Some background: Many electric cars are designed to use a "J1772" charging point. The main signal in the J1772 specification is called the pilot.
Long story short, the car is expected to place a resistor and diode in series across the pilot and ground. A resistor value of 2.74k indicates that the car is present, and a value of 882 ohms indicates that the car wishes to have power.
The spec requires that you make a stopover in state B (2.74k) before engaging state C (882). It does not say how long that stopover must be.
There is a second signal in the specification called "Proximity." It is a signal from the J1772 handle to the car. It is wired to the release button on the handle. When the user is pushing on that button, it is a 450 ohm resistance to ground and when the button is released, it is a 150 ohm resistance.
I am making a device that has a J1772 inlet and derives its power from the charger to which it is connected. Prior to it placing the 882 ohm resistor value in place, it has no power of its own (unlike most electric cars).
The natural thing to do would be to make a circuit such that when that resistance is 150 ohms, the extra 1.3 kOhm resistor between the pilot and ground is in place. When the resistance is, oh, greater than 300 ohms, that resistor is not in place.
The challenge is... how can one do that with merely passive components?
In short, I want to detect whether a particular pin has a resistance of less than 300 ohms to ground, and if it does, I want to change the resistance between another pin and ground from 2.74k to 882 ohms (or put another way, add another 1.3 kohms in parallel). And I would strongly prefer to do this without a power source.
The left half of the diagram below is the proximity circuit that's part of the J1772 charging handle. It cannot be changed. The right hand side is my circuit. The question, again, is when the proximity resistance is less than 300 ohms, I want to close the "mystery" switch.
If I were going to make something active, this is what it would be:
This may be laughable overkill. I'm not sure. But it doesn't solve the power problem. It requires either a set of AA batteries (3 would last around 100 days if my calculations are correct), or a "START" button across the transistor to "bootstrap" the system. Both of those solutions, in my view, are lame.