The Prodigy Keyboard Circuit

The previous posts shows what 1V/oct keyboard CV should look like. So how do we get it? There are two steps. 1.) We need to spit out a CV when a key is depressed. 2.) We need a sample and hold circuit to hold that voltage until the next key is depressed. This is needed because of the "Release" portion of the note's envelope. The pitch must stay constant until after the note fades out (or another note is played). First I will talk about the first step, spitting out the CV when the key is depressed. The Moog Prodigy keyboard circuit/switches are shown here.

Each key is connected to two switches which make contact when the key is depressed (double pole single throw, or DPST). I will concentrate on the bottom switches that are responsible for CV. The top switches are for gate/trigger signals. Each key CV switch is separated by a 100 ohm resistor. When the key is depressed it makes contact with the bus bar labeled "3" in the schematic. this is the output that goes to the sample and hold stage. From Ohms Law we know the resistors will act as voltage dividers. In the last post we figured each key should be 1/12 V higher than the previous key. So let's try to run the CV circuit with a voltage source. This will be the wrong solution for reasons I will explain.

Each of the 31 resistors has 1/12 V. The whole chain should have 31*1/12 = 2.58 V. So we could hook-up this voltage source like so:

Let's say a note an octave above the lowest note is played. Treating this like a voltage divider, the voltage Vout will be 2.58*12(100)/(31(100)) = 1 V. This is the correct voltage. In fact all keys will put out the correct voltage. So what's the problem?

Even though we're talking about a monophonic synth, often two keys are held down at the same time. What happens in that case? Our voltage divider gets screwed up. Let's say the note an octave above the lowest note AND the note an octave above that are held down. The current will bypass the resistors between the notes since they are shorted out.













What does this do to the voltage? In the example I just gave the voltage Vout will be
2.58*12(100)/((31-12)(100)) = 1.63 V. This is not the correct voltage of either note. In fact it's not the correct voltage of ANY note, at least not in the musical/CV scale we're using.

How do we solve this problem? Luckily the synth forefathers already did it for us. Use a current source instead. That will be explained in the next post.