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#1 |
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diyAudio Member
Join Date: Sep 2005
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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? ^_^
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calm down and enjoy your time in experiment |
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#2 |
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diyAudio Moderator
Join Date: Jan 2003
Location: Near London. UK
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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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The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference... |
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#3 | |
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diyAudio Member
Join Date: Sep 2005
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Quote:
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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calm down and enjoy your time in experiment |
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#4 |
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diyAudio Moderator
Join Date: Jan 2003
Location: Near London. UK
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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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The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference... |
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#5 | |
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diyAudio Member
Join Date: Feb 2001
Location: illinois
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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 Quote:
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#6 | |
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diyAudio Member
Join Date: Oct 2008
Location: Vancouver
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Quote:
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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#7 | |
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diyAudio Member
Join Date: Sep 2005
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Quote:
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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calm down and enjoy your time in experiment |
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#8 | |
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diyAudio Member
Join Date: Sep 2005
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Quote:
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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calm down and enjoy your time in experiment |
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#9 |
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diyAudio Moderator
Join Date: Jan 2003
Location: Near London. UK
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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.
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The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference... |
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#10 | |
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diyAudio Member
Join Date: Sep 2005
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Quote:
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