Can Someone Look at My Buffered Mult Schematic Please?

[fredrik]

Now you’ve ended up with the diodes the wrong way (as mentioned above, they’re protection diodes, not LEDs). If the triangle points from + to −, current flows that way, which you don’t want to happen here.

You can fix that by picking the wrong PCB footprint but probably better to fix it in the schematics.

(Also, you don’t need to add +12V net ports at each capacitor. One is sufficient. And when you do the layout, don’t forget that the 10 uFs go near the power connector, the 100 nFs near the IC pins.)

[fredrik]

What’s with the 100 ohm resistors in the feedback loops, btw?

The TL084 inputs have very high resistance, so adding a small resistor in series doesn’t make much sense: 10¹² + 100 = 10¹²

(also, TL074 and TL084 are the same thing, as discussed e.g. here (when they’re not, the more common TL074 is better) so no need to source one if you already have the other)

[twinturbo]

@fredrik

Now you’ve ended up with the diodes the wrong way (as mentioned above, they’re protection diodes, not LEDs). If the triangle points from + to −, current flows that way, which you don’t want to happen here.

You can fix that by picking the wrong PCB footprint but probably better to fix it in the schematics.

(Also, you don’t need to add +12V net ports at each capacitor. One is sufficient. And when you do the layout, don’t forget that the 10 uFs go near the power connector, the 100 nFs near the IC pins.)

Should the 100nf not be tied to GND? presently they are between the + and - rails?

[fredrik]

@twinturbo

Should the 100nf not be tied to GND? presently they are between the + and - rails?

Oh, missed that. Good catch.

(you’ll find some designs out there with a single 100 nF between the rails, but that’s not correct and even if you’d go for that approach anyway putting two capacitors in series is a waste of capacitors )

[popflier]

@fredrik

What’s with the 100 ohm resistors in the feedback loops, btw?

I had it outside of the feedback loop and then moved it inside as I had thought that’s what @analogoutput had told me I needed to do. So I moved it and asked him if that was what he was talking about having me do. It wasn’t. He clarified and I updated the schematic to take them out of the loop but haven’t uploaded a new schematic yet showing the changes.

[fredrik]

@analogoutput

Then output = input voltage regardless of what load you connect. (There can be drawbacks to doing it this way so some like to use it only when preserving V/oct CV is needed.)

The “in-the-loop” pattern is often used to deal with capacitive loads:

Output stage topologies

I have an application note that discusses this somewhere but cannot find it right now, but iirc the version with the output resistor in the loop can deal with capacitive loads (e.g. long patch cables) a bit better. (the resistor itself is also there to protect the output stage and more importantly protect other modules, by limiting the current in and out of this module.)

What are the drawbacks, besides using one more (optional) capacitor? (if you’re aware of any specific ones, post them in the linked thread )

[popflier]

How would I add this switch to my schematic as I have 2 boards stacked? One with all of the pots, leds, and the switch, and the other with all of the components.

How would I route this through the connectors to make this work?

Can someone look over my schematic and tell me if they see any glaring issues please?

[Dud]

something like this

[popflier]

I know that’s where the lines need to end up but I don’t know how I should route the signal on pins 2 & 3 to get there between the connectors? See how the switch is on pots board. I have to route the signal through connectors and not sure how to do that.

[Dud]

Sorry I did not take into account the history of the panel , so if I understand correctly maybe your solution is to add another connector for the switch connection on panel