A balanced (differential) Bottlehead Seduction phono preamp?

Mikey · 9044

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Offline Mikey

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Hello gang,

In my possesion I've got a preamp with a (built-in) differential phono stage.
The RIAA correction is done 'all-at-once', sandwiched between two gain
stages, very similar to the Seduction phono preamp....only balanced.

Looking at the values of some of the components, it seems like it is a high
impedance RIAA circuit, with a series resistor of around 715K or so.

Would there be any benefit to retrofitting a lower impedance RIAA circuit
(a balanced version of the Seduction) in place of the existing circuit?

http://www.kabusa.com/riaa.htm

Mike

Mike Paschetto


Offline Paul Joppa

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Reply #1 on: January 21, 2011, 12:54:03 PM
High impedance keeps the triodes linear and the gain high, and makes the circuit values less dependent on tube parameters (plate resistance). Low impedance reduces hum and noise from electrical fields, and makes the circuit values less sensitive to stray capacitances. Finding the best balance is always a challenge!

Paul Joppa


Offline Mikey

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Reply #2 on: January 22, 2011, 02:50:44 AM
Hmmmmmmmmmm...

I've got gain to spare, so that's not an issue.

Maybe I'll fart around with the RIAA calculator and see
if I can come up with a 'middle of the road' solution.

Thanks PJ!

Mike Paschetto


Offline Mikey

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Reply #3 on: January 24, 2011, 08:15:51 AM
Well, I'll be darned!

Taking the Seduction RIAA section and 'mirroring' it over a virtual GND
results in two capacitors in series for the various corner frequencies:

0.033uF + 0.033uF = 0.0165uF
0.0012uF + 0.0012uF = 0.0006uF

Dang, those are some small values, and I'mm assuming that their precision
has to be pretty good to boot!  Has anyone got any idea where to begin
to search for precision small value capacitors like this?

Mike

Mike Paschetto


Offline HF9

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Reply #4 on: January 24, 2011, 08:27:12 AM
I think 0.0006uF translates to 600pF

I saw this on Mouser: http://www.mouser.com/ProductDetail/Cornell-Dubilier/MCM01-001ED601J-F/?qs=UFm1FN5OnO2LmS4xJzxFYw%3d%3d

it's SMD though, you'd likely have to put in on a little breadboard.

My DIY Audio Electronics Blog: DIYAudioBlog.com


Offline xcortes

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Reply #5 on: January 24, 2011, 08:43:08 AM
I used Auricaps in the pF range when reducing the value of the parafeed caps in my tweeter amps

Xavier Cortes


Offline Grainger49

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Reply #6 on: January 24, 2011, 09:30:09 AM
Dang, those are some small values, and I'm assuming that their precision
has to be pretty good to boot!  Has anyone got any idea where to begin
to search for precision small value capacitors like this?

Mike

For precision buy a number then measure for those closest to the nameplate value.  You will most often see caps are less than the nameplate value.



Offline Paul Joppa

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Reply #7 on: January 24, 2011, 10:11:09 AM
The smaller caps in the Seduction equalization are in parallel, to produce the required 0.0112uF, half of which is 0.0056uF, or 5600pF.

There are really only two caps, and their capacitance ration is 2.9163, close enough to 3.0. For that reason, when we did the Eros I switched to caps of 0.030 and 0.010uF. You can often find 0.010uF caps in precision values even if other values are not available. So you parallel three of them to get 0.030. And usually there's a price break for 10 caps anyhow - by mixing and matching with a cap meter you can get pretty close even if you can't find precision caps.

OR take the Seduction circuit, double the caps and halve the resistors, then mirror it and the caps revert to the Seduction values.

Paul Joppa


Offline Mikey

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Reply #8 on: January 24, 2011, 12:05:44 PM
The smaller caps in the Seduction equalization are in parallel, to produce the required 0.0112uF, half of which is 0.0056uF, or 5600pF.



Whoops!  I 'fat-fingered' the keyboard on that one....my bad!
I'll try to conjure up some resistor values using the 0.030 and 0.010uF capacitors.

Gang, thanks for the feedback so far; I'll follow up on those leads.

Mike

Mike Paschetto


Offline Mikey

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Reply #9 on: January 25, 2011, 04:57:14 AM
OR take the Seduction circuit, double the caps and halve the resistors, then mirror it and the caps revert to the Seduction values.

Well, that certainly makes the calculations easy!

Using Seduction capacitor values of C1=0.033uF X2 (0.066uf) and C2=0.011uF X2 (0.022uF),
I'm calculating resistor values of R1=33.1K and R2=4.81K, just as you predicted.

These resistor values do not take into account the load impedance (RL) seen by the network,
nor the source impedance of the preceding gain stage.  I found another RIAA calculator on the
web that is a bit better than the KAB network computer, available for download here:

http://www.beigebag.com/files/Programs/

It's the TubeCad Journal calculator, and I'm going to fiddle with it for a bit...

Mike

Mike Paschetto


Offline Paul Joppa

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Reply #10 on: January 25, 2011, 06:28:23 AM
Seduction is designed so that the source resistance (the 6922 plate resistance) and the load resistance (the 1 meg grid resistor) cancel each other. You can get very close to this if the series resistance (66.5K in Seduction) equals the geometric mean of the source and load resistors. This also provides the lowest sensitivity to errors in those resistors.

Paul Joppa


Offline Mikey

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Reply #11 on: January 25, 2011, 10:01:34 AM
Seduction is designed so that the source resistance (the 6922 plate resistance) and the load resistance (the 1 meg grid resistor) cancel each other.

Ah ha!  That's very clever....

In my case, I'm 'swagging' a source impedance of 24K, and a load impedance of 10M.
That's roughly a 500K geometric mean....I guess I won't have the luxury of cancellation,
unless I swap out the grid resistor of the next stage.  Maybe I'll revisit that option later...

Mike

Mike Paschetto


Offline Paul Joppa

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Reply #12 on: January 25, 2011, 10:42:24 AM
Seduction is designed so that the source resistance (the 6922 plate resistance) and the load resistance (the 1 meg grid resistor) cancel each other.

Ah ha!  That's very clever....

In my case, I'm 'swagging' a source impedance of 24K, and a load impedance of 10M.
That's roughly a 500K geometric mean....I guess I won't have the luxury of cancellation,
unless I swap out the grid resistor of the next stage.  Maybe I'll revisit that option later...

Mike

OK, now here's the clever part ... pick your geometric mean and 10Meg load, calculate the ideal source impedance (example: 100K mean, 10Meg load, 1K ideal source). Subtract that from the actual source (24K-1K=23K). Now that leftover part, 23K in this example, becomes part of the 100K so you use a 77K resistor.

Makes sense?  :^)

Paul Joppa


Offline Mikey

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Reply #13 on: January 25, 2011, 11:02:06 AM


Makes sense?  :^)

I think so...  ;-)

Didja download a copy of the TubeCad RIAA software yet?  It has the ability for the user to input values for RS and RL, and works out
the 'adjusted' series resistor for ya', no calculator necessary!  I believe it's doing exactly what you just described in your previous post...

<pause>

OK, I just ran the (admittedly rough) numbers thru TubeCad RIAA.  When RS=24K and RL=10M are applied to our differential network,
Rseries drops from 33.1K to about 9K, just as you predicted!

Now, I've read differing opinions on the next tidbit, but I'd like to know:

- How important is it to get the value of the series resistor 'spot-on'?

Is a 1% variance acceptable?  How about 10%?  If Rseries is indeed hyper critical, then I'm assuming the output impedance of the
preceeding stage will have to be calculated and/or measured very closely, no?

Mike

Mike Paschetto


Offline Paul Joppa

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Reply #14 on: January 25, 2011, 12:39:18 PM
The large-value series resistor sets the 50Hz corner frequency, primarily. It's probably more important that they match between channels because the ear is pretty sensitive to phase below 1kHz. In general, 1% is 0.1dB - a few people might hear that in the midrange, less likely in the deep bass.

The output impedance of a triode will increase as it ages, probably as much as 50% before the tube is otherwise obviously near the end of its life. So yes, it's worthwhile to run the eq at high impedances. The problem with high impedance is susceptibility to electric fields and RF interference, and errors due to stray capacitance.

Now run the numbers again with a smaller load resistor, and see how that reduces dependence on the source resistance.

Paul Joppa