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.

Keyboard CV Demystified

There seems to be a lot of misunderstanding about keyboard a.k.a. pitch CV on analog synths. You've probably heard most Moogs and ARPs are 1 Volt/Octave. I've made a video with my Moog Rogue to show what 1V/oct means.

I have the probes of my multimeter connected to the "KYBD CV OUT" and ground. First I play the low F, which puts out close to 0 V. Then I play the F an octave up (1 V) and the F two octaves up (2 V), hence 1V/oct. Note that the CV stays the same until the next note is played.

Next I play F, G#, B, D, F. What should the voltages be? If we have a 1V/oct scale and 12 notes = 1 octave, then 1V/oct = (1/12)V/note. So we get the following:

F = 0 V
G# = 3*(1/12) = 1/4 V
B = 6*(1/12) = 1/2 V
D = 9*(1/12) = 3/4 V
F = 1 V

Next I turn up the portamento. The Rogue's portamento kicks ass and I have yet to hear a softsynth duplicate it. That will be the topic of a future post. With the portamento turned up I play the lowest note and highest note to show the CV slide between the values. The high note incidentally sounded like the slow down part in "Rockafeller Skank."

Then I turn the multimeter to mV. I play the low F and F#. We would expect the voltages to be 0 mV and 83.3 mV, respectively. They are actually 2.7 mV and 85.7 mV. Why the deviation? Analog isn't perfect and that's why it's great.


Measuring Pitch CV for my Moog Rogue from Robert Hume on Vimeo.

First!

This blog will be my investigation into analog synthesizer electronics. How do analog synths work? Why do they act and sound the way they do?

I've decided the best place to start is with monophonic keyboards. Ever wonder why your Moog has low-note priority? Why doesn't it have last-note priority like the Pro-One? Check back here to find out.