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Old 15th December 2012, 08:08 PM   #1
dooper is offline dooper  Canada
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Default Passive mixing network -- calculating total loss?

I have a question about the "New York Dave" balanced mixing network design found here. In a design with four balanced inputs and a single balanced output, the value for Rx would be:

(20000/4)*200/((20000/4)-200) ~ 208Ω.

In the specs, NYD indicates the calculated loss of the network would work out to (about) 40 dB. How would one calculate the exact loss for a circuit with eight 10K input resistors and one 208Ω shunt resistor at the output?

Thanks!
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Old 15th December 2012, 08:53 PM   #2
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I'm not a math whiz, so this is mostly intuitive:
20k * (-40dB) = 200
or Zout/Zin = 0.01
But wouldn't you want to recalculate the formula for 8 inputs?
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Old 15th December 2012, 10:51 PM   #3
dooper is offline dooper  Canada
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Nope, not eight inputs, four. It's balanced circuitry from front to back.

Thanks for the reply,

dooper
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Old 16th December 2012, 02:35 AM   #4
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"Exact" is a pretty big number, because it varies a little with the input level controls and "assign" switching. In general though, each input's 10K Ohm mixing network resistors are loaded by the 208 Ohm resistor and the three other channels (and the next stage's load, which we'll ignore).

Each of the other channels looks like two times 10K Ohm = 20K Ohm plus some amount of their input level controls plus (or not) the "assign" resistors. Worst case is 20K Ohm each for a total of 20K/3 or about 6K67 Ohms. Reasonably ignorable in parallel with 208 Ohms.

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Chris
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Old 16th December 2012, 12:21 PM   #5
AndrewT is offline AndrewT  Scotland
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What is Rx doing?
Does it change when the number of inputs changes, or the source resistances change?
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Old 16th December 2012, 12:45 PM   #6
DF96 is offline DF96  England
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Rx is providing a poor virtual ground, to reduce but not eliminate interchannel issues, and setting the output impedance (roughly) at the required 200 ohms. Why someone would wish to take a line level signal and attenuate it down to microphone level, only to amplify it again, is beyond me. Maybe they like noise?
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Old 16th December 2012, 12:52 PM   #7
AndrewT is offline AndrewT  Scotland
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Maybe Rx should be removed.
What does that do?
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Old 16th December 2012, 02:32 PM   #8
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The exact loss of the network depends on the impedance of the sources at the inputs and the load at the output. E
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Old 16th December 2012, 04:45 PM   #9
DF96 is offline DF96  England
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Without Rx you get less attenuation, but more channel interaction.
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Old 16th December 2012, 05:59 PM   #10
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The summing junction is the trickiest part of mixing. An active summing junction removes all these interaction issues, but requires a fairly powerful amplifier to drive its own feedback resistor, if resistances are scaled to keep noise down. You end up with something between a headphone amplifier size and speaker amplifier size, depending on number of channels.

Paul Stamler wrote the definitive article on homemade mixers in Audio Amateur back in the 1990's. Well worth checking out for an overview.

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