Jack input problems

Hi, everyone. Happy Holidays!

I too have a diy jack question. Could someone please lend some advice?

My project is a diy micro controller module. Whenever I initially insert a cord into my GATE inputs, it makes my CV readings go crazy. The CV goes back to normal after the cord is all the way in. Is there something I can do to stop this? I really don’t want the CV jumping around just because someone is patching a GATE jack.

Here is my circuit so far, attached.

Not sure if I need to decouple my power supply or maybe rewire the jacks?


If you use a jack that has a switch that is opened when the jack is inserted (make it grounded with no jack inserted) then when it starts to go from ground to a positive value add a small delay before acting upon it?

I’m sure there will be better suggestion, but that would be my starting point :slight_smile:

It’s actually a variant of the “switch bounce” problem.


hmm… what happens is : say my patch cord has anywhere from -15V to 15V and when inserted into the jack, the tip initially, accidentally touches the ring (ground). you have a short time where the -15V to 15V signal is connected to ground. since all my opamps, transistors, and pins are using circuits connected to ground, doesn’t this cause havoc thru the ground?

> say my patch cord has anywhere from -15V to 15V

With what kind of output circuitry?

You have design your outputs so that they can withstand a temporary short to ground.


The outputs could be any existing synth module. I heard 1k ohm resistance at the outputs is standard.

> I heard 1k ohm resistance at the outputs is standard.

Yes, or at least an op-amp whose output is not “stiff” enough to behave like an ideal 15V source :slight_smile:

I think my CV input circuits being op amp designed are very sensitive and when this gate tip touches ground the CV value jumps around.

Do you experience the same thing with your Mutable Instruments? I see they use op amp CV input too, right?

In your schematics you show separate analoge and digital ground. These should be connected at a certain point. Is it? Where?

Hi, Picard. They should be already connected somehow with the internal circuitry of my micro-controller. Do they need another external connection?

Another comment: what’s the deal with the “CV pot”. Why not connecting the wiper directly to A0?

I could try that again. I was thinking the op amp helps stabilize the pot reading.

I was also worried that when the pot is turned all the way to ground then the gate tip cord has a direct path to the pin input when it accidentally touches the jack ground. A 14V gate would fry the input pin from this ground short?

When a short occurs, your 14V gate won’t be at 14V.

So the input pin will never see the 14V short? Could you explain what would happen if a + or - 14V signal touches the jack ground? To me it looks like this is a direct path to the input pin thru the potentiometer when turned all the way to ground. I would like to know the electronic theory of why.

Thanks so much!

The answer is in your question :slight_smile:

Not all “+ or - 14V” signals are created equal. A +14V ideal voltage source, a +14V voltage source with a 1k resistor in series, and an op-amp buffer outputting a +14V voltage will all show +14V on a meter, but they are not the same thing.

It’s unlikely to encounter a +14V voltage source with a zero ohm internal resistance in the real world, and it’s unlikely that the path between this source and your module’s ground will have a null resistance. So when you connect the two together, there’s no reason the ground will instantly go up to +14V. What happens is one of the following:

  • Your +14V voltage source stays at +14V, your module’s ground stays at 0V, and some current flows in-between, according to ohm’s law. This current can cause things to catch fire. Example: 1 ohm resistance between an ideal +14V source and ground => 14A of current / 196W of power dissipated through the wire, fire (this won’t happen because your PSU won’t be able to deliver 1A anyway). Example: 1kohm resistance between a +14V source and ground => 14mA of current / 200mW of power dissipated through the resistor. Things get slightly hot.
  • Your +14V voltage source has some current-limiting mechanism, which causes the +14V voltage to drop whenever there is a short. Example: a circuit is designed to deliver +14V with a limit of 100mA. If you try shorting its output through a 10 ohm resistor, its output will just go down to +1V.

Tip: the US symbol for a resistor is a spring-like thing. The lower the resistor value, the stiffer the spring. This should remind you that the two nodes of the circuit (two mechanical parts) connected by a large resistor (loose spring) can be at different voltages (move independently from each other).

Ahh, I see now. Very educational. Thank you!