Electric car "J1772" charger - "pilot" signal sequencing with passive circuitry

Question

J1772

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.

My Device

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.

^{simulate this circuit – Schematic created using CircuitLab}

If I were going to make something active, this is what it would be:

^{simulate this circuit}

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.

Answer 1

You are likely done with this but...

First understand the J1772 Specification. There is no need to place resistors in series or parallel.

The EVSE expects a voltage, which happens to be 3V, so you only need 2AA batteries or a CR123 which fits nicelly inside the enclosure. Since there are no volage dividers there is no current loss from the battery if the cable is unplugued/unused.

When plugged in there is a minor current draw to the EVSE so this battery should last a LONG time. The button operation can be emulated with a voltage divider composed of 2 220R Resistors. At normal operation a 220R resistor is in series with the battery and the proximity pin. Since the current draw is minumun the battery voltage (2.7 - 3.2V) will show up on the EVSE. when the button is pressed another 220R resistor is placed in series with the output from the first and GND, hence the voltage at the output will be ((220R+220R)/220R)x3V which happens to be 1.5V

Some EVSE dont need the buton on the proximity line, as long as you supply 3V to indicate the presence of the circuit they work fine, power is cut using the pilot.

Now to fully answer your question (The above makes it VERY practical, but not "passive") you can use the pilot to draw a very small current to make 3V available. In this case, you can replace the 2.74KOhm resistor with a lower value in series with a 3V9 zenner to generate an additional voltage. You WILL need to adjust this such that the output is still 6V. The zenner voltage should be attached to a diode, and then feed into a small capacitor to remove the ripple from the 50% dutty cycle. 3.9 - 0.7 = 3.2V Voltage can then be used to engage the proximity sensor. If needed you can make the voltage divider the same way as the above, but using 1KOHm resistors not to load the circuit.

Such a circuit will certainly only work depending on the current draw of the EVSE on the proximity line. The one I use more often works fine, but dont always count on this, hence the first option is cheap, very reliable and self contained since the cell can be placed inside the handle and changed every other month or so.

As you say this turns the EVSE off under load which is not very polite, so always have a circuit on the car to switch power ogf before you ask the EVSE to do so. I not only do this but also have a 10S timer to enable the charger and DC/DC after the power is applied. Only the control electronics (30W) are powered directly from the EVSE supply.

Also this circuit should only be used with power up to 15A. Higher power levels require the EVSE to negotiate maximun power with the charger, indicated by the pilot duty cycle.

Source: I own an EV.

http://www.diyelectriccar.com/forums/showthread.php?s=eafa7e1d8df5ebfaa5e4f2d51b79bb7e&t=78701&page=30

Answer 2

In the end, it was pointed out to me that doing it this way means that we're potentially asking the source EVSE to turn the power off while it's potentially under a high load, which is not... polite.

The better solution is to turn the proximity detector into a software input. The software will react by gracefully de-powering the downstream load. With that load off, the source EVSE would only be asked to de-power the handful of watts that run the logic board, which is much more acceptable.

Given that, then the design will be the active version of the circuit powered by the power supply. The "magic" switch in the first diagram will be simply turned into a real, physical one, labeled "power". Of course, turning the power switch off under load will simply move the impoliteness to our own relay(s), but I do think a manual hard shutdown option is a good safety decision.

Answer 3

Why not make your start button a battery connection switch instead (for your active bootstrap circuit)?

That combined with recharging it when it does power up could null and void the issue?