Battery capacities are usually provided in amp-hours or milliamp-hours. A quick search shows that typical car batteries have a capacity of around 45 amp-hours. That means with a one ampere load, it has enough energy to run for 45 hours. This isn't the most accurate way to think of capacity, since the total energy available depends on the manner in which it's used (quickly? slowly?) but it should suffice for some back-of-envelope calculations.
So to estimate the runtime, divide the capacity in amp-hours by the average current requirements of your device. If your device requires 50 mA on average, then:
\$ 45Ah / 50mA = 900 h \approx \text{37 days} \$
Most car batteries are not designed to be deeply discharged as part of normal operation, so avoid that. It's hard to say if your car has any sort of protection against running the battery down enough that it won't have enough to start. It would depend not only on the car, but also on which circuit the device is connected; some may have protection while others may not. I know my car does not offer any protection (even the dome light will run the battery down), but as the car in question gets newer and more expensive it becomes more likely there will be some circuitry that will do something intelligent.
If your device has batteries of its own that will power it should the main power be lost, then you can simply add that runtime to the estimate runtime of the main battery.
It's hard to say how long it will take to recharge the battery. It will depend on the charging method (fast charger? slow charger? the car's alternator?) as well as the type of battery, but something an hour and 2 days is likely. See this page from Interstate Batteries for more information on that.
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
Best Answer
All electrical motors have a high starting current momentarily which is several times higher than the normal running current. This is due to the fact that when a motor is switched on the total applied voltage drives heavy current through the winding depending on its impedance which is quite low. As the motor picks up speed it generates a counter EMF (voltage) which is in opposition to the applied voltage and the value of current (I)drops quickly (I= Applied voltage- counter Voltage/Impedance). Therefore it is seen that if the motor does not pickup required speed quickly the insulation of its winding melts away due to high heat generated by high current and a short circuit occurs which is generally referred to as motor burning in common parlance. Your calculation is perfectly correct. Starting motors (Starters) of cars are designed for heavy current as the load they have to drive is quite high but only for a very short duration. This heavy current is drawn from car battery only for a short duration. If the engine does not start in a short duration we stop the starter for giving rest to the battery. A week battery fails to operate the starter due to its inability to supply requisite current.