Downshifting, not Translating, an AES3/EBU Signal Voltage.

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The voltage specification for AES3/EBU is ±2V to ±7V (revised from ±2V to ±10V), this is needed for diving the cable. How could this be dropped to 0V to ±5V at the receiving end without causing damage to the signal? The drop wouldn't result in data loss, it's still a difference of ±5V, but would be more usable by μCs, FPGAs, and so on. I'm sure this is commonly done, but usually in ICs. I want to make my own receiver because every receiver IC I've used has problems/errors with it. Even the suggested pre-IC circuits are often incorrect. The specification sheets for AES3/EBU are very sparse and incomplete. Maybe a different transformer ratio (1.4?) would do the trick, but would probably need to be custom made, and for now I'm only going to be making about 10 units, so that wouldn't be an option. Maybe there's a different translator that I can use that I'm not aware of? I must use a twisted pair (CAT5) for my application, the other pairs are used for other unrelated signalling, so I can't use something like S/PDIF. Conversion to optical may be an option at the receiving device, but I think in the EBU specification, transformers are mandatory. I wonder if a galvanically isolated digital isolator qualifies as a transformer? Many can handle ±7V inputs. Some are too slow though, so if you use one instead of a transformer, take note of the speed!

I imagine that the same or similar method I'm looking for would also be applicable to "upshift" the signal for a cable driver?

Just a side note, this will be used in both consumer and professional applications (switchable).
 
As I understand the spec, it's not ±7V, it's 7V peak-to-peak. If I understand your question, you want to drop that to 5V peak-to-peak, correct? If so, you could use a balanced voltage divider with 275 Ohms in parallel with the 110 Ohm load, and two 15.7 Ohm resistors (one in each leg) in series with the parallel combination of load and 275 Ohm resistor.
 
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Typical to overlook something as simple as an H-bridge or T-bridge. Both sides are supposed to be the same anyway, even though that wouldn't be the case if the source device was something like a microphone (unless processed that way).

Thanks, I must be getting fatigued!
 
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