Could I just jump into the unpowered section of the breadboard and have multi outs for splitting a clock to multiple inputs?
Actually, some more basic electronics questions:
I’m thinking about using a different device for a clock, I think I can use a sync signal…
I can just connect a female jumper to the out of a mono socket and have basically a eurorack jacket?
Also can I just use a variable resistor with the right resistance to control amplitude? For example, making an lfo affect something less severely?
Ultimately, the answer to these questions are “just try it and see” but the reason I’m hesitant is that I don’t want to break anything - are there any basic rules of thumb to consider so that I can try things out and not worry about frying things?
Yes, but remember: it divides the voltage, so you may get unexpected results. A weak clock might not be enough to trigger what you want.
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Exactly — just use male-to-female pins, and you can split any signal directly using the breadboard itself.
Each output has an impedance of about 100 Ω, and the inputs have an impedance of more than 10 kΩ. So splitting a signal into 2–3 paths is fine and won’t cause a noticeable drop in signal level.
Of course, that’s exactly what the Jacket module is designed for.
Yes — connect your potentiometer between GND and the +5 V rail, and connect the center (wiper) pin to the CV input.
Also, if you have a mixer/VCA module, it provides +5 V and −5 V reference outputs. That’s a better option for this setup, since the +5 V rail on the breadboard can be slightly noisy.
The usable CV range is from −5 V to +5 V, but each module can tolerate up to ±12 V on all inputs (except for the special raw pins on the ESP32/MIDI module).
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I’ve done quite a bit of learning since, but there’s something still confusing me about this.
With audio and eurorack signals, it’s a jack cable which has - and + sides, whereas the signals were routing here are just one wire, I guess that it’s the + with the volts/data/cv and the - just goes through the breadboard - to ground in the power supply module.
So if I’m making my own 3.5mm jack type module, the + of the jack is what I send to a breadboard jumper pin, and the - would need to go to the same ground, which would be into the negative rail on the breadboard with the psu in it.
Am I misunderstanding it here?
To clarify, in audio jacks, you’re dealing with an unbalanced TS (Tip-Sleeve) or Balanced TRS (Tip-Ring-Sleeve). The former is what you typically see in eurorack and microrack. Tip is signal, sleeve is ground.
The signal itself is capable of through-zero.
The sleeve would connect to the gnd pins.