Hi all,

While thinking about ways to speed up the settling of a single-supply single-op-amp RIAA amplifier, see https://www.diyaudio.com/community/...upply-phono-preamp-design.413571/post-7702435 , I found a way to include a second- or third-order Butterworth high-pass filter. As it may be useful outside the context of single-supply circuits, I give it a separate thread.

When you just look at the topology and ignore the component values, this is a rather conventional RIAA amplifier (you could make it even more conventional by connecting R

Normally, C

Note that

With everything ideal, at the value of the Laplace variable

The disadvantage of using C

As an intermediate step, suppose you could add an ideal inductor with a huge value between the output and the negative input of the op-amp, chosen such that it resonates with C

Such an ideal inductor is totally impractical, but it can be approximated with a T network consisting of two resistors with values much smaller than

The transfer from the voltage going into R

I haven't found any simple exact equations for any of the values except

Regarding those approximate calculations:

The DC gain would be 1 +

At

(

At

(

The (theoretical) inductance

Best regards,

Marcel

While thinking about ways to speed up the settling of a single-supply single-op-amp RIAA amplifier, see https://www.diyaudio.com/community/...upply-phono-preamp-design.413571/post-7702435 , I found a way to include a second- or third-order Butterworth high-pass filter. As it may be useful outside the context of single-supply circuits, I give it a separate thread.

When you just look at the topology and ignore the component values, this is a rather conventional RIAA amplifier (you could make it even more conventional by connecting R

_{7}in parallel with C_{5}, that doesn't matter much for the principle):Normally, C

_{8}is used to cause roll-off in the subsonic region and the network R_{7}...R_{9}, C_{5}, C_{6}realizes the RIAA poles and zero. In this case, however, I use C_{8}to realize the lowest RIAA pole at -1/(3.18 ms) and R_{7}to get subsonic roll-off.Note that

*C*_{8}/*C*_{5}= 1000, meaning that without the subsonic roll-off, the DC gain would be 1001, a very ordinary value for a moving-magnet amplifier (1 kHz gain roughly 40 dB).With everything ideal, at the value of the Laplace variable

*s*where the impedance of C_{8}cancels the impedance of R_{12}, the feedback disappears and the gain goes to infinity. This means that there is a pole at exactly -1/(*R*_{12}*C*_{8}), so if this has to be the first RIAA pole, one needs*R*_{12}*C*_{8}= 3.18 ms. It's actually 3.196 ms in the schematic, pretty close.The disadvantage of using C

_{8}for the first RIAA pole is that C_{8}, which has a relatively large value, needs to be accurate to get an accurate first RIAA pole. The advantage is that you can include better subsonic filtering in the loop by adding two more resistors and a capacitor.As an intermediate step, suppose you could add an ideal inductor with a huge value between the output and the negative input of the op-amp, chosen such that it resonates with C

_{5}at the desired subsonic roll-off frequency, and that you chose*R*_{7}such that it damps the LC circuit to a quality factor of 1/2 √2. The subsonic response would then be very close to second-order Butterworth. That's because the gain of the RIAA correction amplifier is one plus the ratio of the feedback impedance to the impedance from the negative op-amp input to ground, and that "one" is quite negligible at low frequencies. Mind you, R_{8}and R_{9}contribute to the damping of the LC circuit, but not by much. You could also choose a quality factor of 1 and design the AC coupling at the input for the same cut-off frequency. The combined response is then third-order Butterworth.Such an ideal inductor is totally impractical, but it can be approximated with a T network consisting of two resistors with values much smaller than

*R*_{7}and a capacitor to ground at the point where they are connected, see this figure:The transfer from the voltage going into R

_{11}to the current coming out of R_{10}rolls off at a first-order rate from some very low frequency onwards, like would be the case with an inductor.I haven't found any simple exact equations for any of the values except

*R*_{12}*C*_{8}= 3.18 ms. In fact, I've been very lazy and just calculated approximate values for the other components, and then used a pole-zero extraction program to fine-tune the values.Regarding those approximate calculations:

*R*_{12}*C*_{8}= 3.18 ms to get the first RIAA pole at the right spot.The DC gain would be 1 +

*C*_{8}/*C*_{5}without subsonic roll-off, so*C*_{8}/*C*_{5}= 1000 gives you a midband gain of roughly 40 dB.At

*s*= -1/((*R*_{8}+*R*_{9})(*C*_{5}+*C*_{6})), the impedance of the network R_{8}, R_{9}, C_{5}, C_{6}goes to zero and the gain of the circuit becomes 1. As a gain of 1 is pretty close to 0, this must be close to the location of the RIAA zero. That is,(

*R*_{8}+*R*_{9})(*C*_{5}+*C*_{6}) ≈ 318 μsAt

*s*= -1/((*R*_{8}+*R*_{9})*C*_{6}), the impedance of the parallel connection of C_{6}and R_{8}& R_{9}goes to infinity. The impedance of the whole feedback network remains finite due to the other branches R_{7}and R_{11}, C_{7}, R_{10}, but it does get pretty large. That means the second RIAA pole must be close, so we get the extra criterion(

*R*_{8}+*R*_{9})*C*_{6}≈ 75 μsThe (theoretical) inductance

*L*is chosen to resonate with C_{5}at the required subsonic roll-off frequency and R_{7}is chosen to get the desired quality factor.*R*_{10}and*R*_{11}get convenient values much smaller than*R*_{7}with*R*_{10}also much greater than*R*_{12}. We then have*C*_{7}=*L*/(*R*_{10}*R*_{11})Best regards,

Marcel

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