IMHO you have only 40% because it doesn’t use the 280° of the rotation. I noticed on my Moog that the modwheel goes from 4 o’clock to 8 o’clock on the pot.
the best is to use 10bits and some algorithmic ( x+ (pot value>> 2 )for example) in order to get a value between 0 and 127.
A 30K pot won’t help, since you’re using the pot as a voltage divider, the range would still be 1.5V - 3.5V (assuming 0V-5V, 40% range at middle)… I’m guessing, you might have one that is only the starting 40% so the voltage is 0V - 2V.
Converting 40% of a 10 bit value into 7 bits would work, but you’d be bypassing the built-in jitter handling. It’s much better to use 0V-5V inputs into the Core.
You could try shifting and amplifying the voltage, but that would involve opamps. I’m not the best person to explain how that would be done… but in theory you could do it with a rail-to-rail opamp… assuming the middle of the range is 2.5V, it could amplify 1.5V-3.5V to 0V-5V.
You could also try modifying the pot so the full range of resistance is only in the middle 40%. That would involve opening the pot and “shorting” the resistant strip at each end… you could use a silver conductive pen (eg. this: http://www.amazon.com/CircuitWriterTM-Precision-Pen-silver-based-grams/dp/B0002BBVQO ). You would have to carefully measure the pot’s resistance at the ends of the modwheel’s range, then with a multimeter, work out where that corresponds on the resistant strip, and “paint” from those points to the ends of the strip. Again I’m assuming it’s the middle 40%, if it’s only the starting 40%, you only need to paint one end.
so I got some mc1458P to do it. This thing has 8 connectors
1 1out
2 1in-
3 1in+
4 Vcc-
8 VCC+
Those are the relevant pins for you.
It does need power. Vcc- needs to be tied to ground and Vcc+ needs to be tied to +5V.
Unless I am completely blind, that’s not a rail-to-rail opamp. The circuit will still “work” but not improve the situation at all, since the voltage will be clipped @ roughly 1V and 4V.