Electronic – Adding multiple differential signal

I think a couple of voltage dividers will fit the bill perfectly. Presuming a 'full-scale' input will produce +1.29 volts on one output and +3.80 volts on the other output (and vise versa for negative full scale), the difference between the two is 3.80 - 1.29 = 2.51 volts. If you scale both output voltages down by a factor of 2.5 to get the differential voltage on the right scale, you get 0.516 and 1.52 volts, both of these within the 0 to 3.3 volt range. So I would suggest simply building two divide-by-2.5 voltage dividers.

Let's assume the output voltages are split into differential and common mode components. In this case, the output voltages Vp and Vn would be Vp = Vc + Vd and Vn = Vc - Vd. If you use decent voltage dividers, the output voltages to the ADC would be Vp,adc = (Vc + Vd) / 2.5 and Vn,adc = (Vc - Vd) / 2.5. Then the converter takes the difference internally, Vadc = Vp,adc - Vn,adc = (Vc + Vd - Vc + Vd) / 2.5 = Vd / 1.25.

Now, if the dividers are not exactly matched, then the result is a little different. Call the divider gains K1 and K2. The ADC output voltage will be Vadc = Vd (K1 + K2) + Vc (K1 - K2). The differential gain is Adm = (K1 + K2) and the common mode to differential gain is Acm,dm = (K1 - K2). In this case, the error caused by the common mode voltage will be proportional to K1 - K2.

This does what the OP wanted, a differential output around a defined output common mode, with no more, and in fact fewer, precision resistors.

^{simulate this circuit – Schematic created using CircuitLab}

If the common mode voltage does not match the input at Vcm, then OA3 drives an input voltage into both inverting inputs, with the same gain, which will cause both output voltages to move the same amount in the same direction, maintaining the existing differential gain, but shifting the common mode until there is no error.

Stability may be an issue, as there are two amps in a feedback loop. I suspect it would be easy to stabilise by clobbering the OA3 bandwidth, and/or speeding up OA1/2 a little with a small C across R3 and R5, which may or may not be desirable from the differential behaviour point of view.

Note that the only resistors that need to be matched are R1 and R2, which set the two output terminals to be equally disposed around Vcm. The differential gain is just (R3+R4+R5+R6)/(R4+R6), it does not need matched resistors, these can be four arbitrary value resistors, subject to getting the correct gain of course. I emphasise that fact by putting 4 unmatched values in the diagram for those resistors. The diff gain is 7 (21k/7k), with the outputs exactly disposed around Vcm because of R1==R2, and OA3. Try it!