The Prodigy Keyboard Circuit part 2

So we determined a voltage source to the keyboard circuit won't work. Here is a current source used instead. The lowest key is still connected to ground. What should be the current I? Each resistor is 100 ohm and should have 1/12 V across it. By Ohms Law I = V/R = 1/(12*100) = .833 mA. What happens when two keys are held down at the same time? Like the previous example with the voltage source, the current will bypass the resistors between the two keys. To solve for Vout, we see the point on the far left is grounded. Vout will just be I*(Resistance from ground to the lowest key held down). Pressing higher keys won't do anything to Vout. This is the reason for low note priority.

It would be just as easy to make a keyboard with high note priority. The highest note could be held at 2.58 V. The current source would be connected the opposite way to the lowest note to sink current. Making a last-note-priority keyboard would be much more difficult to make. Figuring out the last note played really requires digital logic, so it wasn't until the introduction of microprocessors in synths that last note priority became common. The SH-101 and Pro-One come to mind. The SH-101 has a last-note/low-note priority switch so the synth shredders of the day could still use their Moog soloing techniques.

Now that we've established why we need a current source, we can ask why 100 ohm 1% resistors are used. (It looks like .1% on this scan of the schematic. Other scans definitely have 1% and I'm fairly certain .1% aren't manufactured). The 1% is easy. Our ears are very sensitive to pitch. If a note is 5% (5 cents) out of tune, it will be noticeable. Why 100 ohm? Consider this. If the resistors were large, say 100k, we could save power. The power used by the keyboard when no notes are played is V^2/R. We can't change V because we are using 1V/oct. But if we switched from 100 ohm to 100k ohm, it would require 1/1000 the power. What's the problem?

We are using a current source. In a perfect world it wouldn't matter, but real current sources have an output resistance. It is a large resistance in parallel with the current source. If we are driving a load with small resistance, this output resistance won't matter. But if the load is comparable to the output resistance, the current source will start misbehaving.