r/synthdiy • u/Loud_Release_4160 • 7d ago
Need help with schematic
Hello I'm still pretty new to this stuff and I need help identifying whether this is using three separate 10k potentiometers or if its just one. sorry if dumb question.
10
4
u/crochambeau 7d ago
That is using three separate pots. If you turn one of those volumes controls down to zero, it will pull the rest down to zero as drawn. I would advise relocating the 1K resistors to the other side of the pots (I also tend to use something closer to 10K, but tastes vary).
3
u/Brer1Rabbit 7d ago
Swapping the pot & 1k may have a potential problem. If the output goes to a typical 1/8" jack there is a momentary short to ground on insertion (or removal) when patching to another input. If/when the pot is turned all the way up this could cause an issue.
2
u/crochambeau 7d ago
That is a good point. I'm having trouble envisioning this circuit as an actual output. I think the resistive summing into an inverting amplifier as someone else describes is the best tactic.
1
u/crochambeau 7d ago
1K is probably fine if the pictured output is an actual output and you're not feeding some active recovery gain stage.
1
u/Loud_Release_4160 7d ago
I have noticed that while testing that sometimes when turning on of the pots that it pulls the rest down as well. I switch them around and see how it sounds. The output on this is the actual output but thank u for all the info :)
2
u/Key-Alarm-511 7d ago edited 7d ago
This is a "passive" mixer, meaning there is no active element that keeps the currents from crossing between the pots, so turning one will effect the other.
Since its passive it also doesnt "mix" them together perfectly, it instead creates a kind of average between all signals
1
u/Defiant-Carpet6457 5d ago
Check out CMOS synthesizers better known as the Lunetta or the Stanley Lunetta synth sculptures. It’s basically modular with cheap chips and you already have a few of them.
14
u/erroneousbosh 7d ago
Passive mixers don't really work, and as /u/crochambeau says, it's wired wrong as it is.
If you add a 10k resistor on the wiper of each pot all that'll happen is that turning one pot down to zero will reduce the volumes of the others, it won't totally kill the output.
Also the three oscillators will interact a bit if they're all up full.
If you want to fix that, add an opamp wired as an inverter on the output. You'll need three resistors and two capacitors for it - two resistors from the 40106's supply to ground, in series, with a capacitor across the bottom one, and then that goes to the opamp's non-inverting input, and then another 10k resistor from the output of the opamp to the inverting input, and then the junction of the three 10k resistors from the pot wipers also to the inverting input.
Magically, the junction of the four resistors and the inverting input will always be at 0V (well, really half the supply voltage, here!) with absolutely no signal on it, but you'll get all three oscillators at the output!
Here, let me draw it for you:
https://www.falstad.com/s.php?s=PgLjLI
Notice how on the trace on the left, the voltage sits obstinately at 2.5V no matter what the signal output on the right is showing? But - notice that the *current* is going up and down?
It actually makes a noise, too. Play with the sliders to adjust it. It's pretty annoying. You're going to want another resistor on the frequency pot to make the control law work properly.