Phono preamplifier with DC servo: precorrecting RIAA correction poles

Hi all,

I sometimes see people trying to put DC feedback loops around phono preamplifiers. When you put a DC servo loop around a phono preamplifier, the feedback shifts the poles, particularly the lowest one, and may mess up the RIAA correction accuracy. One way around that is to deliberately shift the correction network poles such that they end up where they should be with the DC servo loop closed. I did some calculations about that, see the attachment. I hope it will be of use to someone.

Best regards,
Marcel
 

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  • RIAAwithDCservo.pdf
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Thanks MarcelvdG!
Good to learn that what seems better is not so necessarily.
I read the paper (diagonally), but what is the factual origin of this shift?
Is it the (low) frequency contribution of the servo well above its bandwidth?
And is that the reason for a very low cutoff frequency for the servo, say 100 times lower then 50.05Hz / 3180µsec (0.5Hz in your article)?
Again, much appreciated!
 
Marcel,
Your paper has caused me to check my pre-amp.

I use an integrator time constant of 50s. The DC feedback is injected at the input and the gain is 1000. The -3dB frequency is 3Hz. By 20Hz, the error is -0.1dB. My time constant is a factor of six shorter than your example #1.

The reason is in the criteria:

Design the DC servo to have practically no loop gain left at 50 Hz and above, say magnitude of the loop gain ≤ 0.01 if you are willing to accept an error of the order of 1 %.

This is more conservative than what I do. I take the magnitude of a complex number and ignore the phase. I believe that only the magnitude is relevant.

This is an old pre-amp, but I would have noticed if I had gotten it wrong. ;)
Ed
 
Ed, are you sure you didn't either tweak the value of the correction network a bit, or use higher-order roll-off of the servo, or just calculate the roll-off due to the 3 Hz high-pass without accounting for the shift of the correction filter poles?

The criterion of 1 % pole shift corresponds to about 0.0864 dB of error at frequencies well below 50 Hz (when you normalize to 1 kHz, like everyone always does). It's not that different from your 0.1 dB. As the shift makes the bass increase, there is some compensating effect due to the roll-off of the servo.

With K = 1000, tauDClint = 50 s, the first time constant would have to be corrected by -5.8622431852 % to get the first correction pole at the correct place with the loop closed. Assuming that with the uncorrected network, the time constant ends up at 3.18 ms/(1 - 0.058622431852), which should be approximately true, that results in an error of +0.4474 dB at 20 Hz. Subtract 0.1224 dB because of the roll-off of the servo (taupsubs is about 47.063 ms rather than 50 ms), and you end up at an error of approximately +0.3249 dB at 20 Hz.
 
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Thanks MarcelvdG!
Good to learn that what seems better is not so necessarily.
I read the paper (diagonally), but what is the factual origin of this shift?
Is it the (low) frequency contribution of the servo well above its bandwidth?
And is that the reason for a very low cutoff frequency for the servo, say 100 times lower then 50.05Hz / 3180µsec (0.5Hz in your article)?
Again, much appreciated!

Shifting poles is what feedback always does. It's how you can make an amplifier with a bandwidth in the megahertz range with an op-amp having an open-loop roll-off at 10 Hz.

In this case, the 0 Hz pole of the DC loop integrator is shifted to a higher frequency and the first RIAA correction pole is shifted to a lower frequency. If you don't want the RIAA correction pole to shift, you either have to keep it out of the loop, or to make sure the loop gain is very small at 50 Hz, or to put a zero on top of it that doesn't occur in the input-to-output transfer - which you can do with a zero in the DC feedback circuit.

With a first-order roll-off, a 0.5 Hz bandwidth corresponds to a loop gain of 0.01 at 50 Hz, neglecting the fact that the first correction network pole itself also helps a bit to reduce the loop gain at 50 Hz.

Unlike the poles, the zeros don't shift. For the values of s where the forward path gain is 0, the gain just stays 0 no matter whether you apply feedback.
 
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Marcel,
The 3Hz frequency was hand-calculated without considering the EQ network. I see now that it was too close to 50Hz to ignore the EQ. The EQ is otherwise dead-on accurate.

I may increase the integrator time constant. I recently removed the 100pF loading capacitors once I realized that they were causing an error of +1dB at 10KHz with the AT440ML.
Thanks!
Ed
 
Dear "MarcelvdG",

the discussion thread in question was not addressed to you at all. Strangely enough, however, your name haunted the back of my mind.
I didn't want to trigger you in any way. Nevertheless, I am very pleased that you intervened with an explanation. The first three circuits in my thread are of course nonsense. The idea behind the thread "Haudegen" is both polarization and a little bit of entertainment. A slightly innocent but also defiant approach, initially I was trying to simulate a kind of intuitive approach to building an equalizer. I deliberately switched off my mind and allowed myself to make every mistake.

Perhaps you took this as an opportunity to explain the so-called control loops in a factual manner.
Thank you very much for this, I would personally like to see one or two DIY approaches based on more solid foundations.

I can't stand fiddling with the resulting time constants when everything can be calculated relatively easily (in the control loop).

But I don't want to disturb your great thread with my flippant manner. In my thread, I do follow a rough red thread, but anything can happen there and you shouldn't take it too seriously. It's not really aimed at experts, otherwise I wouldn't use this stylistic device under any circumstances.

Yours sincerely,
HBt.
 
Hi HBt,

I had seen your thread and it was the trigger to start this one, but I've seen many people run into this issue. The first time was in the early 1990's, when a then colleague of mine tried to put a DC servo loop around a phono preamplifier. I don't remember whether he or I or we together found the explanation.

In any case, as it is a quite generic issue, starting a new thread seemed more logical than replying to yours.

Best regards,
Marcel
 
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At the risk of stating the obvious its common for a phono input to have a high-pass filter, aka rumble filter, so that warps and tone-arm resonances are suppressed - with an AC-coupled phono preamp the need for a DC servo is eliminated. The rumble filter would invariably after the preamp RIAA stage and thus not affecting the poles of that stage. (Although encroaching on the bass end overall response).
 
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When you have a straightforward op-amp phono amplifier where you apply DC feedback by simply connecting a big capacitor in series with the resistor from the negative input to ground, you don't run into the issue of the shifting RIAA correction poles. The reason for that is that these poles are then automatically covered by zeros of the feedback network. (In fact the zeros of the feedback network and the near infinite gain of the op-amp define the pole locations.)
 
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We are mixing topologies. ;)

My point was that DC feedback is common in phono pre-amps because of the gain.

You are pointing out that the poles don't get shifted with a large capacitor in the active feedback network.

My pre-amp is in the problematic category because its EQ is passive. The DC feedback comes from a large capacitor that integrates the output.
Ed
 
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Just a couple of plots that follow from the equations in the document. First the required correction in % for the first time constant to get the first RIAA pole where it should be, as a function of the cut-off frequency of the DC servo loop in Hz:

RIAAwithDCloopcorrp1.png


The same for the time constant that sets the second RIAA pole:

RIAAwithDCloopcorrp2.png


These corrections are small enough to neglect for most practical purposes.

Finally the required value of the time constant of the integrator divided by the forward-path DC gain, so 1/(omegaDClintK), in second as a function of the desired servo loop cut-off frequency in hertz. It is not exactly inversely proportional because of the effect of the correction network in the loop.

RIAAwithDClooptauDClintoverK.png
 
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...what is the factual origin of this shift?
Basically, if the EQ network is inside the feedback loop, then the differential input stage will see that the output signal is different from the input signal, so it will produce an error signal to try to correct the output to match the input. Depending on the loop gain at the frequency of interest, the differential input stage can correct what it thinks is a FR error in the amplifier by only so much.
 
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