a lot to verify results for poles but also zeroes locations with a simple dedicated setup. You'll let me know what differences you see/like between the books then : ) Jun 22 at 7:35