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This is problem 6.3 from Vorpérian's Fast Analytical Techniques book. I'm having trouble proving his answer for the numerator of this video amplifier circuit's voltage gain function. The general form of the voltage gain is given by \begin{align} A(s) = A_m \frac{N(s)}{D_L(s) D_H(s)}, \end{align} where \$ D_L(s) \$ and \$ D_H(s) \$ denote the poles for the low and high frequency response, respectively, and \$ N(s) \$ denotes the zeros.

Here's what the circuit looks like simulated in Qucs-S. I generated it from this simulation file. (To open this in Qucs-S, copy all the text in it and save it as a *.sch file. Then open it using Qucs-S.) I realize the simulation is at a ridiculous frequency (up to 100 THz; we should be using transmission line theory at that point), but I wanted to make sure I was able to see all the high frequency poles and zeros. simulated circuit

In the book, the numerator is given by \begin{align} N(s) = \left( 1 + \frac{ s }{ \omega_{oz} Q_{oz} } + \frac{ s^2 }{ \omega_{oz}^2} \right) \left( 1 - s \frac{ C_{bc2} }{ g_{m2} } \right), \end{align} where \begin{align} \omega_{oz} &= \frac{ 1 }{ \sqrt{ C_{bc1} C_{be1} r_{o1} r_{\pi 1} \parallel \frac{ r_{x1} }{ 1 + g_{m1} r_{o1} } } } \\ &\approx \frac{ 1 }{ \sqrt{ C_{bc1} C_{be1} \frac{ r_{x1} }{ g_{m1} } } } \end{align} and \begin{align} Q_{oz} &= \frac{ \sqrt{ C_{bc1} C_{be1} \frac{ r_{o1} }{ r_{\pi 1} \parallel \frac{ r_{x1} }{ 1 + g_{m1} r_{o1} } } } }{ C_{be1} + C_{bc1} \left[ \frac{ r_{o1} }{ r_{\pi 1} } \left( 1 + \beta_1 \right) \right]] } \\ &\approx \frac{ \sqrt{ C_{bc1} C_{be1} \frac{ g_{m1} }{ r_{x1} } } }{ C_{be1} + C_{bc1} g_{m1} r_{o1} }. \end{align}

Now, I managed to figure out where the factor of \$\left( 1 - s \frac{ C_{bc2} }{ g_{m2} } \right)\$ comes from. When the output voltage \$V_o\$ is zero, all of the current from the dependent current source \$g_{m2} v_{\pi 2}\$ has to flow through the two capacitors with which it forms a loop. This allows us to formulate that loop's KVL equation as \begin{align} v_{\pi 2} \left( s C_{bc2} - g_{m2} \right) = 0. \end{align} After two algebra steps, we find the zero at \$\left( 1 - s \frac{ C_{bc2} }{ g_{m2} } \right)\$.

I have been working on deriving \$\omega_{oz}\$ for just over six months' worth of weekends. It's definitely gotten to the point where I can't do this myself.

I figured that since \begin{align} \omega_{oz} &= \frac{ 1 }{ \sqrt{ C_{bc1} C_{be1} \mathcal{R}^{(1)} \mathcal{R}_{(1)}^{(2)}} } \\ &= \frac{ 1 }{ \sqrt{ C_{bc1} C_{be1} \mathcal{R}^{(2)} \mathcal{R}_{(2)}^{(1)}} }, \end{align}

where \$\mathcal{R}_{(j)}^{(i)}\$ denotes a null resistance, I could simply assume that \$\mathcal{R}^{(2)} = r_{o1}\$ and \$\mathcal{R}_{(1)}^{(2)} = r_{\pi 1} \parallel \frac{r_{x1}}{1 + g_{m1} r_{o1}}\$, and try to reproduce the result that way. However, in the process of attempting to derive \$\mathcal{R}^{(2)}\$ and \$\mathcal{R}_{(1)}^{(2)}\$, I found that \begin{align} \mathcal{R}^{(2)} = r_{x1} \parallel r_{\pi 1} \left( 1 + g_{m1} r_{o1} \right) \end{align} and \begin{align} \mathcal{R}_{(2)}^{(1)} = r_{\pi 1} \parallel \frac{ 1 }{ g_{m1} } \parallel r_{o1}. \end{align} I got these results by assuming that \$R_1\$, \$R_{e 1}\$, and the rest of the circuit to the right of \$r_{o1}\$ all do not contribute to the calculations of these null resistances.

The value of \$\omega_{oz}\$ from my results (which is \$ \frac{1}{2 \pi} (31.618 \times 10^9 ~\text{rad/s})\$) and the one from the book's results (which is \$\frac{1}{2 \pi} (31.403 \times 10^9 ~\text{rad/s})\$) agree numerically within an error of about 0.6%. However, I don't think my derivation and the book's derivation agree for all component values.

So here's my question: how would I go about deriving the coefficients \$\frac{ 1 }{\omega_{oz} Q_{oz} }\$ and \$\frac{ 1 }{ \omega_{oz}^2 }\$ from \begin{align} \left( 1 + \frac{ s }{ \omega_{oz} Q_{oz} } + \frac{ s^2 }{ \omega_{oz}^2} \right)? \end{align}

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  • \$\begingroup\$ \$N(s)\$ in your case, is a 3rd-order polynomial factored as a 2nd-order polynomial and a RHP zero. If you develop the whole thing, you end up having \$N(s)=1+a_1s+a_2s^2+a_3s^3\$. Since you use SPICE, you could apply the method of simulating a null double injection (NDI) that I described in my book which would let you independently find and test all a coefficients. Since you have one of the zeroes, you can also concentrate on determining \$1+a_1s+a_2s^2\$ alone with my method \$\endgroup\$ Jan 16 at 11:56
  • \$\begingroup\$ and when you have \$a_1\$ and \$a_2\$ in the second-order polynomial, you determine \$Q=\frac{\sqrt{a_2}}{a_1}\$ and \$\omega_0=\frac{1}{\sqrt{a_2}}\$. You could also consider calculating high-frequency gains as suggested by Ali Hajimiri in his paper. This is a generalized approach capitalizing on time constants already determined in \$D(s)\$. The resulting coefficients are more complicated but it can simplify the analysis by avoiding multiple NDIs. \$\endgroup\$ Jan 16 at 12:01
  • \$\begingroup\$ Thanks Verbal! I read part of your book before trying this; that's actually what gave me the idea to use SPICE. I know that Ali Hajimiri's method exists but haven't tried it yet. I'll work with that next, now that you've clarified the benefits of that method. \$\endgroup\$
    – chips
    Jan 16 at 12:33
  • \$\begingroup\$ Let's say that I want to keep the RHP zero that I already found, and I want to use NDIs for this analysis, just for practice using NDIs. I want to determine \$ a_2 \$ from \$ 1 + a_1 s + a_2 s^2 \$. Would I need to open \$ C_{be1} \$ and short \$ C_{bc1} \$ to find the resistance term \$ r_{\pi 1} \parallel \frac{r_{x1}}{1 + g_{m1} r_{o1}} \$? Or short \$ C_{be1} \$ and open \$ C_{bc1} \$? \$\endgroup\$
    – chips
    Jan 16 at 12:33
  • \$\begingroup\$ When you have two energy-storing elements, you can either reuse the first time constant \$\tau_1\$ and determine \$\tau_{12}\$ where element 1 is in its high-frequency state while you determine the resistance driving 2. In this case, \$a_2=\tau_1\tau_{12}\$. Or, you can reuse the second time constant \$\tau_2\$ and determine \$\tau_{21}\$ where element 2 is in its high-frequency state while you determine the resistance driving 1. In this case, \$a_2=\tau_2\tau_{21}\$. This is the welcome redundancy brought by the method. \$\endgroup\$ Jan 16 at 13:10
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Because the example given in the post is overly complicated and would require many steps, I am going to show how to determine the numerator \$N(s)\$ of a simple 2nd-order \$RLC\$ network as shown below. For this purpose, I will apply several methods that I described in my book on transfer functions and fast analytical circuits techniques (FACTs):

enter image description here

To determine an impedance \$Z\$ - which is a transfer function - you install a test generator \$I_T(s)\$ injecting a current (the stimulus) between the terminals across which you will observe the response \$V_T(s)\$. The impedance is then \$Z(s)=\frac{V_T(s)}{I_T(s)}\$. In the below case, this expression can be expressed as \$Z(s)=R_0\frac{N(s)}{D(s)}\$. The term \$R_0\$ represents the resistance offered by the impedance expression for \$s=0\$. Without writing a line of algebra, you immediately see that this resistance is the parallel combination of \$r_L\$ and \$R_1\$: \$R_0=r_L||R_1\$. The numerator can be written as \$N(s)=1+a_1s+a_2s^2\$ while the denominator is \$D(s)=1+b_1s+b_2s^2\$.

Now, to determine the numerator, we have several choices:

  1. a null double injection or NDI: you keep the current source \$I_T\$ in place and you determine the resistance \$R\$ offered by one of the energy-storing elements while the response is a null. A null means that despite the presence of an ac excitation (the stimulus) it does not propagate in the circuit to generate a response. A bit like if the stimulus was lost in the network for a certain frequency value and the response was 0 V ac. What is good with an impedance determination is that a 0-V response across a current source is similar to replacing the current source by a short circuit. This is called a degenerate case.

  2. Inspection: this is the fastest and most efficient way to determine the zeroes. You observe the circuit and realize that a some branches could cause a transformed short circuit across the current source. And if this branch is a transformed short at a certain frequency (the zero frequency), then the response across the current source \$I_T\$ is 0 V and this is the manifestation of one zero.

  3. generalized approach: by reusing the time constants obtained for the denominator, we can, via the determination of high-frequency gains, reuse these time constants and obtain the numerator without resorting to NDIs. This is a simpler approach sometimes but you obtain a slightly more complicated result than with an NDI. From this technique, we can immediately infer if we have zeroes in a circuit and how many: place one of the energy-storing element in its high-frequency state and check that if you stimulate the circuit (via the current source in this impedance exercise), you have a response \$V_T\$. When \$C_1\$ is replaced by a short (its high-frequency state) and \$L_1\$ is left in its dc state (a short), then if you excite via \$I_T\$, you can measure a voltage \$V_T\$: \$C_1\$ contributes a zero. Should you have the same circuit without \$r_C\$, then shorting \$C_1\$ would cancel \$V_T\$ and there would be no response despite the stimulus presence. So no zero without \$r_C\$. Do the same for \$L_2\$, then when both \$C_1\$ and \$L_2\$ are in their high-frequency state and you have two zeroes in \$N(s)\$.

Let's start with the NDI. As explained, an NDI is performed with the excitation \$I_T\$ brought back in the circuit. The exercise consists of determining the resistance \$R\$ "seen" from the energy-storing element terminals when the voltage \$V_T\$ across the current source is a null or 0 V. 0 V across a current source is similar to having the same current source replaced by a wire:

enter image description here

The time constants in this mode are simply involving the cap and the inductor equivalent series resistances (ESR). With these elements on hand, we can assemble our first coefficient: \$a_1=r_CC_1+\frac{L_2}{r_L}\$. How can we be sure this is the right way to go with this NDI? In other words, is there any other possibility to experimentally verify that it works this way, that I truly cancel the response across the current source? Sure, and this is what I presented in the book with a SPICE simulation involving a voltage-controlled current-source:

enter image description here

The source \$G_1\$ adjusts the injected current in \$C_1\$'s terminals to maintain 0 V across the source. The voltage across the terminals divided by the injected current is computed by the in-line equation \$B_1\$ and you read 120 m after the bias point calculation. This is well the ESR of \$C_1\$. You can do the same exercise with \$L_2\$ of course. The computed bias value should be the exact value found with a solver.

Now that we have these two time constants, we can look for the second-order term obtained when \$C_1\$ is replaced by a short and we "look" through \$L_2\$'s terminals to obtain the resistance we want:

enter image description here

Once we are here, we can write the second-order term as \$a_2=r_CC_1\frac{L_2}{r_L}\$. The complete denominator can thus be expressed as \$N(s)=1+s(r_CC_1+\frac{L_2}{r_L})+s^2(r_CC_1\frac{L_2}{r_L})\$. From this expression, we could then determine a quality factor \$Q_N=\frac{\sqrt{a_2}}{a_1}\$ and a resonant angular frequency \$\omega_0=\frac{1}{\sqrt{a_2}}\$. However, we could also try to rework it as a product of factors and that's where inspection is of great help.

As explained with the NDI, the zero manifests itself by the disappearance of the response when the network is excited at a zero frequency. In our example, it would mean that you tune the current source \$I_T\$ to a frequency where suddenly the voltage across its terminals becomes 0 V. This would happen if the network involving \$C_1\$ and \$r_C\$ or \$L_2\$ and \$r_L\$ would become a transformed short circuit. In this case, the zeroes are immediately determined as shown below:

enter image description here

And, finally, the third method involves determining high-frequency expressions when the energy-storing elements are alternately set in their high-frequency state. In this case, you determine simple gain values and reuse the time constants already obtained when finding \$D(s)\$. This is how it goes for the three expressions \$H^1\$, \$H^2\$ and \$H^{12}\$:

enter image description here

We can then gather all these numbers in a Mathcad sheet and compare the results with the brute-force expression shown in the upper right corner:

enter image description here

If we plot the magnitude and phase responses of all these transfer functions, they are all identical:

enter image description here

So we have seen three different ways to determine the zeroes of a transfer function using the FACTs. The simplest and fastest way is inspection as you can see. Sometimes, unfortunately, it is not possible and you need to resort to NDI. To that respect, SPICE and its bias point calculation are a treasure and helped me find mistakes in many occasions with complicated circuits. And, following Ali Hajimiri method, the high-frequency calculations can be seen as a way to skip NDI but it ends up in more complicated coefficients that you must (or not) simplify. The good thing is that all results are rigorously identical but expressed in different ways.

Acquiring the skill to determine transfer functions using FACTs requires time and efforts but they are worth it in the end. If you start with simple passive examples of 1st- and 2nd-order, then complicate the circuits smoothly, you will succeed solving more complicated active circuits as I have tried to modestly promote in my book.

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  • \$\begingroup\$ I'll accept this answer because generalizes far beyond this problem. However, I did start to get some answers to the specific resistance factors I was looking for, thanks to your encouragement! The key for me was to forget about proving the formulas in the book. That way I could focus on writing down the resistance terms that I saw. I'll post my answer also later this week, in case anybody else was wondering how to do this specific (albeit somewhat ridiculous) problem. \$\endgroup\$
    – chips
    Jan 18 at 10:48
  • \$\begingroup\$ Yes, I have tried to be general in my comments to widen the audience. The problem with the many documents that I've absorbed for my APEC seminar and the book, is that the solved examples are often complicated to tackle and have discouraged many people to pursue. I took a different path with a slower progression together with a SPICE solver. I command your pugnacity to continue the exercise with many controlled sources : ) Chapeau ! or hats off! \$\endgroup\$ Jan 18 at 14:47
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Vorperian problem 6.3

To find the zeros in the numerator you can split the circuit in two parts, like you have it already done.

See images vo314a and vo314b. vo314a

vo314b Zeros occurs if the first sub circuit creates 0V at \$R_{c1} \parallel R_{in2}\$ and the second sub circuit set Vout to 0V.

The sub circuits can be solved with the normal EET (no 4EET).

First sub circuit

Calculation of \$ \mathcal{R}^{(1)} \$ at port 1

First a calculation is done without \$R_{\pi1}\$.

This input impedance is named \$\mathcal{R}^{(1x)}\$.

It follows that \$\mathcal{R}^{(1)} = \mathcal{R}^{(1x)} \parallel R_{\pi1}\$.

See the image vo314a1. (NLR1 is a Nullor and nulls the output in the ndi).

vo314a1

\$R_{\pi1}\$ is removed. Capacitors are opened. Output is nulled.

A current \$i_t\$ is injected in port 1. \$v_t\$ is the voltage at port 1.

\$ \mathcal{R}^{(1x)} = \frac{v_t}{i_t} \$

The current of G1 must flow through \$R_{o1}\$ because the output is nulled:

\$ V(R_{o1}) = - g_{m1} * V_{\pi1} * R_{o1} = - g_{m1} * v_t * R_{o1} \$

\$i_t\$ flows through \$R_{x1}\$:

\$ V(R_{x1}) = R_{x1} * i_t \$

A KVL loop runs from \$0V \rightarrow R_{o1} \rightarrow G1 \rightarrow R_{x1} \rightarrow 0V (GND)\$:

(Check the sign and direction of the voltages)

\$ - V(R_{o1}) + v_t - V(R_{x1}) = 0 \$

\$ g_{m1} * v_t * R_{o1} + v_t - R_{x1} * i_t = 0 \$

\$ \mathcal{R}^{(1x)} = \frac{v_t}{i_t} = \frac{R_{x1}}{1 + g_{m1} * R_{o1}} \$

\$ \mathcal{R}^{(1)} = R_{\pi1} \parallel \mathcal{R}^{(1x)} = R_{\pi1} \parallel \frac{R_{x1}}{1 + g_{m1} * R_{o1}} \$

\$ \mathcal{R}^{(1)} = \frac{R_{x1} \cdot R_{\pi1}}{R_{x1} + R_{\pi1} + g_{m1} \cdot R_{\pi1} \cdot R_{o1}} \$

Note to your check of \$\mathcal{R}^{(1)}\$ with Spice in the comments.

It seems you forgot to null the output, i.e. set the voltage and current of \$R_L\$ to zero with the help of Vin.

Calculation of \$ \mathcal{R}^{(2)} \$ at port 2

See the image vo314a2.

vo31a2

Capacitors are opened. Output is nulled.

A current \$i_t\$ is injected in port 2. \$v_t\$ is the voltage at port 2.

One can do a similiar calculation like in the calculation of \$\mathcal{R}^{(1)}\$.

Another way is to do another EET with \$R_{\pi1}\$ as a shorted Extra Element and \$i_t\$ and \$v_t\$ on port 2 as input and output.

During this calculation the \$\mathcal{R}^{(1x)}\$ circuit and result can be reused.

The result is:

\$ \mathcal{R}^{(2)} = \frac{R_{x1} \cdot (R_{\pi1} + R_{o1} + g_{m1} \cdot R_{\pi1} \cdot R_{o1})}{R_{x1} + R_{\pi1} + g_{m1} \cdot R_{\pi1} \cdot R_{o1}} \$

Calculation of \$ \mathcal{R}^{(2)}_{(1)} \$ at port 2 with port 1 shorted

Simple examination get:

\$ \mathcal{R}^{(2)}_{(1)} = R_{o1} \$

Numerator of the first sub circuit

\$ N = 1 + \frac{\mathcal{R}^{(1)}}{Z1} + \frac{\mathcal{R}^{(2)}}{Z2} + \frac{\mathcal{R}^{(1)}}{Z1} \frac{\mathcal{R}^{(2)}_{(1)}}{Z2} \$

\$ N = 1 + s \cdot C_{be1} \mathcal{R}^{(1)} + s \cdot C_{bc1} \mathcal{R}^{(2)} + s^2 \cdot C_{be1} \cdot C_{bc1} \cdot \mathcal{R}^{(1)} \mathcal{R}^{(2)}_{(1)} \$

By term comparison with

\$ N = 1 + \frac{s}{Q \cdot \omega_0} + \frac{s^2}{\omega_0^2} \$

and some algebraic transformations this matches with the solution in the book.

Second sub circuit

See image vo314b.

vo314b

Find the zero in the output.

\$ V(R_L)=0 \Rightarrow I(R_L) = 0 \$

KCL at top node of \$R_L\$:

\$ I(R_{\pi2}) + I(C_{be2}) + g_{m2} \cdot V_{\pi2} = 0 \$

\$ \Rightarrow V_{\pi2} \cdot ( \frac{1}{R_{\pi2}} + s \cdot C_{be2} + g_{m2}) = 0 \$

\$ \Rightarrow ( 1 + s \cdot \frac{ C_{be2} \cdot R_{\pi2}}{ 1 + g_{m2} \cdot R_{\pi2}} ) = 0 \$

This differs from the solution in the book \$ ( 1 + s \cdot \frac{C_{bc2}}{g_{m2}} ) \$. I checked my calculation on multiple ways and can't find an error.

The book is excellent but has many typos in the formulas.

I get Vorperians result if the output is changed to the collector of \$Q_2\$ and \$R_L\$ is moved to the collector and the emitter is grounded.

I think your check for the factor \$ ( 1 + s \cdot \frac{C_{bc2}}{g_{m2}} ) \$ is wrong because between \$C_{bc2}\$ and \$G1\$ is a GND connection over which can flow a current.

In you Qucs schematic the resistor \$R_{o2}\$ is double used. Explicit as resistor and in \$parcomp\$.

Best regards

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  • \$\begingroup\$ Nice job, +1! Did you try to check the zero resistance you've found which differs from Vorpérian's book using SPICE and my voltage-controlled current-source method effectively nulling the output in the complete two-stage arrangement? If your result is correct, it should match the SPICE dc bias point several digits after the coma. \$\endgroup\$ Jan 24 at 10:11
  • \$\begingroup\$ Thank you. I checked my result with the descriped manual calculation, a 2EET calculation and with a symbolic solver. I checked the Rn_Cbe2 resistance with SPICE in the subcircuit and the complete circuit with removed capacitances and it fits my result. Surprisingly the 2EET is very simple because Rn_Cbe2 is Rp2 || 1/gm2 and all other Rns are zero. I have to compare my Spice solution with your Spice method in a calm moment. \$\endgroup\$
    – JosefC
    Jan 24 at 19:00
  • \$\begingroup\$ +1; thanks JosefC! With regards to your notes: (1) I grounded the output instead of nulling it, so that's probably the source of my problem. (2) I got Vorpérian's solution by assuming the current through \$ r_{\pi 2} \$ was zero. Otherwise, opening \$ r_{\pi 1} \$ to help calculate \$ \mathcal{R}^{(x1)} \$ was pretty surprising, yet intuitive; I wouldn't have thought of that! \$\endgroup\$
    – chips
    Jan 25 at 13:26

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