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
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?
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?
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.
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).
