I am losing my mind over this.

I am trying to design a Deliyannis-Friend section and I am stuck in a loop and don't know what to do. These is the circuit and its equations.

enter image description here

Based on this I have devised a design method.

The design starts by setting equal capacitors, \$ C_1 = C_2 = C \$, and choosing the value of capacitance \$ C \$. Taking the expression of \$ K_p \$ and \$ Q \$, results in

\begin{equation} \frac{|K_p|}{Q} = \frac{ \displaystyle 1+\frac{R_f}{R_g}}{ \displaystyle \sqrt{\frac{R_1}{R_2}}} \,, \label{eq:DFderiv1} \end{equation}

By choosing the ratio \$ \displaystyle \beta = \frac{R_2}{R_1} \$, the ratio \$ \displaystyle \alpha = \frac{R_f}{R_g} \$ comes as

\begin{equation} \alpha = \frac{|K_p|}{Q} \times \frac{1}{\displaystyle \sqrt{\beta}} -1 \,, \label{eq:DFderiv2} \end{equation}

which leads to the design constraint

\begin{equation} 0 < \beta < \left(\frac{|K_p|}{Q}\right)^2 \,. \label{eq:DFderiv3} \end{equation}

Now, from the chosen ratio \$ \beta \$, resistors \$ R_1 \$ and \$ R_2 \$ can be calculated as

\begin{equation} R_1= \frac{1}{\omega_p \times C} \times \frac{1}{\sqrt{\beta}} \,, \label{eq:DFderiv4} \end{equation}


\begin{equation} R_2= \frac{\sqrt{\beta}}{\omega_p \times C} \,. \label{eq:DFderiv5} \end{equation}

Finally, with \$ \alpha \$ and choosing \$R_g\$,

\begin{equation} R_f= \alpha \times R_g \,. \label{eq:DFderiv6} \end{equation}

Ok, into a practical design with \$ K_p = - 0.5 \$ and \$ Q = 5 \$ and \$ \omega_p = 10000 \$

I choose \$ C = 10 \, \text{nF} \$. As \$ 0 < \beta < 0.01 \$ I choose the middle value \$ \beta = 0.005 \$ This allows as to obtain \$ R_1 = 141.42 \, \text{k$\Omega$} \$ and \$ R_2 = 707.11 \, \text{$\Omega$} \$

Plugging this into the \$ \omega_p \$ expression does allow to obtain the \$ \omega_p \$ value of \$ 10000 \$. So until this point we are keeping our sanity.

Now comes \$ \alpha \$. It is equal to \$ \alpha = -1 + \sqrt{2} \$. So choosing \$ R_g = 141.42 \, \text{k$\Omega$} \$ follows that \$ R_f = 58.578 \, \text{k$\Omega$} \$.

And now plugging these values in the gain and Q expressions gives \$ Q = 22.4 \$ and \$ K_p = -0.0035 \$

What is actually going on here? Where was my mistake? Did I accidentally divide by zero or something? Forgot a sign? I am losing my mind I swear...


Ok, so a new attempt. I have decided to actually redefine \$ \alpha \$ as \$ \displaystyle 1 + \frac{R_f}{R_g} \$

So here are the three equations I have

$$ K_p = - \frac{\alpha}{\frac{2}{\beta}- \left(\alpha - 1\right)} $$

$$\frac{1}{Q} = \frac{2}{\sqrt{\beta}} - (\alpha -1) \sqrt{\beta}$$

$$\frac{K_p}{Q} = - \alpha \sqrt{\beta} $$

So replacing the third equation on the second allows to obtain

$$ \frac{1}{Q} = \frac{2}{\sqrt{\beta}} + \frac{K_p}{Q} + \sqrt{\beta} $$

Solving for \$ \sqrt{\beta} \$ gives two solutions:

$$\beta_1 = \left( \frac{1-K_p}{2Q} \times \left( 1 + \sqrt{1 - \frac{8Q^2}{(1-K_p)^2}} \right) \right)^2$$

$$\beta_2 = \left( \frac{1-K_p}{2Q} \times \left( 1 - \sqrt{1 - \frac{8Q^2}{(1-K_p)^2}} \right) \right)^2$$

and \$ \alpha \$ can be calculated with

$$\alpha = -\frac{K_p}{Q \sqrt{\beta}} $$

I have tried to solve this using Mathematica

Clear[Kp, Q]
condbeta1 = (1 - Kp)/(2*Q)*(1 + Sqrt[1 - (8 Q^2)/(1 - Kp)^2]);
condalpha1 = -Kp/Q*1/condbeta1;
condbeta2 = (1 - Kp)/(2*Q)*(1 - Sqrt[1 - (8 Q^2)/(1 - Kp)^2]);
condalpha2 = -Kp/Q*1/condbeta2;
Reduce[condbeta1 > 0 && condalpha1 > 1 && Q > 0 && Kp < 0, Kp]
Reduce[condbeta2 > 0 && condalpha2 > 1 && Q > 0 && Kp < 0, Kp]

My logic was that:

  1. \$ \sqrt{\beta_1} \$ and \$ \sqrt{\beta_2} \$ should be positive numbers. When I did this manually I have arrived with the condition that @Tesla23 arrived: \$ K_p < 1 - \sqrt{8} Q \$.
  2. The second condition is that \$ \alpha \$ should be bigger than 1.
  3. Additionaly the quality factor Q shall be positive and the gain Kp shall be negative.

This allows to obtain

  1. The first solution is only valid for \$ Q>\frac{1}{\sqrt{2}} \$ in which case \$ -2Q^2 < K_p \leq 1 - 2\sqrt{2}Q\$
  2. As for the second solution, it is valid for \$ 0 < Q \leq \frac{1}{\sqrt{2}} \$ in which case \$ K_p < -2Q^2 \$ or for \$ Q > \frac{1}{\sqrt{2}} \$ in which case \$ K_p < 1 - 2\sqrt{2}Q\$.

This seems to solve the problem.

  • \$\begingroup\$ Did you start out by reading "High-Q factor circuit with reduced sensitivity" by T. Deliyannis, 1968, from Electronics Letters, 4(26), 577? (It's what I'd do, anyway.) Note figure 1(c) from the paper. \$\endgroup\$ Commented May 18, 2023 at 0:40
  • \$\begingroup\$ (Hmm. Just noted a mistake by the author. Or, at least it looks like one to me right now.) \$\endgroup\$ Commented May 18, 2023 at 1:03
  • \$\begingroup\$ @periblepsis I am still not being able to figure this out, this is horrible \$\endgroup\$ Commented May 18, 2023 at 2:29
  • \$\begingroup\$ If I set \$\tau_{_0}=R_1\,C_2\$, \$\tau_{_1}=R_2\,C_1\$, \$\tau_{_2}=R_1\,C_1\$, & \$\eta=\frac{\tau_{_0}}{\tau_{_1}}+\frac{\tau_{_2}}{\tau_{_1}}-\frac{R_a}{R_b}\$. Then I get \$K_p=-\frac1{\eta}\left(1+\frac{R_a}{R_b}\right)\$, \$\omega_{_p}=\frac1{\sqrt{\tau_{_0}\,\tau_{_1}}}\$, \$\zeta=\frac12\eta\sqrt{\frac{\tau_{_1}}{\tau_{_0}}}\$, & \$Q=\frac1{\eta}\sqrt{\frac{\tau_{_0}}{\tau_{_1}}}\$. The author then sets \$r=\frac{R_1}{R_2}\$ & \$q=\frac{C_2}{C_1}\$ (using your naming, which is the opposite of his) and then moves on to define similar \$\tau\$ terms but using \$r\$ and \$q\$, instead. \$\endgroup\$ Commented May 18, 2023 at 2:37
  • \$\begingroup\$ By the way, I used the authors nomenclature for \$C_1\$ and \$C_2\$, which is the opposite of your writing. And I use his left-side naming of \$R_a=R_f\$ and \$R_b=R_g\$, where yours is on the right side. So if you want to use any of what I just wrote in the prior comment, you will need to look at his 1(c) schematic that I already posted to you above. I chose not to use your schematic, preferring his. \$\endgroup\$ Commented May 18, 2023 at 2:44

2 Answers 2


The problem you are trying to solve is to find values for \$\alpha\$ and \$\beta\$ that satisfy the two equations for \$K_P\$ and \$Q\$.

The problem you have solved is to find values for \$\alpha\$ and \$\beta\$ that satisfy the equation for \$\frac{|K_P|}{Q}\$, but not the separate equations for \$K_P\$ and \$Q\$. You need to add one of these in as a requirement.

I suggest you write these two equations out, eliminate one variable (\$\alpha\$ or \$\beta\$) and solve. If I have done it correctly, I think you will find a separate condition \$|K_P| > \sqrt{8} Q - 1\$, which your attempted design fails.

Also, your equations don't match your schematic - the capacitor values are swapped. This doesn't really matter as you assume they are equal, but it's still wrong.

  • \$\begingroup\$ Hi! Thank you for your comment! I am trying to go for your approach I might update my answer with that later. Can you also please provide me some steps you have taken? Also yes I did realize my capacitors are swapped and have updated the expressions with that taken into account. \$\endgroup\$ Commented May 20, 2023 at 8:38
  • \$\begingroup\$ You have two equations in two unknowns, they are nonlinear and you have to figure out how to get solutions. I have given you a hint, it's your thesis.. \$\endgroup\$
    – Tesla23
    Commented May 20, 2023 at 9:57
  • \$\begingroup\$ Hi again. Thank you I think I have figured it out, can you please check my update? I think there are no further constraints to the design to be taken into account. \$\endgroup\$ Commented May 20, 2023 at 12:31
  • \$\begingroup\$ I haven't worked through it, but it looks much better. I also got a quadratic in \$\sqrt{\beta}\$. Well done. \$\endgroup\$
    – Tesla23
    Commented May 20, 2023 at 21:41

@Granger Obliviate, I did not check your derivations in detail. But some time ago I have analyzed this filter structure in detail. I have chosen another approach - perhaps it can help a bit?

Here is my transfer function:


with: m=Rf/Rg ; k2=R2/R ; R=R1 ; C1=C2=C.

This leads to:

  • Midband gain Ao=Q(1+m)SQRT(k2)
  • Pole frequency wp=1/[RCSQRT(k2)]
  • Pole-Q Qp=SQRT(k2)/(2-mk2)

Approach: Choose Qp, wp, C and k2.

Hint: My analyses have shown that the best trade-off between a moderate Ao value and a low componenet spread can be reached for k2=1.

EDIT From one of the comments above I have learned that you are interested in the sensitivity of Qp against passive tolerances. This a rather simple task.

The definition of the sensitivity figure "S_Qp" against a parameter "x" is S_Qp=(x/Qp)(dQp/dx). So you have nothing to do than to find the differential quotient - and to find S according to the definition. Then, you can analyze the expression in order to see under which conditions the value of S would be at a minimum.

Example: It is the purpose of the Deliyannis modifikation (pos. feedback) to enhance the Q-value. Therefore, we can expect that any tolerance of the factor "m" will have a comparable large influence of Qp. As an example, the product mk2 must not assume a value of "2" (infinite Qp).

Let us finde the sensitivit S_Qp against m. Applying the above definition, the calculation gives

S_Qp (against m)=(m/Qp)(dQp/dm)=mk2/(2-mk2).

From this, it is clear that we should try to make the product mk2 as small as possible.

  • \$\begingroup\$ Thank you for your contribution. Is Qp the same as Q in your design? I can't "choose" Qp though. I am supposed to impose the gain, the quality factor and the frequency and design the components knowing that. \$\endgroup\$ Commented May 20, 2023 at 8:37
  • \$\begingroup\$ For a second-order bandpass the Q-value (midfrequency divided by the 3-dB bandwidth) is identical to the pole-Q (Qp=1/2d with d=damping ratio). Remember that the pole location is decribed by wp and Qp only - and that both pole parameters (wp, Qp) do appear in the transfer functions denominator for lowpass, highpass and bandpass. \$\endgroup\$
    – LvW
    Commented May 20, 2023 at 8:57

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