The Phono Preamplifier

Design

The phono preamplifier amplifies the audio signal of a MC^2.1 phono pickup with a sensitivity of 150$\mu$V to line level and also equalizes the frequency response according to the RIAA deemphasis curve (see [Hü97]). Additional equalization according to the IEC deemphasis curve is not done as my turntable does not produce significant drive noise and the electronics following in the signal chain is able to handle all signals produced by the MC pickup. Usually there is at least one highpass somewhere within the signal chain between pickup and speaker which has enough suppression of sub bass frequencies.

The preamplifier consists of several coupled stages (see the schematics ):

1. an ultra low noise linear amplifier stage offering 40 dB amplification,
2. a passive lowpass with a corner frequency of 2122 Hz (75 $\mu$s),
3. a second amplifier stage which also works as active filter for the poles 3180 $\mu$s and 318 $\mu$s,
4. and finally an offset compensation circuit which eliminates the DC offset of the first amplifier stage.

You must use at least two amplifying stages because an amplification of 64 dB at 1kHz (84 dB at 20 Hz!) cannot be handled by one single stage in high quality. The THD ratio would be high because a single stage has not enough noise gain to compensate the inherent non linearities.

I choose two separate filters instead of one composite filter for the RIAA equalization because it is very difficult to construct a three pole filter where the poles are very close together and influences each other. If you look into the excellent paper in [Lip79] you'll understand what I mean.

Abbildung 2.1: Schematics of the Phono Preamplifier Stage

The First Amplifier Stage

Nowadays you can get extremely ultra low noise operational amplifiers which compare with transistor pairs regarding price but are more easily available. Furthermore a discrete amplifier using a transistor pair needs very high precision resistors and a very good stabilized supply voltage. Operational amplifiers on the other hand have all these parts built in. Examples of ultra low noise operational amplifiers are LT1028 by Linear Technology and AD797 by Analog Devices.

A none inverting amplifier is built with the feedback resistors $R_{8}=1k\Omega$ and $R_{12}=10\Omega$ to get an amplification^2.2 of about 40 dB. Usually I do not use feedback resistors with such low resistance to prevent overloading the output of the operational amplifier (it gets warm and does not sound very good). This case is an exception because the input noise current of the operational amplifier flows through $R_{12}$ which means that the output noise increases with the resistance of $R_{12}$. Therefore its resistance should be so low that its residual noise is small compared to the input voltage noise of the amplifier.

The input impedance is set by putting $R_{2}$ in parallel to the input. Its value should be selected according to the specifications of the MC pickup. Usually this is 100-150$\Omega$, for low output MC pickups, high output pickups might need up to 1000$\Omega$ (check the owner's manual). I chose $R_{2}=100\Omega$.

This means that the none inverting input sees a source impedance of $100\Omega\Vert6\Omega$ and the inverting input of $1000\Omega\Vert10\Omega$. As the input offset current (which is relatively high) flows through both input impedances the imbalance leads to an additional input offset voltage which is then amplified by 40 dB. Usually this can be prevented by adding an additional resistor at the inverting input to recreate the balance. I did not do this because the input noise current also flows through this additional resistor, thereby adding unwanted voltage noise. The resulting offset voltage of the first stage is later eliminated by the offset compensation circuit.

At some later time I replaced $R_{8}$ in one channel by the serial connection of a 750$\Omega$ resistor and a 500$\Omega$ adjustable resistor. This allows to balance pickups with unequal output voltage.

The Passive Lowpass Filter

I chose a passive filter to realize the 75 $\mu$s lowpass for the following reasons:

1. The 75 $\mu$s lowpass must be the first filter to prevent clipping of the following amplifier stages by high frequency transient signals (e.g. due to scratches).
2. It is not advisable to realize the first amplifier stage as an inverting stage due to the high noise contribution of the input resistor. This input resistor of typical 100$\Omega$ would be added to the output impedance of the pickup-approximately 6$\Omega$ at my MC pickup-which would increase the input noise voltage by a factor of
   
   $$\sqrt{\frac{100\Omega+6\Omega}{6\Omega}}=4.2=12.5dB$$
   
   
   This cannot be accepted.
3. It is not possible to construct the 75 $\mu$s lowpass as active filter using a noninverting amplifier, because the minimum amplification is always 1 (0 dB) which inserts a second pole into the frequency response. This second pole can be compensated by adding a second lowpass to the output of the active filter. Unfortunately the second pole depends on the amplification of the filter, which means that an adjustment of the amplification (which is necessary as output voltage and source impedance of moving coil pickups vary considerable, e.g. 50 $\mu$V to 2 mV and 2$\Omega$ to 100$\Omega$) requires an adjustment of the compensation filter. This is very impractical to implement.

The passive lowpass consists of a simple RC filter. It is calculated as follows:

$$From\quad\omega=\frac{1}{2\pi f_{g}}\quad and\quad f_{g}=\frac{1}{2\pi R_{20}C_{35}}\quad results:$$

$$\omega=R_{20}C_{35}\quad therefore\quad C_{35}=\frac{\omega}{R_{20}}\quad or\quad R_{20}=\frac{\omega}{C_{35}}$$ (2.1)

The capacitor should be of the highest quality both in precision and sound. The only choice is a styroflex capacitor, which is available with 1% tolerance. The biggest value is just about 15nF. I chose $C_{35}=10nF$ which leads to a resitance of $R_{20}=7,5k\Omega$ according to equation 2.1, and this part is available with 1% tolerance. I got hold of a large bunch of 10nF capacitors and used a capacitance meter to select two capacitors of equal value (to improve channel balance) as close to the specification as possible.

The Second Amplifier and Filter Stage

The second amplifer stage also serves as filter for the poles 318$\mu$s and 3180$\mu$s. It is possible to use a passive filter-however that makes no sense as then the signal must be amplified much more which leads to increased distortion, and this is not our target.

Therefore we need an active filter where the amplification is constant between DC and 50,05 Hz, decreases by 6 dB/octave up to 500,5 Hz and stays constant again for all higher frequencies. This is easily achieved by adding the serial connection of resistor $R_{24}$ and capacitor $C_{36}$ parallel to resistor $R_{26}$ in the feedback loop. At very low frequencies the impedance of $C_{36}$ is very high so that the amplification depends on $R_{26}$ alone. At very high frequencies the impedance of $C_{36}$ is close to a short cut so that the amplification depends on $R_{26}\Vert R_{24}$.

To be able to calculate the filter exactly we must find the transfer equation of the circuit. Having done this we can compare the coefficients of the amplification equation with those of the filter equation. Let's do this assuming an ideal operational amplifier (having infinite difference amplification and input impedance and zero output impedance)^2.3.

The transfer equation of the active filter is:

$$A_{s}=1+\frac{R_{26}\Vert(R_{24}+C_{36})}{R3}$$

Replacing the filter components by the complex impedances results in:

$$A_{s}=1+\frac{1+j\omega C_{36}R_{24}}{1+j\omega C_{36}(R_{26}+R_{24})}*\frac{R_{26}}{R_{22}}$$

We reduce the equation to the lowest common denominator

$$A_{s}=\frac{1+j\omega C_{36}(R_{26}+R_{24})}{1+j\omega C_{36... ...}R_{24}}{1+j\omega C_{36}(R_{26}+R_{24})}*\frac{R_{26}}{R_{22}}$$

and work it out to:

$$A_{s}=\frac{R_{26}+R_{22}}{R_{22}}*\frac{1+j\omega C_{36}\fr... ...R_{26}R_{24}}{R_{26}+R_{22}}}{1+j\omega C_{36}(R_{26}+R_{24})}$$ (2.2)

The transfer equation of both poles 318$\mu$s and 3180$\mu$s results of the combination of the corresponding equations of the single poles (see [Hü97, equations (3) and (4)]) and a constant amplification $k$:

$$A_{f}=k*\frac{1+j\omega318\mu s}{1+j\omega3180\mu s}$$ (2.3)

If we compare the cofficients of both equations (2.2) and (2.3) we find the linear equation system:

$$k = \frac{R_{26}+R_{22}}{R_{22}}=1+\frac{R_{26}}{R_{22}}=\; DC-amplification$$ (2.4)

$$318\mu s = C_{36}\frac{(R_{26}+R_{24})R_{22}+R_{26}R_{24}}{R_{26}+R_{22}}$$ (2.5)

$$3180\mu s = C_{36}(R_{26}+R_{24})$$ (2.6)

If we assume $C_{36}$ and $k$ as known or required, we can solve the equation system in a way that results in the equations for the resistors $R_{26}$, $R_{24}$ and $R_{22}$^2.4:

$$R_{26} = \frac{3180\mu s}{\frac{10C_{36}}{9}(1-\frac{1}{k})}$$ (2.10)

$$R_{24} = \frac{R_{26}}{9}(1-\frac{10}{k})$$ (2.11)

$$R_{22} = \frac{R_{26}}{k-1}$$ (2.12)

Again I chose a $10nF$ styroflex capacitor for $C_{36}$, and $k=171$ for the DC amplification. The resistors turn out to be:

$$\begin{eqnarray*} R_{26} & = & 287883\Omega\\ R_{24} & = & 30116\Omega\\ R_{22} & = & 1693\Omega\end{eqnarray*}$$

Regarding the qualitiy of both capacitor and resistors please refer to 2.1.2 on page .

Subjective listening tests comparing the new preamplifier with my old one^2.5 by connecting the output of $U_{1}$ to the line level amplifier showed that the sound of the new phono preamplifier was more clean, had much more resolution and was much better at the frequency extremes, but was also more cool, sterile and metallic and not that musical. After countless tests I found a solution which added the missing characteristics: a class A output stage within the feedback loop using a power Mosfet (credits go to Nelson Pass).

$Q_{1}$ is a source follower using the source resistance $R_{28}+R_{29}$. The quiescent current running through $Q_{1}$ is calculated according to the equation

$$I_{r}=\frac{-(-U_{B})}{R_{28}+R_{29}}$$

as being 22,7 mA. $Q_{1}$ dissipates a power of

$$P_{tot}=(+U_{B})I_{r}$$

equal 340 mW and $R_{28}$ and $R_{29}$ each dissipate 170 mW. Therefore $Q_{1}$ needs a small heatsink, otherwise it gets very hot^2.6.

The signal is coupled to both the line level amplifier and the tape output via a serial resistor of $50\Omega$ each, because I use $50\Omega$ coax cables (RG58 / RG214U) both internal and external. This serial resistor prevents reflections^2.7.

The DC Offset Compensation

Both stages of the phono preamplifier together have a DC amplification of approximately $85dB$. Even when using the best precision operational amplifiers money can buy today it is not possible to get rid of the DC offset without reducing the DC amplification-which makes sense anyway in a phono preamplifier. There are several possiblities how to realize this:

1. The most simple way is to use an output capacitor. The disadvantage is that the output capacitor and the input impedance of the following stage (don't forget the tape recorder, if there is one as there is no buffer amplifier) form a highpass. To prevent that the corner frequency of this highpass gets too high you must use a big capacitor which usually leads to bad sonic performance.
2. It is possible to use capacitive coupling between the first and the second stage. In this case it is necessary to add a resistor parallel to $C_{35}$ to ground the input offset current of $U_{1}$. This resistor must be much bigger than $R_{20}$ if it should not influence the filter characteristic, and this leads to an increased DC output voltage of $U_{1}$.
3. You can reduce the DC amplification of the second stage to 0 dB by adding a capacitor in series to R22. Then the DC output voltage originates of the first stage only.

   Approaching a corner frequency below 20 Hz requires a capacitance of 4,7 $\mu$F. Subjective listening tests using a capacitance of 6,8 $\mu$F (leading to a corner frequency of 13 Hz) gave the impression of reduced bass dynamic (punch) compared to a version with full DC amplification. For a corner frequency of 1 Hz a capacitance of 82 $\mu$F is required, and a good one is very big in size and very expensive too. You cannot use a cheap electrolytic capacitor as there is no DC voltage applied which would lead to significant distortion.

4. It is possible to compensate the DC offset of the first stage so that just the DC offset of the second stage appears at the output of the phono preamplifier. The solution is to connect the DC voltage of the first stage to the inverting input of the operational amplifier of the second stage. In practice you connect the output signal of the first stage to a lowpass filter, amplify the filtered DC by a precision operational amplifier and connect the resulting voltage via a large resistor to the feedback resistor of the operational amplifier of the second stage. If the damping of the resulting voltage divider has the same value as the amplification of the precision operational amplifier the offset compensation is perfectly adjusted.

I chose solution 4. $R_{14}$ and $C_{33}$ form a lowpass with a corner frequency of approximately 1 Hz. The amplification of the precison operational amplifier $U_{5}$ is calculated according to the equation

$$A=1+\frac{RVAR_{1}+R_{18}}{R_{16}}$$

and turns out to be $119..128$ (41,5 .. 42,1 dB), depending on the adjustable resistor $RVAR_{1}$ ($1k\Omega$). $R_{36}$ and $R_{22}$ form a voltage divider the damping of which is calculated according to the equation

$$A=\frac{R_{22}\Vert R_{26}}{(R_{22}\Vert R_{26})+R_{36}}$$

and turns out to be approximately 122,6 (41,8 dB). You can see that amplification and damping are almost identical, and the trim pot $RVAR_{1}$ is used to adjust the values to absolute identity.

A top view of the printed circuit board is in figure 2.2 on page .

Abbildung 2.2: Top View of the Phono PCB