Need circuit, 6,12,18,24db attenuator - Page 2 - diyAudio
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Old 28th October 2010, 06:19 PM   #11
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get yourself one of those wall mounted volume controls. 11 steps. -3,-6,-9, -12, 15db...etc
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Old 17th March 2013, 03:29 PM   #12
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Rousho, How do I make this balanced?
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Old 19th March 2013, 09:39 AM   #13
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Use two of them but do not connect the ground to any ground. Connect it between them. You will have to match the resistors to a good level, between both.

Gajanan Phadte
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Old 19th March 2013, 10:14 AM   #14
Roushon is offline Roushon  India
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Thanks Gajanan!
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Old 19th March 2013, 10:15 AM   #15
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Thanks for your reply. So I don't need to halve the resistor values and place one resistor/2 in hot and one resistor /2 in cold? Im only talking about the series resistors, not the bridge resistors.
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Old 19th March 2013, 10:23 AM   #16
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by Roushon View Post
The attenuator circuit has input and output impedances 10K...........
No.
The diagram is showing that when the output is loaded with 10k that the input impedance seen by the source is also 10k.

If the load were changed from the suggested 10k then the input impedance would no longer be consistently 10k.

But when looked at from the receiver end, the output impedance of this attenuator is NOT 10k.
Set Rs for the source to 200r. Set all the relays to 0dB. The load is still 10k. But looking back from the load to the output of the attenuator, the receiver sees a source impedance of 200r. The load sees Rs when set to 0dB.
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Last edited by AndrewT; 19th March 2013 at 10:44 AM.
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Old 19th March 2013, 10:28 AM   #17
AndrewT is offline AndrewT  Scotland
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Be careful.
To get the benefit of balanced impedance connections, the impedances matches MUST be VERY accurate. You must aim for much better than 1%. Usually 0.1% is accepted as the maximum tolerance, but benefits are to be found by achieving <0.01%.

It would be worth looking at typical switch resistances and how they vary, to see what effect, if any, these inconsistent resistances would have on the precision of your very accurately matched resistances and capacitances.
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Old 19th March 2013, 12:28 PM   #18
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If I place this between a CD player (200Ohm output Z) and a line input (100K or higher Z) Will the attenuation be accurate within.. halve a dB? Between every step or in total, How should I look at this.. Thanks
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Old 20th March 2013, 04:49 AM   #19
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This matching requirement will make it costly if low tolerance resistors are used. The audio source, if it is not a fully differential design, can be modded to accomodate a unipolar volume of your choice. This will give better results to maintain the CMRR of the setup.

Gajanan Phadte
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Old 22nd March 2013, 08:54 PM   #20
GoatGuy is offline GoatGuy  United States
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Quote:
Originally Posted by AndrewT View Post
Be careful.
To get the benefit of balanced impedance connections, the impedances matches MUST be VERY accurate. You must aim for much better than 1%. Usually 0.1% is accepted as the maximum tolerance, but benefits are to be found by achieving <0.01%.

It would be worth looking at typical switch resistances and how they vary, to see what effect, if any, these inconsistent resistances would have on the precision of your very accurately matched resistances and capacitances.
Not really... the original poster has an unbalanced source (which could start a whole different discussion) I like the russian design, though it really doesn't quite accomplish what it says it does. (just did the spreadsheet with combinatorial attenuation characteristics, using the Thévenin equivalent resistance and attenuated source voltage as the feed...) But its in the right direction.

With only 4 switches ... The resistor matrix would be much simpler, as only ONE switch at a time could be selected for attenuation.

OP: the formula you're looking for starts here:

dB = 20 log10( ratio ) therefore ratio = 10 ^ (dB/20)

You wanted ... 6, 12, 18, 24 db
The ratios would be 0.501, 0.251, 0.126 and 0.063 respectively

These in turn correspond to the following resistor pairs (to "kind of" match the 10K sink)

1K:1K -6 dB
1K:335R -12 dB
1K:144R -18 dB
1K:67R -24 db

Even figuring the 10K load into it (which will reduce output a bit more), the worst case is the -6 dB case, which becomes -6.4 dB... hardly worth worrying about. Its not as if you're building a precision instrumentation pad. You're building an attenuator for your CD player!

Which of course ... brings up the idea of just having a volume-pot (audio-taper) in a small project box, with the unbalanced IN, and a balanced OUT. Use a quality pot. Ground the MINUS pin of the balanced jack. The Pot should be relatively low value (1K, 2.5K or thereabouts) so as to minimize the effect of the 10K load on its attenuation. A nice knob and a carefully measured set of attenuation settings, and you have the best of both worlds - totally variable attenuation, and, calibrated set-points for your rig. Heck... once you got that together, you can then lay out (in Illustrator or equiv.) the reticle scale and use crafty DIYaudio techniques to transfer it to your project box.

GoatGuy
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