Zener diode for overvoltage protection and earthing

Question

I am designing a small circuit where the goal is to measure the current. The CT measures the current, which passes through a shunt resistor. The ADC convert the signal so it can be further processed.

I am thinking about using jack port for the input of the CT. However, someone might think it is a power supply input and might plug in a 230 V cable.

So I want to protect my circuit against high voltage. Apparently, placing a zener diode is the simplest solution.

However it looks like my circuit is wrong. The zener diode will be connected to itself. Can someone tell me what I am doing wrong?

Answer 1

First, that's not the symbol for a Zener diode. That may be the fault of your layout software.

For a Zener to protect the IC it would go between the pin and ground, with the cathode (line end) to the pin, something like this:

^{simulate this circuit – Schematic created using CircuitLab}

The positive voltage on the pin will be limited to the Zener voltage \$V_Z\$ (more or less) and a negative voltage on the pin will be limited to the Zener's forward voltage drop (approx. 0.7 V) with respect to ground.

The current will need to be limited, otherwise a voltage higher than \$V_Z\$ will cause excess current in the Zener, likely burning it out. This can be done by using a series resistor after the shunt like this:

^{simulate this circuit}

If you are worried about someone plugging 230 VAC into the jack you should use a jack that doesn't accept a 230 VAC plug, using something that can be mistaken for an AC receptacle for anything else is just asking for trouble.

Answer 2

Just cut the connection upward from "1" on the Zener to break the loop. You also need to make sure that both resistor and diode are good for the resulting current from applying 230V if that is your concern.