for someone who's not an electrical engineer, you're pretty good at picking this up!
The measure from the datasheet you're interested in is called dropout voltage, i.e. the minimum voltage difference between in- and output.
The LM317, being old enough to be of archaeological interest, has an enormous dropout voltage of up to 2.5V, depending on the input voltage and output current. That's what's biting you here – your input voltage is too close to the desired output voltage for the LM317 to regulate sufficiently well.
Generally, the LM317 is probably not the regulator you'd want in 2022 to regulate an output voltage well, especially for oscillator purposes: It's precision is kind of bad, as well as it's stability – it changes with output current, with temperature, to a lesser degree, but still, with input voltage. Any regulator will do that, no real-world system is perfect, but the tolerances (up to multiple percent of relative error under normal conditions!) here are really worse than what you could buy for very little more. (There's more aspects that make the LM317 undesirable here, such as the bad rejection of audio-frequency input and load noise, which can actually become problematic for voltage-controlled oscillators, but I don't have enough insight into the design and application of this circuit to make a quantitative statement here.)
Working with nearly no margin with a regulator that's borderline unsuitable to the task does call into question the design of the circuit you've bought, but I guess this is easy enough to fix that we don't need to worry about the rest for now.
Since you need a low dropout voltage, a so-called Low-dropout regulator, or LDO in short, will be what you want.
What you need to make sure is that the current it can provide fulfills (or over-fulfills) your circuitry's needs. You should definitely check voltage regulation in the data sheets to be better than the LM317 during the same operation – but I promise you, it's not going to be easy to find a worse regulator. (OK, aside from the LM7805 that is used as voltage regulator for the CV-12 chip itself, but since that IC is a digital one, and probably has an internal voltage regulator anyways, that won't matter much.)
Now, 10.6 V is a voltage too specific to find a fixed-output voltage regulator that outputs exactly that. So, just as the LM317, your LDO will need to be an adjustable model.
As said, you'll be able to supply the necessary current. Now, I don't have the design insight that the original designers have, so I'll have to go by what I can see to find a safe upper bound for the current the whole 10.6 V supply might draw.
From the schematic:
We can see that the only thing driven by the 10.6 V supply are the two level shifters U2 and U3, which are MC14504 according to the schematic, and CD4504 (in a variant made by Texas Instruments) according to the photograph. So, let's look up the CD4504 and see how much current it can draw, at most!
The datasheet breathes pure unadulterated 1970s, but that's fine with me. The table "Static Electrical Characteristics" tells us the maximum quiescent current (i.e. what it uses without doing anything) it will draw is 5 mA. The maximum output current is 6.8 mA per output. But it can't source that, since these output are connected to 100 kΩ resistors (R3 through R14), so even taking a bit of resonance into account, that's at most 10.6 V / 100 kΩ = 0.106 mA per channel. There's 12 channels, so 1.2 mA, plus two chips' 5 mA quiescent current, making for maybe because we're bad at math and to make things easier, 20 mA of current needed at 10.6 V. (also, we don't know what you will attach to the external 10.6 V pin.)
Armed with that knowledge, we go and find the linear voltage regulators at our electronic distributor of choice – farnell, mouser, rs components, or digikey, and enter the properties we need into the table filter there:
- needs to be available,
- needs to have a positive output voltage,
- needs to be adjustable,
- needs to have a max input voltage higher or equal to 12 V, and
- a min output voltage lower than 10.7 V and
- a max output higher than 10.5 V,
- an output current of at least 20 mA
I prepared such a digikey filter for you here; the first results you find are, to little surprise, regulators from the ubuiquitous xx1117x-ADJ family. (These are very popular, not as old as the LM317 or LM78xx series of devices, and there's a archload of manufacturers that make them.) That would actually work. However, let's look for a regulator with a lower dropout voltage, so that slight supply variations hurt nobody.
That would lead us to the AP2202K-ADJ.
So. Buy:
- AP2202K-ADJ (buy two or more. You might lose or break one...)
- a 1 µF electrolytic capacitor. Rated voltage >= 16V, better 25V. (you could reuse C3)
- a 2.2 µF or up to 10 µF electrolytic capacitor, rated voltage 12V or higher.
- Precision (that is, 0.1% tolerance or better) resistors; you'll want a 0805 or 1206 package (smaller works as well, but it's harder to solder, mount)
- 68 kΩ
- 9.1 kΩ
- (optional) a ceramic chip capacitor of 100 pF to 1 nF for lower output noise. Make it the same package as the 9.1 kΩ resistor.
So, unsolder IC2, R1 and R2, as well as C3. Solder in a 1 µF electrolytic capacitor between the +12V (of the now unused IC2 contact holes) and GND (you can get that GND contact at the top end of R2, if you look at the board).
This would have been good in the original circuit as well, and the lower the dropout, the more important adding a decoupling capacitor on the input becomes.
Solder in your 2.2 to 10 µF capacitor in place of C3.
Glue (or don't, I'm not your mother; a small amount of superglue does the trick, e.g. take it up with the tip of a nail, deposit on board, drop chip on the droplet using tweezers) the AP2202K-ADJ "upside down" next to the old place of IC2, so that you can, with short pieces of wire (e.g. from resistor "leg" snipoffs), connect its Vin to 12 V. The datasheet tells you how:
From the output pin Vout to the ADJ pin, we'll connect the 68 kΩ precision resistor, and from ADJ to GND, the 9.1 kΩ resistor.
The ADJ and Vout are the two pins on the side with only two pins of the AP2202K-ADJ. So, same idea, drop a droplet of glue, place 68 kΩ so that it's not too far from ADJ and Vout to these contacts.
Do the same for the 9.1 kΩ resistor, placing one of its contacts right next to the ADJ-adjacent contact of the 68 kΩ resistor. If you've bought the "additional" chip capacitor, solder it atop and in parallel of the 9.1 kΩ resistor.
Now, add all the necessary connection as shown in the datasheet schematic above using short, manageably thin pieces of wire (if you kept the snipoffs from the resistors of the original kits, these work well, but so does "magnet wire").
