Electronic – Second order all-pass filter using a single op-amp

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Is it possible to build an active second order all-pass filter using a single op-amp and no inductors? After Googling it I've found no less than three different circuit topologies, but when simulating them they all have a non-flat frequency response. I've also tried analysing them using some simple Laplace transforms and some algebra, but have failed to get anything similar to the transfer function that a second order all-pass filter should have. This might be because the algebra does get slightly messy, and I don't handle messy algebra very well when tired.

It would be great if such a circuit does exist, as I'm (purely for fun) designing a phasing network to obtain a fairly flat 90 degrees phase difference (quadrature) in the output over a fairly wide range of frequencies, to be used in a phasing SSB receiver for side-band rejection. Currently I'm using a software (called QuadNet) that outputs a phasing network for me, but it uses first order segments, which results in a whole lotta op-amps. The goal is to halve the necessary number of op-amps.

Just for reference; the transfer function of an all-pass filter takes on the following form $$\frac{s^2-As+B}{s^2+As+B}.$$

To be clear, I'm simply looking for a circuit topology that provides this transfer function (2nd order all-pass filter) using a single op-amp and no inductors, and nothing else. Assuming ideal components is totally fine for my purposes.

I'm eagerly awaiting enlightenment!

Best Answer

One option for a second order all-pass filter with one OpAmp is the Delyiannis structure, as shown here (p.2):

enter image description here

The transfer function of this filter is given by

$$H(s)=\frac{R_4}{R_3+R_4}\cdot \frac{s^2-s\frac{2}{R_2C}+\frac{1}{R_1R_2C^2}}{s^2+s\frac{2}{R_2C}+\frac{1}{R_1R_2C^2}}\tag{1}$$

where

$$\frac{R_2R_3}{R_1R_4}=4\tag{2}$$

must be satisfied.

For a given center frequency \$\omega_0\$, a given quality factor \$Q\$, and a chosen value of \$C\$, the resistors \$R_1\$ and \$R_2\$ must be chosen as

$$R_1=\frac{1}{2Q\omega_0C},\quad R_2=\frac{2Q}{\omega_0C}\tag{3}$$

and \$R_3\$ and \$R_4\$ must be chosen to satisfy \$(2)\$. One possibility (as suggested in above document) is \$R_1=R_3\$ and \$R_4=R_2/4\$. (Note there is an error in the definition of \$R_4\$ in the document cited above).

From Eqs \$(2)\$ and \$(3)\$ it is straightforward to show that the filter gain is related to the quality factor \$Q\$ by

$$g=\frac{R_4}{R_3+R_4}=\frac{Q^2}{1+Q^2}\tag{4}$$