Sawtooth VCO and current source for modular synthesizer
I am confused about how to derive an expression for the frequency 
  of the sawtooth wave at output at U3.


The frequency is determined by the current coming out of the collector of Q2. This current charges C3 until Vcompref is reached. Then the comparator will trigger and discharge C3. 

Voltage across C3 is determined by 

  V = Ic(Q2)*t/C3.


Since you discharge at Vcompref the frequency is: 

  f = Ic(Q2) / (VcompRef * C3)


With the values from your schematic Ic(Q2) will peak somewhere at VCC/Rref or 120µA. Plugging this into the equation above gives a peak frequency of:

  f = 120µA / (1.5V * 10nF) = 8000Hz.


This is the frequency you should get with Vcon at 0V. With increasing Vcon the frequency will drop because more current will flow through Q1 than Q2. The inverting nature of U1 makes sure that you get the positive response you want for synthesizer applications.

To reduce the frequency you have two options:


Increase C3. If you double C3 the frequency will drop by factor two. The value you've choosen is reasonable though and I'd leave it as it. 
Decrease the current at Q2 collector. This is done by increasing Rref. Try 1 Megaohm as a start.


More important is a design problem around your comparator though: 

You're working without any hysteresis. That means that it is not guaranteed that the comparator will completely discharge your timing capacitor C3. 

The comparator will turn on (and discharge C3) as soon as Vcompref is reached and will turn off as soon as the signal drops below Vcompref. What you want is to discharge C3 down to 0V or so.

The only reason why your circuit somewhat works is the propagation delay of your comparator along with R3 and the parasitic comparator input capacitance working as a small delay.

Add a resistor from the comparator output back to the comparator non-inverting input. This gives positive feedback and will make sure that the comparator stays turned on until the voltage of C3 drops by a significant amount. The ratio of this resistor and R3 will define the amount of hysteresis. R3 seem to be way to small btw, I'd expect a value of 10k or so.

Fixing this will likely solve your high frequency problem as well. I would expect that the frequency you're getting is not the VCO but the comparator oscillating around Vcompref.


  Also, the oscillator tends to decrease in frequency over
  time, and I have not detected yet what component is 
  responsible for this. Do you have any thoughts on this?


Yep! Temperature dependence. 

This is quite hard to get under control and is one of the more difficult part in VCO design.

You're using a CA3086 array for the expo converter: This chip does not guarantee matching of the transistor pair also it could be matched. A matched pair is only guaranteed in the CA3046 array. Same pinout! If you later find out that you're not able to dial in a good 1 Volt/octave response change the chip to CA3046. 

Oh, have you connected the substrate pin of the CA3086 (pin 13) correctly? It should be somewhere at -2 to -5V.

Furthermore, you've mentioned that one of your resistors is temperature dependent. Which one, and what kind of temperature coefficient does it has?

Since you're using the CA30x6 array already you may want to have a look how Bob Moog solved the problem: He used one of the transistors as a temperature sensor and another transistor as a heater to keep the silicon die at a constant temperature above ambient (45°C or so). This solves the problem without temperature compensation resistors (Google Moog Prodigy Schematic).

And finally: In your VCO I'm missing the Franco compensation resistor. This resistor, typically 680 ohm, is in series with your timing capacitor C3 and compensates for the finite time it takes to discharge C3. You want this resistor, otherwise your oscillator will go flat at higher octaves.


  I bet many of you have worked with similar designs.


Yep! A CA3046 expo-converter based VCO is on my desk right now doing tuning stability tests over night :-)