How to deal with inductive proximity sensor current leakage?

Thread Starter


Joined Jun 3, 2019
I'm using a 3 wire inductive proximity sensor to trigger a 24vdc SPDT relay. The relay has a minimum pickup current of 8mA with an unknown dropout current (the mfg. cannot provide a spec for that for some reason). The problem is my sensor has a leakage of 4 mA when off. What's happening is when I initialize my system, the 4 mA is not strong enough to trigger the relay, when the sensor does turn on, the current jumps to 17 mA and the relay picks up; however when the sensor turns off, apparently the 4 mA is enough to prevent the relay from dropping out.

I've looked on line and while there is a lot of information about bleed resistors for 2 wires sensors, there's nothing for 3 wire, so I don't know if a bleed resistor is feasible. If it is, does anyone know if there is a formula for figuring out what the resistance should be? I've also tried a socket with an RC network that didn't fix the problem, and have experimented with different resistances in parallel to the coil and running the the 24vdc return, but either the sensor doesn't trip, or it's the same problem with leakage current keeping the relay on after actuation. Any suggestions would be appreciated.


Joined Mar 14, 2008
Try adding a bleed resistor across the relay coil to shunt away some of the current.
Start with about 2kΩ, and go up or down from there until the relay works properly.

If that doesn't work, we could go to a two transistor current-mirror circuit to bleed the current, which won't reduce the maximum current as much as a resistor would.

P.S. It sounds like the sensor has a standard 4-20mA current loop output.
Last edited:


Joined Mar 14, 2008
I looked at one general purpose relay and it stated the dropout voltage/current was just 5% of the rated (or 0.4mA for your relay).
If that's true for your relay, then we likely will need to use something like the current-mirror bleed circuit to get it to work.

What supply voltage(s) do you have available?

Below is LTspice simulation of such a circuit.
Q1 and Q2 form a current-mirror circuit.
Q1 is biased at ≈4ma through R1, which generates a 4mA mirror current in Q2, thus bleeding 4mA from the signal.
This is shown in the simulation, where the 4mA bleed current (green trace) means there is no current flowing into the relay coil (yellow trace) when the signal is a 4mA (blue trace).
The relay current thus tracks the sensor current, but at 4mA less.

If your power supply is other than 24V, then you just need to change the value of R1 to give 4mA at the supply voltage.



Joined Apr 7, 2016
Like @crutschow , I can't really get the description to fit a standard digital three-wire sensor.

On a three-wire sensor, the output leakage current must be down to a few μA.

Two-conductor sensors, on the other hand, must have a "leakage current" to supply the electronics.

What type numbers are sensor and relay?
What is the resistance of the relay coil?


Joined Jul 18, 2013
You normally should have no problem with a 3 wire prox sensor as they have an open collector output.
Typically this mode is often used and should not need a bleed resistor.
4ma is quite a high leakage, either the Prox is the wrong type or there is some other issue with it that is not the norm.


Joined Feb 20, 2016
I agree with Crutschow. It does sound like you have an analog 4-20mA sensor, not an open collector switching one.
As Kjeldgaard has asked, what is the sensor type number?