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Old 10th January 2003, 04:52 PM   #1
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Default Interesting norwegian RIAA amp

http://home.no.net/andiha/articles/a...afig/riaa1.gif

Interesting RIAA amp with descrete design and DC-servo.
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Old 10th January 2003, 05:13 PM   #2
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The riaa network appears to be fed by the current mirrors (transconductance amp). Is it possible to calculate the driving impedance with sufficient accuracy?

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Old 10th January 2003, 06:02 PM   #3
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Default And there is more...

here..:
http://home.no.net/andiha/articles/audio/index.htm

ArneK
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Old 11th January 2003, 06:14 AM   #4
Electrons are yellow and more is better!
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Quote:
Originally posted by analog_sa
The riaa network appears to be fed by the current mirrors (transconductance amp). Is it possible to calculate the driving impedance with sufficient accuracy?
It's not kind of taste of driving RIAA networks but I think that the output impedance is rather "resistive" in the audio band, therefore the gain wil be predictable.

It's also not my kind of taste to have parameters that give varying results. The gain (I suspect) will differ between different amps because the transistor parameters will variate.
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Old 11th January 2003, 06:31 AM   #5
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Peranders

I was not worried about the gain but rather about RIAA accuracy. Don't you need to know very accurately the driving impedance in this configuration? There doesn't seem to be a resistor between the transconductance amps and the 'passive' RIAA.



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Old 11th January 2003, 08:32 AM   #6
Electrons are yellow and more is better!
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Quote:
Originally posted by analog_sa
Peranders

I was not worried about the gain but rather about RIAA accuracy. Don't you need to know very accurately the driving impedance in this configuration? There doesn't seem to be a resistor between the transconductance amps and the 'passive' RIAA.
The "resistor" is the output impedance and the gain will variate but, not the much the RIAA curve looks, just a little maybe.
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Old 11th January 2003, 10:58 AM   #7
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It's obvious the output impedance is the 'resistor'. But how do you calculate its value without knowing the current gain of the transistors? That's why normally there is a resistor with value a lot higher than the preceding stage output impedance. Is this circuit supposed to be tuned to the selection of active devices?

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Old 11th January 2003, 01:30 PM   #8
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When I've taken a closer look I believe that the DC-gain is (R23/R17)/2, quite predictable.
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Old 11th January 2003, 09:10 PM   #9
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Quote:
Originally posted by peranders
When I've taken a closer look I believe that the DC-gain is (R23/R17)/2, quite predictable.
If I was to judge by the formulas at Andiha Homepage Andiha Audio
I would say he probably knows what he is doing.

This page for example, makes me DIZZY just by a quick glance:

The Complete Amplifier

And this is what he is talking about
a typical voltage feedback operational amplifier
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