Switching jacks (how to prevent shorting)

Technology Synth-DIY
#1

Hi guys/girls,

I ran into a very simple design issue that I have not been able to find a nice solution for. I was hoping for some input/experience from others?

I want to design a module with CV inputs that are normalized with switching jacks. If a plug is inserted it takes the voltages that’s on there. If there’s no plug inserted, the switch in the socket closes and connects to an internal +5V reference voltage. However… with most jack sockets there are two ways you can momentarily short what’s connected to the switch terminal (my +5V reference in this case). Either:

(1) It shorts to ground when the tip that is slowly inserted, it touches the ground of the socket while it also touches the tip contact that is still conducting to the switch contact. Hence my +5V reference on the switch terminal is shorted to ground.

or

(2) the tip of the plug (carrying let’s say a maximum Eurorack voltage of +12V) touches the tip contact inside the socket that’s still conducting to the switch contact (when you push in further, it disconnects). Here my +5V reference on the switch terminal is connected to the external +12V, which (depending on impedance) may cause a reverse current through my +5V reference regulator.

How do you overcome that?

Using a current limiting resistor affects the level of my reference voltage (it becomes 4.9V or so). And using a diode may not be enough (and cause a voltage drop as well). Any other ideas? Perhaps buffering the voltage in an opamp circuit that recover from being shorted?

Best,

Rutger

#2

Use a 10V voltage reference and a 100k resistor between the voltage reference and your CV input.

Assuming the input impedance of your input stage is 100k (and it really should be 100k!), you’ll sense a source equivalent to 100 / (100 + 100) x 10 = +5V

Use 0.1% resistors if precision matters.

#3

Frames uses another approach - the voltage reference is buffered by TL07x whose output can be shorted to any voltage between Vee and Vcc for a short amount of time. If you want to be more robust, you can have a current limiting resistor in the buffer’s feedback loop.

I still prefer my first approach in which the current limiting resistor is part of a divider :slight_smile:

#4

Great, thanks Olivier!