David Rogoff sent the following into the Yamaha CS80 list. I asked him if I could put it up and he gave me the OK.
"This touches on a big, somewhat technical, issue of what kind of VCOs the CS80 uses. The VCO III chip is a linear VCO, sometimes called Hz/Volt, as opposed to the more common exponential (Volts/Octave) VCOs (e.g. MiniMoog, Curtis & SSM chips in SCI and Oberheim polys).
Here's a pretty good explanation:
linkHere's a (I hope) quick one:
The most basic VCO is a sawtooth one, which can be a capacitor charged by a current. For non-EE types, here's my modified toilet analog (and you though the Metasonix vacuum-tube VCO was weird) : The capacitor is like the water tank of a toilet. The water filling it up is the current. The height of the water is like the voltage across the capacitor. Now, modify the float valve so that when the tank is full it automatically flushes. Then the cycle starts again. If you double the water filling rate ( = double the current), you double the frequency of the flush cycles.
The is a basic, linear VCO (actually Water-CO). It shows a couple of things. First, it's not actually voltage controlled, but current controlled. Ignore that for now. Also, the filling time is adjustable, but the discharge/flushing time is fixed. This is an issue with all sawtooth VCOs and is why many (e.g. Moog) VCOs have a high-frequency-tracking adjustment, which helps cancel this out. Here's the CS80 VCO:
linkOk, so why don't all synths use linear VCOs? As the above link explains, human ears don't hear frequency linearly. A above middle C is 440Hz. An octave about is 880Hz, or double the frequency. The next octave would be 1760Hz: double that. If you graph this, it's an exponential curve. So, the space (in Hertz) between two notes keeps getting bigger as we get to high pitches. If you had a modular synth with linear VCOs (like that old Paia), the top key might output 5 volts. One octave down would be 2.5volts. The next 1.25volts, followed by 0.625v and 0.3125v. This is a pain to generate. Also, as you get to lower notes, smaller voltage inaccuracies start becoming bigger pitch errors to our ears.
To avoid all this, someone (anyone know who? Dr. Bob? Tom Oberheim? Don Buchla?) came up with exponential VCOs. Basically, they're just a linear VCO with a circuit in front of them called (big surprise) an exponential converter. This is just a circuit that takes a linear input (1volt/octave) and outputs the doubling voltage (actually current...) that the VCO wants. Now, everything is simple.
So, why did Yamaha go for the linear? Two reasons, I'd guess. First, adding the exponential converter to each VCO adds more cost to the chips, since there's more circuitry. A bigger issue is temperature stability. As we've been talking about lately, all circuits are affected (i.e. knocked out of tuning) by temperature changes. The exponential converter, for reasons I won't go into, is really sensitive to this. People have been complaining about the tuning stability of the CS80, but it's rock solid compared to any poly-synth with exponential VCOs (P5, OBX, A6, etc). They all need computer-controlled auto-tuning routines to have any chance of staying in tune.
So, what issues/problems/advantages does the CS80 having linear VCOs create?
Good things:
1) modulation - linear vibrato sounds a bit different than v/oct vibrato, probably closer to acoustic vibrato (e.g. violin). Also, as the modulation speed increases, you start getting into F.M. land, which requires linear modulation (you don't want to know the math!). This is why some modular VCOs have linear FM inputs in addition to the normal v/oct controls.
2) sweep to D.C. - my favorite. If you start a pitch bend at the right end of the ribbon and slide all the way to the left, the pitch of the VCOs all go down to 0Hz / D.C. / flat-line. This is because the input to the VCOs goes to 0 volts and the frequency equals the voltage times a constant. With a exponential VCO this is impossible. Going 1 volt less on the control input goes down one octave. Mathematically, you can't get to zero Hz. You'd need to input -infinity volts! Also, many other limitations in the circuit block the VCO from even getting close. Big win for linear VCOs!
Bad things:
1) Keyboard voltages - as I wrote above, the keyboard has to generate exponential voltages. This is a big pain. In a digitally-controlled analog (like the CS80, P5, etc), the keyboard voltage comes from a DAC (digital-analog-converter). 99.99% of DACs are linear. The CS50/60/80 (and others in the family) have bizarre, custom exponential DACs. This makes interfacing the CS80 to other synths and/or MIDI-CV converters a pain.
2) CV mixing. Finally, we get to the original question of adding a pitch-bend input to the CS80. In the volts/octave world, everything is easy: you just add voltages together. Adding voltages is simple to do - just an op-amp and a few resistors. Let's say you had the following voltages come out of a v/oct keyboard: 1v, 2v, 4v. This could represent a low C (c1), C one octave up (c2), and C two octave above that (c4). To make it simple, let's say we have a pitch wheel or pedal add 1 volt to this (2v, 3v, 5v). This would be c2, c3, c5, so we've just transposed the sequence up an octave.
Ok, what happens if we try this with a linear voltage. For the same c1, c2, c4 notes, we might have 1volt, 2volt, 8volt. Adding one volt gives 2volt, 3volt, 9volt. The first note is correctly up an octave, but the next is only up about a 5th and the third note is only transposed up about a semitone. This, obviously, doesn't work. What we need to do, instead, is multiply the voltages. To transpose up an octave, double the voltages. To transpose down an octave, halve them. This is easy for a fixed transpose, but if you want a variable, like a pitch-bend pedal input, you need to multiply voltages. Just like it's much, much easier for people to add and subtract than multiply and divide, so it is for analog (and digital) circuitry.
If you follow the schematics or block diagram of the CS80, you can see that the voltage to the VCOs comes through a long chain of multiplications. The ribbon is actually the initial voltage source for the whole instrument. If the ribbon isn't pressed it outputs some fixed voltage (not sure the actual value - call it 2 volts). If the ribbon is slid up, all the way, from the left to the right, it would output double this voltage, which corresponds to one octave up. If the ribbon is slid the other way, it outputs zero volts, as mentioned above. Next, the voltage is sent through the concentric pitch knobs. Any normal potentiometer is a voltage multiplier, which can multiply the input by anything from zero to one.
This voltage then becomes the reference input to the exponential DAC on the KAS board, which multiplies it by it's exponential resistor network to create the CVs for each of the either voices. These voltages go to the VCO chips on the M-Boards. Are we done - nope - one more CS80 weirdness. In a v/oct synth, the octave/foot switches would just generate a voltage that would be added to the keyboard CV (e.g. MiniMoog). The CS80 VCO, instead, has a special footage input that needs an exponential current for each feet setting. Because this is difficult to do accurately over a wide range, we end up with the wonderful VR4, VR5, and VR6 trimmers to get the feet switching calibrated separately for each of the 16 VCOs. Yuch!
Getting back to the original question (remember Alice? There's a song about Alice...), a pitch bend input would need to control a voltage multiplier. This could be an added circuit, after the ribbon circuit, or could probably be merged with the ribbon voltage. I haven't figured out the details, but it's not rocket science. However, it is a lot more work than it would be on something like a Prophet 5.
Ok, I guess that wasn't quick, but at least I didn't have an graphs or get into transistor curves or Bessell functions.
David"
via this auction
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The Power Supply regulator board is missing. There are three main options:
