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For my diy synth, I have a Meanwell RT-65B PSU that delivers +11.5V, +5V and -12V. The +11.5 should be 12, but the difference is within the tolerance, according to the datasheet.

I also have a CV-12 module from midimuso that requires 12V to operate. On that module, a LM317 regulator is used to produce +10.6V, which has to be exact because the module is used for tuning other oscillators. However, the +11.5 V don't seem enough for this, and the regulator won't go beyond +9.9V, making the module out-of-tune.

I'm not an electrical engineer and hardly know which part does what, and I'm curious which options I have left, other than buying a different PSU. Are there any regulators that are more suitable for this purpose? I have looked at the data sheet for the LM317 regulator, but I'm not sure which part of it refers to the situation that I am in right now.

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  • \$\begingroup\$ Is changing your LM317 to an LDO an option? \$\endgroup\$
    – winny
    Commented Oct 20, 2022 at 11:48
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    \$\begingroup\$ if precision is a necessity, the LM317 is a design mistake... so I duly hope that replacing it with a much better precision and stability is an option! \$\endgroup\$ Commented Oct 20, 2022 at 11:50
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    \$\begingroup\$ Voltage regulator used for tuning? Hmm. 1% error (which is good) would be an unacceptable (in my book) 17 cents of tuning error... \$\endgroup\$
    – TypeIA
    Commented Oct 20, 2022 at 11:57
  • \$\begingroup\$ I've ordered the module as-is from the manufacturer, and that's what was in the package. Running it with a different, off-the-shelf 12V DC power supply at 12.4V worked fine, but I'd like to use the one that's already in my synth. Also, today I'm learning something new! :) \$\endgroup\$
    – bmurauer
    Commented Oct 20, 2022 at 12:07
  • \$\begingroup\$ Do you have an over 10% load on the 5v rail? What about the 12V load? Under the Minimum load, it can cause the voltage to not be steady. Can you adjust the 5v rail up using the adjustment pot or remote sense? That should help improve the 12V in proportion. \$\endgroup\$
    – Passerby
    Commented Oct 20, 2022 at 14:18

3 Answers 3

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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:

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:

  1. AP2202K-ADJ (buy two or more. You might lose or break one...)
  2. a 1 µF electrolytic capacitor. Rated voltage >= 16V, better 25V. (you could reuse C3)
  3. a 2.2 µF or up to 10 µF electrolytic capacitor, rated voltage 12V or higher.
  4. 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)
  5. 68 kΩ
  6. 9.1 kΩ
  7. (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:

Schematic from datasheet

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").

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    \$\begingroup\$ Silly question: Aren't all series regulators adjustable if you connect the ground pin to a feedback voltage divider? some even call the ground pin ADJ for this exact reason \$\endgroup\$ Commented Oct 21, 2022 at 19:59
  • \$\begingroup\$ @bmurauer By the way, the pinout on the first page of the datasheet is looking from the top of the device. If you mount it "dead-bug-style" on its back, don't forget to flip it over (I found making a drawing on paper to be sure what to connect where is a good help) \$\endgroup\$ Commented Oct 22, 2022 at 13:46
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    \$\begingroup\$ @user253751 not a silly question at all! Maybe there's a question on that already somewhere around here; if it's not, I think this would be a good thing to ask about! Generally, the downside would be that you'd be making assumptions on the ground current. If that changes during operation, it will affect the current through the lower half of your voltage divider. \$\endgroup\$ Commented Oct 22, 2022 at 13:51
  • \$\begingroup\$ @user253751 In the case of this regulator, it varies with the pass transistor base current, i.e. it's load-dependent and would for our loads vary between 150 µA at 0 output current and up to 1.5 mA. So, to not affect the accuracy by more than ca 1% of the voltage divider-based based solution, you'd need to make the current through the top of the voltage divider 100× as large – 150 mA. That would be much more than the current used by the actual load! (and remember, we're originally aiming for better than 1% for the whole system!) \$\endgroup\$ Commented Oct 22, 2022 at 13:59
  • \$\begingroup\$ now there's a question about it: electronics.stackexchange.com/questions/639980/… \$\endgroup\$ Commented Oct 26, 2022 at 3:12
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You have 3 solid options

  1. Follow Marcus Müller recommendation from anouther answer - replace LM317 with proper LDO on CV-12 module
  2. Replace Meanwell RT-65B PSU for anouther one which has at least 12,2V on 12V output (why 12,2V you may ask - your particular LM317 chip at your particular load provides 9,9V instead of 10,6V - 0,7V deficit when fed with 11,5V so it needs at least 0,7V higher input for desired output)
  3. Add a boost converter beetween Meanwell PSU and CV-12. MT3608 adjustable booster boards are readily avaible and one of those can output 12,5V at 1A when fed with 11,5V (lets give LM317 a 0,3V extra headroom)
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For completeness' sake, I'll add another important piece of information:

The RT-65B PSU has a trimmer to adjust the +5V output. However, it seems that the other voltages are referenced relative to the +5V rail, so adjusting it also adjusts the other rails. This is not stated on the data sheet, but a kind Redditor informed me that this is the case. This way, I managed to get +12.0V and +5.3V, which makes the board behave as it should.

Nevertheless, this whole endeavor has taught me a lot, and I now understand some flaws in the design of the board.

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