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Old 12th December 2009, 07:32 AM   #1
bbp is offline bbp  China
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Default Ultra high precision Volume module like Mark Levinsion's

Finally, I collect enough information to design such a REAL R-2R attenuator.

Fully balanced attenuation through 16 bit R-2R resistor network.
65535 steps current attenuations available, two Ultra low-noise low distortion OPAMP AD797AR were used as I/V convertor, the same structure as the module in Mark Levinson's No.32 ^_^

Considerate LAYOUT, 4-LAYERS PCB with 2 Oz copper,thanks Mark ^_^
Considerate POWER MANAGEMENT, onboard ultra-high performance linear regulator
High-precison SMT chip resistors
Simple data interface for micro controller
dimension:5545mil*2135mil


current switches: DG413DY, ultra precision analog switches
I/V convertor : AD797AR, ultra low-noise,ultra low distortion OP from ADI
Onboard regulator:LT1355
Resistors : YAGEO RT series @ 1% tolerance


The PCB will arrive next week, I will assemble the whole module and test then.
Anyone interest in it? ^_^
Attached Images
File Type: jpg R-2R D1.jpg (105.7 KB, 1694 views)
File Type: jpg R-2RR.jpg (78.2 KB, 1644 views)
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Old 12th December 2009, 07:44 AM   #2
EC8010 is offline EC8010  United Kingdom
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If you have 65535 steps and an R-2R ladder, you need far better than 1% tolerance resistors. You need 0.001%. I've no idea where you'd get such things.
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Old 12th December 2009, 07:55 AM   #3
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Originally Posted by EC8010 View Post
If you have 65535 steps and an R-2R ladder, you need far better than 1% tolerance resistors. You need 0.001%. I've no idea where you'd get such things.

we don't need such high precision resistors up to 0.001%, 1% is enough in this application. even though, Mark Levinsion's product uses 5% resistors in their module in No.32, 1% resistors in No.320s/326s ^_^

We don't need a tolerance = 1/65535=0.00001526 in each step, just think about the R-2R DACs like PCM1704U-K ^_^
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Old 12th December 2009, 04:13 PM   #4
EC8010 is offline EC8010  United Kingdom
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You might not need 0.001%, but you do need better than 1% if you want to keep the errors down and make the accuracy reflect the precision. I'm assuming you don't actually make all those 65535 steps available to the user, but actually code for a logarithmic law and pick codes that give 1dB steps. At 6dB of attenuation, 1% resistors give an uncertainty in attenuation of 1.4%, and between channels of 2%, or 0.2dB. Stepped balance controls typically go in steps of 0.25dB, so those 1% resistors could cause balance error of one balance step. I don't see that a stepped attenuator is entitled to any balance errors, so whilst 0.001% isn't necessary after all, I think you need better than 1%, and 5% is certainly not good enough.
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Old 12th December 2009, 05:45 PM   #5
mlloyd1 is offline mlloyd1  United States
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i find it rather strange that mark levinson would use 5% tolerance resistors in the precision attenuator in such a flagship product as the no.32 preamp.

are you sure about your source of information?

mlloyd1

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Originally Posted by bbp View Post
we don't need such high precision resistors up to 0.001%, 1% is enough in this application. even though, Mark Levinsion's product uses 5% resistors in their module in No.32, 1% resistors in No.320s/326s ^_^

We don't need a tolerance = 1/65535=0.00001526 in each step, just think about the R-2R DACs like PCM1704U-K ^_^
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Old 12th December 2009, 07:03 PM   #6
cbdb is offline cbdb  Canada
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At 6dB of attenuation, 1% resistors give an uncertainty in attenuation of 1.4%, and between channels of 2%, or 0.2dB.

How did you figure that? I get more like .01db.


10 log 100= 20, for 2% error: 10 log 102=20.09 or .09db error, for 5% 10 log 105=20.21 or .21db.


So the 1% tolerances are fine and even the 5% would be fairly inaudible.
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Old 13th December 2009, 12:32 AM   #7
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Quote:
Originally Posted by mlloyd1 View Post
i find it rather strange that mark levinson would use 5% tolerance resistors in the precision attenuator in such a flagship product as the no.32 preamp.

are you sure about your source of information?

mlloyd1

Hi mollyd1,

I just download the image from Mark's website and found this >_< But I'm not sure all the version of different date manufactured module uses the same 5% resistors


regards
bbp
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File Type: jpg MARK 32.jpg (149.9 KB, 1482 views)
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Old 13th December 2009, 12:49 AM   #8
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Quote:
Originally Posted by EC8010 View Post
You might not need 0.001%, but you do need better than 1% if you want to keep the errors down and make the accuracy reflect the precision. I'm assuming you don't actually make all those 65535 steps available to the user, but actually code for a logarithmic law and pick codes that give 1dB steps. At 6dB of attenuation, 1% resistors give an uncertainty in attenuation of 1.4%, and between channels of 2%, or 0.2dB. Stepped balance controls typically go in steps of 0.25dB, so those 1% resistors could cause balance error of one balance step. I don't see that a stepped attenuator is entitled to any balance errors, so whilst 0.001% isn't necessary after all, I think you need better than 1%, and 5% is certainly not good enough.
Of course we don't need all the 65535 steps in the actual application,I plan to make the attenuations adding 0.1 dB/step from -96dB to -46dB and 1 dB/step from -46 to 0 dB.
That means I need to calculate the relationship for decimal/dB >_<

For the tolerance, 0.1dB is equal to 1.14%
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Old 13th December 2009, 09:56 AM   #9
EC8010 is offline EC8010  United Kingdom
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Quote:
Originally Posted by cbdb View Post
How did you figure that? I get more like .01db.


10 log 100= 20, for 2% error: 10 log 102=20.09 or .09db error, for 5% 10 log 105=20.21 or .21db.


So the 1% tolerances are fine and even the 5% would be fairly inaudible.
No, I'm afraid your calculations above are incorrect. A 1% error would mean that the attenuation was 0.99 or 1.01 of what it ought to be. Further, 10 log is for powers, but potential dividers deal with voltages, so we use 20 log. Taking the 0.99 case, 20 * log(0.99) = -0.087dB - near enough 0.1dB.
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Old 13th December 2009, 12:27 PM   #10
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No, I'm afraid your calculations above are incorrect. A 1% error would mean that the attenuation was 0.99 or 1.01 of what it ought to be. Further, 10 log is for powers, but potential dividers deal with voltages, so we use 20 log. Taking the 0.99 case, 20 * log(0.99) = -0.087dB - near enough 0.1dB.
Acutally divides current ^_^
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