Deliyannis-Friend design: where is my mistake?

Question

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.

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}

and

\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...

UPDATE MAY 20

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.

Answer 1

@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:

H(s)=-[s(1+m)k2RC]/[1+sRC(2-mk2)+s²k2R²C²]

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.

Answer 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.