JamJar: an HPA-1-inspired power amp

Getting back to the puzzle Mark left for me:

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This chart says (to me) that the opamp I'm using in my FE shunt regulator has poorer noise rejection on its negative supply than its positive supply.

Mark also gives the hint:

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I'm not sure if those are "put something here" boxes, or ISO resistor symbols. But the location has me stumped too. The box is shown on the positive supply of the positive regulator's opamp, and on the negative supply of the negative regulator's opamp.

Perhaps Mark didn't notice that the positive and negative regulators are not symmetrical?

But even if they were, I'm not sure what this addition would do....
 
On your shunt regulator either throw a 33,000uF cap after the input resistor on each rail, alternatively and probably even better you could build up the total resistance at the input of the shunt with CRCRCRCRC, eg 2200uF, 27Ohms, 2200uF, 27Ohms, 2200uF 27 Ohms, 2200uF, 27Ohms, 2200uF
That should improve things significantly, once you have implemented that the PSRR of your opamps will no longer even matter, ie you will have eliminated any issue in the performance of the opamps being a problem.
 
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I'm not sure if those are "put something here" boxes, or ISO resistor symbols. But the location has me stumped too. The box is shown on the positive supply of the positive regulator's opamp, and on the negative supply of the negative regulator's opamp.

..

RC Filtering is what he is most likely referring to, but you should also use CRCRCRCRC at the input as mentioned in my post above
 
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Jeff, make a few assumptions and use them to do a few calculations, to figure out how good your power supply needs to be. Then design to that target. Boom, done.

Someone (not you, just a hypothetical strawman example hobbyist) might, for example, assume that their power supply needs to be constant and ripple free and noise free enough, that the supply imperfections, times the signal handling circuit's PSRR, is less than or equal to (2 to the power -18). The supply-created imperfections in the signal circuitry's output, are the same magnitude as one LSB of an 18-bit D-to-A. {my calculator claims that equals 3.8 parts per million.} Now this "someone" can work backwards to find out how good his supply must be. Design to that target, slather on some extra safety margin if possible, Boom. Done.
 
I tried these two ideas out: RC filtering on the opamp supply, and CRCRC pre-filtering.

RC filtering on the opamp is interesting. Sticking with a C0G cap you can't really get enough of a filter to make any difference. Moving to an X7R allows for an improvement of a bit over 2dB at 300kHz. Using a tantalum doesn't help as the ESR takes away everything the increased capacitance gives. A big film cap can produce significant results (in the region of 10dB).

Moving on to pre-filtering, the opamp doesn't like CRCRC (or CRC) filtering (it puts a big spike in the attenuation curve). CRCR does produce a huge win, increasing the attenuation by something like 50dB(!).

This begs the question that while this is all interesting (and instructional), is it material? Will there be any audible difference between 75dB and 125dB of attenuation at 20kHz? I know it's not much of a bar, but the venerable L78XX can't even get within a country mile of the 75dB spec.

(Mark posted after I had written this, so I'm now off to do his suggested math....)
 
OK, Mark's LSB (out of 18) is 54dB. So let's add 11 ('cause my volume control goes to 11) and say I want my regulator-attenuation + amp-PSRR to be 65dB. If my regulator's attenuation is 75dB, then my PSRR can be... well, completely non-existent.

Now to measure output impedance.

SPICE says my FE/VAS draws 96mA +/- 3mA. If I put that load on the regulator (at 20kHz), then I get a 20kHz PtP ripple of 38uV. R = V/I so my effective output impedance (assuming I've done this right) is 0.038/6 or 0.006 ohms. Whoo hoo!

I look pretty good on the Linear Audio chart, too, though I suspect most of those were far more efficient than my paltry 37%. (Fortunately for me this is the Pass forum and none of us are afraid of a little heat. :cool:)
 
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Jeff you did the math incorrectly.

2 raised to the power (-18.0) equals 10 raised to the power (-5.419). Since we're talking about voltage not power, that's (20 * -5.419) = -108.4 dB.

A nice aid to memory is: -120dB is 1ppm. -100dB is 10ppm.

Since we're talking about 4 ppm, the number of dB must be somewhere between -120dB and -100dB. And it is. Double check: yes!