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.

  • 3
    \$\begingroup\$ There's a lot of jargon in this question that is diluting what you want - can you explain it more simply. What is "stopover" all about? Is "pilot" a type of connection? What is a J1772 handle"? What is "EVSE"? Please don't supply a detailed answer to any of these questions but try and make your question simpler without the jargon that I suspect a few folk here would not understand. \$\endgroup\$
    – Andy aka
    Jul 15, 2013 at 16:55
  • 1
    \$\begingroup\$ J1772 is the specification for electric car charger connections, if anybody is confused (Like I was). OP wants to create a device which emulates an electric car and draws its operating power from the charger. Wiki has some good overview which should make this question's objectives a bit clearer. en.wikipedia.org/wiki/SAE_J1772 \$\endgroup\$
    – dext0rb
    Jul 15, 2013 at 18:12
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    \$\begingroup\$ @nsayer Not suggesting you add anything to the question, but that you remove stuff from the question to simplify it to the bare concept(s) that need solving. The protocol name, the specific resistance values, and so on, do not help with that, they only prevent the salient question from standing out. \$\endgroup\$ Jul 15, 2013 at 20:56
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    \$\begingroup\$ What voltage is placed by the charger on the "pilot" signal. It is dissipating power through R3. Maybe that power can be used to run some low power circuit that senses the voltage between R1 & R2 and controls your "mystery" switch. A minimal microcontroller may be feasible. It is possible the J1772 committee anticipated the possibility of a flat battery. \$\endgroup\$ Jul 16, 2013 at 7:45
  • 1
    \$\begingroup\$ @RedGrittyBrick Now that I've said that, though... I wonder if tying the anode of the diode to, like, a 100 uF electrolytic cap would be able to keep a stable enough voltage for the amp to operate properly. It would have to potentially work with a duty cycle of as little as 10%, but the current requirements of the amp are only .5 mA. The worry I have is that either the voltage would be too unstable or that the draw would drag the supplied voltage down too far. I may just have to try it and see. \$\endgroup\$
    – nsayer
    Jul 16, 2013 at 17:03

3 Answers 3


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.



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.


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?

  • \$\begingroup\$ What battery? The device in question is not an electric car. It has no power source of its own. Adding either a start switch (it would only need to be across C and E of the transistor. As soon as the pilot resistance drops to 800 ohms the EVSE will kick on the AC power and everything is fine from then on) or a battery is, as I said in the question, a lame solution. \$\endgroup\$
    – nsayer
    Jul 18, 2013 at 16:00
  • \$\begingroup\$ You mentioned a battery in your initial question as an option (albeit an option you wanted to avoid) \$\endgroup\$
    – back_ache
    Jul 19, 2013 at 10:10
  • \$\begingroup\$ You mentioned a battery in your initial question as an option (albeit an option you wanted to avoid). What I am firstly saying is to generate the PWM signal J1772 needs, there is no other way to reliably power the circuit than a battery, my second point was to counter your worry about only having a 100 hours battery life by only connecting the circuit to the battery when the start button is held down. The only real draw back in this approach is following a powercut it won't restart till you press and hold the button. \$\endgroup\$
    – back_ache
    Jul 19, 2013 at 10:18
  • \$\begingroup\$ An document that explains the signal is quite well is open-evse.googlecode.com/files/J1772.pdf \$\endgroup\$
    – back_ache
    Jul 19, 2013 at 10:36
  • \$\begingroup\$ I see your point now. But if you simply put the START button across the transistor, then you can power the rest of the circuit from the power supply. No batteries required. You push the button and it makes the pilot resistance 880 ohms and the EVSE will kick on the AC power. The active circuit will then hold the power until the proximity button is pushed, then commit suicide. \$\endgroup\$
    – nsayer
    Jul 19, 2013 at 14:24

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