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In Reply to: RE: How to devise a single-ended output from a balanced preamplfier posted by Lew on July 11, 2024 at 12:36:20
Just put the same load on the negative as the plus. 20K from the negative to ground and a 20k pot to positive will not disrupt your balanced output.Alternatively, the way I do it, is to use a 5K pot off of the amplifier output to drive a subwoofer amp.
Edits: 07/12/24Follow Ups:
Thank you very much for the response. It reminds me that I once talked to the late Bill Thalmann about this problem, and he suggested 20K resistors, too. But why a 20K pot vs a resistor on the positive leg?
In any case, the circuit if I tap off the linestage would be...
20K between the negative phase of the balanced output and ground.
20K potentiometer or resistor between the positive phase of the balanced output and positive of the SE output.
Is that correct? I guess the 20K pot on the positive leg would permit control of gain, but the subwoofers I plan on using have built in attenuators.
If it is 100K, then you just need to take the signal off the positive and put a 100K resistor from the negative leg to ground. One resistor
I haven't purchased the subwoofs yet, but I can certainly find out the input impedance of the built-in amplifiers. I hate to reveal my ignorance or at least my insecurity, but I assume you mean the negative phase of the balanced output, when you refer to the "negative leg". Then connect the ground lug on the RCA SE output jack to ground per usual.
That couldn't be simpler.
Not all balanced outputs are referenced to ground. This should be confirmed by reviewing the specs or the schematic for the preamp before proceeding.
Exactly.
In a pure balanced line, there is no ground. There is only a shield, which is NOT the signal ground of components at either end. This is best implemented with a transformer. In that case, you can ground one end of the secondary to get a single-ended connection.
Many (most?) modern balanced components are more like differential lines, where there are two signal paths in opposite phase, which share a common signal ground. This is a decent approximation as long as the impedance to the source's common signal ground is much, much smaller than the receiver's impedance to ground. That means you can't just connect one side to ground.
Many problems are caused by the confusion between signal and shield grounds ... :^)
Yes. Ground is floated at the balanced outputs. Which is why I originally thought I needed to use a transformer, in the way that you mention which is to ground one side of the secondary while taking the SE signal off the other side of the secondary.
Edits: 07/13/24 07/13/24
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