Passive mixing network -- calculating total loss?

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dooper

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Joined 2004
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!
 
Chris Hornbeck

Chris Hornbeck

Member
Joined 2011
"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.

All good fortune,
Chris
 
Chris Hornbeck

Chris Hornbeck

Member
Joined 2011
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.

All good fortune,
Chris
 
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KoksRox21

Member
Joined 2017
dooper said:
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!
Click to expand...

Did you by any chance figure out the exact formula for calculating the loss?

I would be very grateful if you did :)
 
PRR

PRR

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Joined 2003
Paid Member
> how would I calculate that?

20K loaded in 200r, find the dB of that.

Why "exact"? Your sources have faders. Your sources are dynamic, fading and swelling. The "loss" is calculated for one source but if "mixing" then obviously you have multiple sources which will add-up more or less.

Build as shown. (NYD was not a fool.)
 
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